Embed Size (px)
DESCRIPTION
GEN CHEM MCAT
Citation preview
J ''$1qctilrn VICasesby Todd Bennett
o = Lighter gas
g = Heavier'gas
eoo9o
^6O@vOg
o
-9o" @oo
oe
o o
oe
oo
o
eooo
Gas mixtffe is richer inthe lighter gas on fheriglrt side of the tube
Terminology and Conceptsa) Cas,Phase Definitions'b) Kinetic Molecular Theory of Casesc) Boltzlann s Distvibutiond) ldeal Casese) Reil Qases
..1
Gas Lau'sa) Applying Cas Laws
b) Avogadro's Law
c) Boyle's Lawd) Charles's Law
Gas $ystem Pfopertieg " t' ,
a) Partial Pressure
b) Manometers
Gas Motiona) Qas Speedland Veloci$rb) Diffusionc) 0ffusion and Infusion ,
,
d) Isotopic Dnrichment i ,
e ecleoo
rooo, " l': i. '.o ooulj " utt'o o lT@ o
G.as mixture is richerin'the heavier gas,, onthe le{t side of the tube
Gases Section GoalsKnow the definition of a tras and the terms associated with qases.
E?
You must know what an ideal gas and a real gas are. Understand the following terms used to describea gas system: pressure, volume, concentration, density, temper-ature, moles, molecules, collisionfrequency, and mean free path.
flave a solid understanding of the kinetic theory of gases.Be able to recite a viable description of gases at the microscopic level using the kinetic theory ofgases. Know that gases move about their bnr.ironment in a random fashion, and that they frequentlycollide with other gas particies and the wails of the container. Know that the momentum of heaviergases is greater than the momentum of lighter gases. Know that the rrelocities of each gas particlesum to zero for a gas in a stationary, closed container.
Know the proper applications of the various gas laws.Know hen to use Avogaclro's, Boyle's, and Charles's laws from the r.aliabies presentecl in the question.Have an understanding of the
-experimental conditions required to use 6ach law. li.ecog'nize the
shape of the graphs thai show the relationships betlveen the variables oi each gas 1aw.
Be able to distinquish between a real gas and an idgal qas.An ideal gas is composed of molecules that occupy no volume and exhibit no intermolecular forces.A real gas is just what the name implies, a gas made up of real molecules that do occupy volumeand exhibit intermolecular forces. The pressure and volurne of a real gas are therefore different fromthose of an ideal gas. The Van der Waals equation (w hich you shoLrld"understand but not mcmorizc)shows the relationship betweenideal variablcs alrd rcal variables ftrr a gas.
Be able to determine the partial pressure of a gas r,{ithin a rnixture.The total pressure of a system is the sum of the individual pressures (partial presstrres) of all of thecomponent gas particles. The partial pressurc due to one componerr{ irr the mirture can be determincdby multiplying the mole fraction of the component by the total pressure of the system. You mustbe able to convert between different pressure units.
Understand the relationship between speed, mass, and temperature f,or a gas.You must know the effect of a particle's mass o1r iis velocity. Lighter gases travel at a faster rate thanheavier gases. The velocity of a gas particle is ini crsely proportional trr the squ.rre root ul iis mass.You musl know the effect of iemperature on r elocity. At higher temperatures, gases travel at a Fasterrate. The velocity of a gas particle is directly proportional to the square root of the temperature.
Understand the concepts of effusion and effusion rate, infusion, and diffrrsion.Effttsiotr is the process by which gas molecules pJss from within a container to the ouLside lhroughpores irr the container wall. The rate of elfusion depends on thc r clocity of thc gas molccules, thcnumber of pores in the material, and the size of tire pore relative to the size of the molecule. Whengases entera coniainer through a pore, the process is rcferred to a5 1i rrlci{)ir. When a gas particlemoves from a region of higher cohcentration to a ltrwer concentratir;n, the process is known as,liffusion.
sv
@?
'vov
G'
E,:.-- .j
- ='
'
::=::i:
:.:-:: _ _-;
_'. :t":: "::
. ;.] ir:l:
a, a,, a
I -,,_
lr'll l-t "::
.1.,.1 :
:'i.
: - 1^,-
-.;*
!
r_r:
._t",.Lr ,
!:_::
-. ,-," :
""t* :; l
t_.
. iL:- f
:.i:1"
- ,i:il:-.l :'": li. :
,-,,-t: ' .1::
:t i.r :
rr -,: r'l:l
_ ;:, ":
General Chemistry Gases and Gas Laws Introduction
The gas phase, from the macroscopic perspective, is defined as the state of aclosed system in which molecules have no definite shape and no definite volume.From the microscopic perspective, the molecules are freely moving particlestraveling through space, where the kinetic energy associated with each particle isgreater than the potential energy of intermolecular forces. The gas phase is thefirst of three phases of matter that we shall address. Gases differ from liquids inthat liquids have a definite volume, and the molecules in a liquid are incontinuous contact with neighboring molecules. Gases differ from solids in thatsolids have a definite volume and definite shape, and the molecules in a solidundergo no translational motion and are in continuous contact with neighboringmolecules. Of the three common phases of matter (solid, liquid, and gas), the gasphase has the largest amount of kinetic energy and the greatest entropy. Toreach the gas phase, energy must be added to the molecules within a system thatis not already in the gas phase. It is essential to understand a gas from both themacroscopic and microscopic perspectives.
Macroscopic View of a Gas: A gas assumes the shape and volume of its container.Gases are compressible and must be contained on all sides to hold them in place.Gases are described by the macroscopic variables pressure, concentration,temperature, and volume. Chemists treat gases as if they are composed of inertspheres that occupy no molecular volume. This is a basic assumption of the idealgas law. Most calculations involving gases are based on the ideal gas equationiPV = nRT). There are two common types of calculation questions about gasesthat you may encounter on the MCAT 1) the before-and-after questions about theeffect of changes on a gas system and 2) the straightforward type using PV =rRT. Calculation questions requiring you to find a precise value are few, but;onceptual relationships can be determined from calculations as well. The firstsivle of question is better understood as an application of one of the gas laws-\vogadro's, Boyle's or Charles's laws). For instance, if the temperature increases
-n a sealed, rigid container, what changes? This is another way of asking how aiange in temperafure affects the pressure of a gas, assuming constant volume:nd moles. Rather than memorize gas laws, know qualitatively how one variablen the ideal gas equation affects another.
Microscopic View of a Gas: Gases consist of individual molecules or atoms that arerandomly moving about the space within a container. The particles are in contactonly during collisions. This concept is known as the kinetic theory of gases. Gasescan be described by the microscopic variables of collision frequency, mean freepath, and mean velocity. The fact that microscopic behavior involves interactionsretween gas particles implies that the molecules do in fact occupy a small'i'olume and they are capable of exhibiting intermolecular forces (both attractiveand repulsive). The strength of these forces varies with distance. Gases exist as:ea1 gases, which are approximated as obeying ideal behavior when they do notnteract. Ideal gas behavior is best simulated at low pressure (fewer collisionsretween molecules and minimal forces between molecules) and highlemperature (at high temperature, molecules have the energy to overcomentermolecular forces). Under these conditions, the forces acting between gasnolecules become negligible. This is the basis of the kinetic theory of gases.
lt.e behavior of a gas is predictable in terms of the variables that describe the gas:\'stem. We use the ideal gas law to predict the effects of changes on gas systems.-{lthough there are no ideal gases, the approximation is still rather close.
Copyright O by The Berkeley Review Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Terminology and Concepts Ge
fcffiiffidtbffi',fifid C'ficbptsGas Phase DefinitionsThe terms used to describe a gas can be broken down into categories ofmacroscopic and microscopic. We can consider either the system as a whole(macroscopic perspective), or the individual particles that constitute the system(microscopic perspective). Each of the macroscopic properties has a microscopicequivalent. Table 6.1 lists the macroscopic and corresponding microscopicproperties that describe the state of a gas system.
Table 6.1
It is imperative to have a well-developed glossary in your memory by the timeyou take the MCAT. Organize the terms in a manner such that one definitionhelps you to recall additional definitions. The definitions of selected terms fromTable 6.1 are:
Gas Pressttre: Gas pressure is defined as the force per unit area exerted by agas through collisions against a defined area of the container wall. As thegas moiecules collide more frequently with the container walls or increasetheir force during collisions with the walls, the gas pressure increases. Thegas pressure depends on the number of gas particles, the volume of thecontainer, and the temperature of the gas system. The standard unit forpressure is atmospheres.
Collision Freqtrcncy: The collision frequency is defined as the rate at whichmolecules in the gas system collide with each other and with the wall of thecontainer. The collision frequency can be increased in one of three ways:increasing the temperature (energy of the gas system), increasing theconcentration of gas particles, or reducing the mean free path. All of thesechanges result in an increase in the number of collisions experienced by a
molecule within the system in a given period of time.
Collision Force: The collision force is defined as the force exerted by a gas
particle during a collision between it and the container wa1l. It is an impulse,so both greater momentum and a shorter time of contact increase the force ofimpact. The collision force can be increased by increasing the temperature(energy of the gas system), because greater temperature imparts greatervelocity, and thus greater momentum, to each particle.
Volume: The volume of a gas is defined as the region within the walls of acontainer. The actual volume that a gas molecule can occupy (the realvolume) is the volume of the container minus the volume of the other gasmolecules, because no two gas molecules can occupy the same volume at thesame time. The volume of a gas in an open environment is undefined,because the container holding the gas is also undefined. The standard unit ofvolume is liters.
:'i-
:iL
;l*
IE
;ii:
:r[*:s
:.. .-
4r:
:..
::
:rll|lllttr];lli
T*run5
finpur*
;;tilr';
}c:r U:
I illlillir
J,*rn]uu-
Llin,* u
t,'rfun i Ll.LtiI
[wr Afi[|:
lrtrIw]t"lPljt]lnf,
tr iiLi,lil
l4 I llllllt
luu
:
I
: l:
t_ i
Macroscopic Measurements Microscopic Measurements
Pressure (P; standard unit is atm.) Collision frequency and collision force
Volume (V; standard unit is liters) Mean free path
Moles (n; standard unit is moles) Molecules
Temperature (T; standard unit is K) Average kinetic energy
Copyright @ by The Berkeley Review The Berkeley Review fT|t ll
ts General Chemistry Gases and Gas Laws Terminology and Concepts
ofrle
mric
ric
Concentrntion: The concentration of a gas is the number of gas particles perunit volume in a container. A gas is assumed to be homogeneous, so asample from anywhere in the container may be used to determine the gasconcentration. As more molecules of gas are added to a system, or as thevolume of the container is decreased, the gas becomes more concentrated.This means that as a gas is compressed, it becomes more concentrated.
Mean Free Psth: The mean free path is defined as the average distance aparticle can travel before colliding with another particle. Although it isn't thesame thing, it can be thought of as the average distance between gas particlesat any given time. It is the microscopic equivalent of concentration. If theconcentration of a gas remains constant, then the average distance betweenany two particles within the container also remains constant.
Temperature: Temperature is a measure of the total kinetic energy of asvstem. The greater the energy of each particle in the system, the greater thetotal energy of the system, and thus the greater the temperature of thesvstem. Temperature can be measured in degrees Celsius or Kelvin,although Kelvin is the better measurement to use when working with gases.
Auerage Knetic Energy: The average kinetic energy of a system refers to themean energy of a particle in that system. As the energy of each particle in thesystem increases, the average kinetic energy of the system increases, therebyresulting in an increased temperafure.
Example 6.1\\'trich statement below accurately describes the relationship betweentemperature and energy?
A. Temperature quantifies the energy of a system.B. Energy quantifies the temperature of a system.C. Temperature is independent of energy.D. Temperature is variable, while energy is constant.
SolutionTemperature is a measurement of the average kinetic energy of a system. Thehest answer is choice A. A thermometer is used to determine the temperature ofa svstem. The way a thermometer works is based on the kinetic theory of gases.The thermometer is an evacuated closed column that is partially filled with apure liquid, preferably of low volatility (such as mercury). Gases collide with the,rutside of the evacuated glass casing that contains the non-volatile liquid. Theenergy from these collisions is transferred through the glass walls and into theniquid. A sufficient number of collisions can increase the vibrational energy ofihe liquid in the container and force the liquid to expand, thereby causing theheight of the column of liquid to rise. This implies that the density of a liquid isinversely proportional to its temperature, because as the temperature rises, ther-olume of a liquid increases (the liquid is expanding). Because the density oflvater does not change uniformly (it increases from 0"C to 4'C and then decreasestrom 4"C to 100"C), it cannot be used in thermometers. In addition, the range oftemperatures over which water exists as a liquid is too small. The liquid chosento fill a thermometer must have a large temperature range between its freezingpoint and boiling point. In most thermometers, the liquid used is mercury. Inother thermometers, a non-volatile alcohol tinted with a red dye is used.
ne
)n(rl
Copyright @ by The Berkeley Review Bxclusive MCAT Preparationgw b
General Chemistry Gases and Gas Laws Terminology and Concepts Ge
Kinetic Molecular Theory of Gases
The kinetic molecular theorv of gases takes a microscopic view of the componentmolecules (or atoms) that make up a gas. Four assumptions associated with thistheory with which we shall concern ourselves are the following:
1. Particles are so smail compared to the distances between particles (inter-nuclear distances) that their volumes are negligible (assumed to be zero).
2. Particles move in straight lines. The direction of a particle's motion ischanged only by its collision with either another molecule or the walls of thecontainer. The collisions are said to be elastic (no energy is dissipated), andmomenfum is conserved.
3. Particles are in constant random motion. Gas pressure is caused by collisionsof the particles against the walls of the container.
4. Gas molecules exhibit no intermolecular forces. This is to sav that theparticies neither attract nor repel one another.
Figure 6-1 shows the random pathway of one gas particle over time, according tothe rules of the kinetic molecular theory of gases. If the kinetic energy of theparticle increases, the particle's speed increases, so it collides more frequentlywith the wall. Because it is moving faster, it collides with greater momentum, soimpulse increases. The result on the macroscopic level is that the force per unitarea exerted against the walls increase, meaning pressure is greater. The kinetictheory of gases explains macroscopic observations using principles derived froma microscopic model. Figure 6-1 shows a random particle in motion.
The particle moves until it collideswith the wal1. In a real system, theparticle would also collide with otherparticles present in the container.
Figure 6-1
When there are many gas particles in the container, collisions between particlesbecome more common than collisions with the wall. However, more particles inthe container also results in more collisions with the walls, so the pressureincreases as particles are added to the system. When there are particles ofdifferent masses in a mixed gas, heavier particles move more slowly, hence theyexhibit lower coilision frequencies. However, because they have a greater massand only slightly reduced speed, they collide with greater force (momentum). Asa general rule, lighter gas molecules have greater average speeds (and greatercollision frequencies) than heavier ones, but less momentum (and thus lesscollision force). Because pressure depends on both collision frequency andcollision force, gas particles of different masses exert the same pressure. On themacroscopic level, this means that pressure is the same under identicalconditions for all ideal gases, independent of their molecular mass. A goodexample is to compare helium and nitrogen. The reason they have the samepressure at the same temperature is because they have the same kinetic energy(mv2 term). The molecule with greater mass has less speed. The average speedis inversely proportional to the square root of the mass.
$,lrr$,u
xri{iflrlfi[:r Tll,:ll,lf
Tfrt5-al:
- trfr
#il,igoitni'
a, LULr i-tit
6Copyright @ by The Berkeley Review The Berkeley Review I.m.r
Its General Chemistry Gases and Gas Laws Terminology and Concepts
ntris
lr-
ishend
,nS
he
tohetlySO
nittic)m
Example 6.2ln a rigid, closed container, how does an increase in temperature affect the gasparticles in the system?
A. The mean free path increases.B. The collision force decreases.C. The collision frequency increases.D. The particle momentum decreases.
SolutionThe fact that the container is rigid means that the volume does not change.Because it is a closed system, the moles of gas do not change (matter can neitherenter nor exit a closed system). The concentration remains constant, so the meanrree path does not change. This eliminates choice A. The mean free path is thea\-erage distance a particle travels between collisions. It can also be thought of asthe average distance between particles at any given instant. Because the volumerf the container does not change, and the moles of gas do not change, theconcentration (density of gas) does not change. This means that the particles arethe same distance apart, so the mean free path does not change. According to theequation PV = nRT, if the temperature of the system increases with volume and if:Loles remain constant, the pressure must increase. This means that collisionIorce, collision frequency, or both must have increased. As temperatureicreases, the velocity of the particles increases (the kinetic energy increases in:erms of velocity). With an increase in velocity, the particles travel the distance'ietu'een collisions at a greater rate, so they collide more frequently. Because the:arficles have greater velocity, they have greater momentum, so the collision:orce increases. The result is that the collision force and collision frequency bothrcrease, so choice B is eliminated. At greater velocity, the particles have greater:romentum, so choice D is eliminated. Choice C is the best answer.
B oltzmann's DistributionLn a gas system, not all particles have the same kinetic energy. There is instead a:andom distribution of energies, known as Boltzmqnn's distribution. Figure 6-2 isa Boltzmann's distribution of kinetic energy for the particles in a gas system. The::rean kinetic energy does not correspond to the apex of the curve, but to a point
=-rghtly to the right of the apex. This is because the energy distribution graph is
=<en'ed to the right. As temperature increases, each particle gains kinetic energy,
=nirting the distribution to the right. Figure 6-2 shows a Boltzmann's distributionat tu'o different temperatures, where T2 is greater than T1.
Kineticenergy +
Figure 6-2
lesin
rreof
rey
ISS
AsterESS
ndthecal
'odme:wrcd
TztTt
olrCJ
-o-.
z
Copyright @ by The Berkeley Review Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Terminology and ConcePts
Ideal Gases
An ideal gas is a theoretical gas which obeys the following three conditions:
1. The molecules exhibit no intermolecular forces.
2. The molecules occupy no microscopic volume (are point masses)'
3. All collisions are perfectly elastic.
Because the constraints of the ideal gas cannot be met by real molecules, an ideal
gas is strictly theoretical. Gases are closest to ideal at high temperature (having
more kineiic energy to overcome intermolecular forces), low pressure
(interacting minimafy with one another), and when the system is composed of
small, ineit particles. The most ideal gas is helium (which has the smallest
particles of any substance known and which exhibits negligible intermolecular?orces). to veiify these conditions, consider a phase diagram. When pressure is
small and temperature is large, the state of matter is gaseous, far from the other
phases in the phase diagram.
The ideal gas law was the result of empirical observations, such as noting that the
volume oi u gut is inversely proportional to its pfessure, volume is directlyproportional io temperature, and volume is inversely proportional to moles.
From these relationships, Equation 6.1 (a composite formula) was derived.
(6.1)
Equation 6.L can be converted into Equation6.2by introducing a constant.
V = kn'TP
Equation 6.2 can be manipulated to yield: P'VEquation 6.3, where the constant is R, rather than
6,3, the ideal gas equation'
PV = nRT (6.3)
where P is pressure, V is volume, n is moles, T is temperature, and R is the ideal
gas constani of 0.0821 L'atm.'mole-1'IC1.
In the best of all possible worlds, all gases would be ideal. But real gases make
up 100% of all gises, so there are no ideal gases. F{owever, while there are no
ideal gases, *e .ur', make calculations for gases that are ideal and then adjust for
orr. uiorr. The ideal gas law produces answers close enough to the exact values
for real gases that *" .utt .ttu it it everyday practice. Deviations from ideality
are appr:oximated, and rough colrections ale made to determine the conditions
for a ieal gas. For example, as volume decreases, gases behave less ideally' This
is because*the actual size of the molecules does not change, so the percent of the
volume occupied by the molecules increases. The molecules interact more and
are limited in the free space they can occupy.
Ideal gas problems are straightforward, plug-and-chug in their purest form-
Ideal g:as qlestions also encompass the before-and-after case questions associated
with gas liws. As you approach these problems, isolate the value that you are
lookiig for and solve foi it in terms o{ the other variables. Be sure to use the
correciunits. It is easy to forget to use kelvins. What you must do is observe
what remains constant when you consider a system'
--nr]i::r ::'i
tr,.- *-- al lr
n.T
P
SL
sl.
LD
Snrrir:l
*.:rU
lhm
Enaxd
.h
s
T\
:r@u
fif,fi..u!fi.
Jl,s
lhemrs
(6'z)
= k.n'T, which in essence isk. Ideal gases obey Equation
PV = nRT .'. R = PV, ,o PlVl =P2Y2nT nlTl nZT2
Copyright @ by The BerkeleY Review The Berkeley Keviewa
General Chemistry Gases and Gas Laws Terminology and Concepts
Cancel out the terms that are constant, and isolate the variables. Three examplesof commonly used variations of the gas laws are:
If PandTconstant: Yr =Y2.n1 n2
If n and T are constant: P1V1 = P2YZ.
If nandPconstant: Vr =Y2 .
T1 T2
Lxample 6.3Given that the pressure of a twentieth of a mole of gas is 0.82 atm. at 27'C, whal.-. its volume?
\. 1.50 litersB. 7.72litersC. 2.4litersD. 50.0liters
5,crlution
-,. - latrng the volume term (V) in the ideal gas equation yields: V =:::stituting given values yields:
.,- _ 0.O5moles x 0.0821 L.atm.mole-1' K-1 x 300K _ 0.05 x0.0821 L x 300
0.82 atm. 0.82
V = 0.05 x0.1 x300 L = 0.05x 30L = 0.5 x 3L = 1.5 L.
,]: best answer choice is choice A.
Lr,"ample 6.4
-: : r:s occupies 618 mL at STP, what is its volume at 50.4"C and 1.22 atm?
{.. l;: mLS -31mL:. :-,- mLD 1-l mL
nRTP
5,rlulbion-- " I rF. ',he pressure is 1.00 atm. and the temperature is 0'C. This means that the::=s..-i:e increases from 1.00 atm. to 7.22 atrn., and the temperature increases::-r- l-l K to 323.4 K. An increase in pressure reduces the volume of the gases..--: ::.;:ease in temperature increases the volume. The change in pressure andr-: -*;. rn temperature have opposite effects on the volume in this question, sor:*E :itar'-ge in volume should be minimal. This makes choice C the most likelyli-*." \:::. Flon'ever, to be certain, we need to do the math.
-,.. -.^. PlV1 P'tY't . . ,.- PtVt PtY2Lrlven: r r =-W/ftconstant= !n1T1 nZTZ T1 T2
"' =(#l+)'' = (#\#)ors-r-= (a[9ora-r-
Vt = 2 x 618 mL= 600 mL-55l-': :'=s: ans\l'er choice is choice C.
I,::r-:-:h: e br- Ihe Berkeley Review Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Terminology and Concepts Gr
Real Gases
There is no such thing as an ideal gas; all gases are real gases. An ideal gas issaid to have molecules that occupy no space and exert no force lrpori o."another. A real gas exhibits intermolecular forces (i.e., the molecules ittract orrepel one another), has particles of microscopic volume that transfer energy uponcollision. The degree to which a real gas deviates from ideal behavior t-ris to aowith the magnitude of the intermolecular forces and the size of the particles. TheDutch chemist J. D. van der Waals studied real gases and developed correctionsto the ideal gas law to explain real gas behavior. His name wis given to theweak intermolecular forces between particles of a gas.
Deviations in pressure are due mainly to the intermolecular forces. For instance,if the gas particles exhibit attractive forces, the system implodes to a smalldegree. The result is that gas particles coilide less frequently with the walls ofthe container, so the observed pressure (Pobs) is less than ideal pressure. Thismeans that a correction term must be added to the observed pressure to make itequal the ideal pressure. This is shown as Equation 6.4 below:
Pideal =Pobserved * ua1v2
where c is an empirical value for each gas, r is the number of moles of gas, and vis the volume of the container. It is easiest to think of a as being an attractioncoefficient. \44een the particles repel, the value of a is negative.
Example 6.5\Alhich of the following types of gases has a negative a term?A. PolarB. Non-polarC. HydrophobicD. lonized
SolutionThe first step is to determine what this question seeks. it is asking for a negativea tetm, which is associated with particles in the gas phase that repel one another.The question can be reworded to read, "In what type of gas do th! particles repelone another?" choices B and C are the same answer worded differently, so bothshould be eliminated. Polar particles have attractive forces, so choice A iseliminated. An ionized gas is generated when an electron is removed from thevalence shell of the gas atom or molecule. The particles of an ionized gas carry apositive charge, so they repel one another. This makes choice D the best answer.
Deviations in volume are due to the fact the molecules have volume. Becausethey have volume on the microscopic level, the actual free space (space notoccupied by molecules) is less than the volume of the container. The bigger themolecules, the greater the volume they occupy, thus the greater the deviation.The more molecules, the greater the volume they occupy, thus the greater thedeviation. The free space (ideal volume) is found by taking the container volumeand subtracting the volume of the molecules. This means that the ideal volume(volume of empty space) is equal to the difference between the volume of thecontainer and the volume of the particles. This is shown as Equation 6.5 below.
Videal=V66n1sin61 -nb (6.s)
where b is an empirical value for each gas and n is the number of moles of gas.
-:li
;i*$i
4r
m
rL,
D
E:mcr
(6.4)
5I{rUr
,E
: T:lii
'i_.4,,frilifj:T
i6li;tlljut
"lfl,"rt't,:
* ru:rL
rrs."a]lt i
ruL i
llhltll:llrtuury
;' ITfItrKJI
lfitne n-il frlrtdll
:Jru,,rr1,1"111,66;
:rInJrIml
ltil;nlrigt
.-J-"
:[xrrygs
rlg$lL
r iflulJll{*
-"T-rfi r
tLf
r;" &
-a
Copyright @ by The Berkeley Review lo The Berkeley Review
General Chemistry Gases and Gas Laws Terminology and Concepts
s
e
rn
D
e
s
e
ItS
t
The greater the number of gas particles, the more volume the molecules occupy.This accounts for the n in the equation. Think of b as beinga bigness coefficient(despite there being no such word as "bigness"). A11 particles have some volume,so b values are always positive.
Example 5.5Which of the following gases has the greatest b term?
A. MethaneB. EthaneC. EtheneD. Ethyne
SolutionThe first step is to determine what the question is asking for. It is asking for thelargest b terrn, which is associated with the largest molecule. Methane has onlyone carbon, so choice A is eliminated. Choice C is eliminated, because ethene(C2Ha) is in the middle of a sequence that starts with ethane (CzHd and endsrvith ethyne (CzHil. The largest molecule can never be in the middle of as€quence like this. Because ethane has the most hydrogen atoms and has sp3hvbridization, it is the largest molecule of the choices. Choice B is the best.
Combining the deviation in pressure and the deviation in volume leads to thereal gas equation, also known as the aan der Waals equation. The van der Waalsequation, Equation 6.6, is derived by substituting the corrected terms for pressureand volume into the ideal gas equation.
(Pour"*"a
The value of the a term can be either positive or negative, because intermolecularforces are both attractive and repulsive. If the molecules attract one another, thepressure is reduced from ideal behavior, so a correction term must be added toPobserved. This means that the sign of a must be positive for attraction. The b
term is always positive, because molecules have positive volume. Understandthe deviation and be able to predict its effect on a real gas. Attractive forces, forilutance, make the volume of a real gas less than the volume of an ideal gas.
Example 6.7l\'hat are the a and b terms for an ideal gas?
-d. a=1;b=IB. a=1;b=-1 /C a=0;b=1D. a=0;b=0
Solutionr,\lren both a and b are zero, both of the correction terms in Equation 6.6 are zero,and thus drop out. This leaves us with PV = nRT, the ideal gas equation,indicating that choice D is the best answer. An ideal gas has no intermoleculartorces, so the attraction coefficient is zero (the a value is zero.) The particles of anr,Ceal gas occupy no microscopic volume, so the bigness coefficient is zero (the b
ralue is zero.) This confirms that choice D is the best answer.
* utllV.ontainer - nb) = gq1 (6.6)vzl L\);,t oo
P'-'\
,\) u-"'l L\) ;"s ''"
Copyright @ by The Berkeley Review Exclusive MCAT Preparationll
General Chemistry Gases and Gas Laws Terminology and Concepts
Example 6.8For an inert real gas, if you were to reduce the pressure to half of its originalvalue, then what is the final volume (Vi) relative to the initial volume (V1)?
A. lV' - a little bit2'B. lVr +alittlebit2'C. 2Yr- a little bitD. 2V1+alittlebit
SolutionWhen pressure is cut in half, the ideal gas law predicts that volume shoulddouble. Because only the space between molecules increases, while themolecules remain the same size, the increase in volume is not as large aspredicted by the ideal gas law. This makes choice C,2Yi- a little bit, the bestanswer. The "little bit" term is attributed to the size of the molecules. Figure 6-3shows this.
Volume = 1.55 LEmpty space 1.45 LMolecules 0.1 L
Volume = 3.0 LEmpty space2.9 LMolecules 0.1 L
Figure 6-3
Volume = 5.9 LEmpty space 5.8 LMolecules 0.1 L
ffirueryq,llr*re*@#
iiErl'w -:nar,ceui
,mc adItl,,mr:h
mluryhd
lnum .larnir
llrr"':dhr d.lh
"uumrl:n a
Trn{sffillul
ffrresffiru[
triLllli lon,,e
re"ffannlTlm,'grurririls
iinn *ilhhman;1im$,
.4tF mrsnir
,Wmll 16 1
Tkns retil
#ms-dmlr:e ld'JtgFrnurmf
tltlur,r :mm
iilllJ:llrE
lftrre :r:r.rqr
iw
ir- r
}c -*u, :r.k..t -ir ii:mrtsmrtrFT nrtcfl;M ruA:
Pressure isdoubled
-
Pressure isreduced bv half
-4-'>>
Assume we have a 3.0 L container filled with a gas. AIso assume that if theparticles of the gas were compacted together, then their volume would be 0.10 L.When the dimensions of the container change, only the empty space changes, notthe actual size (volume) of the molecules. When the container expands, the sizeof the molecules will still be 0.1 L. This means that when the pressure is cut inhalf, the volume doesn't quite double, but is 2Yi - a little bit. The value for ourhypothetical system is 5.9 L. If the pressure were doubled, the volume wouldn'tbe reduced by exactly one-half, either. The new volume would be fV, + a littlebit, as shown in Figure 6-3. The best answer is choice C.
Figure 6-3 shows that the microscopic volume of the molecules does not change.The size of an atom can change only when its radius is increased (by exciting itselectrons to occupy a higher energy orbital with a greater radius). Note also inour example that when the pressure was increased (doubled), the volumedecreased, but not all the way down to one-half of its original value. VVhen thepressure was reduced by one-half, the volume increased, but not all the way upto double the original value. It may be easier to remember that the change isnever as large as you think it is (i.e., what it is predicted to be according to theideal gas law) when dealing with a real gas. Corrections to the ideal gas lawshould be intuitive for the most part.
Copyright @ by The Berkeley Review t2 The Berkeley Keview
General Chemistry Gases and Gas Laws Gas LawsPts
inal
ruldthe
3asbest:6-3
Lt.8 LL
f the10 L.
', notsize
ut int ourIdn'tlittle
-mge.rg itsso inlumen thery uprge is:o thes law
Applying Gas LawsThere are three common gas laws to know: Avogadro's law, Boyle's law, andCharles's law--the A, B, and C laws of gases. They all stem from the ideal gaslaw. It is possible to deduce them by solving for R in the equation PV = nRT,canceling out any terms that remain constant, and equating the values for beforeand after the change in the system. If you find yourself getting confused aboutwhich gas law refers to what, try creating a simple story about how each scientistmight have made his discovery: Avogadro was into counting big numbers, sohis law focuses on the number of molecules. Therefore, Avogadro's law dealsn ith the relationship between moles of gas and volume. Big Boy Boyle sat on hislunch and smashed it (decreased the volume of his sandwich), by increasing thepressure on it. Therefore, Boyle's law deals with the relationship betweenpressure and volume. Good ol' Chuck overheated his popcorn and it scatteredall over (increased its volume). Therefore, Charles's law deals with therelationship between temperature and volume. These descriptions may seemluvenile, but if they help you recall the gas laws by name, then they are worth it.For solving any gas questions, it is critical that you use temperature in terms ofkelvins; otherwise, you will encounter much sadness and dismay.
-{s mentioned earlier, all of the gas laws, in conjunction with the ideal gas law,iead to the following relationship:
P1v1 =
n1T1
PzYznzTz
This relationship is ihe foundation in physics for the operation of any system thatEenerates air flow, whether it is a human lung or a mechanical ventilator. One:-;pe of ventilator is an accordion-like apparatus used to maintain uniform:espiration (circulation of air) through the tung of patients during their recovery:rom many kinds of medical problems that affect breathing, such as a collapsed-ung.
lhe four steps in the normal cycle of a human lung are shown in Figure 6-4.
Step 1: The diaphragm contracts and the thoracic cavity expands,increasing the volume of the lung.
Vlung t, Plung J so that Pintemal ( Pexternal
Step 2: Air flows into the lung through the trachea.
nair T, ptrr,'g t until Pir.,1".rral = Pexternal
Step 3:The diaphragm relaxes and the thoracic cavity contracts,decreasing the volume of the lung.
Vlung J, Plung t so that Pinternal ) Pexternai
Step 4: Air flows out from the lung through the trachea.
nair J, Pt.r.rg J until P61".r.,a1 = Pexternal
Figure 6-4
lhe rvorkings of a ventilator are similar to these steps, but it has one-way valves.-{.:r flows in both direction through the trachea, but ventilators typically have:-et and outlet tubes. (A ventilator is discussed in one of the passages in ther:,ses section of your in-class general chemistry booklet.)
:Ylew -mrlight O by The Berkeley Review l5 Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Gas Laws
Avogadro's LawAccording to Avogadro's law, the empirical relationship between the volume of agas system and moles of the gas within the system, is that as the moles of the gasincrease, the volume of the gas increases, if the temperature and the pressureremain constant. This really is just a common-sense observation. For instance, ifyou add more gas to a balloon, the balloon gets larger. You prove Avogadro'slaw every time you blow up a balloon. Figure 6-5 depicts this phenomenon in aschematic fashion.
$,nwin
gq:rri r,[j
riu Tr':
: tflnft:rn!.ss
illlT Ll ff-a Ld
'"tr ffr
,'L*PiUi]l
rl,,.1ilfg
]r rilll.lum
*.] t$de..
,;lliIriiliil
lmcSrt."llltt:r :.niLnliti,.:
'illfir,'i ir:li,ffrLut
$,rrtq,m -
5'llr: .l
ill i :Tr:
I r l,.lult
T:rr t'sft*i.i"
ripi iplll tl\ii r 1m .l
0.33 moles+added
0.33 moles_---:>added
Initialconditions1.00 molesV =22.47L
After firstaddition
1.33 molesV = 29.88 L
Figure 6-5
After secondaddition
1.67 molesY =37.35L
Example 6.9When 0.15 moles of helium gas are added to a piston containing 0.82 moles ofanother gas,by what percent does the total volume increase?
A. 6.4%B. 74.4%c. 1,8.3%
D. 25.0%
SolutionMoles of gas are being added to a closed system that can expand, so the molesincrease, and the initial intemal pressure equals the final internal pressure. Thismeans that the addition of moles of gas increases the volume, makingAvogadro's law applicable. The conclusion of Avogadro's law is that V = k.n.The conditions are isothermal and isobaric, so the following relationship applies:
v1 =v2n1 n2
In this question, the moles increase from n1 = 0.82 to n2 = 0.97. The question asksfor the percent increase in volume, which is equal to the percent increase inmoles. The math is shown below:
vl =
v2 ...Y2 =nz =y2 =0.97 ,son2 -ogZn1 n2 v1 n1 v1 0.82 n1 0.82
0.97 -0.97 +0.194 -7.764 = 1.180.82 0.82 + 0.164 0.984
Or to approximate the answer:
"'4 .97 > 1.15 = 1.15,soincrease >rs"/o .97 < 1'.00 = T.2s,soincrease <2s%
:t t-* .82 1.00 .82 0.80
The increase is between 15% and 25o/o, so the best answer is choice C.
fi
:lrrt!T
/ _ @ o @\ae @@ o"o
/u on"t^oao ^-Oo@oO"Oo- l,o o ooooo o @/
Copyright O by The Berkeley Review t4 The Berkeley Review ] t[.n;r,,:
fls General Chemistry Gases and Gas Laws Gas Laws
fa;as
re,if)'sta
Boyle's LawBoyle's law applies under isothermal conditions in a closed container. Accordingto Boyle's law, when the external pressure is increased on a gas within a flexiblecontainer, the volume decreases in a manner that is inversely proportional to thepressure increase. That is, voiume and pressure are inversely proportional to oneanother. This is true for ideal and real gases, although real gases show somedeviation as the conditions become more extreme. Once the volume stopschanging, the internal pressure again equals the external pressure. The overallresult is that as the volume of the container decreases, the pressure of the gaswithin the container increases. Boyle demonstrated the relationship between gasvolume and gas pressure with an experiment that varied the volume of a gas in aclosed, glass tube. He poured mercury into the open end of a J-shaped tube, asshown in Figure 6-6. As more liquid is added in, the height of the column ofmercury rose, so that it exerted a greater pressure on the gas in the closed end ofthe tube, causing the gas to compress. Boyle observed that it became moredifficult to compress the gas as its volume continued to decrease. This is becauseihe repulsion between molecules increases as they are packed more closely.Figure 6-6 shows three stages of the experiment.
Initial system After first addition After second addition
of
Pru,
+ pgh'
+ pgh"
Les
risn8-n.
5:
P*u,
Dt gut Pat.r,
;ks
in
gw
Figure 6-6
::ep 1: Initially, just enough of the mercury is added to isolate the two sides ofthe J-shaped tube. Mercury is ideal for this purpose, because gases aregenerally insoluble i. Hg. Initially: Pgas = Patm
::ep 2: Additional mercury is poured into the J-shaped tube. The mercurydistributes itself unequally due to gas pressure now being greater thanatmospheric pressure. After first addition: Pgas' = P26 + pgh'
More mercury is added to the ]-shaped tube. The difference between gaspressure and atmospheric pressure is even more significant. h is larger.After second addition: Pgas" - P3i6 + pgh"
.-' : hole were poked in the glass at the apex of the closed-end side of the tube- --crving the last addition, gas would escape from the hole, because the internal
::::sure exceeds the external pressure. The mercury would flow from the sideri higher nrercury to the side with less mercury, until both sides of the tube.:e leve1. This would result in the equilibrating of gas pressures.
- - rr right @ by The Berkeley Review l5 Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Gas Laws C(
An experiment you can conduct to demonstrate Boyle's law for yourself is toobserve gas bubbles as they rise through water. At lower depths, water pressureon the bubble is greater due to the force exerted by the water above. As a bubbleascends, the external pressure acting on it is reduced (because the total volume ofwater above it is steadily decreasing), and consequently the bubble expands as itrises. Consider a 1.O0-liter air-filled rubber ball that is submerged below thesurface of a lake to a depth of thirty feet. At thirty feet, the pressure due to thecolumn of water is approximately one atmosphere. With each subsequent dropof thirty feet, the total pressure on the ball increases by one atmosphere. Table6.2 shows the relationship between volume, depth, and pressure.
Volume (liters) Depth (feet) Pressure (atm)
1.00 at surface 1.00
0.50 -30 2.00
0.33 -60 3.00
0.25 -90 4.00
Table 6.2
Notice that the pressure changes by uniform increments, while the volume doesnot. This means that the volume change varies with pressure. It can beconcluded that a gas becomes less compressible as the pressure increases. Thismeans that a gas expands most when it is at low pressure. Scuba divers, forinstance, are at greatest risk for developing air embolisms during ascent, as theynear the surface. During their entire ascent, this is where the volume of gasundergoes the greatest change with changing pressure. Figure 6-7 is a graphrelating pressure and volume at constant temperature, for two distincttemperatures. The graph shows two asymptotic curves.
Both curves are isothermal, butT2 is greater than T1. The resultis that P2Y2is greater than P1V1.
Pressure (atm.)
::mflnm-uII a.fi.'."l1|l@
*.qt :
fr.:(_-
tD. :
SMfre
l}ns .
Xlmilr
P--n:"
m i.5
hsmrrtlfiuutldmdm
hm'wx
fhire ru
mrrofurdme,m
dflmffiirl#mt'w
S-&di'iArU mwmffidhmlisi'u
mm,r"mW
ffiumm
lL $-
LAc"LD" iil-
tic)
AJd.
o
Copyright @ by The Berkeley Review
Figure 6-7
The Berkeley Review Crmptiwtr
General Chemistry Gases and Gas Laws Gas Laws
toJE
rle
ofit
heherprie
Lxample 6.1.0-: the pressure of a gas in a 1.250-L container::essure if the volume of the container is:e=rperature does not change?
-\. 0.667 atrn.B. 0.726 atrn.C. 1.046 atm.D. 1.333 atm.
is initially 0.872 atrn., what is theincreased to 1.500 L, assuming
)€s
behisiorrey
tasphrct
Solution-]e conclusion one draws from Boyle's experiment is that PV = k. This means::at if the system remains closed and isothermal (constant temperature), then:- r'-1 = P2Y2. Il is given in the question that Pl is 0.872 atm., V1 is 1.250 L, and V2*. 1.300 L. The answer is found as follows:
P1v1 = P2Y2 :.P2 = Plvt - P2 =PtlYf)Y2 \V2l
:*.ause the volume increases, the pressure must decrease. This eliminates-:.,ices C and D. The initial pressure (P1) must be multiplied by a factor lessr:::, 1.0 to obtain final pressure (P2). Some calculation is necessary to decide::,1\-een ChOiCeS A and B.
v, =(0.872)(1.25) _(0.872)(s) _ (0.eX5) = 0.1s(5) = 0.75-1.5066
lle r-alue is just less than 0.75, so the best answer is choice B.
:'nample 6.11lle ','olume of a ballast bulb at sea level is 1.00 liters. If you dive 66 feet below:: surface with the bulb, and the temperature of the water surrounding you::es not change, what is the new volume of the bulb? (For every 33 feet you::-:end, water pressure increases by 1.00 atmospheres)..q- 1.00 litersE 1.00 liters
- -.10 litersL.33liters
5 r,.ution--: -ea level, air pressure on the bulb is 1.00 atm. At 66 feet below the surface, the::.'ssure is 3.00 atm., which includes 2.00 atm. for the column of water directly::---,'e the bulb and 1.00 atm. for the column of air directly above the water,lo:ru-se the pressure has tripled, the volume must decrease by a factor of three.
---. irLeans that the final volume is 0.33 liters, choice D.
lew t7 Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Gas Laws &Charles's LawCharles's law states that if the temperature of a gas is increased at constantpressure (isobaric conditions) and constant moles (closed system), then itsvolume increases proportionally. This means that volume is directlyproportional to temperature. Just as with Boyle's law, Charles's law is true forboth ideal and real gases. The deviation from ideal behavior is more substantialas the volume of the gas decreases, because the molecules interact to a greaterextent at short distances. Charles's law can be demonstrated rather easily bycarrying out an experiment with a piston holding a known quantity of gas. Apiston is chosen because the volume change is predictable and easily measured.With a piston, volume changes in only one dimension. Demonstrating Charles'slaw using a balloon has its problems, because the volume of a balloon changes inall directions, making the change in volume difficult to measure. Also, a balloonhas a different restoring force at different volumes. If the volume of the balloonsystem were to be measured using displacement of a liquid into which it is fullysubmerged, the pressure on the balloon would vary.According to Charles's law, the relationship between volume and temperature isthat as the temperature of the gas increases, the volume of the gas increases, if thesystem remains closed and the pressure remains constant. Figure 6-8 illustratesthe system experiment that demonstrates Charles's law.
"J0rffi5-6>
-hffimrlrnD"trf,n.{lt[{
Applied heat
Initial h: 6.00 cm
Initial Y:6nr2
Initial T:25'C
After expansion
New h: 6.50 cm (not 121)
New V: 6.5nr2
New T:50"C (= 323K)
Figure 6-8
It is critical that you consider the temperature in kelvins when applying Charles'slaw. The applied heat can be from many different sources. Typical sourcesinclude a heating coil (thermal energy generated by resistance of electrical flowthrough a wire), a Bunsen burner (thermal energy generated by combustion), andIR radiation (thermal energy generated by the release of an IR photon uponrelaxation). In the example in Figure 6-8, the new volume is found bymultiplying the initial volume times the ratio of the new temperature (323) to theinitial temperature (298). The factor is roughly 325 divided by 300, which isequal to 13 divided by 12. If the temperature of the gas in the piston werelowered to 0"C, then the new height of the piston would be 5.50 cm. Figure 6-9shows the graph of volume as a function of temperature over a range oftemperatures that spans the three common phases of matter. The graph ceases tobe linear at lower temperatures, because the gas undergoes a phase change tobecome a liquid, and the liquid becomes a solid. As Figure 6-9 indicates,expansion of a gas is more significant than expansion of either a liquid or a solid.When heated by small increments, liquids and solids expand only slightly.
Copyright @ by The Berkeley Review la The Berkeley Review
Iit/S General Chemistry Gases and Gas Laws Gas Laws
antitst1yfortialterby
A
ed.
)S'S
in,on
,on
llv
r1s
:he
tes
SS
CS
rdln)yhe
isre,-9
oftotorq
d.
Temperature (K)
Figure 5-9
E:.ample 6.12
' re volume of a piston filled with an inert gas is 4.31 L at25"C, then what is the-.',',- r'olume occupied by the gas, after it is heated to 50"C, assuming the system::.:eriences no net change in pressure?
{. 3.98 LB. +.31 Lf,. 1.67 LD. 3.62 L
5olution-:: conclusion of Charles's law is that V = k.T. This means that if the system:-rains closed and isobaric (constant pressure), then T2V1 = T1V2 It is given in--: question, that V1 is 4.31 L, T1 is 298K, and T2 is 323 K. The answer is found:.s :ollows:
vl =
V2 \/r - ViT2 = v, = vr(Tz\
T1 -Tr"vz- 1, - ''-"'\Tt/le;ause the temperature increases, the volume must increase. This eliminates
-:.-ices A and B. The initial volume (V1) must be multiplied by a factor greater-::r 1.0 to obtain final volume (VZ). Math is necessary to decide between choicesC and D.
Y2= (323 KX4.31 L)
298K
8.62L > 13 (4.31L) > 4.31L72
fte final volume is greater than 4.31 L, but far less than 8.62 L, so the best answer
-. choice C. Forgetting to convert degrees Celsius into kelvins would erroneously-"ad you to select choice D.
-323@srL1 =Wgsi L) = 1314.a1L)298 300 t2
f opyright @ by The Berkeley Review Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Gas System Properties fre
qt$::::Sff$ft,ffi: ffiib gS
Partial Pressure
A gas system can be composed of one pure gas or it can be composed of amixture of gases. within a mixture of gases, each component gas can be treatedseparately from other components. The particles can migrate to any positionwithin the system, so each component gas occupies the same volume. But theyare independent particles. The notion of partial pressure stems from the conceptthat you may treat different components of a gaseous mixture independently.Partial pressure is the independent pressure exerted by a gas within a mixture.For instance, each gas in an equal molar mixture of three gases has a partialpressure due to collisions of that component against the walls of the container.In that case, the partial pressure of any component equals one-third that of thetotal pressure of the system. Consider the system in Figure 6-10, where the threecomponents are theoretical gases X,Y, and Z.
rc.7)
:|ruiluu
Fllls;m'S"lidr'
:t:mnqi {ii ]tttll il
lil,,.- i1*
lht*mlltrn* :r
-.i ",JH
lhme ni.
ili@nud
ilL8ttrfiI.
ldn',ff,e[[
Uimrlftru
Is imiuru
F rm'm$umr
:ffim(mWlffindl
TM
Ar 1il-
m336. IhM' S,il
$nilii'ndit!
il:nm m,
MmeM
llil,iirme mWr
W0SF4mru
Lrnm -lihfl,rlirfinnmrmrn
momili1ifimi'
@ o o o ooo ,. ;.* o.*too
r .o "o" @o ^o-@@ co o (J o
O GasX
@ GasY
O GasZ
Figure 6-L0
The idea that the mixture is a composition of the individual gases leads to theidea that the total number of moles of gas in a mixture is the sum of theindividual moles of each separate gas. This is shown in Equation 6.7.
ntotal = I ni = na + nb + nc + ...
(where i refers to each component gas within themixture, and a, b, and c are arbitrary components)
The gases are in the same container, so they are at the same temperature and inthe same volume. Because each gas is only a fraction of the total moles, it mustalso be a fraction of the total pressure. The total pressure of gas is the sum of allthe individual partial pressures of each gas. Equation 6.8 shows this relationship,which is similar to the relationship for moles.
Ptotul = I Pi = Pu + P6 + P" + ... (5.8)
Equation 6.8 can be used to find the total pressure of a mixture of gases from thepartial pressures of its constituents. Example 6.13 illustrates how this is done.
Example 6.13What is the total pressure of a mixture created by adding 0.15 moles He(g) to a
5.0-L flask that contains 570 torr of N2(g) and 0.20 atm. Ar(g) at 100"C?
A. 1.10 atm.B. 1.33 atm.C. 1.61atm.D. 1.93 atm.
llRrWmri
u,lli$ d,ll
Copyright @ by The Berkeley Review The Berkeley Review Cm$llmu
ties General Chemistry Gases and Gas Laws Gas System Properties
ofaated:tiontheyceptrtly.hlIe.rtialner.' the:lree
the:he
5.7)
lin',is tall
:.P,
i.8)
-1e
.,4
-lr torr- : torr-: - :orr: :orr
! rlution- :.t off, everything must be converted to the same units. The units in the::-iiver choices are atm; so we should convert everything to atm. 570 torr cleanly.:rlerts to 0.75 atm; so there are 0.75 atm. NZ(S). To get the pressure of He(g),, -. must employ the formula PV = nRT.
;. = n'R'T _ (0.15x0.082)(4oo)
V 5.00
= total pressure is: P1s121 = PHe +
--5est answer is choice D.
(15)(0.082X4) . ^=
-
= t1x0.082 = 0.984atm.5
PA. + PN2 = 0.98 + 0.20 + 0.75 = 1.93 atm.
. -retermine the partial pressure of a component in the mixture, you must know. - . ::role percentage of the gas in the vessel (referred to as the mole fraction). The: .::-a1 pressure of a component is directly proportional to its mole fraction. This" . ,:::onship is shown in Figure 6-7\, for the three gases in Figure 6-10.
ny=uyntotal Ptotal Ptotal
Figure 6-11
=:r. that the pressure fraction of a component in the system equals its mole' : ::--rn, the partial pressure of one component can be determined from knowing- ,-- -.ie fraction and the total pressure of the system. Equation 6.9 below is used
. r,:,.ilate the partial pressure from the total pressure.
Pi _ ni p:_ niptotal =
4"* .'' Pi =
-#Ptotal = Xi Ptotal (6'9)
, ,ne partial pressure of an arbitrary component i, ni is the moles of an-" .:r:-r\-component i, and X1 is the mole fraction of that component.
: . - .:-on 6.9 is used to obtain a partial pressure from the total pressure. Example= '-rstrates how this is done.
:::,ple 6.14' .: -s the partial pressure due to nitrogen in a balloon at STP that contains 1.00.=s relium, 1.25 moles nitrogen, and 1.75 moles argon?
nx=P*ntotal Ptotal
rlz -ntotalPz
-.:
- -:':ation 6.9 to solve this problem, because the partial pressure of nitrogen is- : :'. multiplying the mole fraction of nitrogen by the total pressure. At STP,: ::- pressure of the system is 760 torr. The number of total moles of gas in: : .:em is 1.00 + 7.25 + 7.75, which is 4.00. The mole fraction due to nitrogen-: -:ii ided by 4, which is 31.25 7o. One-fourth of 760 is 190 torr, so the partial
. -.,.:: due to nitrogen is greater than 190. This eliminates choice A. One-third- -s 253.3 torr, so the partial pressure due to nitrogen is less than 253.3. This
: .:.:es choices C and D. The correct answer is choice B. If you want to- :::-: this by math, 31.25% of 760 = 76 + 76 + 76 + 7.6 + 7.9 = 237.5 torr.
t&' :rt O by The Berkeley Review 2l Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Gas System Properties Gr
ManometersBecause a gas has no definite shape and no definite volume, it assumes the shapeand volume of the container in which it is present. As the dimensions of thecontainer are altered, so are the properties of the gas. A gas may be compressedor expanded by moving the walls of the container. Likewise, the walls of thecontainer can be moved by adding or removing gas (think of blowing up aballoon). If the container is not flexible, then the pressure increases or decreases.Gas pressure is measured in units called torrs (named after EvangelistaTonicelli). A column of mercury is used to measure gas pressure, becausemercury has a high density and it doesn't compress easily. This means that onlya significant change in gas pressure on a column of mercury produces anoticeable and measurable difference in the height of the column.
Figure 6-12 shows measurements of gas pressure using a manometer. In eachcase, atmospheric pressure is compared to the pressure within a column. Insystems I and II, the column is evacuated, so the initial pressure inside thecolumn before mercury is added to either system is zero. Atmospheric pressureforces the mercury up into the evacuated column. Because there is lessatmospheric pressure in the mountains than at sea level, the mercury does notrise as high in column II as it does in column I. In each case, the mercury in thetube remains ai a fixed point once the force of gravity cancels the force exerted bythe atmospheric pressure. In system III, the column is open to the atmosphere, sothe pressure in the column equals the pressure exerted on the base.
Tl n
;;i;fiE
- Irlrai
i::iirt:
System IColumn evacuated
Base at sea level
System IIColumn evacuatedBase in mountains
System IIIColumn open to air
Base is anywhere
Iitilrue :iliithmillttue :aEryuli
sFiH|Iril'rili
Mr.[JLltm
rilmilm:.1
*.[|'rgruujillTr
udfffi:
TflM Ilq,iltflifimtfi
,mlttttmi -!:Jr
flNmHffin
'uunltErumm
dmdffum
Liil0MLEIUlIil
LiimmMIIlm
,Ugmmmmtl
ltm mrul
Atmosphere at sea level Atmosphere in mountains Tube open to atmospherePcolum.=0.'. Patmosp6"r"=760 Pcolrmn=0.'. Parmospl,,"r" =520 .'. Pcolumn=Patmosphere
Figure 6-12
Systems I and II of Figure 6-12 show the pressure difference between a closedsystem in the column and the atmosphere. Manometers may also be used wherethe two sides of a mercury-filled U-tube are connected to two different closedcontainers. When both columns of the manometer are exposed to differentpressures, there is a height difference between the two columns of mercury.
Copyright @ by The Berkeley Review 22 The Berkeley Keview tffiroqmr'mq
General Chemistry Gases and Gas Laws Gas System Propertiesties
lapef thessed: therpa15ES.
iistalUSe
onlyesa
:achIn
thesurelessnotthe
lbvl, so
,,",,":".-T"lPsystem < Patmosphere
alre
J-:-:re 6-13 shows a manometer under different conditions with respect to:' ::.:.r pressure. one end of the manometer is connected to a closed system.' = ,':her end is open to the atmosphere. The pressure difference between the::; -11 the system and the atmosphere can be measured. Knowing thei: -.pheric pressure allows us to determine the pressure of the system.
Psystem > Patmosphere
Figure 6-13
- , :'-::=:=:-:e n heights between the two columns can be used to measure the. " :ri--r :: -:. :iessures, if the columns have an equal bore size (internal radius).'^" .::=-::-:r'..letrveen column height and preisure can be cierived from the.i:- : :- *j: :=-aconship between gravitational fot"e and the force exerted on the* " -'- -, -ie rressure difference. The force pulling the mercury down is
"":r'' - : - =-: . The force pushing the mercury up is the pressure of the gas'- -. :" ,:': :-. --:,e cross sectional area of the mercury F = pA). As long m ih"::r:-:.: >tationary, then the two forces are equal, meaning th;t pA =- : -: ::.. :i'ae mercury in the column is the density of the irercury (p)' : :: :-" ::.- '.'olume of the mercury (V). Thus, column pressuie isr' -r- "-r : : :'. : i*ation 6.10, derived as follows:
: - tr. -
DA---: =rjo*'n = l-A = mg =pvg ... P -O{ = OSh
AP = pg& (6.10)
. :r::.:e.ce betn'een any connected columns can be determined
.: : - , '..'here Ah is the difference in height between the columns,L u[rErrrtLs ur rlctSrrl ucLwcelt ute columns/: -::.:rn.e 1n pressure between the columns. For instance, if the gas-- -'. . :o: one of two columns (we usually know that ur, op1nr :'rr-€iii Fressure exerted on it), then the pressure of the second
:.:=::r.:recl bl' adding the difference in pressure between the two
tr.:- ii ji
" 1",1 l]t ::i-
-: ,.i- --.1':. :.essure (atmospheric pressure in this case.) This is a::i I -: i-- q2c nh:co rpanlinnc anrl raen+i^-c +L-+ -^^,,.:-^ ^- :-^..!:::- ::: gas phase reactions and reactions that require an inert
- :: :> .!::ogen gas or argon gas). Manometers are an easy way* rl'll'
'.ere
::e
rederetedent
elt' !,.:ie-e-,' Review 23 Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Gas System nnoPertir
In Figure 6-1,4beIow, the application of Equation 6.1'0 to a manometer svs{en fu
demonstrated.
Pr>Pz
AP=Pr-PzAh=hz-hrlP = pg&
Figure 5-14
The pressure applied to the left side of the manometer (P1) is grelter than iltc
pr"rJrrru appliJto the right side of the manometer (P2), forcing the liquid to ri*irlgt,"r in it e column on th" right. The difference in pressure_between the tr+o
sides of the manometer is prolortional to the difference in the heights of the
liquid in each column. When the tiquid is mercury, each mm difference is
referred to as a torr. Mercury is chosen, because it is the densest liquid, so SiIr
does not readily diffuse through mercury. A closed system connected to a
mercury manometer remains a closed system'
Example 6.15
Whatls the pressure in atmospheres of a column of gas in a closed tube abole
*"..rry if tile height difference at sea level between a connected column of
*"r"rrry open to tlie atmosphere and the closed column above mercury is 317
mm?
A. 317 176gaan.
B. 443 /zeOun.C. 7601,Lnut^.
D. 317 144gann.
SolutionThe height difference of 317 mm means th{ lhe pressul€ difference is 31-7_torr'
The atriospheric pressure at sea level is 760 tirr, so the gas pressurg *-dtocolumn is i+s torriz60 - 317). The conversion from torr to atm. involves dividing
by a factor of 760 torr per atm. This ryul"t choice B the correct answer. As I
qlestlo' is worded, ih" pr"*r-rre difference is provided, but the relati
pr""rrlr", is not mentioned. The pressure could also be 1077 torr (760 + 317),
ihis value is not listed as an answer choice'
Copyright @ by The BerkeleY Review 24 The BerkeleY Keview
ES General Chemistry Gases and Gas Laws Gas Motion
is
IC
>e
'o
re
is
IS
a
e
rf
7
g
e
t
ifias ffifi,i ffrii+ iir$iiiiDrl r1i.$;111ffi,.,,,. ' ,,,:,,,,.iri.i,ii.,',,i:,,, ilil-if$$$l+tiiffi
-.as Speed and Velocity*:e .,-elocity of a particle is dependent on both its mass and the temperature. The*r:: at which a gas particle travels is directly proportional to the square root of:rre lemperature and inversely proportional to the square root of its mass. By::":iliflg the equations for kinetic energy of a gas to one another (lmvZ to |nf;,i: *' lossible to determine the root mean square velocity of a monatomic gas, bothi- = absolute manner and relative to another gas. The derivation of Equation: -l, used to calculated the root mean square speed of a particle, is shown below:
Kinetic energy = lmv2 and also = ,enf :. !mv2 = !.nf
Because lmu2 = r"*f "t = T, so prms2 = 3Tr-
Prms = /F, where Frms is the root mean square speed
l-trms = 1@vm
Lrample 6.16
"'1at is the root mean square speed of neon atoms at27"C?{" 19.3m/s3 211mls: 672m/sD. 1018 m/s
5'olutionlir-. question requires the use of Equation 6.11. Because scientists employ thej'KS system, the mass must be in terms of kilograms, and the temperature must:e rn units of kelvins. The mass of neon is 0.020 kg and the temperature is 300 K.lire solution is as follows:
(6.11)
trruVmz
(6.12),
t,
-
_ "fuRr _ _ /3(8.314X300)-:rns- V * -'V 0.020
= {G*-o4 = 6x ro2 - 600?
-he speed of neon at27'C is slightly more than 600 meters per second, so the best::swer is choice C.
-: is also possible to determine the relative velocities of two gases using the:nergy relationship. The velocity of a particle is inversely proportional to the-;quare root of its mass, so the lighter the gas, the faster it travels at a given,emperature. This leads to Graham's law for gas flow, which is listed below asEquation 6.12
v2v1
8x 9
2
Copyright @ by The Berkeley Review Exclusive MCAT Preparation25
General Chemistry Gases and Gas Laws Gas Motion
It is also possible to determine the relative speeds for particles of the same gas attwo different temperatures. The speed of a particle is directly proportional to thesquare root of the temperature, so the greater the temperature, the faster a gastravels. This is summarized in Equation 6.13
(6.13)
Example 6.17\A/hat is the speed of a gas particle at 725'C, if it has a speed of 100 rn/ s at25"C?
A. 500 m/sB. 223m/sC. L33 m/sD. 1.L4m/s
SolutionThe speed is greater at higher temperatures. This doesn't help, because all of theanswer choices are greater than 100 m/s. To determine the exact value, thetemperature must be converted to kelvins. The temperature increase is from 298to 398, which means the temperature is 1.33 times greater. Because it is a squareroot function, the speed increases by a factor of t[T33, which when multiplied by100 m/s yields a value less than 133 m/s. Only choice D is possible.
DiffusionWith diffusion, gas particles exhibit net flow from a region of higherconcentration to a region of lower concentration until the concentration isessentially uniform throughout the container. A gas can diffuse as quickly as thegas travels. This means that lighter gases diffuse faster than heavier gases.Diffusion involves flow in all directions, until the pathway is impeded by a
barrier. Unlike effusion and infusion, it involves no pores through which gases
pass. Diffusion describes the dispersion of a particular gas through a containerand is concentration-dependent, proceeding most rapidly when the mean freepath is larger and the average kinetic energy is greater.
Example 6.18Which of the following gases stinks up a room the fastest, if they are all releasedsimultaneously?
A. SO3B. C2H6SC. CaHlsSD. H5C6NCS
SolutionStinking up the room the fastest results from having the greatest rate of diffusion.The rate of diffusion depends on the speed of each gas. They are all under sameconditions of temperature, mean free path, and volume, so the only factoraffecting their speeds that differs between the answer choices is their molecularmass. The lightest gas has the greatest velocity, and thus stinks up the roomfastest. The lightest gas is C2H6S, choice B.
/=-v,, I 1.,
v1 VTr
MWffinmGmmriba
fu@miffiffiilmilF.@
dlffimdamffirtulititill[Ii
W&pmnnemmm!@
h{lUfturrffidndlr M0nffifimwd
mrg
m
Copyright @ by The Berkeley Review 26 The Berkeley Review
ron General Chemistry Gases and Gas Laws Gas Motion
sattheJ:q5*'',
,13)
.?
thethe298:are
1by
:hernis, theISCS.
ASES
inerfree
ased
s10n.
;amelctor:u1ar'oom
vrew
::--sion and Infusion- ,-.': is the process of gas escaping from the region within a container to the- :,-rment outside of the container through pores (often microscopic pores) in. : -: tairer's walls. The faster the molecules move, the more often they collide-- ::.e n'alls, and thus the more often they can pass through the pores. Ther: :rlncentrated the species, the more frequently they interact with the pores,
. :. -:- js the faster they effuse. The pore size has some effect on the effusion rate, -, rtained gas, when it is about the same size as the diameter of the gas
= -:-e. The relative effusion rates of two or more gases can be determined by*: r::ng the relative concentrations and speeds of the molecules.
- ,': is the reverse of effusion and involves a gas entering a container through: - - r€S of its walls. The faster the molecules move, the more often the
:--r-es pass through the pores and thus the faster the molecules enter. The.:' = :.
'les that apply to effusion are also valid for infusion.
::"lLe 6.19.:ie of effusion for nitrogen gas is initially 17.5 mL/mtn, what is the initial. .::usion for carbon dioxide under identical conditions?
- . rllmin: , :rl/min:
= rl/min, : rllmin
- _-n
:-::\'€ effusion rates for the two gases are equal to the relative average:- :,-l the two gases. Nitrogen is lighter than carbon dioxide, so it has a
:: ,'.-..erdg€ speed and thus rate of effusion, resulting in an effusion rate for. . : -lran 77 .5 mL / rntn This eliminates choices C and D. The effusion rates: :,.,-c gases are compared in the following way:
---^:lon Rate l\Jii.o*"r",
: -'::: of effusion for N2 is 1.25 times greater than the rate of effusion for CO2.
-: ::..ans that the rate of effusion for CO2 is 0,80 times the rate of effusion for',',::rch is (0.80)(17.5) = 14.0 mllmin, making choice A the best answer.
: ,,.-.-n and infusion are defined in terms of direction rather than concentration,., ::.e rates of both effusion and infusion are affected by concentration.
!'- --.-on and infusion can occur with or against a concentration gradient. When'. -"::Jn rate exceeds infusion rate, there is a net flow of gas out of a system.- =r infusion rate exceeds effusion rate, there is a net flow into the system.
-: ias flow speed affects the rate at which particles migrate through their-:l.:1er. Migration is a general term for motion in any direction, which takes
. -, e:count diffusion, effusion, and infusion. Diffusion, effusion, and infusion---:-t similar trends. The rate of diffusion, effusion, or infusion for any gas is
: :. --:,',' proportional to its average speed, which is inversely proportional to the: -::. root of its mass. We assume that the pore size of a gas container has an, -:- often minimal) effect on the effusion rate or infusion rate for a gas.
^l 3RTf-_ u u mN2 =
v*co2 =^[44 =\tr.sn = r.2s-'rB - 't'Jt L
V /mCOr
:rght @ by The Berkeley Review 27 Exclusive MCAT Preparation
General Chemistry Gases and Gas Laws Gas Motion
Isotopic EnrichmentIsotopic enrichment is a process by which isotopes are distributed in a non-uniform manner. Relative effusion rates can be exploited to separate isotopesfrom one another. Because the rate of effusion for a gas varies with its atomicmass, the different masses associated with different isotopes allow for isotopes tobe separated after migration through several pores. A typical example involvesenrichment of deuterium from hydrogen, as shown in Figure 6-15.
Gas mixture Gas mixtureevacuated outadded
Figure 5-15
The lightest molecule is H2, so it travels the fastest. It is in highest concentrationin cells to the right. The heaviest molecule is D2, so it travels the slowest. It is inlowest concentration in cells to the right. To obtain an enriched sample o{
deuterium, the last gas to escape from the tube is collected. Figure 6-16 showsthe distribution of speeds for a hydrogen-deuterium mixture that has equal molefractions of H2, HD, and D2.
Figure 6-16
tU)0)
)Uq)
ozlr
-otr
z
stwth
ilm
mm
fi4
ooo %^o@ o".!o.atv|
o@
a
o@
o
^a-@
e o"^Oo oa"^O@o "aoo
o'o %
";". ;-1O
o o ooo
:';'. ;o i^o
Copvright @ by The Berkeley Review The Berkeley Revieu
tion
:IOn-
)Pesrmic-.s tolves
bion
sin:of)wsrole
Passage
The btwhere P is:he tempen:he behavi:elationshil
lases to stu
'pecificalllihows the r
I. Real Qas Data
II. Kinetic Theory of Gases
III. Submerged Ball ExPeriment
IV. Charles's Law
V. Boyle's Law
VI. Lung Function
(1 -6)
(7 - 14)
(t5 - 22)
(25 - 28)
(2e - 35)
(56 - 43)
(44 - 50)
(5r - 57)
(58 - 65)
(64 - 70)
(7r - 78)
(7e - 86)
(87 - e3)
(e4 - lOO)
VII.
VIII.
IX.
X.
XI.
XII.
XIII.
Isotopic Enrichment of Uranium
Air Bag
Effusion Spheres ExPeriment
Closed-system Reactions
Weather-Sampling Balloon
Keal Qas Equation
Buoyancy and Effusion
Q,uestions Not Based on a Descriptive Passage
Cases Scoring Scale
Raw Score MCAT Score
84 - 100 15-15
66-85 to-1247 ,65 7 -9
34-46 4-6t-33 t-3
The da:elatir.elv,h attracti:s PV rs-'ride gas-
Pressur
1.00
25.4
50.0
100.0
200.0
Deviaticeme con,
lme forle 2 list(1.00 at
r.right (
Passage | (Question 1 - 6)
The behavior of an ideal gas obeys the law PV = nRT,
' :ere P is pressure, V is volume, n is the moles, and T is-: remperature (in kelvins). The equation only approximates: behavior of real gases, but well enough to predict
:, =,ionships between real gases. A researcher chooses three:::iS to study the effect of varying pressure on gas behavior,:.:iticaliy the relationship between PV and P. Figure 1
- '',,. s the results obtained in the experiment.
0.20 0.40 0.60 0.80 1.00
Pressure (atm)
-
HzG) ' Oz(g)
-
COz(g)
Figure 1
.: data show slight deviations from ideal gas behavior: ::,',elv standard pressures. Deviations are attributed to' -:.:l.tive and repulsive forces in the gas phase. Table 1
l'n' r's. P for helium gas, oxygen gas, and carbon-: :]s. respectively, obtained in a second experiment.
l'ressure PH"vH" PorVo, Pcozvcozrrl 22.44 22.39 22.26
..< 22.89 21.10 8.16
r" il 23.40 20.97 10.1 1
r r_l 24.32 20.01 11.25
-i I 26.36 20.13 13.58it r 30.27 23.01 18.23
Table 1
- : '.jirons from ideal behavior are compounded under'] - :i i , nditions, but near standard conditions, the molar.-r: r r almost any gas is approximately 22.41 liters., - .,'-s the molar volume of some common gases at
t .rm and 273.15K5.
Gas Molar Volume (L),\r 22.395Coz 22.261
Hr 22.435
He 22.436N2 22.404
\H: 22.081
O2 22.396
Table 2
* r:: I bv The Berkeley Review@
tv
fr -1/'f Which of these e
behavior as prelz- range?P
. A. Llelium only
1 . Which of the following does NOT account fordeviations from ideal behavior in a real gas?
.X: A real gas is composed of molecules that have ameasurable volume.
-8". " There are attractive forces between molecules.
--e.'ihere are repulsive forces between molecules.-.
D L Gases are not uniform, so the pressure against thewalls ofthe container does not accurately reflect thepressure of the system.
2 . What is true about the interactions between gasmolecules at higher pressures?
y6. The interactions are reduced, because the moleculescollide less frequently.
.B: The interactions are reduced, because the moleculescollide more frequently.
-gatne interactions are increased, because themolecules collide less frequently.
,'DlThe interactions are increased, because the\€ molecules coltide more frequently.
aN r- ,JUWhich of these experimental gases shows no change inbehavior as pressure is increased across an extreme
f-*l -{ &-Lle :-+ t#o w rt q. *i { 'r* -"-
tF-Bl--.Oxvsen onlv_j)' ""ta-" ""'r a tC. Both oxygen and helium
FBoth oxygen and carbon dioxide
4. How does the molar volume of ammonia gas compareto that of an ideal gas?
,"'A: lt is larger than ideal, due to attractive forces.
.-8". It is larger than ideal, due to repulsive forces.
ff) lt is smaller than ideal, due to attractive forces..p-< It is smaller than ideal, due to repulsive forces.
y. What is the expected' gas at 75 atm.?
Ql ;-"i;' 'o.oe7 riters
B. 0.142 liters.-Ci 10.63liters{,' n.zztit"r,
molar volume of carbon dioxide
F(i -' v' re -tr
1p "t\-)
Under which of the following conditions does a real gas
behave MOST like an ideal gas?
-#F gntemperature and high pressure
G-. Higtr temperature and low pressure
..e'. Low temperature and high pressure
..4- Low temperature and low pressure
5l
6.
GO ON TO THE NEXT PAGE
Passage ll (Questions 7 - 14)
The kinetic molecular theory of gases is employed to
explain the behavior of an ideal gas. It is, in essence, the
theoretical perspective of an ideal gas on the microscopic
level. The postulates of the kinetic molecular theory of gases
as they relate to the particles of an ideal gas are as follows:
1. The particles are so small, compared to the distances
between them, that the volume of the individualparticles can be assumed to be negligible.
2. The particles are in constant random motion. The
collisions of the particles with the walls of the
container are responsible for the pressure exerted by
the gas.
3. The particles are assumed to exert no force upon
each other; they are assumed neither to attract nor
repel one another.
4. The average kinetic energy of the particles is
assumed to be directly proportional to the
temperature of the gas in units of kelvins'
The kinetic theory of gases serves to explain temperature
and pressure on the microscopic level. While it does not hold
true for real gases, it is a good model for an ideal gas. Real
gases exert force upon one another, and their particles have a
finite volume.
As the properties of an ideal gas change, the particles are
assumed to remain uniform. For instance, the average kinetic
energy of a gas increases as it is heated, but the size of each
atom remains the same and the bonds remain in place. This
is to say that the particles are not altered by physical changes.
7 . For a closed system in a 1.00-L sealed piston, what is
observed when the temperature of the system in"."3lll "
-A . The number of molecules of gas increasat. /. .. ,.
-Y..a-\he average kinetic energy remains constant. ":...,f. Cu The mean free path increases.
-rD. The average molecular momentum remarns
constant.
PJ t t'-r i'''"1
8. For an inert gas system in a sealed rigid glass container,
what occurs when the average kinetic energy ofmolecules increases?
9.<' Collision frequency increases, while collision force' decreases.
B<-. Collision force increases, while collision frequency
decreases.
El"Bothmean free path and collision force increase.
Zf)F,otn collision frequency and collision force(-'/rncrease.
. /| ,t 1 *"..--* * ,. \.,nttt- j \- )
Copyright @ by The Berkeley Review@
Number of particles
9. As temperature of a gas system in a closed flas
decreases from 20"C to 10'C, the average speed of the
component gas molecules:
7dl' increases slightly.
-ffiemains the same.
--;Q - -&creases slightly.
D. decreases substantiallY.
10. What is observed over time if a mixture of H2 and
initially at 50Vo by moles H2, undergoes effusio
fr[tr !A.Tl-n
d.:.*gtft
D-td
through a small pore in the container?
A . PH2 decreases; Xp, decreases.
B ._ PHz increases; Xp, decreases./)(S/ Pordecreases: XD, increases.
1
D. PD2 increases; Xp, increases. \ ,- -)t J
j'*- lo{o. .i_r-. 'ta,.1 ', ...*t -.. ,"'-..,'-.^/ |
e6rm
t"TTtn
t:,
11.under identical conditions
A. Helium
--B: Neon"€. Nitrogen
{ft Sulturdioxidet\
'e* s'r
12. Which of the following graphs best describes
collision frequency as the number of particles in a
B.
ho
Number of partic
Number of
Wfrictt of the followingmomentum, if they are all
temperature and pressure?
gases has the GREA
UoU
I
i
\
irli
o
oO
.tfQU:*'" Ll\ \'"r Y
r /EPz" ,- -gcooNTorHENExrI ,._. ,.. e.,-a \
(,),- ,-ir'l
Number of particles
32 PA
I flask i -:.
i of the
Trhich of the following is an expected behavior of a; . s. according to the kinetic molecular theory of gases?
{, The gas has an infinitely high boiling point.ts. \folecular gases show different behavior than
atomic gases.
r- . The average velocity of a particle doubles when ,
temperaturedoubles. :I v J *.*..'" 31.-,..,..:
LI . The pressure exerted by a gas is independent of the
=hape of its container. : ri,:i 1
"ili: ,:;. -\-z tr'v A= \v-
. : .s differs from a liquid in all of the following ways:
"rlEPT:-+: . liquid has a definite volume while a gas does not.
!. ::rticles in a liquid are always in contact with one.rother, while particles in a gas are never in contact.; rth one another.
:** . _:as is more compressible than a liquid.ir' .: object is more buoyant in a gas than a liquid.
r;(l,V
. t]
5l/ II
urd D2.
ffusion
[;
The Berkeley Review@ 33 GO ON TO THE NEXT PAGE
Passage lll (Questions 15 - 22)
A researcher sets up a salt water tank (see Figure l,below) with a depth of fifty feet to study Boyle's Law. Anair-filled rubber ball is fitted with a thin metal ring around itscircumference, positioned slightly on the lower side of theball. The ball is placed in the tank of salt water, which ismaintained at25"C. At the base of the tank is an adjustablemagnetic field that is used to submerge the ball to selecteddepths. The displacement of the water in the tank ismeasured to determine the volume of the ball.
Floating on the surface, the ball is 50Vo submergedbefore the magnetic field is applied. The initial volume ofthe ball is 36.00 liters, but as it descends towards the bottomof the tank, its volume gradually decreases. The volume ofthe elastic ball is measured at the several depths. Table 1
lists the values for the ball's changing gas volume.
Depth (ft) Pressure (atm) Volume (L)
5 1.16 31.14
10 1.31 21.41
15 1.41 24.43
20 1.63 22.01
25 1.78 20.21
30 t.94 18.54
35 2.09 r1.2040 2.25 16.02
45 2.41 14.91
50 2.56 t4.ltTable 1
Note: A pressure of 1.00 atm. is required to raise a
column of pure water 32.6 ft.
1 5. The greatest change in volume is experienced by the ,
ball during which segment of the descent? , I --, L*A ., From 0 to 7 feet below the surFace "
: : v
. Bl From I to 14 feet below the surface
_-r€. ,From 14 to 2l feet below the surface
,",D . 'From 21 to 28 feet below the surface
/ te . lt what depth will the ball neither rise nor sink if the
magnetic field is turned off?
A . Between 0 and 10 feet below the surface
B. Between l0 and 20 feet below the surface
C . Between 20 and 30 feet below the surface
4!,'Between 30 and 40 feet below the surface
17 . If fresh water were used rather than salt water, how
would the results be affected?
-.N{Tn" change in volume of the ball would be greater
when submerged in salt water than fresh water'
-, because salt water is less dense than fresh water'
,nf. fne change in volume of the ball would be less
when submerged in salt water than fresh water,
...,-.because salt water is less dense than fresh water.
C .' The change in volume of the ball would be greater
when submerged in salt water than fresh water,because salt water is denser than fresh water.
,t)\/ )1. fot a ball that is initially 607o submerged, what is
1 observed when compared to the ball in Figure 1, under
the same experimental conditions?
A. The depth at which the density of the ball equals
the density of surrounding water is below -32 feet,
because the ball is less dense than the original ball
^ in the expcriment.
, b./ fn. depth at which the density of the ball equals\-'/ the density of surrounding water is above -32 feet,
because the ball is less dense than the original ballin the experiment.The depth at which the density of the ball equals
the density of surrounding water is below -32 feet,
because the ball is denser than the original ball inthe experiment.
The depth at which the density of the ball equals
the density of surrounding water is above -32 feet,
because the ball is denser than the original ball inthe experiment.
4
,d . tt't" change in volume of the ball would be less
when submerged in salt water than fresh water,hecause salt water is denser than fresh water'
18. For a ball that is filled with an ideal gas and immersed
in fresh water, and that is initially 25Vo submerged
before the magnetic field is applied, at what depth willthe ball no longer be buoYant?
A . At depths between 0 and32 feet below the surface
B . At depths between 32 and 64 feetbelow the surface
C . At depths between 64 and 96 feet below the surface
D. At depths greater than 96 feet below the surface
Copyright O by The Berkeley Review@ 34 GO ON TO THE NEXT PAG
,--/2 0. What has no direct effect on the volume of the ball?
t,4' 4, The external pressure.' 6. The moles of gas in the ball
.*eJ The temperature of the water
D., The magnetic field strength
21. It more salt were added to the water in the tank
Figure 1, what would be observed?
,"f) tne ball would rise, and a stronger magnetic,i---' field would be required to submerge the ball bel
the surface.
B. The ball would rise, and a weaker magnetic, B, fie
would be required to submerge the ball belowsurface.
d'The ball would sink, and a stronger magnetic,lield would be required to submerge the ball
-fhe surface.
O. ttre ball would sink, and a weaker magnetic'field would be required to submerge the ball bel
the surface.
CI
2,fu me compressibility of a gas is GREATEST when:
'? A. the moles are high and the volume is low'
" 8.. the moles are high and the volume is high'
:&.rthe moles are low and the volume is high'
LDr'th" moles are low and the volume is low.
Chdqma qln
F0tmf,
il-rpffiimil
ffitu
fuoffis
ru!um
ball?
e tank in
;netic, B.,a11 below
;, B, fieldrelow thr
lnetic, B';
'a11 belowll
hen:
Pa,ssage lV (Questions 23 - 28)
,-arles's law states that when the temperature of a gas;r --,' r closed system is increased, the volume of the gas:: -":::es proportionally, if the moles of gas and pressure"l'- - . Jonstant. To demonstrate this law, a student sets outiL ::.r\ e the change in volume of a gas within a cylinder.l-i: -',.rnder contains a heating coil that can be set byiilrLr - ,-.: the voltage. As the voltage is increased, the cunent:'r', -il :he wire increases, causing more heat to be releasedi' - ,,-: coil. The change in volume of the piston isIi ; - -::l by the change in its height. Table I shows the data,ril : - :: tV the student for an unknown gas.
Temp lleight Volume10'c 6.00 cm 50.0t2 cmJ20'c 6.21 cm 55.J'[ cmJ30"c 6.42 cm 60.6n cmj40'c 6.64 cm 65 9n cmJ50'c 6.85 cm '11.2n cm360'c 7.06 cm 16 5n cmJ70'c '7.27 cm 81 .8n cmJ80'c 7.48 cm 8'/.ln cmJ90'c 7.70 cm 92.4n cm3100'c 7.91 cm 97.]n cm'110"C 8.12 cm 103.0n cmj120'c 8.33 cm 208.3n cmr
Table 1
': r::r from the experiment show little deviation from; ' :,,rr. Equation 1 below represents Charles's law:
Vinitiul =
Vfinal
Tinitiul Tfinat
Equation 1
"' ,r -.:ne can be expressed in any metric units, but theri'--i.-:: nust be measured in kelvins,
| ,-.,.'. .?
' !. '" ..,
;l'"i"-r r! rhe volume of the unknown gas at 150'C?1 r'i ' + ir- - : YIICm' V<:u-L4
. ? -t , '(.j, ___ l;i cmJ .. ,
^)--: -: cmJ {. i '"1i,;: '' {..
i, '.:. : .:e volume of a heated gas in a closed, n#G,r " ri l-'C. if it had an initial volume of 7.0 liters at
-" : ii the heating of the gas was carried out under" ;: :,rnditions.
] .::CIS
)l , - ',:ers' nt at: ).,.-rs \ll. -l\_.4 \ aiP{ :,,- The Berkeley Review@ cit
2 5. As you drive, what happens to the pressure in your cartire?
.-#It decreases, because your tire cools down andpressure decreases with decreasing temperature.
-Wlt increases, because your tire cools down andpressure increases with decreasing temperature.
' C . It decreases, because your tire heats up and pressure
decreases with increasing temperature.
Cn, :It increases, because your tire heats up and pressureincreases with increasing temperature.
1', ',
a-4 ) ..i; )
,r ,
t
2 6. Charles's law states which of the following?
.ffiTconstant volume, as the pressure of a gas
increases, the temperature decreases.
-B-'-Al constant volume, as the temperature of a gas
' increases. the pressure increases.
.-t:*it constant temperature, as the pressure of a gasincreases. the volume decreases.
(''D)At constant pressure, as the temperature of a gasincreases, the volume increases.
-{r)r':.^(a\
. .1. " i*- -'f-
27 . How can it be explained that the volume of the particlesin the experiment at -15"C is a value less than 20 cm3?
/. ldletd behavior applies only to gases between 0"Cand 100"C.
The gas condenses into a
slightly greater than -15"C.
The gas condenses into a
slightly Iess than -15'C.
liquid at a temperature
liquid at a temperature
When the temperature of a gas is very low, itsmolecules get small enough to effuse out of thecontainer.
28. In a system of three gas components (water, oxygen,and nitrogen), where water has a partial pressure of 221ton' and the percentage of oxygen gas is 20.9Va, what isthe partial pressure due to the third component(nitrogen), if the total pressure is I 62 torr?
! --l-t-l c,
B.
C.
A;'159.3 ton
'' 8. 444.4 torr(
(C) sso.o torr
D; 719.0 torrl*.,
1 i' -..| ': r-
I L i <; a-)I \ ',--' -*
GO ON TO THE NEXT PAGE
Passage V (Questions 29 - 35)
Boyle's law is based on the observation of a closedgaseous system at constant temperature. The conclusion isthat when the system is closed so that moles of gas are heldconstant, the pressure varies inversely with the volume, if thetemperature remains constant. Boyle's law is stated below:
As the pressure exerted on a gas in a closed system isincreased, the volume of the system decreases in a linearfashion, if the temperature and moles of gas are heldconstant.
Summary: P f : V I with ng and T.
To prove this to be true, a researcher directs a constantstream of argon gas into a bent glass tube sealed at one end.
The argon stream displaces the air in the tube, because argonis heavier than air. The system remains relatively pure inargon even though the tube is open at the other end to the air.The researcher then pours enough mercury into the tubebarely to fill the bend in the tube. Once the mercury reachesa level where the gas in the left side of the tube is isolated,the height of the gas is measured to be 10 cm in the left sideof the tube. The researcher next adds enough mercury to theopen end of the tube to create a difference of 76 cm betweenthe heights of the columns on the left and right side (this isthe first addition of mercury to the tube). The height of thegas in the left side is then measured to be 5 cm. Mercury isagain added to the right side, so that the difference in heightsis now 152 cm. The height of the gas in the left side of thetube is found to be 3.3 cm. The three stages are drawn inFigure 1 below:
Initial
Figure 1
Every 760 mm of height in a mercury column exerts a
gravitational force that when applied over a unit area equals1.00 atm of pressure. This is independent of the radius of thecolumn. If a less dense liquid is used, a greater height in thecolumn is required to exert 1.00 atm of pressure.
Copyright @ by The Berkeley Review@
{10.0 cm
f
-78"C 0'c 59"C
P (atm) V (L) P (atm) V (L) P (atm) V (L)
0.25 2.029 0.25 2.841 0.25 3.455
0.50 1.o14 0.50 r.420 0.50 1.127
0.7 5 0.6788 o.15 o.9466 o.75 1.139
1.00 0.5068 1.00 0.7095 1.00 0.8628
2.O0 0.2532 2.00 0.3545 2.00 0.4311
3.00 0.1685 3.00 0.2359 3.00 0.2869
4.00 0.1264 4.00 0.1769 4.00 0.2151
5.00 0.1010 5.00 0.14r4 5.00 0.1120
In a second experiment, the researcher adds 1.000 gramsofpure oxygen gas to a cylinder that can expand. The systemis a closed system and the temperature is held constant bysubmerging the unit into a oil bath. Changes in the volumeof the cylinder are determined by observing the level of a
plunger connected to the top wall of the cylinder. Thevolume of the cylinder is proportional to its height, so thevolume can be determined from the height of the plungeraccordingtoV=fir2h.
Table I shows the data collected at -78'C, 0'C, and59'C. Figure 2 is a graphical representation of the data listedin Table 1.
Table 1
190 380 510
Pressure (torr)
Figure 2
29. According to the graph in Figure 2, what can be
about the volume of 1.000 gram of oxygen gas at
temperature and 0.50 atm. of pressure?
A. The volume is less than l.0l liters.
B. The volume is greater than 1.01 liters, butthan 1.42 liters.
C . The volume is greater than 1.42 liters, butthan 1.71 liters.
D. The volume is greater than 1.71 liters.
It- ln'mJI
D.
ffinrdffiudim;
ilr-l,G*l-
!o
oE
o
36 GO ON TO THE NEXT PA
C
)0 grams ie syster.rstant br
: r'olume
:r'el of a
er. Th:)t. so thrr plunger
D'C, and
ata listed
v (L)
3.455
T.J2 ,1
1 .139
0.8628
0.4311
0.2869
0.215.
0.17:i
'ilil
,iliilllllul it" ' :ilr It _
:ullt ulilt i
';-':at would be observed if a liquid less dense than-;iJufy were used in the J-tube in the first experiment?
r" . The argon gas in the left side of the tube wouldrompress less, and thus the height of the column of:es would be greater than what was observed with:rercury in the manometer.
!, lhe argon gas in the left side of the tube would: rmpress more, and thus the height of the column,: eas would be less than what was observed withrL3rcury in the manometer.
- l-: argon gas in the left side of the tube would. :lpress the same amount as with mercury, and-.i rhe height of the column of gas would be the:::3 as what was observed with mercury in the
- ::ametef., -": rrson gas in the left side of the tube would
. r:ress more, and thus the height of the column' :,. r-,ould be greater than what was observed
.- :r-rcury in the manometer.
. : -, --:ie up of molecules rather than atoms werer' . - :: -,rr argon gas in the first experiment, whatli1 : -r :sert'ed for the heights of the gas in the leftr iir .-. . -b,e in each of the trials?
*- . i3 mm, 35 mmll * . 53 mm,31 mm
-- J mm,35mm- : -l- mm, 31 mm
-r
::: I-rndings of the first experiment, one. ::;.sure when applied to an open column
." ,. j :aise the height of the mercury in the- :. :lltrUlit']
. --1-
P. -.suaaApplvPressure
ri[ii,,,
lillln
llllrui
llillil rnllll -:,:-.
: - - , :'= ::. = -:',- Rgf igw@tTP GO ON TO THE NEXT PAGEcl
3 3 . What would the volume of 2.000 grams of oxygen gas
be at 0'C, if the pressure were 1.00 atmosphere?
A. 0.71 liters
B. 1.00 liters
C. 1.42 liters
D. 2.84 liters
3 4 . If the temperature of a gas enclosed in a flexiblecontainer were increased from 25"C to 50'C while thepressure on the system were increased from 0.5 atm. to1.0 atm., then what would happen to the volume of thesystem?
A. The volume would increase, because the effect ofincreasing pressure is greater than the effect ofincreasing temperature in this example.
B. The volume would increase, because the effect ofincreasing temperature is greater than the effect ofincreasing pressure in this example.
C . The volume would decrease, because the effect ofincreasing pressure is greater than the effect ofincreasing temperature in this example.
D. The volume would decrease, because the effect ofincreasing temperature is greater than the effect ofincreasing pressure in this example.
3 5. When mercury is added to a J-tube containing argon gas
in uniform 100.0-mL increments, the change in volumedecreases with each subsequent addition. How is thisbest explained?
A . The argon gas becomes more dense with each
addition of mercury, so the particles become harder
to compress.
B. The argon gas becomes less dense with each
addition of mercury, so the particles become harder
to compress.
C. The mercury becomes more dense with each
addition, so the pressure exerted by mercuryincreases with each addition.
D. The mercury becomes less dense with each addition,so the pressure exerted by mercury increases witheach addition.
Passage Vl (Questions 36 - 43)
The circulation of vital gases through the body starts in
the lung, where large volumes of gas are exchanged between
the body and the atmosphere. The volume of the lung can be
broken down into four distinct volumes and dead space.
There is the tidat volume, which is the volume of gas
inspired or expired in a normal breath. There is the
inspiratory reserve volume, which is the volume that can be
inspired beyond the tidal volume, invoked during periods ofexercise and strained breathing. There is rhe expiratoryreserve volume, which is the volume of air that can be
expired after the expiration of the tidal volume. There is the
residual volume, which is the volume of gas that remains in
the lungs after maximal expiration. Finally, there is dead
space. There is the physiotogical dead space, which is the
volume of the lung that does not eliminate carbon dioxide,
and there is anatomic dead space, which is the volume of the
conducting airways (about 150 mL in the average adult).
The breathing cycle is regulated by changes in the
internal pressure of the lung. Operation of the lung can be
explained by the ideal gas law. The pressure change is caused
by expansion and relaxation of the thoracic cavity. At rest'
alveolar pressure is equal to atmospheric pressure. As defined
in physiology, this is zero pressure. The interpleuralpressure is negative, due to the tendency for the lung tocollapse and chest cavity to expand. The onset of normal
inspiration is a result of contraction of the inspiratory
muscles, in particular the diaphragm. When the diaphragm
contracts, the abdominal contents are forced downward, and
the thorax expands. This increases the volume of the 1ung,
making the alveolar pressure negative (sub-atmospheric).
Expiration is normally a passive process, resulting from
the relaxation of inspiratory muscles back to rest. The lung
and chest system is elastic, so after active inspiration, where
it is expanded from its resting state, the system relaxes back
to its resting state. This decreases lung volume, making the
alveolar pressure positive (super-atmospheric). Air flows out
of the lung as a result of this pressure difference. The
amount of gas exchanged in one cycle of normal breathing is
the tidal volume. As some of the definitions imply, there is
also facilitated breathing, where inspiration is greater than
normal and expiration is an active process.
3 6. Inspiration into the lung is BEST explained by:
A. Archimedes' PrinciPle.B. Bernoulli's princiPle.
C. Boyle's law.
D. Charles's law.
3 7. What is true during exPiration?
A . Lung pressure is less than atmospheric pressure.
B. Lung pressure is greater than atmospheric pressure'
C . The thoracic cavity is contracted more than normal'
D. Lung volume is at its smallest point.
3 8. At a low altitude with a low ambient temperature,
is true of atmospheric air?
A. It has a lower partial pressure of O2(g), but
same mole fraction of OZ(g) as normal air.
B. It has a higher partial pressure of O2(g), but
same mole fraction of OZG) as normal air.
C . It has both a lower partial pressure and lowerfraction of O2(g) as normal air.
D. It has both a higher partial pressure and
mole fraction of OZ(g) as normal air.
3 9 . What is observed during normal expiration?
I. Total lung volume is less than tidal volume.
II. Internal pressure exceeds external pressure.
m. Moles of oxygen exceed moles of carbon dioxide'
A. I only
B. II only
C. I andll only'D . II and III only
4 0. For lungs with a tidal volume of 400 mL and a
volume following normal expiration of 1200 mL,
does internal pressure change to cause inspiration?
A. Itincreasesby33Vo.B. It increasesby 25Vo.
C. It decreasesby 25Vo.
D. Itdecreasesby33Vo.
41. What is NOT true as the diaphragm contracts?
A. Internal pressure is less than external pressure'
B. Partial pressure ofcarbon dioxide decreases.
C . Lung volume is less than it is at rest'
D. Air temperature in the lungs remains relativconstant.
42. Why do scuba divers experience different breathi
patterns underwater than on land?
A . Air is more concentrated, so they breathe more 01
B. Air underwater behaves more like an ideal gas'
C . Air viscosity is greatly reduced underwater.
D . Air in the anatomic dead space has a reduced parti
pressure of CO2 underwater.
4 3. The air flow associated with the lung moves:
A . at a decreasing rate during a notmal breath.
B. at an increasing rate during a normal breath'
C . at a constant rate during a norrnal breath.
D. in a sinusoidal fashion during a notmal breath'
Milffi x[
'"' @
qfu.
ffihmc
lt-'
M
[,r
ffiiffilllt
dr-
D"
Copyright @ by The Berkeley Review@ 3a GO ON TO THE NEXT PA
ro-
fe, wha ;:ssage Vll (Questions 44 - 50)
but th, \uclear reactors requiring uranium use the isotope 235U
- --:1. Naturally occurring uranium is only 0.l2To 235tJ,
L.-1 r,-_ 'ne majority of the uranium being 11-t" 238lJ isotope.Dut tn'
- ::actor to run efficiently, the uranium should be at least
_ -::9. To enrich the sample with 235U, the uranium iser mol, - .ed to the hexafluoro species (UFO), a white solid at
' .emperature. The uranium hexafluoride compound isihighe; :r io its sublimation point, and the gas is all,owed to
, :: :hrough tiny pores between a series of connected' - -3rs. With each effusion, the vapor becomes enriched'-' l35g species, because the lighter compound effuses
" .:cording to Graham's law. The relative rates of. ,nd 238UF6 are detemined as fbllows:
= I masszrs, ,r_ I '-= 4/' - "''",n0 =qf 3)2 =y'|.00859 = 1.004
'rrde. V mass2r5l rFb V :+S
Figure 1
: -: 'use the average speed of the U-235 species js 0.4Vo: - i.ln the U-238 species, the effusion rate of the U-235, .: 0.1Vo faster. This is assuming that the diameters, - "L F6 and 238UF6 molecules are equivalent, which is
a tota -;. ruse the only difference between the two isotopes ofI, ho.; : .- .: three neutrons. Only charged particles affect ther :n atom. By carrying out successive effusions, the
-.:. the UF6 is converted back to uranium metal.
d -.: effect would the diameter of the molecule of a gas, - ,rn its effusion rate?
, The greater the diameter, the slower the effusion:"re, because the molecule is hindered when passing:rough the pore.'e : - he greater the diameter, the slower the effusion:.ie. because the molecule collides with the pore--,tre often.
atr ve.,- le greater the diameter, the faster the effusion rate,::Jause the molecule is hindered when passing-.:'ru_eh the pore.
- -e greater the diameter, the faster the effusion rate,-:Jause the molecule collides with the pore more
..an.
--:n be said about the reactivity of fluorine gas. n U-235 and U-238?
-orine gas reacts more readily with U-238,- : - fuse it is heavier.
= -,rrine gas reacts more readily with U-235,-:::use it is lighter.
-.riine gas reacts with U-235 and U-238 equally,-:::use the extra neutrons do not significantly- --, the chemical reactiviry.
-':rne gas reacts with U-235 and U-238 equally,- : -:.rse the extra protons do not significantly affect-. -'hemical reactivity.
rathinl
,re o:*i, n
. partJLi
h.
)-\Gn I bi The Berkeley Review@ GO ON TO THE NEXT PAGE
46. Assuming that the molecular diameters areinsignificant, which of the following compounds wouldeffuse roughly twice as fast as UF6?
A. SO2
B. SO3
C. PF3
D. Xe
All of the following affect the effusion rate of a gasEXCEPT:
A . the size of the molecule relative to the pore.
B . the temperature of the gas.
C . the presence of isotopes in the compounds.
D. the concentration of the species outside of thecontainer.
What is the mass percent of uranium-235 in 23sgp6r
A . Less than 50Va
B. Between 50Vo and 60Vo
C . Between 60Vo and'|}Vo
D . Greater than 707c
Had chlorine been used instead of fluorine to formUCl6, what differences would have been observed?
A . The difference in effusion rates for the two uraniumisotopes would be greater than 0.4Vo, so feworfiltering chambers (pores) would be needed to enrichthe sample.
B. The difference in effusion rates for the two uraniumisotopes would be less than 0.4Vo, so fewerfiltering chambers (pores) would be needed to enrichthe sample.
C. The difference in effusion rates for the two uraniumisotopes would be greater Ihan 0.4V0, so morefiltering chambers (pores) would be needed to enrichthe sample.
D . The difference in effusion rates for the two uraniumisotopes would be less than 0.4Vo, so more filteringchambers (pores) would be needed to enrich thesample.
Which of the following pairs of gas molecules would bethe MOST difficult to separate using the successiveelfusjon technique?
A. OZ and N2
B. Hz and Ne
C. CO2 and SO2
D. CO andC2H4
47.
48.
49.
50.
Passage Vlll (Questions 51 - 57)
Air safety bags are standard features on new cars. These
bags function through rapid gas generation. Because they are
safety features, air bags must use a non-toxic gas that does
not oxidize readily. Nitrogen gas is ideal because of its inertchemical behavior. The Reaction 1 is employed by airbagsto produce nitrogen gas (N2) rapidly.
2NaN3(s) + 2 Na(l) + 3 N2(g)
Reaction 1
Sodium azide decomposes quickly to fill a 60.0 liter airbag in approximately 20 milliseconds. To proceed at this
rate, the reaction must be run at 350"C. To maintain this
temperature, Reaction 2 is employed.
6Na(l) +FezOr(s) + 3 Na2O(s) + 2Fe(s)
Reaction 2
Once the air bag is filled, it must deflate rapidly. To
accommodate this, there are small pores in the material to
allow the gas to effuse out. If the bag remained inflated,vision would be hindered. The rate ofeffusion depends on the
average speed ofthe molecules. The root mean square speed,
Frms, of a gas is calculated using Equation 1.
Frms = 1EvmEquation 1
where k is Boltzmann's constant, T is for temperature(measured in kelvins), and m is for mass (measured in kg).
5 1 . What can be said about the thermodynamics of Reaction
1 and Reaction 2?
A. Entropy increases in Reaction 1; Reaction 2 isendotherrnic.
B. Entropy increases in Reaction 1; Reaction 2 isexothermic.
C. Entropy decreases in Reaction 1; Reaction 2 isendothermic.
D. Entropy decreases in Reaction 1; Reaction 2 isexothermic.
52. As the nitrogen gas cools from 350'C to roomtemperature:
A . its pressure increases, and the kinetic energy of the
gas increases.
B. its pressure increases, and the kinetic energy of the
gas decreases.
C . its pressure decreases, and the kinetic energy of the
gas increases.
D . its pressure decreases, and the kinetic energy of the
gas decreases.
Copyright @ by The Berkeley Review@ GO ON TO THE NEXT PAG
5 3. An air safety bag fills at a rate of 100 moles/sec. Af10 milliseconds, what is the volume of the gas in t
bag at 31'C? (At 31'C, gas is 25 L/mole)
A. 2.5 liters
B . 22.4litersC . 25 liters
D. 250 liters
54. If the average speed of a gas at 25'C is exactlym/sec, then at 125'C the approximate average
would be which of the following values?
A. 115 m/sec
B . 133 m/sec
C . 224 tnlsec
D. 500 m/sec
55. If an air bag were filled with argon gas, helium gas,
nitrogen gas, which of the following relationsaccurately describes the relative rates of effusion forgas from the air bag?
A. He>N2>ArB. Nz>He>ArC. He>Ar>NzD. Ar>N2>He
56. In a mixture of 50 grams nitrogen gas with 50 gr
carbon dioxide at STP, the partial pressure attributedcarbon dioxide would be which of the following values
A. 50 torrR. 296 torr
C . 380 torrD. 464 torr
5 7. What can be said about the average speed of neon gas
25"C compared to argon gas at 25"C?
(Ne has mass = 20 g/mole)
(Ar has mass = 40 g/mole)
A. Ne gas is twice as fast as Ar gas.
B. Ne gas is 1.4 times as fast as Ar gas.
C . Ar gas is 1.4 times as fast as Ne gas.
D. Ar gas is twice as fast as Ne gas.
milffimr
tu${fr
,dlffi
100 ,il\mNE
MMmi
tuiiffiffi
40
)'1CS
lX (Questions 58 - 63)
lr*ii. : ir is defined as the flow of a gas from the inside ofr r' --.iL"rE: :'r the outside of a container through a pore in the
' .:: :ontainer. The size of the pore can range fromri" ,.,:f -: ibarely greater then the diameter of the gas
, l:r:, :-, :racroscopic (as is the case with a valve). The
,riL i, T,:: a gas effuses depends on the velocity of the gas'
,rrt i::. of the pores relative to the surface area of the-:i- .r.l Ihe concantration of the gas in the contarner'
','r,i:::her sets out to study effusion using inert gases,
riiLlL r ::rolecular forces are minimal. The apparatus,, - lnequal spheres connected by a tube with a
r : The left sphere (Region I) has a radius' 11,*r -::: sphere (Region II) has a radius, 12. Region II
ri uti {r il1{t , :,lume of Region I, as shown in Figure 1
Figure L
,ii1,,;pr 1 I .. -ritially charged with equal moles of helium,
inr,: ::-ion. aI 25'C. Region II is evacuated. The,li : .s set to allow gas exchange. The vessels are
T{: ,1er that is essentially pore-free for the lifetime'tlil1tttrllr i;fld-l:':.-rt. Figure 2 shows the pressure of each gas
lLlfillrtllirrql'r : 3r itme.
hilttt
Time (seconds)
Figure 2
;lti i'lq;p; 111 :--: lnce equilibrium has been established?
q -T,: --:le fraction of helium is lowest in Region I.
[ *ie ::,.: liaction of argon is greatest in Region II.I -re :.:hs of nitrogen are greatest in Region I.
I -ru *,:.es of helium are greatest in Region II.
gas.
ton
lor
i0
ibutoi: r'al
t
let valve I Inlet valve
Gas exchange valve
rgrln , :-, The Berkeley Review@ GO ON TO THE NEXT PAGE4t
5 9. What occurs in Region II, once the gas exchange valve
is opened?
A. Helium partial pressure grows rapidly, reaching the
greatest partial pressure once at equilibrium.
B. Helium partial pressure grows rapidly, reaching the
smallest partial pressure once at equilibrium.
C. Argon partial pressure grows slowly, but reaches
the same partial pressure as helium once at
equilibrium.D. Argon partial pressure grows rapidly, but reaches
the same partial pressure as helium once at
equilibrium.
6 0 . Which of the following statements must be TRUE with
regard to the system shown in Figure 1?
I. The final pressure in Region II is three times as
great as in Region I.II. At equilibrium, the mole fraction of helium is the
same in Region I as the mole fraction of argon inRegion II.
ru. Increasing the temperature would increase the rate
of effusion and increase the equilibrium partialpressures, but it would not change the molefractions at equilibrium.
A. II onlyB. III only
C . I and II only
D . II and III only
61 . All of the following can increase the effusion rate
EXCEPT:
A. opening the valve to a greater area.
B . increasing the temperature of the system.
C . increasing the volume of the system.
D. increasing the partial pressure of a component.
6 2. How would hydrogen compare to the gases in Figure 2?
A. It would have the steepest drop initially, reachingequilibrium before the other gases.
B. It would have the flattest drop initially, reaching
equilibrium after the other gases.
C . It would be identical to the graph for helium'
D. It would be between the graphs for nitrogen and
argon.
6 3. How does the rate of diffusion change with time?
A. It decreases for helium and increases for argon'
B. It increases for helium and decreases for argon'
C . It decreases for all three gases.
D . It increases for all three gases.
Passage X (Questions 64 - 70)
A researcher carries out a series of reactions understandard conditions, in a round- bottom flask connected to amanometer. The line from the flask to the manometer isfitted with a valve that opens to the environment. Thefollowing four reactions take place in the reaction vessel:
HI(g) + Ba(NO3)2(aq) + BaI2(s) + HNO3(aq)
Reaction IKHCO3(aq) + HCI(aq) -+ KCI(aq) + H2O(l) + COz(g)
Reaction 2
AgNO3(aq) + NaCl(aq) -+ NaNO3(aq) + Agcl(s)
Reaction 3
CIZ(aq) + Zn(s) -+ ZnCl2@q)
Reaction 4
The change in pressure is found by observing thedifference in heights of mercury in the two columns of themanometer. The two columns have the same internal radius.The reaction apparatus is shown in Figure 1 below:
Manometer
Figure 1
The data for each reaction are calculated using the idealgas equation for the vapor in the system. The vapor pressureof water at 25"C is 27 .2 torr. The pressure for each reactionis recorded along with the resting volume of the system. Thevolume of the gas space following the reaction isapproximated as the original volume plus any positive ornegative correction for the manometer. For a reaction inwhich the volume increases, for instance, the change involume is the cross-sectional area of the manometer coretimes the height change.
6 4 . Which of the reactions would result in an increase in the
height of the mercury column on the right of side of theU{ube?
A. Reactions 1 and 2
B. Reaction 2 only
C. Reactions 1, 3, and 4
D. Reaction 4 only
Burette
Copyright @ by The Berkeley Review@ 42 GO ON TO THE NEXT PA
6 5. As the bore radius of both tubes in a manometerreduced, how is height of the fluid in the column in tatrnospheric side affected?
A. If the radius decreases by half, the height iby a factor of four.
B. If the radius decreases by ha1f, the height iby a factor of two.
C . If the radius decreases by hali the height does
change.
D . If the radius decreases by half, the height decby a factor of four.
6 6. For a reaction that results in only a small changevolume, what type of tube in the atmospheric sidethe manometer yields the MOST accurate reading?
A . One with a very dense liquid and a large bore radi
B. One with a liquid of low density and a largeradius
C . One with a very dense liquid and a small bore
D. One with a liquid of low density and a smallradius
6 7 . What is the role of the stopcock in the middle line?
A. To equilibrate the two sides of the manobefore the reaction begins
B . To measure the pressure inside of the flask
C . To measure the pressure outside of the flask
D. To prevent pressure buildup inside the flthrough one-way venting
Wrhrc
nnnmu
d.gro
-il- D
l-1c-ED-C
uffi
ffim,r6mfuct- T
m
tu
t-uffi
O;urft
ffi
6 8. Once the systemof the manometermanometer, then:
is at equilibrium,is higher than the
if the right colleft column of
A' Pry.t.* at equilibriumthe stopcock is opened,
B' P.yrt.t at equilibriumthe stopcock is opened,
C. Psystem at equilibriumthe stopcock is opened,
D. Prytt.rn at equilibriumthe stopcock is opened,
> Psyste. initiallyithe net flow of gas is ou
( Pryrt.rn initiallylthe net flow of gas is
) Psystem initiallyithe net flow of gas is in.( Psystem initially; $
the net flow of gas is in
"-\tich of the following is the BEST liquid to use in the:renometer for a system that produces a large quantity:: las?
\. Mercury
B. Water
C. Ethanol
D. Glycerol
--:i an endothermic reaction that neither consumes nor:r-:cuces gases, what is observed after reaction?
d. The left column rises initially until the solutiontemperature falls to 4'C, then the left columnhegins to drop.
ts. The left column rises initially until the solutiontieezes, then the left column remains constant.
[ - The left column drops initially until the solutionIemperature falls to 4"C, then the left columnbegins to rise.
D. The left column drops initially until the solutiontreezes, then the left column remains constant.
:hangeic sideng?
)re
arge
ore radi
mall
i
;k
the fl
llyi ws is out
tlyl ws is oulllyi w$ rs ln.
llv;l515ln
IPA ,r,r"'::Eht @ by The Berkeley Review@ 43 GO ON TO THE NEXT PAGE
Passage Xl (Questions 71 - 78)
A researcher fills a balloon with a mixture of nitrogengas (ZOVo by volume) and helium gas (80Vo by volume) and
then seals it. The balloon is used to transport samplingdevices into the atmosphere to collect gases at variousaltitudes. It is filled to a total internal pressure of 760.0 torrat2'7"C and a volume of exactly 24.63 liters. The ballooncontains exactly one mole of total gases. It is allowed to rise
to a height of 20,000 feet, where the atmospheric pressure is
measured to be 507 torr (0.67 atm.) and the temperature is-3'C (270 K). The balloon carries a basket aloft containingequipment to monitor the atmospheric conditions. Theballoon is released early in the morning and returns to the
ground at night, where it is 7'C and the pressure is 760 torr.
In addition to sealed balloons, hot-air balloons may also
be used. Drawn in Figure 1 below is a hot-air balloon withan attached basket.
Sand bags(ballast)
Figure 1
Hot-air balloons demonstrate the principle of buoyancy.Hot air is less dense than the surrounding air, so when hot airdisplaces cold air in the balloon, it becomes less dense than
the surrounding medium. Hot air rises to the top of the
balloon, as shown in Figure 2. This lowers the center ofmass for the system.
Figure 2
7 1 . What is the volume of the sealed balloon at 20,000 feet,
where the temperature is -3'C and pressure is 507 torr?
A. 14.78 liters
B. 18.24 liters
C . 33.25 liters
D. 41.05liters
7 2, What is the density of helium gas at STP?
A. 0.18 grams/liter
B. 0.36 grams/liter
C. 1.00 grams/liter
D. 4.00 grams/liter
73. As a hot-air balloon is lifting off, which point indrawing below accurately reflects its center of mass?
A. Point a
B. Point b
C. Point cD. Point d
7 4. What is the partial pressure of N2(g) at 507 torr and
7'C?
A. 102 torr
B. 133 tonC. 157 tonD. 608 torr
7 5. Which of the following statements BEST describes thebuoyancy of a hot-air balloon under varying conditions?
A. The balloon is most buoyant on a hot day, becausethe surrounding air is denser than on a cold day.
B. The balloon is most buoyant on a hot day, becausethe surrounding air is less dense than on a cold day.
C. The balloon is most buoyant on a cold day, becausethe surrounding air is denser than on a hot day.
D. The balloon is most buoyant on a cold day, becausethe surrounding air is less dense than on a hot day.
the
Copyright @ by The Berkeley Review@ 44 GO ON TO THE NEXT PA
76. The relative rate of effusion from a balloon (withHe and 507o N2) for helium compared to nitrogenbe which of the following?
A . 7 times faster for helium
B. {1 times faster for helium
C . 1[1 times faster for nitrogen
D. 7 times faster for nitrogen
77. The molecular weight of a gas, where the massgrams is represented by (g), can be calculated byof the following formulas?
A. gPV
RT
B. gVT
PR
C. gRP
TV
D. gRT
PV
78. How many moles of helium gas are in thethe pressure is reduced to .507 atm. and theis reduced to -3"C?
A. 0.48 moles He
B. 0.80 moles He
C. 1.08 moles He
D. 1.20 moles He
Cffirlomqa
!@irol:fuffis[
mlnwfi pnam hil
d
P;ilsd
LfiIfiffih c
mtu
0,r!
frftnn
NASS I1
'. whici
loon.ler
rh 509i Fassage Xll (Questions 79 - 86)i wouk
Chlorocarbons are a class of compounds composed ofrne. carbon, and hydrogen, andfluorocarbons Ne a class
- -npounds composed of fluorine, carbon, and hydrogen.:r ' :iasses of molecules consist of polar compounds, for
: - rst p&rt. Their polarity affects their physical properties,, - ,. boiling point, so they are often used as refrigerants.
;l. , . . -S3 of their strong intermolecular forces, neither class of- rrnds acts like an ideal gas. An ideal gas obeys-_ _rn 1:
PV = nRT
Equation 1
,-' P is pressure in atm., V is volume in liters, n is the--. of moles of particles, R is the ideal gas constant
' - L atm..mole-1.K, and T is temperature in kelvins.
',::n comparing a real gas to an ideal gas, deviations in- : and volume are taken into account. The corrected
,: :en substituted into Equation l, yield Equation 2:
I(P + an-) x (V - nh) = nP1
v2Equation 2
': -,S &r empirical value related to the strength of the: rntermolecular forces, and b is an empirical value' :he size of the molecule.
- .:rerai, bigger molecules have greater b values." -. irom ideal behavior decrease as the mean free path
:: This deviation is attributed to reduced interactions, :. the increased average distance between particles.: lme of the container increases, the space that is
: - r". the particles becomes insignificant.
. irsk initially filled with equal moles of methane,
- . : ,nethane, chloromethane, and bromomethane,
" -:. gas has the GREATEST mole fraction after a' r iuration of time, if each gas is allowed to effuse* .:: flask?
t, 1,1:thane
! -.:oromethane. - r-oromethane
, ::'tmomethane
* :: rlue for 1,1-difluoroethane is:
" - . : of the b-value for 1 ,1 -dichloroethane.
: :rtly less than the b-value for I ,1 -
: :..-oroethane.
_.r.rtly more than the b-value for 1,1-- - -.loroethane.
:: as large as the b-value for 1,1-dichloroethane.
" I bv The Berkeley Review@ GO ON TO THE NEXT PAGE
8 1 . The a-value for a halocarbon increases proportionallywith:
A. molecular mass.
B. nucleophilicity.
C. boiling point.
D. polarity.
8 2. Which of the followingvalue?
A. Chloromethane
B. Fluoromethane
C. Methane
D. 1,2-Difluoroethane
gases has the GREATEST a-
45
83. What is the volume of 1.00
and 1.00 atm.?
A. 20.t'7 LB. 22.41 Lc. 24.24LD. 32.11 L
moles of CHrF at 25"C
8 4 . What does NOT decrease as the pressure exerted on a gas
system in a piston increases?
A. Volume
B. Compressibility
C . Mean free path
D. Gas concentration
85. What is TRUE when comparing CH2F2to CH2Br2under identical conditions?
L The diffusion rate of CH2F2 is greater than the
diffusion rate of CH2BI2.
[. The collision impulse of CH2F2 is greater than the
collision impulse of CH2Br2.
m. The average kinetic energy of CH2F2 is greater
than the average kinetic energy of CH2Br2.
A. I only
B. I and II only
C. I andIII only
D. I, il, and III
8 6. Which of the following gases is MOST ideal?
A. I,l-Difluoroethane
B. Difluoromethane
C . Fluorine
D. l,2-Difluoropropane
Passage Xlll (Questions 87 - 93)
A balloon filled with enough helium floats in the airsurrounding it, because its countermass exceeds its mass.The countermass is the mass of the surrounding medium thatis displaced by the balloon, and it is found by multiplyingthe volume of air displaced by the balloon times the densityof the medium. Helium is lighter than all gases excepthydrogen, and it has the smallest molecular size of any gas
known. Because of its low mass and small molecular size,helium effuses rapidly from a balloon. The restoring force ofthe rubber from which the balloon is made provides a drivingforce for effusion to exceed infusion. As the helium effuses,the volume of the balloon is decreased, reducing thecountermass. Eventually, the mass of the balloon exceedsthe countermass, and the balloon is no longer buoyant.
The rate of effusion depends on the temperature and massof the gas, and on the size of the gas particles relative to thesize of the pores in the container's walls. The mass andtemperature affect effusion rate, because they affect theaverage speed of the gas particles. Equation 1 shows therelationship of gas speed (v) to temperature (T) and mass (m):
Equation ITo study the average speed of a gas, particle flow in onedirection through a vacuum is observed. The apparatusshown in Figure 1 is used to generate such conditions.
Valve
Figure IThe system is completely evacuated, and a gas sample is
injected into Chamber 1. The cylinder is then set in motionand Valve 2 is opened briefly, allowing gas to flow intoChamber 2. A sensor in Valve 3 detects collisions.Chamber 2 is re-evacuated, and the process is repeated withthe cylinder spinning at a different rate. The goal of each
adjustment is to determine the spin rate that causes thegreatest collision frequency at the detector. It is assumed thatthe gas is passing straight through Chamber 2 at this spinrate. Under such conditions, the diameter and period of thespin are used to determine the speed ofthe gas particles.
87. A balloon filled with which of the following gases
would require the GREATEST minimum volume tofloat in air?
A. Ammonia
B. Helium
C. Hydrogen
D. Methane
, * ^[TVm
Chamber I
Spinningcylinder
Copyright @ by The Berkeley Review@ 46 GO ON TO THE NEXT PA
8 8. In the experiment illustrated by Figure l, how must the
spin rate of the cylinder be adjusted when methane (16
grams/mole) is replaced by nitrous oxide (44grams/mole) to get the same results?
A . It must be slowed by a factor of 2.15.
B. It must be slowed by a factor of 1.61.
C . It must be increased by a factor of 1 .61 .
D . It must be increased by a factor of 2.15.
89. Which of these balloons has the GREATEST buoyan.force in air?
A. A 4.O-liter methane filled balloon
B. A 3.0-liter helium filled balloon
C . A 2.0-liter neon filled balloon
D . A 1.O-liter nitrogen filled balloon
9 0. If a larger cylinder were used in Chamber 2, how mus.
the spin rate be adjusted to keep the gas passing througrthe pore after half of a rotation?
A . The rate must increase by a factor of 2.
B . The rate must decrease by a lactor of 18.
C . The rate must increase by a iactor of y'7.
D. The rate does not change.
91. Why is Chamber 2 evacuated before gas is releathrough Yalve2?
A . To minimize collisions between particles withe cylinder
B. To prevent the effusion of gas
C. To prevent the infusion of gas
D . To maximize the buoyancy of Chamber 2
Questlrrldescrip::'
;.1 . \\-hn1.Cr,
lite;s
A.B.C.D,
A.B.C.D.
|&- l 'liri;li
ry I lllllif
ll ri; ;Iliu
{.tti
Ir
r<
El r
-1"li F
9 2. How would the speed of the gas
system shown in Figure 1, if theincreased from 25'C to 100'C?
change within
A . It would increase by a factor of 4.
B. It would increase by a factor of 2.
C . It would increase by a factor of 1.25.
D. It would increase by a factor of 1.12.
9 3 . How do the speeds of N2O, CO, F2, and Ne compare
A. vNzO ) vp, ) VCO > VN"
B. vNe > vCO ) Vr, > VNzO
C. uNzO ) vCO > Vp, > VN"
D. vN" ) vp, ) VCO > VIVzO
temperature we
iilillb,ur
'nwllliu]n
sitfl:ff,, ,*
D l(
nustlane (lde(
buo,v
rparea
", :..r.: is the final pressure in a balloon that occupies
,'" liter at S.T.P. after the volume is decreased to 0.75
::: rt constant temperature (isothermally)?
s" , .l13 tonB ! -1.-1 torrI -50 torr
fx l-0 torr
1"":,:: rs the partial pressure of argon in a mixture that is
'- - 'rr moles argon at STP?
q. lji.2 torr3 --lrl.0 torr
- i,r-).0 torrIt :15.8 torr
r ;.r ::lloon that occupies 67.2: rn '..'i11 occupy what volumer l -.cP=0.90atm.)?", it.l iiters
! :5 1 liters
- :' -t liters
I i -.-i liters
liters at 25'C and 1.0
in the mountains (T =
'" :--r,es a balloon filled with argon gas 23.6 minutesl,:';:ease its volume by 0.100 liters due to effusion at
:,-"':: temperature and constant pressure, how long,, :- : ri take if the balloon is filled with sulfur trioxide-, ,:r::l .Ar = 39.95 gim, SO3 = 80.06 g/m)
q, -- - minutes
il :-- I minutes'" : - minutes
I i minutes
qi rL :-. ri the followingir - i .i i,ast as water?
gases moves approximately
I:-- -LI - ^tl1
,--: 3 br. The Berkeley Review@ 47 YOU'VE PASSED GASES!
i:i-.rnS 94 through 100 are NOT based on a
Li '::1:.-\ e passage.
9 9. Select the sequence which corresponds to the infusionrates for the four gases indicated below in ascending
order:
A. CHa < CO < SO2 < C12
B. SO2 < Clz < CO < CH4
C. CO < Clz < CH4 < SO2
D. Cl2 < SOz < CO < CH4
L00. The relative rate of effusion from a balloon formethane compared to helium would be which of the
following?
A . 4.00 times faster than helium
B. 2.00 times faster than helium
C , 2.00 times slower than helium
D. 4.00 times slower than helium
1. D 2.D 3. A6.8 1.C 8.D
11. D 12. A 13. D16. D 11. C 18. D21. A 22. C 23. C26. D 21. B 28. C31. A 32. C 33. C36. C 37. B 38. B41. C 42. A 43. D46. C 41. D 48. C51. B 52. D 53. C56. B 57. B 58. D61. C 62. A 63. C66. D 67. A 68. A71. C 72. A 73. C16. B 11. D 78. B81. D 82. B 83. C86. C 87. A 88. C91. A 92. D 93. B96. C 91. B 98. A
4.C 5.B9.C 10.C
14. D 15. A19. D 20. D24. C 25. D29. C 30. A34. C 35. A39. B 40. C44. A 45. C49. D 50. Ds4. A 55. As9. c 60. D64. B 65. A69. A 70. A14. A 15. C79. D 80. B84. D 85. A89. A 90. D94. A 95. A99. D 100. C
@T"C
g
i1
1I
-;ue:r
C
1..
Gases Passage Answers
Choice D is correct. Because the particles of a real gases have a molecular volume, the volume of the containerdoes not reflect the actual volume that the gas can occupy. This is based on the idea that in a real gas, no twoparticles can occupy the same point at the same time. As the container volume decreases, the space inoccupiedby gas molecules is decreased, so the molecules collide more often. This increases the interaciions and causesdeviations from ideal behavior. Choice A is eliminated. Forces between molecules (attractive or repulsive)result in deviations in the behavior of the gas. The deviations can be attributed to changes in the collisionfrequency of the particles with the walls. For instance, particles that experience attractive forces have agreater tendency to collide with one another, decreasing the frequency of their collisions against the walls.This eliminates choices B and C. Choices A, B, and C represent the assumptions of the ideal gas law, whichunder real conditions do not hold true. The gas is assumed to be uniform ilthe flask, so choice"D is not astatement. In addition, this would not necessarily account for deviations from ideal gas behavior.collisions would simply be asymmetrically distributed against the walls. Pick D to be totally hip and now.
Choice D is correct. Pressure is a measure of the collisions of the gas molecules with the container walls. AsPressure increases, the gas particles strike the wall more frequently. It can be inferred. that molecules acollide with one another more frequently, eliminating choices A and C. Because they collide more frequentlytheir time in contact, and thus their interactions are increased, making choice D the best answer.
Choice A is correct. Because the question asks for the trend over a large pressure range, we need Table 1.observed from Table 1, the pressure'volume product for helium gas showi a uniform increase over the periduring which the pressu,reis increased up to 400 atm. During this same period, the pressure.volume prodlctgxygen gas decreases slightly before beginning to increase. This means that oxygen gas shows a chang"behavior as pressure is increased, which eliminates choices B, C, and D. Althougn w"e tnow the answer mustchoice A at this point, we'Il analyze the trend for carbon dioxide to be certain. ih" p."rrure.volume productcarbon dioxide shows a drastic drop until some pressure around 25 atrn., then the pressure.volume prodrapidly increases. The best answer is choice A.
Choice C is correct. This is really two questions combined into one. "Is the volume larger or smallerexpected" and "Is this due to attractive or repulsive forces?" The value for the molar volume of an idealunder nearly standard conditions is listed as 22.4L liters in the passage. The value for the molar volumeammonia (NHe) is listed in Table 2 as 22.087liters. This means that the molar volume is smaller than ifor ammonia (NH3). Choices A and B are eliminated. The volume of a gas is reduced due to attractive forceIn the case of ammonia, the attraction can be attributed to hydrogen bonding. The best answer is choice C. Bechampion and choose C.
Choice B is correct. Table 1 lists values for the pressure.volume product. The value for the pressure.voluproduct at 75 atm. should be somewhere between 10.11 and 11.25. A good approximation of lhe value forpressure'volume product at 75 atm. is 10.63 (an average of 1011 and 11.25). To determine the molar volumecarbon dioxide at75 atm., the pressure.volume product (10.63 L.atm.) is divided by the pressure (25 atm.). Tlyields a number far less than one. Choices C and D are eliminated, because those ,rair"t better approximathe product of the Pressure and the volume at75 atm.,not the volume alone. The correct value is greater tl0.1333 (the value obtained when 10 is divided by 75), because 10.63 dividedby 75 is greater than 10 divided75. Choice A should be eliminated, because it is too small (less than 0.100, which is tO divlaed by 100).best answer is choice B.
6. Choice B is correct. A gas is most ideal when there are minimal intermolecular forces between theThis occurs when the gases do not contact one another as frequently. At low pressure, the gases do not colliwith one another (or the walls) as often, so they have fewer intermoleculir interactions. This makes kpressure more ideal than high pressure. At high temperature, gases have the necessary kinetic energyovercome the intermolecular forces, so they do not stay in contact for as long a period of time. This makes-hitemperature more ideal than low temperature. The best choice is B, high temperature and low pressure. If yconsider a phase diagram, at high temperafure and low pressure, the material is in the gas phise, far from 1
other phases, making it most gas-like.
c.;
i"{;5{
dff
FfgE
2.
3.
C[WM
'dm&q
4.
5.
0d6ffimtu
&m
Copyright O by The Berkeley Review@ 4a Section VI Detailed
'n
Choice C is correct. In a closed system, the number of particles (moles of gas) cannot change. This means that::oles are constant during the process. If the piston is motionless, then the internal gas pressure equals the
=-.temal pressure. This means that overall, given that the piston starts and finishes at rest, the pressure shows::: r-ret change. Only the temperature and volume exhibit a net change. When the temperature of the system is:.:reased, the average kinetic energy of the molecules increases. The number of molecules does not change, so
-:,:ice A is eliminated. Because the average kinetic energy increases, rather than remains constant, choice B is:.::uinated. The concentration decreases as the container expands, so the average distance between particles:::eases, meaning that the mean free path increases. This makes choice C true. As the average kinetic energy:i::eases, the average particle speed increases, so momentum (mv) increases. This eliminates choice D. Choicet[ -*. the best answer.
,Cioice D is correct. Because the container is sealed, the system is closed, so the number of molecules does not-:::ige. The container is rigid, so the volume of the system does not change. Because neither the moles of:,:-cles nor the volume changes, the concentration does not change, so mean free path cannot change. Mean::e lath is the microscopic equivalent of concentration. The particles may collide more frequently, but that is;l*'::ilse they cover the distance between molecules faster. The molecules on average are the same distancer:':t" Because mean free path does not change, choice C is eliminated. With greater average kinetic energyr j thus greater temperature), both the collision frequency (based on velocity) and the collision force (based:r: =omentum) increase. This makes choice D the best answer. This question addresses both the chemist's and:.- -, sl;Lst's perspective of gases. The correlation between macroscopic measurements and microscopic ideas ofirs.": are listed below:
Temperature correlates to Average Kinetic Energy (as T increases, KE increases)
Concentration correlates to Mean Free Path (as n/V increases, d.r,"u., free path decreases)
I:=s=ure correlates to Collision Force and Collision Frequency (Pressure increases when collisions increase)
Moles correlates to Molecules (hopefully this relationship is a freebie)
Volume correlates to Container Walls (the container walls dictate the volume of the system)
l:r,:i;e C is correct. As the temperature is decreased, the speeds of the gas particles decrease, so the averagelrq*: :f the particles in the gas system decreases. This eliminates choices A and B. The important thing to',lr':i-rere is that the change is considered in terms of the Kelvin temperature scale, not the Celsius scale. The:'.i:.3e is only from293 K to 283 K, and the speed is proportional to square root of T. Because the temperaturerl:::ease is only about3.4"k, the speed decrease is even less than 3.4ok, so the decrease in particle speed is only;-lrt. Choice C is the best answer.
Chroice C is correct. This question is asking for the effect of effusion on the absolute pressure (P) and the:.lative abundance (X) of component gases. Because both H2 and D2 are escaping from the container (due to::tusion through the pore), the moles of each gas decrease over time. This causes the partial pressure of each
:a-. to decrease as well. This eliminates choices B and D. However, because H2 is lighter than Dz, Hz escaPes
:.ster than D2, so the relative amount of D2 (Xpr) increases. This makes choice C the best answer.
Choice D is correct. Under identical "orlaiUojr,
the gases are all at the same temperature, so they have the:ame average kinetic energy. The equation for kinetic energy is K.E. = 1f 2mvz. This means that if two:articles have different masses, they must have different average velocities. Because velocity, v, is squared in--:e kinetic energy relationship, it varies inversely with the square root of the mass. In other words, if one:article is four times as massive as another, it has half the velocity of the lighter particle. Momentum is the:roduct of mass and velocity, so particles with greater mass have greater momentum. This means that the gas-,r-ith the greatest particle momentum is the heaviest gas. Sulfur dioxide is the heaviest gas of the choices, sochoice D is the best answer.
Choice A is correct. Collision frequency affects the pressure. The number of molecules affects the moles.
-{ccording to the ideal gas equation, PV = nRT, moles and pressure are directly proportional, if all otherconditions are held constant. The relationship of pressure and moles is linear, so collision frequency with the'rvalls as a function of the number of particles should also be linear. Twice as many particles results in twice as
many collisions. The best answer is choice A.
nolcu
rulsiollisihave)wa,w.tar.lw.
As.es
uen
luctngenustluctrrod
er
umenlfo
C. Be
er.ded
eal
D.
les loergyes hiIf
'om
:;ight @ by The Berkeley Review@ 49 Section VI Detailed Explanations
13. Choice D is correct. According to the kinetic molecular theory of gases, particles exert no force on one another.If there are no intermolecular forces, the particles cannot be held iogether, so they can never form a solid or aliquid. This results in a boiling point that is extremely small, ,roiir",fir,it"1y large. Choice A is eliminated.According to the kinetic molecular theory of gases, particles have a negligible vol,ime, whether they are atomsor molecules. This means that no matter what the particles may be ot-t i-n" microscopic level, thef ail behavethe same. Choice B is eliminated. When the temperature doubles, the average kinetic energy of ine particlesdoubles. The mass of the particles remains the sime, so a change in velocity is attributable'to the change inkinetic energy. However, because the equation is K.E. = 1 / 2mv2, the viocity doesn't double when thetemperature doubles; the velocity increases by square root of two. This eliminaies choice C. The pressure,according to the kinetic theory of gases, is attributed to coliisions with the walls of the container. If the shapeof the container changes, as long as the volume is the same, there is the same density. This results in the samenumber of collisions per unit area against the walls, so choice D is valid.
Choice D is correct. A gas has no definite volume, because its particles are in contact with one another onlybriefly during collisions. A liquid has a definite volume, becauie its particles are always in contact with oneanother, eliminating choices A and B. Because particles in a gas do not remain in constant contact with oneanother, gases are more compressible and less dense than liquids. In other words, the particles of a gas can bepushed closer together, while particles in a liquid are alieady touching, so they aie hard to pJsh closertogether. Because gases are less dense than liquids, objects in gases are less 6uoyant (due to the lower density ofthe medium). This makes choice D invalid, so choice D is the best answer.
Choice A is correct. From Table 1, it can be seen that the greatest change in volume takes place during the firstfive feet of descent. This is attributed to the inverse relationship betwJen pressure and volume. The"graph oivolume as a function of pressure shown below, demonstrates that as the pressure increases in uniform jncrements,the volume changes by a smaller increment each time.
Volume change is in decreasing increments:AYtAVztAV3>AV4
Pressure change is in uniform increments:AP1 =AP2=fS=AP+
C
ijF:
t{$
J]It
fl;-
W,CiMl
Tdffi
riiil
lXJr
:f.,r
M
M
14.
lmil
L.jfr
ffi
fli1'11:
ill:
inr
:[lr r
\_*'
JTili
llilfi{
:lmr
15.
Ct]
5o
AVz
AV.
AV,
I
III
I I I \\
-t-- - - - - -F - - - - - -f -- - -
t--
ruflr
nnnI
tttltr
APr APz AP' AP,J+
Pressure
The greatest volume change is observed at relatively low pressures, because the percent change in pressurecaused by an incremental increase in pressure is greatest at low pressures. This trend would be observed in datafor the first seven feet of descent into water, also resulting in the greatest volume change of any increment ofseven feet of the descent. The best answer is therefore choice A. This same pheno*"r.J^ explains the pain ir.your ears you may have experienced, if you ever descended into the deep end of a swimmrng pool.
16. Choice I) is correct. This question is asking for the point at which the ball used in the experiment experiencesno net force (where the buoyant force equals the weight). At this point, the density of ttre ball must lqual thedensity of the surrounding water. The ball is initially 50% submerged, so the density of the ball is initiallr-50% that of the_surrounding water. This means that the density of the ball must increase by a factor of two tcreach a point where it is no longer buoyant. The mass of the ball is not changing, so the increase in density mus:result from a decreasing volume. The volume must decrease to half of its oiiginal value (36.0 L at the surface)in order to have a density equal to that of the surrounding water. From Table 1, the volume of the ball equals18.0 liters at a depth of somewhere around 32 feet. This makes choice D correct.
.T
C!*lI
.!rn'
'[]' i I
, z.-
rmlil
5()Copyright O by The Berkeley Revier.r'@ Section VI Detailed Explanations
@
mother.rlid or a
rinated.e atomsbehaverarticlesange inren the:essure/e shapere same
er onlyith oneith onecan becloser
rsity of
he firstaph ofments,
rssurer dataent ofain in
encesal thetially,vo tomustface),quals
Choice C is correct. Fresh water is less dense than salt water, which causes the ball to be less buoyant in freshwater, although that is not the focus of this question. Because salt water is more dense than fresh water,choices A and B are eliminated. A greater volume of the ball would initially be submerged in fresh water.However, as the ball is submerged below the surface, the pressure exerted by the less massive fresh water isless than the pressure exerted by the more massive salt water. The greater the pressure, the more the volumedecreases, so salt water reduces the volume of the ball more than fresh water. This means that the change involume of the ball is greater when submerged in salt water than in fresh water, making choice C correct.
Choice D is correct. If the ball is initially only 25% submerged, then the density of the ball is 25"h that ofwater. To cease to be buoyant in the water, the density of the ball must equal that of the surrounding water.This means that the density of the ball must increase by a factor of four. The mass of the ball is not changing, sothe increase in density must result from a decreasing volume. The volume must decrease to 25"h of its originalvalue (at the surface), in order to have a density equal to that of the surrounding water. The pressure musttherefore be four times its initial value (at the surface). The pressure is one atmosphere at the surface, so theball must be submerged to a depth where the pressure is 4 atmospheres (3 atmospheres of which are due to thewater). The passage says that 32.6 feet of water exerts one atmosphere of pressure, so to have 3 atmosphere ofrvater pressure a depth of 97.8 (3 x 32.6) feet is required. This makes choice D correct.
Choice D is correct. Because the ball that is initially 607o submerged is denser than the ball in the experiment,it need not be submerged to as great a depth to have a density equal to that of the surrounding water. Thegreater density can be inferred from the greater submerged volume. This makes choice D the correct answer.
Choice D is correct. According to the ideal gas law, PV = nRT, the volume of the ball is directly affected bychanges in the pressure (Boyle's law), the moles of gas (Avogadro's law), and the temperature of the container(Charles's law). The magnetic field strength should have no direct effect on the volume, as long as the metalring on the ball is attached at the greatest circumference of the ball. The best (but not perfect) answer is choiceD. If the magnetic ring were attached at a point other than the circumference, the ball could elongate due tothe force, and thus the volume would change with the asymmetric deformation (elongation) of the ball.
Choice A is correct. If salt is added to the water, the density of the solution increases, thus making the ballmore buoyant in the solution. If the ball became more buoyant, more of the ball would rise above the surface ofthe water, resulting in less of the volume of the ball being submerged. This eliminates both choice C and choiceD. Because the ball is more buoyant, a stronger applied force is necessary to overcome the increased buoyantforce, so the B field (magnetic field) must be increased. Pick A.
Choice C is correct. The compressibility of a gas is greatest when the particles are farthest apart from oneanother. This is true when the mean free path is largest. To maximize the mean free path, the concentrationmust be low, which equates to a low number of moles and a large volume. This makes choice C the best answer.
Choice C is correct. This is answered by following the trend in the data. Table 1 shows that volume increasesby 5.3 every ten degrees, so at 150"C, the volume should be 15.9 greater than 208.3n cm3, the volume at 720"C.
This makes the volume approximatety 224.2n cm3, so that choice C is the best answer.
Choice C is correct. The bulb temperature increases fuom7"C to 27 'C, so the bulb expands. This eliminateschoice A. Three answer choices remain, making a calculation necessary. The question can be solved intuitivelyin the following manner. Before calculating, be sure to convert from Celsius into Kelvin. If you fail to do this,vou will incorrectly choose answer choice D.
Vi = -T?y',whereVi>Vi .'. Vf = 300 x 7.0L = b x7.OL = -15*L = 7.50LTz 280 t4 2
The key on these types of questions is to plug the numbers into the equation in such a manner that the value isreasonable (greater or less than the original value). Knowing that the volume increases with temperatureimplies that the final volume is greater tltan7.O, so the temperature ratio mustbe greater than 1.0. Choice C isa terrific answer in a situation such as this.
3X
-:rr-right @ by The Berkeley Review@ 5l Section VI Detailed Explanations
25.
26.
Choice D is correct. As you drive, your car tire heats up due to friction from contact with the road surface. Thieliminates choices A and B. The volume of the tire stays roughly constant, so the only significant chacaused by the increasing temperature involves pressure. As the temperature of a gas at constant voluincreases, the pressure must increase, according to the ideal gas law. This makes choice D the best answer.
Choice D is correct. This is a case of memorization. Charles's law equates changes in volume to changestemperature, at constant pressure. This eliminates all of the choices except D. As temperature increases, tvolume increases according to both Charles's law and the idea gas law. Choice D is the best answer.
an Choice B is correct. According to the trend in the data, the volume at -15'C should be about 13.3n less than tvolume at 10'C. This means that we expect the volume to be L36.7n cm3 at -15"C. Choice A is eliminabecause there is no reason for the ideality suddenly to deviate that much. The size of the particles doeschange with temperature, so choice D should be eliminated. The best explanation for the drastic dropvolume is condensation of the gas. A liquid occupies less volume than a gas. At -15"C, the compound is a liquiso its boiling point is greater than -15"C. At 10"C it is a gas. This means that the boiling point is slightgreater than -15"C. The best answer is choice B.
28. Choice C is correct. The partial pressure due to nitrogen is found by subtracting the partial pressures of waand oxygen from the total pressurc,762 torr. The partial pressure due to oxygen is 20.9% of 762 torr. This vis just over 20"h of 760, which is 152 torr (the exact value is 159.3 torr). The partial pressure of nitrogen isminus the sum of 22.7 and a little more than 152. This leads to a value in the high 500s, so choice C is theanswer. The exact answer is found as follows:
Ptotul=PHzO+Pgr+P111,
762=22.1,+159.3+Pp,
PNz= 762 - 787.4 = 580.6 torr
29. Choice C is correct. The volume of a gas increases with temperature, so the volume of 1.000 gram of oxygenat room temperature is greater than the volume of 1.000 gram of oxygen gas at 0"C. The volume of 1.000 gramoxygen gas at room temperature is less than the volume of 1.000 gram of oxygen gas at 59"C, becausetemperature is less than 59"C. This means that the volume of 1.000 gram of oxygen gas at room temperaturebetween 1.42 liters and 1,.71.liters. Pick choice C.
30. Choice A is correct. If a liquid less dense than mercury were chosen, then the mass of the liquid (and thusforce exerted by the liquid) would be less in the right side of the column as compared to the mass and {associated with mercury. This results in lower pressure being applied to the gas in the left side of the colso the gas in the left side of the manometer would be compressed less, and the height measured wouldgreater than the value obtained using mercury. In layman's terms, the less dense liquid 'squishes' the gasthan mercury. Pick choice A.
31. Choice A is correct. Because molecules are larger than atoms, they are less compressible than atoms. In owords, as you compress molecules, they interact (collide and repel) more than atoms interact, socannot be compressed as easily as atoms. This means that the volume of a molecular gas does not decreasemuch as it does for argon gas when a pressure of equal magnitude is applied to both systems. This meansthe volume of the molecular gas is greater than the volume of the argon gas. The radius of the tube is uniso for the volume to be greater, the height must be greater. The heights when using a molecular gas arethan 50 mm in the second manometer and greater than 33 mm in the third manometer. This is best describedchoice A. Choices B and C should have been eliminated based on trend recognition. If the value is greater50 mm after the first addition, it must be greater than 33 mm after the second addition. Likewise, if the vis less than 50 mm after the first addition, it must be less than 33 mm after the second addition.
Choice C is correct. According to both the passage and the middle of the three manometers, a differencecolumnheights of T6cmisindicativeof apressuredifferenceof 1.00atm. Thismeansthatwhenl.00atm.pressure is applied to an open column of mercury, it will rise 76 crn. The best answer is choice C.
32.
52Copyright @ by The Berkeley Review@ Section VI Detailed Ex
e. Thischangei'olumelr.
nges ir.,ces, th€
han the:inatedoes no:lrop irliquid
slighth'
f wate:.s valuet ts 761
he
gen
ram oj
Choice C is correct. According to the table, the volume of 1.000 gram of oxygen gas at 0"C and 1.00 atm. pressureis 0.7095 liters. If the mass were doubled from 1.000 gram to 2.000 grams, then the moles would also double, so
the volume should likewise double. The volume would be 1.4190 liters, which is close enough to 7.42 to makechoiceCasafechoice.
Choice C is correct. When the pressure of the system is increased from 0.5 atm. to 1.0 atm., the volume of the
system is reduced to half of its original value. When the temperature increases from 25"C to 50"C, it has
actually increased from 298 K to 323 K, so the volume of the system increases, but it won't double. This means
that the overall effect is that the volume decreases, because the effect of pressure is more significant than the
effect of temperature on the system. The best answer is therefore choice C.
Choice A is correct. The greatest change in volume is experienced when the first aliquot of mercury is added.
-A,s more and more mercury is added, the change in the total mass of mercury is less (percentage-wise), so the
change in volume is less each time. This trend can be observed when comparing the difference between the firstand iecond manometers (5.0 cm) and between the second and third manometers (1.7 cm). This change intompressibility is attributed to the fact that as the gas is compressed, the particles get closer together (more
Jense), so there is less room to compress them further. The best answer is choice A. Choices C and D should be
eliminated, because the density of mercury does not change with addition of more mercury. The densities ofliquids and solids are most significantly affected by temperature, not by pressure.
Choice C is correct. The operations of the lung depends on the expansion of the thoracic cavity (defined as the
region above the abdomen inside of the rib cage) and contraction of the diaphragm, which result in an increase
rn the volume of the lung. This reduces the internal pressure (pressure within the lung). This demonstrates
Boyle's law, which states that pressure and volume are inversely proportional under isothermal conditions.Ciroice C is best. Archimedes' principle has to do with buoyancy, which plays no role in lung expansion.
Bernoulli's principle deals with the flow rate of a fluid (like air), pressure differences, and the radius of a
:ube. While it is true that air is flowing within a lung, and Bernoulli's law may be applied to explain air flow:henomenon, the lung does not operate because of Bernoulli's principle. Charles's law states that volume and
iemperature are directly proportional under isobaric conditions. The lung is thermoregulated, so a temperature:}'range cannot be responsible for the success or failure of its operation.
Choice B is correct. During expiration, the lung contracts, causing a decrease in volume and therefore an
-rlcrease in internal pressure. This means that internal pressure is greater than external pressure (referred to as
,.. positive pressure according to physiologists), which accounts for air flow out of the lung. This eliminates::roice R and makes choice B the best answer. During expiration, the thoracic cavity and lung are relaxing back
:r iheir normal (smaller) size. Although they are shrinking (their volume is decreasing), they are still larger-:an their normal resting state, so choices C and D are eliminated.
Choice B is correct. At high altitudes, the gas is less concentrated (due to the lower atmospheric pressure). At,rn'altitudes, the gas is more concentrated (due to the higher atmospheric pressure). As temperature
iecreases, gases become denser (and thus more concentrated). This means that at a low altitude and low::mperature, there will be more of all gases, oxygen included. But each gas increases by the same proportionate.:nount, so as far as relative amounts are concerned, there is the same percentage of all gases. This means that
:,\\zgen has the same mole fraction, but a higher partial pressure than standard conditions. This makes choice
B ihe best answer.
Choice B is correct. The tidal volume is the volume of air entering or leaving during a normal breath, while the
.,.ta1 lung volume includes residual air, dead space, and tidal volume. This means that lung volume must
:.,rvays be greater than tidal volume. This makes statement I invalid. Air flows out from the lung during...piration, so the pressure inside the lung must be greater than the external Pressure. Air flow is from higher
.."rrrrr" to lower pressure. Air flow gradually lessens until it stops, once the internal pressure and external
::essures u." "qnui.
This makes statement II valid. Expiration occurs to displace the carbon dioxide, so it must
--a\-e a greater mole fraction of carbon dioxide than oxygen. This makes statement III invalid and makes choice
ts the best answer.
ature
hus tdfolumlruldras I
e
no,lecu'ease
Lns thnifgrea:ibed:er th;e val
:::ht @ by The Berkeley Review@ DC Section VI Detailed ExPlanations
42.
40. Choice C is correct. According to the laws of fluid dynamics, gas flows from a region of higher pressure toregion of lower pressure. This is true of any fluid that experiences a pressure difference across its surInspiration results from internal pressure being less than external pressure, so choices A and B are eliminatedDuring inspiration, the volume increases from 1200 mL to 1600 mL, which results in a value that is 1.33 tgreater than its original value. Because PV = k at constant temperature, the pressure must decrease to 0.75 ioriginal value, assuming that temperature and moles of gas are constant. Starting at P11i1iu1 and finishing0.75 P61621 is a25"h decrease in pressure, so choice C is the best answer.
4't. Choice C is correct. When the diaphragm contracts, the lung is pulled downward, which causes it to exThe expansion of the thoracic cavity also plays a role in the expansion of the lung. Because the lung expanits volume becomes greater than it is at rest, which makes choice C an invalid statement. The question asks fwhat is NOT true, so choice C is the best answer. Upon expansion, lung volume increases, internal pdecreases, so internal pressure (the pressure inside the lung) is less than external pressure. Choice A is vbecause the internal pressure is less than the external pressure (which ultimately causes air to flow intolung). Under these conditions, the lung is referred to as having subatmospheric pressure (whichphysiologists refer to as negative pressure). Choice B is valid, because all gas concentrations, including cadioxide, decrease as the lung expands. This is because the moles of carbon dioxide gas remain constant whithe volume increases. Choice C is invalid, and thus the best choice, because lung volume increases asdiaphragm contracts. Lung temperature is around 37"C and does not vary drastically as the lung expands.means that air temperature is relatively constant, making choice D valid.
Choice A is correct. A scuba diver underwater exists in an environment of high external pressure. Whenthoracic cavity expands, because it is working against a much greater external pressure, more moles of airneeded to fill the lung than are needed on land. Choice A is valid, because the air is more concentraresulting in higher concentrations of all gases, including oxygen. At great enough depths, the gases arewith helium, to reduce the mole fraction of oxygen, helping the diver to avoid breathing excessive amountsoxygen. Too much intake of oxygen can result in oxygen poisoning. Underwater, because of the greater extepressure, air is significantly denser (more concentrated). Consequently, the viscosity is much greater than itat sea level and it deviates from ideal gas behavior. This eliminates choices B and C. Because all gasesmore concentrated underwater, there are more moles of CO2 throughout the lung, including the anatomitalspace, eliminating choice D.
43. Choice D is correct. During a normal breath, air flow is driven by pressure differences between the exenvironment and the intemal pressure within the lung. At the very start of a breath, there is no air flow.air does not flow until a pressure difference is created. So initially, air flow increases from zero to some vat the start of a normal breath. However, because the system is open, a large pressure differencedevelops. Air flows into the lung while it is still expanding, so the pressure difference is relativelyduring the middle of the breath. This causes the flow rate to be relatively constant during the middle ofbreath. At the end of a breath, once the lung stops expanding, the air flow gradually slows and comes toThis means that during a normal inspiration, air flow rate goes from zero, to some positive value, and thento zero. This describes a sinusoidal function. Choice D is the best, albeit not a flawless, answer.
44. Choice A is correct. The collision frequency of the gas particles depends on the concentration of the moleculesthe container and the temperature of the system, not on the diameter of the molecules. This eliminates chB and D. The greater the diameter of the molecule, the harder it is for the molecule to fit through the porethe wall of the container. For instance, golf balls can easily pass through the circumference of a basketballbut basketballs cannot pass through the circumference of a golf hole. This makes choice A the best answer.may recall that this is the principle behind the operation of molecular sieves.
45. Choice C is correct. The extra neutrons do not significantly affect the reactivity of an atom, because the orbelectrons are responsible for the reactivity. The neutrons are in the nucleus, so they have little to no effectreactivity. Only when the molecular velocity is involved do the isotopes make a difference. This holdsfor all isotopes except hydrogen, deuterium, and tritium. There is an isotope effect that correlates todifferent bond lengths associated with bonds to hydrogen, deuterium, or tritium. This is not important inquestion, however. Pick choice C.
Copyright @ by The Berkeley Review@ s4 Section VI Detailed Explana
toa[ace.
rted.mes5 itsrg at
and.nds,; forsurerlid,the
cme'bon'hi1e
theIhis
Choice C is correct. At the same average kinetic energy (same temperature), velocity of a particle is inverselyproportional to the square root of its mass. To travel twice as fast as uranium hexafluoride, a molecule must befour times lighter than uranium hexafluoride. The mass of the uranium hexafluoride compound is roughly 350grams per mole, so the correct answer must be a molecule with a mass just under 90 grams per mole. Sulfurdioxide has a mass of 64 grams per mole, so choice A is too light, and thus eliminated. Sulfur trioxide has amass of 80 grams per mole, so choice B is too light, and thus eliminated. Phosphorus trifluoride has a mass of 88grams per mole, so choice C is the best so far. Xenon has a mass of 131.3 grams per mole, so choice D is tooheavy, and thus eliminated. The best answer is therefore choice C.
Choice D is correct. The size of the molecule affects its ability to escape through a pore in the wall of thecontainer. The temperature of a gas affects the average kinetic energy of the system. The average kineticenergy of the system in turn affects the velocity of the gas molecules, so the effusion rate of a gas is affected bythe temperafure. As mentioned in the passage, isotopes have different masses, and therefore different effusionrates. The concentration of a species outside of a container affects the backflow of the gas into the container, butit does not affect the effusion rate. Backflow affects the net flow of gas, which is known as diffusion, noteffusion. The best choice is answer D.
Choice C is correct. The mass of the uranium isotope is 235 amu, and the mass of the uranium hexafluoridecompound is given as 349. The mass percent is thus 235 divided by 349. This is a little less than 235 divided by350, which is equal to 47 dividedby 70. 47 divided by 70 is less than 49 divided by 70, but more than 42 dividedb1. 70. This means that the answer is between 60 and 70 percent, making choice C correct.
-47 -70"h=49 >47 >Q=60o/,70 70 70 70
Choice D is correct. Had chlorine been used instead of fluorine, then the mass of the uranium hexahalide:ompounds would have been greater. The greater the mass of the two isotopic compounds, the lower the ratio of+Jee masses of the two isotopic compounds. The less the relative difference in masses, the less the difference in:ffusion rates, and therefore the harder it would be to separate the isotopic compounds from one another. Moreiltering would be required with the chlorine compounds than the fluorine compounds, making choice D correct.
lhe mass o5 235gpu is 349 grams per mole, and the mass of 238gpu is 352 grams per mole. The mass o1235g66'.235 + 6(35.5) = 235 + 213= 448 grams per mole. The mass o1238966 is 23b + 6(35.5) =238 +213 = 457 grams perno1e. The difference in relative effusion rates is greater for the fluorine compounds than for the chlorine
:rmpoundsrbecause 1@ , 1@. In addition, chlorine has two major isotopes while fluorine has only oneY s+s V sss
:rajor isotope. The isotopic impurity associated with two majors isotopes chlorine will affect the distribution,:.aking^it_harder to isolate the uranium isotopes. The UCl6 species can have a mass anywhere in the range of:a; 6o,23sg35clO to 460 for238g37g1u.
Choice D is correct. It would be hardest to separate molecules with the same effusion rate. To have the same
=:rusion rate, the molecules must have the same molecular mass. For choice A, 02 has a mass of 32, while N2:as a mass of 28. Choice A is thus eliminated. For choice B, H2 has a mass of 2, while Ne has a mass of 20.lhoice B is thus eliminated. For choice C, CO2 has a mass of 44, while SO2 has a mass of 64. Choice C is thus=rminated. For choice D, CO has a mass of 28 and CZHS has a mass of 28, so they both have the same:,olecular velocity and thus the same effusion rate. They do not separate by this successive effusion technique.lhoice D is the best answer. It should be noted that of the choices, it would be easiest to separate hydrogen gasIil) from neon (Ne) because of the large difference in their masses. CO could be separated from C2H4 by using:.olecular sieves, which distinguish compounds by molecular size.
Choice B is correct. In Reaction I, two liquids and three gases are formed from two solid reactants. This:=:resents a large increase in entropy, so choices C and D are eliminated. Reaction II is employed to generate-:e heat that is necessary for Reaction I to proceed. The generation of heat implies that the reaction (Reaction* must be exothermic. The best answer is therefore choice B.
the' areted,Lxed
so{rnalrt
235 = 235349 350
r,i -:::-nt @ by The Berkeley Review@ DD Section VI Detaited Explanations
52.
53.
Choice D is correct. As any gas cools (whether it be nitrogen at 350"C or any other gas), the pressure of thesystem and the total kinetic energy of the system and the average kinetic energy of the gas particles decrease.The temperature is a measure of the average (and therefore total) kinetic energy of the system, so a lowertemperature is indicative of decreased kinetic energy for the system. The best answer is therefore choice D.
Choice C is correct. If the bag fills at a rate of 100 moles,/sec and the total inflation time is 0.01 seconds (10
milliseconds), then one mole of gas has filled the bag in that period of time (100 moles/sec x 0.01 seconds = 1
mole). At 31'C, one mole of gas occupies 25 liters, as stated in the question. Had the question not stated thatfact, rt wouid have been possible to compare the volume of a gas at 0"C to the volume at 31'C. At STP (0"C and1 atm.), one mole of gas occupies 22.4 liters. At 31'C, the temperature has increased on the Kelvin scale onlrslightly from 0'C (from 273 K to 304 K). The volume therefore increases oniy slightly above 22.4liters. Theonly answer that is in the range of "slightly above 22.4" is choice C, your best answer.
Choice A is correct. The key to this question is using the Kelain temperature scale. Because average speed (theroot mean square speed) is directly proportional to the square root of the temperature, an increase intemperature fuom 298 K (25'C) to 398 K (125'C) increases the average speed of the gas by a factor of roughlv1fu3. This eliminates all of the answer choices except choice A, because 100 xfi33 < 133 < 224 <500. Theexact mathematical solution for the problem is as follows:
@fr0 ch
-=:__:::r.
ir.L:
si54.
55.
v125 -'[732 so v1,2s = 1001/132 < 133100
Because the speed is less than 133 m/s, choices B, C, and D are incorrect (too large), making choice A the bes:answer. Be sure that all of the temperature values that you use are in terms of kelvins. False answer choices orquestions like this one may take advantage of the notion that you will forget to convert degrees Celsiuskelvins.
Choice A is correct. The lighter the molecular mass of a gas, the greater the average speed of the gas at a gitemperature. The greater the average speed of the gas, the faster the rate at which the gas effuses frompores within the container's walls. The ranking of the relative effusion rates follows the trend lighter is faste:than heavier. The best choice is He (4 g/mo1e) > N2 (28 g/mole) > Ar (40 g/mole), making choice A thechoice. This assumes there are equal portions of the three gases within the container.
56. Choice B is correct. At STP, the total pressure of the system is 760 torr. The pressure due to carbon dioxi(COZ) is the mole fraction of carbon dioxide (CCOr) times the total pressure of the system (P16121). The
fraction of carbon dioxide can be found by dividing the moles of carbon dioxide by the total moles of the systernThis is done as follows:
I zp.(zssl Ivt25 - ll ^ fzues54 = \[3eE =,[t3zv25 V / -* u298
50 5044 - 44 _1
(#.#) (#.#) 2
The mole fraction of CO2 is less than one-half, which implies that less than half of the moles of gas are COiThis means that the partial pressure due to CO2 is less than half of the total pressure. The partial pressureCO2 is less then 380 torr, which eliminates choices C and D. Choice A is considerably too small. For choice.to be true, the mole fraction of CO2 would have to be less than 0.033. The mole fraction of CO2 is notsmal1. The best answer is therefore choice B. You should do all of your calculations like this, zeroing in onbest answer without spending time finding an exact answer. Remember, you do not get points on the MCATshowing your work!
57. Choice B is correct. Because neon is lighter than argon, neon has a greater average speed than argon.relati-n'e average speeds of the gases are inversely proportional to the relative square roots of their mThe mass of argon is roughly twice the mass of neon, so the average speed of neon should be roughly {2 tifaster than the average speed of argon. The best answer is choice B, where B is for BEST choice on tparticular question.
[hi,r;rl{:i
t*-- "ln';-L"ILE
the"iru:
*:_r fr__
.5 irtuffr.. -.;:-d.*
:rillli:lhre :
il] ,,,,,,,,,,,,,,,r,r
tr4 '
Jlriyilffii1ru
lL-lln Hlr,'
mr[itrx
nFrne
4 - "11
.u'& -
Ch@rffi
ffir*,,ru1
all "'
.WrrfxI
mgru1
5(,Copvright O by The Berkeley Review@ Section VI Detailed Dxplana
)f therease.
lower).
ls (10
is=1i thatI andr only
The
t (therS€ inughly
The
e best:es ons into
givenm thefastere best
.oxidemole
'stem.
Coz.ure ofrice Art thatrn the\T for
. TheIASSES.
timesn this
Choice D is correct. Figure 2 shows the progression of the experiment as it approaches equilibrium. All of thepartial pressures are equal at the start of the reaction, so the mole fraction for each gas is one-third. Onceequilibrium has been established, the partial pressure of each gas in region I has dropped to about one-fourth ofits original value, but all three gases have the same partial pressure once again. This means that atequilibrium, the mole fraction for each gas is still one-third. This eliminates choices A and B. The molefraction doesn't change, but the moles of each gas do. The loss of moles of gas from Region I causes the drop intotal pressure. A1l of the gases show a net movement from Region I to Region II, so the moles of gas in Region Idecrease, while the moles of gas in Region II increase. This eliminates choice C and makes choice D the bestatswer.
Choice C is correct. Helium is the lightest of the three gases in the mixture, so it travels the fastest. This:neans that initially, helium effuses from Region I into Region II faster than argon and nitrogen. As a result,rrior to equilibrium, Region II is richest in helium and poorest in argon (the slowest gas). Flowever, once
relium has equilibrated between the two regions, argon continues to exchange, until it too reaches equilibrium.Once at equilibrium, the rate of effusion (movement from Region I to Region II) equals the rate of infusionmovement from Region II to Region I). The overall change from initial conditions to equilibrium is that thetotal volume of the system has increased, causing the total pressure of the system to decrease and the partialpressures of all gases to decrease equally. The mole fractions at equilibrium, however, are equal for all of the
iases at any point in the system, Region I or Region II. Because the total pressure is uniform and the mole::actions are equal, the partial pressure of each gas is equal at equilibrium. Choice C is your answer.
Choice D is correct. Once a gas system is at equilibrium, the pressure is uniform throughout the container, so the:ressure in Region II is equal to the pressure in Region I. Region II has three times the volume of Region I, but at:quilibrium it also has three times the moles of gas. The result is that the mole fraction and the partial:ressure of any component gases are equal in each region, once equilibrium is established. This makes statement
- rnvalid. The system is initial charged with equal moles of all three gases, so the moie fraction of helium and::.e mole fraction of argon are both one-third at the start of the experiment. Once equilibrium is established,--:re gases are mixed evenly once again, so despite the total moles in Region I decreasing, the relative amount of-ach gas is the same. This also means that the relative amount of each gas is the same in Region II. Once at
-qurlibrium, the mole fraction of helium and argon are each one-third, at any point in the system. The mole
-:action of helium and the mole fraction of argon are equal, making statement II valid. When the temperature:- increased, the average speed of each gas increases. This results in a greater rate of effusion and infusion,:-though the net flow may not be affected. At the higher temperature, the pressure of the system is greater, so
=lch partial pressure exhibits a proportional increase. The increase is attributed to a greater frequency of: llision and increased momentum for particles at higher velocity. However, the moles have not changed, so--:le mole fractions have not changed. This makes statement III valid. The best answer is choice D.
Choice C is correct. By opening the valve to a greater area, there is more space through which gases may:avel from Region I into Region II. The flow rate increases, so effusion rate increases. Choice A is valid'-:,.reasing the temperature of the system increases the average speed of each gas, so they effuse faster, infuse:aster, and diffuse faster. This makes choice B valid. Increasing the volume of the system reduces the pressure
rumber of collisions against the walls and with the pores), which results in less gas traveling from Region I
-:.to Region II. A decrease in concentration (caused by an increase in volume) would decrease, not increase, the
:-ite of effusion. Choice C is an invalid statement, and thus the best answer. Increasing the partial pressure ofa .omponent increases the number of collisions with a pore experienced by that component. An increase in mole
-action (partial pressure) increases the rate of effusion, which makes choice D a valid statement.
Choice A is correct. Hydrogen , H2, has a molecular mass of 2 g/rnole. Helium, He, has an atomic mass of 4: mole; nitrogen, N2, has a molecular mass of 28 g/rnole; and argon, Ar, has an atomic mass of 40.1 g/mole.iecause hydrogen is lighter than all of the component gases in the system, it has a greater average speed than:-1 of the other gases. This means that hydrogen effuses faster than the other gases, ancl reaches equilibrium:iior to the other gases. The graph for hydrogen would have a steeper initial drop than helium, nitrogen, or:rgon. This makes choice A the best answer.
: ,:::ht @ by The Berkeley Review@ 5/ Section VI Detailed ExPlanations
64.
63. Choice C is correct. Diffusion is the net flow of gas. Initiall/, there is high pressure in Region I and no pressure
in Region II. This means that the net flow of gas is from Region I to Region Ii, Over time, as pressure decrease
in Region I and grows in Region II, the net flow of gas diminishes until it reaches zero once at equilibrium. Thi
1-r-r"ur,, that the rate of diffuiion decreases with time until it reaches equilibrium. On a more specific level,
decreases for all three of the component gases. This makes choice C the best answer.
Choice B is correct. The fluid in the right side of the manometer rises when the internal pressure exceeds
external pressure. This will occur when the internal pressure increases or the external pressure decreases.
likelihood of external pressure (environmental pressure) decreasing under controlled conditions is minimal,
the change in mercur;rheight must be due to an increase in the internal pressure of the system' In order Jor I
internal fr"rrrrt" to increase, the volume of gas in the reaction vessel must increase. Only Reaction II produces
gas product, so choice B is the correct answer.
65. Choice A is correct. Because the bore size is reduced equally in both tubes, the pressure in both tubes rem
equal. Because there is no change in pressure difference between the two sides, the height of the columns is
same in each tube. The manometer measures relative pressure difference between the two sides, not abso
pressure. The question asks for the height, however, not the height difference. Because the volume depends
ihe radius squared, if the radius is cut in half, then the volume is reduced by a factor of four. The amount
mercury has not changed, so the total length of tube filled with mercury must increase by four. The heig
changes by a value close to four, so the best answer is choice A.
Choice D is correct. A small change in volume must be stretched out to be made more visible. A less de
manometer fluid increases the observed change in volume. The smaller bore radius in the atmospheric t(assuming that the inner tube keeps the same bore size) results in the liquid climbing the atmospheric tube
the manJmeter (the column "*porld
to the atmosphere) by a greater amount (resulting in an increased heigh
The greater the change in column heights, the more accurately the difference between the two sides of
manometer can be measured. Pick D to get the tingly sensation of correctness.
Choice A is correct. This middle stopcock is the only port into and out of the system for a gas, and it mremain closed during the experimentlor the system to remain closed. It is a valve, so it has no capacity
measuring pr"rr.rr",--eaning that choices B and C can be eliminated. The stopcock is a two-way valve,
choice D is eliminated. The best answer is therefore choice A. The pressure must be equal in both sides of
manometer initially to get an accurate reading of the change in pressure during the course of the reacti
Hairline differences betireen the two columns initially make for errors in the measurement before the reacti
Venting the system initially assures that the heights of the fluids in both sides of the manometer are equal.
Choice A is correct. The right column of the manometer is higher, because the left side of the manometer
"pushing" the liquid o.r"r. -Thi, means that the Psystem is greater than the Pexternal. Considering that t
"*t"rr1uiprurrrrru do", change during the reaction, thre system pressure must have increased as the reaction
to equilibrium. The Psystem at equilibrium is thus greater than the Psystem initially', H"1 tl: :t"p:::l^ts:^lthe system vents this "ri;;;;
p;#"." L/ uttot"ing the net flow of gasio be outward. The best answer is choice '
Choice A is correct. A large quantity of gas results in an increased internal Pressure, causing the liquid inleft tube to descend and thJ hquid inthe iight tube to ascend. The less dense the liquid, the greater the cha
in heights (and thus the greater the chance"for spillover of the manometer fluid). To avoid overflow, a de
liquid, such as mercury (cioice A), should be used. Less dense manometer fluids are chosen for hairline readi
oismali pressure differences, where more accuracy is needed. The less dense fluid results in a greater change
height per unit difference in pressure. The greater the change in height, the easier it is to read the manomei
and thus the less significant small errors in the reading will be'
5il" Cr&:l,de
[q
ryffinr
wi0t
de
mn
ft]["'i
MIr
ftrutilSlnnrm
tuftm
frft'ma
frturutu
anm
ilhu
ilm6
66.
67.
68.
59.
Copyright @ by The Berkeley Review@ 58 Section VI Detailed
Pressureecreasesm. Thislevel, it
eeds the
-1. Choice A is correct. An endothermic reaction results in a cooler solution. The cooler solution contracts, creatinga smaller liquid volume and thus a greater air space volume. In addition, the gas is cooled to some extent,depending on the degree of insulation provided by the glass. Either way, the inteinal pressure decreases. Thisresults in the left column of manometer fluid rising initially. We eliminate choices C ind D. Once at 4.C, thea-queous solution begins to expand (water is densest at 4"C). This assumes the behavior of pure water. Assumingthat the gas temperature remains equilibrated with the environment, as the solution volume increases, the gasvolume decreases, so the internal pressure begins to rise again. The fluid in the left side of the column begins todescend. The best answer is choice A.
;es.
imal,r forrduces a
remars is
Choice C is correct. This question requires rearranging the gas law, PV = nRT, to isolate R. The gas constantnever changes (thus the name constant), so we can work from PrYr
= PzYz, which reduces to PtVr
= I&, given
that no gas was added or removed (nr = nz). From here, t, iit]il",rtl*f isolating v2. We ""t1or"ol?
that thereduced pressure increases volume and the reduced temperature decreases volume, buithis doesnit help much.
rz^ - PrVrTr = (fr)(fr)", = *ffi ,, ffif x24.63L = ?. #, 24.6sL = z, 24.63Lv2 --
PzTtThe best answer is choice C.
Choice A is correct. At STP (0"C, 1 atm.), 1 mole of a gas (such as helium) occupies a volume of approximately2.4L. The question requires you to find the density (mass per volume) of helium. The mass of 1 mole of heliumrs 1 molea4 Srams
= 4 grams He. Density = :+=0.18 qu*t. Select A for optimal satisfaction.mole 22.4L liter
Choice C is correct. The center of mass for the balloon lowers as the balloon is filled with helium. This isstated in the Passage- This lowering of the center of mass takes place because helium is lighter than air andihus the mass of the balloon is not as great as it is with the helium in it. In any event, the i'est answer placesthe center of mass in the basket, where the greatest portion of the mass lies. Beiause most of the mass is in thelower half, choice B is eliminated. Choices A and D are throwaway answers. The best answer is choice C.
Choice A is correct. The easiest way to do this problem is to recall that N2(g) is 20% of the balloon. This means:he partial pressure of nitrogen is 20% of the total pressure (the equation is: Pry, = (XNz)(ptotat)). Twentyrercent of 507 is just slightly more than 100 torr, so choice A is the best answer.
Choice C is correct. The buoyancy of the balloon depends on the density of the surrounding air, so maximumiensity requires that the surrounding air be as dense as possible. This eliminates choices S ana D, because theystate that the surrounding air is less dense. As air is cooled (temperature decreases), the volume decreases, soie density of the air increases. This means that on a cold day, the surrounding air is densest. This is bestlescribed in choice C.
Choice B is correct. The relative rates of effusion of two gases are given by the inverse ratio of the square roots':f the masses of the gas particles (Graham's law of effusion): MW of He = 4 g/mole, MW of N2 = 28 gfmole.
Rateof effusionHe = @ =@- = y'7 where B ={i >.[7 >,[r =2Rate of effusion N2 {MWH" ,14
The effusion rate for helium is y'7 fimes faster than for nitrogen, so choice B is the best answer.
Choice D is correct. It is known that molecular mass is mass per moles, which has units of grams per mole. In--:-ris problem, we are told that mass is represented by g (grams), s_o we need only find the reciprocal of moles.lhe units for n = mole, so the reciprocal of moles has units mole-l, which is what we need in the best answer
:-troice. tt = S, ro! = ItI, and the formula for molecular mas is MW = 8RT, choi.u D.RT N PV PV'
rbsoluteends onLount: hei
sdric tutubeheight).;of
it mucrtyrlve,sofracti:actiual.
neterhat trn
)pe
oice A
lin
Lderad
mge
r';r--ght @ by The Berkeley Review@ 59 Section VI Detailed Explanations
78. Choice B is correct. From the passage, we know that there is 1 .0 mole of gas in the balloon at 27'C and 1 atm.The number of moles does not change with temperature or pressure, so to solve this question, you must determineonly what percentage of the 1 mole of gas is due to He. He is 80% of the gas by moles, so nHe = (0.80)(1 mole) =
0.8 moles He. Pick B if you crave correctness.
Choice D is correct. If all components start with an identical mole fraction, then after a short time, the
component with the greatest molecular mass, and thus the lowest average speed, will be enriched. The correctanswer is the heaviest of the molecules. Bromine is heavier than hydrogen, fluorine, and chlorine, so choice Dis the correct answer.
Choice B is correct. The b-value for a gas describes its size (deviation from ideal volume). The molecular size
of 1,1-dichloroethane is greater than the molecular size of 1,1-difluoroethane, because chlorine atoms are
bigger than fluorine atoms. The non-halogen portion of each molecule is equal in size in both molecules, so 1,1-
dichloroethane is only slightly greater in molecular size than 1,1-difluoroethane, so choice B is correct.
Choice D is correct. The a-value correlates with attractive intermolecular forces. This eliminates choice ANucleophilicity depends on molecular size of a gas and its ability to share electrons with an electrophile. Thr:
eliminates choice B. While boiling point depends on intermolecular forces, it also depends on molecular mass
so the a-value and boiling point do not always correlate. This eliminates choice C. As polarity increases, the
strength of the intermolecular forces increases, resulting in a greater a-value. Choice D is the best answer.
Choice B is correct. The greatest a-value is found in the compound that exhibits the strongest intermoleculeforces. Choices C and D are non-polar, so they are eliminated. Fluorine is more electronegative than chlorineso fluoromethane is more polar than chloromethane, making choice B the best answer.
ChoiceCiscorrect. AtSTP,anidealgashasavolume of 22.4 literspermole. At25"Cand1.00atm.,anidea,gas has a volume of 24.5 liters per mole. While fluoromethane, CH3F, is not an ideal gas, the correlation Lvolume between that of an ideal gas and that of fluoromethane is very high. This means that the volume c
1.00 moles of fluoromethane at 25"C and 1.00 atm. is close to 24.5 liters. The best answer is choice C. Th
reduced volume compared to ideal behavior can be attributed to the attractive forces of the CH3F molecules.
84. Choice D is correct. As the pressure exerted on the gas within a piston increases, the volume of the
decreases, forcing the particles closer together. The reduction in volume makes choice A a valid statemen:
which eliminates choice A. As the particles get closer together, they become less compressible, eliminatir-choice B. As the particles get closer together, the mean free path decreases, eliminating choice C. And as th
particles get closer together, they become more concentrated, not less concentrated, so choice D is an invahstatement, making it the best answer.
85. Choice A is correct. Under conditions of identical temperature, concentration, volume, and total pressu
differences between CH2F2 and CH2Br2 are due to differences in their respective molecular masses. Becau
CH2F2 is lighter than CH2Br2, it has a greater average speed and therefore diffuses faster. This mak
statement I valid. Because CH2F2 is lighter than CH2BI2, it has a lower average momentum and theref
exhibits lower collision impulse. This makes statement II invalid. At the same temperature (identic
conditions), all gases have the same average kinetic energy. This makes statement III invalid, which mak
choice A the correct answer.
85. Choice C is correct. The gas that is closest to an ideal gas has the smallest a-value and b-value. This is true
the gas with the smallest size and the fewest intermolecular forces. Fluorine gas, F2, is the smallest and
only non-polar gas of the answer choices. This makes Fr the closest to ideal of the answer choices, and ma
choice C the best answer.
79.
80.
81..
::::
(C
u:
Cr
's.ri:
3l L
'[ni*:
Affrl :
-[tl[4i'1.]l:]
,nhi
15lril
lrrr_ I.]!Llr
lff
illi4. j rl
I gr]:
,3m'TtrLo,
dbrqHt:
rtl f
f:lX ll]Li:l
4,tfrll,u-n
* S,',dllti' ; i:
LJ:ri
i15
dt_*
,Tlu*i
82.
83.
Copyright @ by The Berkeley Review@ 60 Section VI Detailed Ex
1 atm.:rminetole) =
re, the:orrectoice D
ar sizens areso 1,1-
mass,ES,
eculIori
ridionune
Choice A is correct. For a balloon to float in air, its buoyant force must equal its weight. The buoyant force is:roportional the volume of the balloon, so the greatest minimum volume ii associated with the balioon svstem:equiring the greatest buoy_ant{orce to float. The greatest buoyant force is required with the balloon ryrt'"- of-eatest weight (the one filled with a gas of highest density). Ammonia is the heaviest of the choices, so},oice A is the best answer.
Choice C is correct. Methane is lighter than nitrous oxide, so methane molecules have a greater average speed-:'an nitrous oxide molecules. This means that the cylinder's spin rate must be increased tJ Ue in phare"with the::ssing methane molecules. This eliminates choices A and B. The ratio of the masses of the two gases is 44 to-:, rvhich means that nitrous oxide is 2.75 times heavier than methane. Because relative ,p"uii, inversely::oportional to the square root of the relative difference in mass, the speed of methane is {IiE times the speed:: nitrous oxide. This eliminates choice D and makes choice c the besianswer.
Choice A is correct. The buoyant force of a gas-filled object depends on the density of the surrounding medium-':'d the volume of the object. The buoyant force does not depend on the contents of the balloon, alttough the' =:ght (and thus net force) do depend on the particles in the balloon. The greatest buoyant force belongs-to th"-:iect with the greatest volume, which makes choice A the best answer.
CLroice D is correct. If a larger cylinder is used, then the diameter and circumference are both increased, albeit:'' the same amount. The gas must travel farther to pass through the cylinder, but the hole in the cylinder must'-*'o travel a greater distance around the circumf"tu.rc" with Jach revolution. The two effects cancel out onel-r-'ther, so the spin rate does not need to change if the size of the cylinder increases. Choice D is the best::-._s 1\'er.
Cnoice A is correct- \Mhether Chamber 2 contains gas or not, the gas from Chamber 1 can effuse into Chamber 2,:-- '' then infuse into the cylinder. This means that evacuating Chamber 2 does not limit the ability of a gas to''-'-'er effuse or infuse' Choices B and C are eliminated. Buoyancy is irrelevant in this experiment, so chJice D::--1]6f be eliminated. In addition, upon its evacuation, the density of the gas in Chamber 2 becomes zero, so:-=:e is no buoyant force (no gas) within Chamber 2. Chamber 2 is evacuateJ to reduce the number of collisions,I :-'r-ing the gas to travel at a greater speed through the cylinder. Choice A is the best answer.
f iroice D is correct. If the temperature of the system increases from 25"C to 100'C, then the speeds of all gases-:hin the system also increase. The increase in temperature is from 298 K to 373K, so the increase is lessthan
: ::.tor of 2. This eliminates choices A and B. The speed of a gas depends on the square root of the temperature.r.:Leasured in kelvins), so the relative speed is equal totlgTz/298, which is roughly \TE. The value otttL2S
:r -ess than 1.25, so choice C is eliminated. The best answer is choice D.
lhoice B is correct. The relative speeds of gases under identical conditions depend on the relative masses of the' '::'pounds. N2O has a molecular mass of 44 grarns per mole, CO has a molecular mass of 28 grams per mole, F2-;= a molecular mass of 38 grams per mole, and Ne has an atomic mass of 20 grams per mole. This means that- - -'n has the greatest velocity, eliminating choices A and C. The next lightest gui it carbon monoxide, the.-lcity of CO must be second in the sequence. This makes choice B the best inr*"r.
Ciroice A is correct. Intuitively, as the volume decreases, the pressure increases, because the collisions increase.liis eliminates choices C and D, because the pressure must be greater than760 torr. It now becomes a question of:::ermining how much the pressure changes. The equatior,
"ur-t be set up intuitively. The final urlr*"i is in torr,
'-- it must be P1 = U, / rr'nr, where f is final, i is initial, and ? is yet to be determined. We know the pressure:ust increase, so insert the larger volume in the numerator and the smaller volume in the denominator. This.":ratestoP6=''00t/o.rSy'760torr. Thissimplifiesto4f gxT60tortwhich is760+7601u. Thevalue tor760f 3-< qreater than 250, so the answer is greater than 760 + 250 which best describes choice A. This may seem like a- : of work on paper, but this should all be done in your head. Do not write math, think math!
6l:-l::t O by The Berkeley Review@ Section VI Detailed Explanations
95.
96.
Choice A is correct. At STP, the total pressure is 760 torr. A system with 32% argon by moles corresponds to a
mole fraction of 0.32 for argon. The partial pressure of argon is 0.32(760 torr), which is roughly one third o1760.This is roughly 250 torr, which makes choice A the best answer. Choices C and D are all clearly too high.
Choice C is correct. The pressure decreases from 1.00 atm. to 0.90 atm., and the temperature decreases from 296K to 277 K. The decrease in pressure increases the volume. The decrease in temperafure decreases the volume.The change in pressure and change in temperature have opposite effects in this question. This means that thevolume should be roughly the same. This eliminates choices A and D. The math, carried out as follows,that choice C is the best answer.
lightest compound. This question is in essence asking for the relative molecular masses of the answer choiMethane is the lightest, so choices A and C are eliminated. Chlorine gas is the heaviest, so choice D is cor
100. Choice C is correct. The relative rates of effusion can be determined from the relatives masses of the twoaccording to Equation 6.12. Methane has a molecular mass of 1,6 grams per mole, while helium has anmass of 4 grams per mole. The relative effusion rate is found as follows:
EffusionRateH"Irr- = vHe _ 1@ru =r[16 =r[i =2
Effusion RateN4s1hl1e vCH+ /mHe ,t4
Because helium is lighter, the rate for helium is greater than the rate for methane (twice as fast, accordingthe math). Select choice C for optimum results in a situation such as this.
97. Choice B is correct. SO3 has a molecular mass of 80.06 grams per mole. Argon has an atomic mass of 39.95 graper mole. Sulfur trioxide is approximately twice as heavy as argon, so argon has a velocity that is roughly 1
(square root of 2) times greater than the velocity of SO3. This means it takes longer for the SO3 moleculeseffuse out of the balloon. The exact amount of time is greater by a factor of 7.4. 23.6 minutes x 1.4 is greaterboth choices C and D. The value is less than choice A, so the best choice is choice B.
98. Choice A is correct. The average speed of a gas can be found by using the equation " = /3kf . For this p
the relative rates of one gas to another is all that is asked, so 3, k (Boltzmann's constant), and T all can
Girrer' P1V1 = P2Yz w/ nconstant = PrVr
= P2Y2n1T1 nZT2 T1 T2
* = (fl(F, )u' = (ot-nJry\ozz ^t = (#)ezz,or > 67 2 m1
canceled out. The ratio of the velocities of the gases is: vunknown = -fil3*-"* = 2. This can be checked" Vwater fiassunknown
making sure that both ratios are greater than 1.0. It can be simplified by squaring the equation; - se9lwate.
' massunknown
4, so rn.,',Lr.,o*., = ffidss*utet
= 18 - 4.5 grams per mole. This is close to the molecular mass of He (4.0 grams44
mole), so choice A is the best answer. ]ust by looking at the choices given, one can see the other choices harmolecular weights that are well over 4.5 grams per mole.
99. Choice D is correct. The greatest infusion rate is observed with the fastest compound, which corresponds to
g
Copyright @ by The Berkeley Review@ 62 Section VI Detailed Ex
m 295,1ume.
at theihows
grarn:r1v 1.4
iiesrrt
tohoicorrec:
gas
atorr
Section VIIPhases
and
Phase Changes
Phasesa) Definitions of Solids and Liquidsb) Properties of a Solidc) Properties of a Liquid
Phase Change processesa) Terminology and Energeticsb) lypical Phase Diagramsc) Atypical Phase Diagramsd) Water Triviae) Supercritical Fluidsf) Heating Curves
Liquid-Gas Dguilibriuma) Vapor Pressure
b) Vapor Pressure eraphsc) Boiling Pointd) Boiling Point Variations
i. Increasing Atmospheric pressureii. DecreasingAtmospheric pressure
e) Raoult's Lawf) Distillation
Colli gative Propertiesa) Concentration Effectsbi Boiling Point Elevationc) Freezing Point Depressiond) Osmotic Pressuree) Conductance
WSpeciahztng in MCAT Preparation
Generic fleating Curve
cs = l.o tu'1n n
Cr = 2.otu'/noCs: 1.stut/sx
Solid to Liquid
Phases Section GoalsKnow the definitions for a solid and a liquid.
You-must klow yh-ere the solid, liquid, ul4 g?t sections of the phase diagram are. You must alsoDe aore ro rcrennry rne rrrPre Pornr/ normal p(Jllrng Pornr/ anq tne crlllcal pomt. De aole to loentllyatypical regions of the diagram, such as the area where the material exisls as a supercritical fluid.Be lble to iecoenize the ohase diaeram of water bv focusine on the nesative slooe of the barrierBe lble to iecognize the phase diagram of water by focusbetween the solid and liquid regions of the phase diagram.
of water by focusing on the negative slope of the barrier
Understand vapor pressure and boiling point.
Gr
I
E.:
'*j,', :'
|r
"-;I _: L
@?
You must know the differences between a solid and a liquid at the microscopic and macroscopiclevels. You must understand lattices, intermolecular forcei, and volume. ln boih liquids and soli'ds,the molecules are in contact with their neighbors, so they have a defined volume. '
l{now the terms and definitions of the phase change processes.You must know the terms and conditions for phase change processes. The phase changes include:evaporation, melting, sublimation, condensation, freezing, and deposition. You must also know theterms that describe the conditions under which these processes oicur, such as isothermal, isobaric,isochoric, and adiabatic condjtions. -You must, know the energetics associated with each phase changeprocess in terms of free energy, enthalpy, and entropy
Be able to read a heating,curve. ,.
You must be able to read a heating curve to extract information about enthalpv of fusion, enthalpvo[ vaporization, and heat capacities for the three phases. You should be able io identify the phJ#at different points on the hedting curve.
Kecoqnize the kev points and features of the phase diaqram.
g?,r*.*j 1..1..:':
"J .-.r '
il! ri u,;y:
il(i:: _
- lift *....
x::
ililI .! ._
rilii,iitit^ti:
illllfittil "t: .:. ;
Hlil:iidl't .:
,lllil r, : .
llfitlri,ry';;.
'IMlllliliii:uti:rt:-ii-
mitlhxl:i: :
*"i:@? rou must Know wnere tne sollcl/ llqulcl/ ano 8as sectlons or tne pnase ctlagram are. You must alsbe able to identify the triple point, irormal bo-iling point, and thb critical p'oint. Be able to identifatypical regions of the diheram, such as the area where the material exists as a supercritical fluic
t?
The conversion between liquid and gas occurs over a wide range of temperatures. At the boilingpoint, the vapor pressure i's equal td the atmospheric pressurel At temieratures lower than thEboiling point, vapor pressure still exists, but at a value that is lower than atmospheric pressure. Theeasierlf is to vapbrizb a compound (in terms of lower enthalpy), the greater its vapor^pressure.
Understand the principle behind Kaoult's law.Raoult's law describes vapor pressure above a mixture of two or more liquids. According to Raoult'slaw, in an ideal mixture, the vapor pressure of any given component depends on the niole fractionof that component inthe solutioh and itslormal vapor pressurb under the system conditions. Vaporpressure increases with temperature and mole fraction.
Understand the theory behind distillation.Distillation is a purification method you normally consider to be an organic chemisLry laboratorytechnique. On tihe theoretical level, it is based on the relative vapor pr"essures of the iomponentsand on Raoult's law. The vapor above a solution is richer in the rirore volatile component than thesolution. This difference in relative mole fraction can be exploited bv repeatedly evaporating andcondensing the mixture. With each cycle, Jt gets becomes riiher in th6 leds volati'le .oti.po.'*t-r?.
Understand the colligative properties of matter.Colligative properties are properties of a solution that depend on the number of solute particles inthe solution, not on the nature of the solute particles. Exahples include: freezing, point depression,boiling point elevation, osmotic pressure, andconductance. The colligative properties can be bxploitedto convert salt water into a more ion-frce state.
o?
@?
General Chemistry Phases and Phase Changes Introduction
-ie three common phases of matter are solid, liquid, and gas. The fundamental:-iference between the three phases is how the molecules (or atoms) interact with-:-e another. Solids have the least kinetic energy of the three common phases.-:.eu kinetic energy is in the vibrational and rotational forms. The molecules of a;, -jc do not change their positions. Liquids are more energetic than solids, yet-: as energetic as gases. Their kinetic energy is in the translational, vibrational,
.- i rotational forms, although the amount of their kinetic energy is less than that:: a gas. Gases have the most kinetic energy of the three common phases, and it.,-<ts ir the translational, vibrational, and rotational forms.
, : nore interest than the phases themselves are the phase change processes.-': -:rg from a less energetic phase to a more energetic phase is both an---:othermic and endergonic process. Changing phases requires changing the:r-, -iorfil€fltal conditions acting on a system. Each material has a unique phase;-i:ram that summarizes this information. we shall look at several phase:-iirarns in this section, noting key points and features of generic and specific: -.se diagrams. The MCAT test writers to date have focused on the atypical, r::cts of phase diagrams. Terminology for both conditions and phase changes' make up much of the phase change portion of this section.
- :.:t amounts of energy can be stored and released via phase change processes.*- = Jquid-gas phase change process is the source of the energy that drives
'* r'-:-res and makes refrigeration possible. It is important to understand that- - : se changes occur with changes in tempera ture or pressure, not just changes in:::erature. The idea of vapor pressure is not a complicated one, but it certainly":.,i :omplicated applications. The study of vapor pressure includes Raoult's law,
: = llausius-Clapyeron equation, and the theoretical perspective on distillation.--
' -argest portion of this section shall address the nuances of the liquid-gas. - 'brium.
. .:,a11 also address the behavior of solute particles in solution. Addition of' -:t affects certain physical properties of a solution, including boiling point,::nJlng point, osmotic pressure, and conductivity. These are known as
:;iiae properties, and they vary with the concentration of the solution.' ::=:sing the amount of a soluble material (solute) increases the interactions
'': *", €€r molecules in the liquid phase only, not in the gas or solid phase. Ice that, rirZ€r from a solution of salt water is relatively pure, with the occasional::::ed ion in its lattice strucfure. Equally, steam formed from evaporation from
.:-: rvater solution consists of pure water vapor. Purification processes are=::ed by colligative properties.
*'. :-iost important features of the phase change section are applications of
- ":.e changes. These often make up the bulk of the material from this section" :: :ppears on the MCAT. Understanding water as both a pure material and as
. - --,'ent is critical. water has many unique features, due to its extensive" ::-rgen bonding. Water has the highest density of hydrogen bonding of any-.::=:ial, which drastically affects its physical properties, such as boiling point,*.,-::tg point, and solubility.
;', nght @ by The Berkeley Review 65 Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Phases
PhasesDefinitions of Solids and LiquidsSolids and liquids are two of the three common phases in which we observe
matter. In boih cases, molecules are in contact at the microscopic level with one
another, which accounts for solids and liquids having a definite volume. The
difference between the two phases lies in the ability of their molecules to move.
In solids, molecules are in a lattice structure, and they experience vibrational and
rotational motion only. The atoms are not displaced, so there is no translational
motion. This results in solids having definite shapes on the macroscopic level. In
liquids, molecules are free to move, so they experience translational motion,
although there is no net motion of the system. The molecules also experience
vibrational and rotational motion. The result of translational motion of itscomponent molecules is that liquids have no definite shape on the macroscopic
levei. Definitions for both a solid and a liquid are listed below:
Solid: A state of matter having both a definite shape and a definite volume.
The molecules do not change position, resulting in a fixed structure,
whose molecules are in contact with neighboring molecules at alltimes.
Liquid: A state of matter having a definite volume, but no definite shape.
The molecules change position, resulting in continuous randommotion, and molecules that remain in contact with neighboringmolecules at all times.
Properties of a SolidBecause solids are defined as having both a definite shape and a definite volume,
they have dimensions on the macroscopic level. The molecules of a solid are
arringed in a fixed lattice structure, on the microscopic level, they have repeating
structural subunits referred to as unit cells. The macroscopic shape of a solidobject can vary, but the microscopic arrangement (lattice structure) is constant.
This is often referred to as the crystal structure (or lattice structure) of the solid.
Lattice structures exist for both elements and compounds. Figure 7-1 shows two
lattice structures and unit cells for atomic solids.
Unit cell is a face centered cubicwith four atoms per cell. Unit cell is a simple cubic
with one atom per cell.
Gt
-:E
:.-"'"; -li:::.:
i,:--
t--t-J
;f:.1
::h:: -_--ts-:
,": --_ lts-:
:e:-;=jqr-:- -.E
:[:k
fr-
./$.-
m"
D
Copyright @ by The Berkeley Review 66
Figure 7-1,
The Berkeley Keview
General Chemistry Phases and Phase Changes Phases Change Processes
The three most common unit cells are the simple cubic with one atom perrepeating unit cell, the body-centered cubic with two atoms per repeating unitcell, and the face-centered cubic (which forms the cubic closest packed structure)rvith four atoms per repeating unit cell. Alkali metals often pack to form thesimple cubic. Gold and silver pack to form a face-centered cubic.
Solids have density (which is often used as a physical measurement in theidentification of an unknown solid). A common procedure for determining thedensity of a solid is to place the solid into a volumetric container filled with aknown liquid. The solid's volume is determined by measuring the displacementr-olume of the liquid. It is assumed that the displaced volume is due strictly tothe volume of the solid. The mass is measured separately, and from this, thedensity of the solid may be calculated. This technique should not be mistakenn ith displacement weight, as in buoyancy. If the solid is buoyant in the liquid,then the density of the solid is less than the density of the liquid, and the relativedensity of the two materials is equal to the percent of the solid that is submerged.It should be intuitively obvious that in an immiscible mixture of a solid and-rquid, the less dense material rises to the top. \A/hen the solid is less dense thanlhe liquid, the mass of the floating solid is equal to the mass of the liquidrccupying the same volume as the submerged portion of the floating solid.These topics are more likely to be found in the physics section of the exam.
Example 7.L
"\hich of the following statements about solids is (are) NOT true?
I: Solids change molecular structure constantly.
II: Solids have less kinetic energy than gases and liquids.III: Solids have more entropy than gases or liquids.
-\. I onlyB. III onlyC. I and II onlyD. I and III only
5olutionSolids have organized lattice structures, so they do not change moleculars:ructure, making statement i an invalid statement. The question is asking fornvalid statements, so the correct answer choice must contain I. This eliminatesloice B. Because atoms in solids exhibit no translational motion, solids have the,east kinetic energy of the three common phases. This makes statement II valid,;nd eliminates choice C. Because solids form lattice structures, their atoms:-raintain position, resulting in lower entropy than liquids and gases (where theitoms are free to move and thus change position). This makes statement III,ni'alid. From this, it can be concluded that choice D is the best answer.
Properties of a LiquidLiquids have no definite shape, but they do have a definite volume.-nderstanding liquids is important in both chemistry and physics. Liquids are:'luids, so they are fundamental to all discussions of hydraulics, buoyancy, and:re Bernoulli effect. From a chemistry point of view, however, it is more:nportant to know the molecular structure and the solvent properties of liquids.'r ou must consider intermolecular forces such as hydrogen bonding, polar-rteractions, dipole moments and van der Waals forces. The strength of these:lrces is in the same order as just listed in the preceding sentence.
Jopyright @ by The Berkeley Review ot Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Phases:
It is important to recall that the strength of these forces is comparable up to theboiling point. Above the boiling point, molecules are no longer in contact.Certain intermolecular forces also affect solubility properties, as often studied inorganic chemistry. Figure 7-2is a schematic representation of a liquid.
All molecules are in contact with one another,but no definite lattice structure exists.
FigareT-2
Because their molecules are in motion, liquids can flow, which defines them as a
fluid. Incidental facts about liquids that you should know are that liquidstypically have the highest heat capacity of the three common phases, and liquidsare compressible. The density of a liquid decreases (because the volumeincreases) with increasing temperature, with the exception of water from 0"C to4"C. It should be noted that a compound is a liquid at room temperature, if itsmelting point is less than room temperature and its boiling point is greater thanroom temperature. A liquid also has other physical properties, such as surface
tension andaiscosity. These tetms are defined below.
Surface tension: The resistance of a liquid to an increase in its surface area. Itgenerally increases as the intermolecular forces increase.
Viscosity: The resistance of a liquid to flow. It can also be observed as
the resistance to flow by an object through the liquid.
ExampleT.2If an object floats on the surface of a liquid in such a manner that sixty percent ofthe object is submerged, which of the following sets of density values accuratelydescribes the liquid and solid?
A. Liquid: 0.90 grams per mL; solid: 1.50 grams per mLB. Liquid: 1.25 grams per mL; solid: 0.75 grams per mLC. Liquid: L.00 grams per mL; solid: 1.67 grams per mLD. Liquid: 1.50 grams per mL; solid: 1.00 grams per mL
SolutionBecause the solid floats on the liquid, the buoyant force upward isthe gravitational force (weight) downward. This means that thedense than the liquid and eliminates choices A and C. According to
greater than
principle, the percent of the solid submerged is equal to the density of the solidivided by the density of the liquid. This means that choice D is eliminabecause the ratio of the solid density to the liquid density is2/3, which0.67, not 0.60. This would make the solid 67"h submerged. The correct ans(and only choice remaining) is choice B. The ratio of 0.75 to 1.25 is equal to 0.60-
ffi
TeASrah,
PTIdrisolcbtpdnuwhptrmtllfre*q
solid is
Copyright @by The Berkeley Review 68 The Berkeley
General Chemistry Phases and Phase Changes Phases Change processes
P. :i1ffiffiffigGlffi$ffi$$$bsTerminology and Energetics-\ phase change process is a physical process, not a chemical process, by whichthe state of matter changes. ln a phase change, the molecules themselves do notchange, but the interactions between molecules do change. A phase changeprocess is reversible. For instance, if a solid is converted into a liquid bychanging the external conditions, then the liquid can be converted back into asolid upon reestablishing the original conditions. All phase changes involve a:hange in the enthalpy (heat energy of the system) and entropy (organizational:otential energy of the system, considered as randomness of the molecules).Phase changes are either endothermic or exothermic processes, depending onl'hether the final state is of higher or lower energy than the original state. If alarticular phase change is an exothermic process, then the addition of heat is:equired for the reverse process because it must be an endothermic process. On:he macroscopic level, phase changes result in changes in volume and changes inshape. The six typical phase change processes are listed in Table 7.1 below.
Phase Change Term Heat Term Entropy Term
Liquid to gas Vaporization Endothermic (AH,ruo) (AS"u")
Gas to liquid Condensation Exothermic (- AHuup) (- AS.,uo)
Solid to liquid Melting Endothermic (AH6rrsien) (ASfnrion)
Liquid to solid Freezing Exothermic C AHf,,ri.,n) C ASrnsior,)
Solid to gas Sublimation Endothermic (AHrrr6) (ASt,rb)
Gas to solid Deposition Exothermic C AHs"b) (- ASsuu)
Table 7.1
-\s a point of interest, you should know that the relative energetics for the threerhase change processes are: AHrr6limation > AHrrapo.ization > AH1,rs1or., andlSsublimation ) ASrruporization > ASfusion
Example 7.3lhe conversion from a solid to a gas represents what kind of phase change and:,as what kind of energy change is associated with it?
{. It is known as sublimation, and it is exothermic.B. It is known as deposition, and it is exothermic.C. It is known as vaporization, and it is endothermic.D. It is known as sublimation, and it is endothermic.
SolutionThis question focuses on either memorization or your ability to move your eyesup this page slightly to read Table 7.1. (For your exam, you may prefer thenemorization route, because if you have Table 7.1 in sight, you are probably:heating.) According to Table 7.1, conversion from solid to gas is sublimation,and the process is endothermic (i.e., because the particles finish at a higherenergy level, then energy must have been added to the system.) Choice D is the'rest
answer.
Copyright @ by The Berkeley Review 69 Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Phase Change Processes
Example 7.4Which of the following statements is NOT true with regard to phase changesstarting from the solid phase?
A. Sublimation results in a system with increased entropy.B. During melting, the lattice structure of solid matter is broken down.C. Following sublimation, the space between molecules is increased.D. Following melting, atoms lose their ability to migrate freely.
SolutionSublimation changes a material from the solid phase into the gas phase. In thesoiid phase, molecules are held in a rigid lattice structure and are not free tomove. Once in the gas phase, the molecules are free to move in any directionuntil a collision occurs. Because of the increased mobility associated with the gas
phase, there is more entropy in the gas phase than the solid phase' This makeschoice A a valid statement. During melting, the material transforms from thesolid phase into the liquid phase. Any lattice structure that may exist in the solidphase is lost in the liquid phase, as the molecules are free to migrate. This makes
choice B a valid statement. Because a gas is less dense than a solid, more fr'ee
space must exist between the molecules of a gas than the molecules of a solid. lnthe gas phase, the molecules are independent and exhibit no intermolecularforces, so they are far apart. In the solid phase, the molecules exhibitintermolecular forces, and they are touching one another. This makes choice C avalid statement. In the liquid phase, molecules have the ability to migrate freelythrough a solution, so when the phases changes from a solid (with no freemigration of molecules) to a liquid (which has free migration of molecules),molecules are able to go in any direction through solution. This makes choice Dan invalid statement, and thus the best answer.
In addition to the phase change terminology, you should know the terms thatdescribe the conditions of a system. There are four terms that describe staticsystem conditions with which you should be familiar. They are listed below:
Isothermal Conditions where the temperature of the system does not change(sometimes referred to as T6661nn1).
Isobaric Conditions where the pressure of the system does not change(sometimes referred to as P.6r-rs1211).
Isochoric Conditions where the volume of the system does not change(sometimes referred to as Vs6ns1611).
Adiabatic Conditions where the system is perfectly insulated, so that heatneither enters nor exits the system (referred to as qq61s16.1).
Isochoric is not likely to be seen, but nonetheless it is a relevant term. You can
often distinguish whether a term is important or not by its familiarity. If it was
never mentioned during your course lectures on campus (e'g', words likeisochoric and deposition), it is less likely that you will need it memorized for the
MCAT. They will likely define it for you. That said, a perfect example is asupercritical fluid, whichhas appeared on several different versions of the MCAT.This is a great example of how the MCAT presents an unfamiliar term to describe
an interesting concept, which can be deciphered from the parts of its name.Beyond the critical point exists a supercritical fluid. It is supercritical because itexists beyond the critical point, and it is a fluid because it is part gas and partliquid, both of which flow (are amorphous).
Ge
TypicA ptt;g1\zen
phaseinterfdiagrthe x-the gr
The ftripX*is ttuBel'sso ittemp
ExatrIn rrl
A. I
B. l
c. l
D..
SoltIhe-\n r
1-aPavaihat
GCopyright O by The Berkeley Review 70 The Berkeley Review
General Chemistry Phases and Phase Changes Phase Change Pnocesses
Typical Phase DiagramsA phase diagram is a graph that summarizes the different states of matter for agiven compound or element with respect to temperature and pressure. From a
phase diagram, you can determine the phase that is most favorable or theinterface of phases for a material under a given set of conditions. Reading phase
diagrams accurately requires drawing lines from a given point on the diagram tothe x-axis and the y-axis. The point where the lines intersect is in the region ofthe graph for the most favorable phase. Figure 7-3 is a generic phase diagram.
Figure 7-3
The triple point, critical point, and normal boiling point are shown in Figure 7-3. The
triple point is where all three phase can coexist simultaneously The critical pointis the highest temperature and pressure at which a liquid may be observed'Beyond the critical point, it is impossible to distinguish between a gas and liquid,so it is referred to as a supercritical fluid. The normal boiling point is thetemperature at which a material boils when the pressure is 1'00 atm.
Example 7.5In what phase (or phases) does the material exist at point e on the graph below?
Temperature
A. Liquid onlyB. Liquid with a small amount of solidC. Liquid with a small amount of gas
D. Liquid with a small amount of vapor
SolutionThe middle region of a phase diagram is where the compound exists as a liquid.An equilibrium between evaporation and condensation exists, so liquids exhibitvapor pressure. Choice D is the best answer. Choice C looks tempting (because
a vapor is a gas), but vapor is the better word. The term aapor desctibes a gas
that exists in equilibrium with a liquid at a temperature below the boiling point.
.}
;Lr
(nog
ot+
(,U)CJtsr
Critical point
Temperature (K) ----+
Copyright @ by The Berkeley Review 7l Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Phase Change Processes
Temperature(K) +
It should be noted that the phase diagram does not show the only phase that isobserved under specific conditions, but the most abundant (and stable) phaseunder those conditions. The most challenging part of phase diagrams is workingthrough the terminology.
Example 7.5If a compound that exists primarily as a liquid at 39"C and 1..72 atm. is heated at
constant pressure, it converts into a gas. Upon cooling the compound back to39'C and 7.12 atm., the compound remains a gas. which of the followingexplanations BEST explains this observation?
A. The boiling point for the compound at 1.12 atm. must be 39'C.B. The molecules rearranged so that the bonds are different from those in the
original compound.C. The compound must have two liquid phases.
D. The compound was heated beyond its critical temperature and thus cannot
become a liquid again.
SolutionA compound can exist in either of two phases when it is at its boiling point(liquid/gas), melting point (solid/liquid), or sublimation point (solid/gas)' For
the compound to exist as either a liquid or gas at a given temperature and
pressure, that temperature must be the boiling point at that pressure. There is no
indication that a chemical reaction took place, so choice B is not the best choice.
Having two unique liquid phases would not explain the observation, so choice C
is eliminated. Even after a compound is heated beyond its critical temperature, itis capable of changing back to its original phase once cooled, so choice D iseliminated. The best answer is choice A.
ExalrrpleT.TWhich arrow is INCORRECTLY identified below?
j
t
ll
^d
"I
M
d@
Imftsm
,T
3CJk)30)k
4{Arrow a is isobaric condensation.Br-.Arrow b is isothermal deposition'
,fr..1lrrow c is isothermal freezing'H Arcow d is isobaric sublimation.
Copyright Oby The Berkeley Review 72 The Berkeley Review
General Chemistry Phases and Phase Changes Phase Change Processes
SolutionIsobaric condensation is where the material is cooled at constant pressure from agas into a liquid. Because pressure is constant, the arrow musl be horizontal,which it is. Because it is condensation, the arrow must start in the gas region andfinish in the liquid region, which it does. Arrow a is correctlyldentlfied, sochoice A is eliminated. Isothermal deposition is where the material iscompressed at constant temperature from a gas into a solid. Becausetemperature is constant, the arrow must be vertical, which it is. Because it isdeposition, the arrow must start in the gas region and finish in the solid region,which it does. Arrow b is correctly identified, so choice B is eliminited.Isothermal freezing is where the material is compressed at constant temperaturefrom a liquid into a solid. Because temperature is constant, the arrow must bevertical, which it is. Because it is fueezing, the arrow must start in the liquidregion and finish in the solid region, which it does not. Arrow c is incorrictlyidentified, so choice C is the best answer. Isobaric sublimation is where thlmaterial is heated at constant pressure from a solid into a gas. Because pressureis constant, the arrow must be horizontal, which it is. Because it is subfimation,the arrow must start in the solid region and finish in the gas region, which itdoes. Arrow d is correctly identified, so choice D is eliminated.
-{typical Phase DiagramsThere is a difference between a typical phase diagram for most compounds andthe phase diagram for water. The difference between the two phase diagrams isJre negative slope associated with the line separating liquid ?rom solid. Mostcompounds can be compressed from a liquid into a solid at constanttemperature. The unusual thing about water is that an isothermal increase inDressure compresses the solid (ice) into a liquid (water), resulting in the liquidberng denser than the solid. The properties of water can be referrbd to as wiirdlFigure 7-4 shows the phase diagram for water and a typical compound.
Fhase Diagram for Typical Compound phase Diagram for water (Hro)
Figure 7-4
1,.}
;ti
(,]aqJs<ei
Liquid
Temperature (Kelvin) --*
Phase Diagram for Water (HzO)
Temperature (Kelvin) --+
- :rr,-right @ by The Berkeley Review tc Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Phase Change hocesses
Temperature
Example 7.8Point a on the phase diagram below is which of the following?
A. The critical point for carbon dioxideB. The triple point for waterC. The triphasic point for carbon dioxideD. The plasma point for water
SolutionPoint a is where all three phase lines intersect, so it is the triple point. The triplepoint is defined as the temperature and pressure at which a compound may existi. uny or all of the three phases sirnultaneously. The negative slope of the liquid-solid interface line confirms that the compound is water. The correct answer ischoice B.
Water TriviaRandom, unrelated facts about water have been tested on the MCAT on morethan one occasion. Two key facts about water are that its liquid form is denser
than its solid form, and it is densest at 4"C. The solid can be compressed into a
liquid under relatively mild conditions. The ramifications of this include ourability to ice skate upon frozen water. The fact that water is densest at 4'C resultsin the presence of the warmest water being at the bottom of a frozen lake (useful
information on youf next ice fishing trip). Water also has an extremely highboiling point for a compound with such a low molecular mass. Ammonia (NHS)
is of roughly comparable mass, and it has a boiling point of -33.4"C' Methane(CHa) is also of roughly comparable mass, and it has a boiling point of -1'62.6'C.
This means that for water, hydrogen bonding has a substantially more significanteffect on the boiling point than mass or other intermolecular forces. Water has
the densest hydrogen bonding of any compound'
Supercritical FluidsA supercritical fluid exists when the conditions exceed the critical point. Super-
critical fluids take on properties that are a compromise between gases and
liquids. Molecules are in free-moving clumps, where the density is greatest at
the bottom of the container (like a liquid), but the material fills the entirecontainer (like a gas). Supercritical fluids have a density between that of a liquidand gas. Specific examples of supercritical fluids that are used industrially are
carbon dioxide and oxygen. Supercritical fluid carbon dioxide is used as adecaffeinating solvent for coffee. It is a safe solvent for dry cleaning, whichmakes it a great alternative to the more toxic organic solvents like methylenechloride. Supercritical fluid oxygen chambers are used to completely oxidizematerials that do not burn cleanly under standard conditions with oxygen gas'
I
fl
il
fr
0
,0
Copyright @by The Berkeley Review 74 The Berkeley Review
General Chemistry Phases and Phase Changes Phase Change Processes
Heating CurvesTo generate a heating curve, a material is heated or cooled at a constant rateunder isobaric conditions over a broad temperature range, and the temperatureis recorded as a function of the heat added (or removed). A heating curve showsthe same features as a horizontal line in a standard phase diagram, but insignificantly more detail than a phase diagram. Figure 7-5 shows a typicalheating curve, drawn according to convention. As a good practice, draw graphswith the axes interchanged, so that you recognize non-conventional views.
During phase changes: AHuuo > AH6,r, and
During heating: Cphur" - #
v0lfr
6fr
Liquid to gas.phase change
Total heat added at constant rate
Figure 7-5
Figure 7-5 depicts the overall view of the phase change process at constant:ressure from an energy perspective for a generic material. The graph of:emperature as a function of heat added shows the phases, phases changes, andreat capacity of each phase. As labeled, plateaus in the graph represent phase
,Jrange processes, because as a material changes phase, the temperafure remains;onstant. During a phase change, the temperature does not change, because theenergy added to the system is being used to break intermolecular forces, rather'han to increase the average kinetic energy of the molecules. The enthalpy of'"'aporization (represented by the second plateau) is greater in magnitude than:he enthalpy of fusion (represented by the first plateau). This is because more:nergy is necessary to break the intermolecular forces (as observed with the:onl'ersion from a liquid into a gas) than is necessary to weaken the,ntermolecular forces (as observed with the conversion from a solid into a liquid).Perhaps it is more obvious that the boiling point is greater than the melting point,--ran it is that AHvaporirati.r., is greater than AHfrrrio.r, but both concepts are
:ooted in the same supporting idea.
\on-horizontal lines represent where the material is being heated without any:hase change transpiring. In these regions, the temperature of the material is-:creasing. This means that the slope of each line correlates with the ratio of:ernperature change to the heat input. The heat capacity of a material is the heat:equired to raise one glam of the material by one degree Celsius, so the units of:.eat capacity (C) are calories per gram'kelvins. The slope of the line in the-emperature change regions of a heating curve is temperature change per heat:Jded, which has units of kelvins per calories. The slope of the line is inversely::oportional to the heat capacity of the material in that particular phase. This is::, say that the flatter the line, the greater the heat capacity.
ts
-c)
-opyright @ by The Berkeley Review tc Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Phase Change Processes
In Figure 7-5, the slopes for each of the three phases are different, implying thatthe heat capacities for each of the three phases are different. The heat capacity ofthe liquid is the greatest of the three phases. This is not unusual, given thatliquids can absorb intermolecular vibration energy as well as translational kineticenergy. Solids can only vibrate, and gases can only move translationally. This isthe reasoning behind choosing a liquid as the medium to absorb heat in a heat-exchange system, such as an automobile radiator. Heat-exchange systems are
always set so that heat is transferred in a counter-current fashion. This is also
seen with ion exchange (a good example is the kidney).
Example 7.9
If melting a solid requires exactly 20 k] per mole, then which of the followingstatements is true?
A. Sublimation requires less than 20 kJ per mole.B. Vaporization requires less than 20 kj per mole.C. Freezing releases more than 20 k] per mole.D. Condensing releases more than 20 kJ per mole.
SolutionGiven that AHrrr6limation is equal to AH6ssi6r', + AHrru'orization, and thatAHvaoorizatior., is greater than AH6rrrion, then AHsublimatio. must be greater thanAHf,rrio., (which equals 20 kJ per mole). This eliminates choices A and B'
Freezing releases the same amount of heat that melting absorbs, so choice C is
eliminated. Because vaporizing absorbs more heat than melting, condensationreleases more heat than freezing. The correct answer is choice D.
u
.N
Idil
m
'!t
qffi
d,d
Copyright O by The Berkeley Review The Berkeley Review
General Chemistry Phases and Phase Changes Liquid-Gas Dquilibrium
ffi $ffi;ffi u*iiiircHfiiIf; ffi ffi ffi .
Vapor PressureAbove every liquid and solid (although we mostly consider liquids), there is afinite amount of vapor formed when molecules at the surface layer escape. Avapor is composed of gas particles that are in equilibrium with the liquid phase,so they temporarily exist a gas, until they condense back into their morefavorable phase. For example, we refer to oxygen as "oxygen gas" at roomtemperature, because it exists as a gas in its most stable form. On the other hand,we refer to water in the gas phase as "water vapor," because the most favorableform of water at room temperature is a liquid. We refer to steam as "water gas"only when it exists at a temperature above the boiling point of water (100'C).Gaseous water molecules at temperatures lower than 100'C are in a state ofequilibrium between evaporation and condensation. The formal definition ofvapor pressure above a liquid is:
Force per unit area aboae the surface of a liquid exerted by molecules formedupon eaaporation of the liquid.
The vapor pressure of a liquid is simply the partial pressure exerted by the gasmolecules formed by evaporation from the surface of the liquid, when it is inequilibrium with the gas molecules condensing back into the liquid. The vaporFressure above a pure liquid depends on the temperature of the liquid and theSvaporizatior., of the liquid. Figure 7-6 shows the surface of a liquid and therost favorable points from which molecules can evaporate.
Molecules can evaporate only from the surface,and tend to evaporate from the comers.
Figure 7-6
--. the liquid phase, molecules are not packed in the same orderly fashion as they;:e in a solid. The picture in Figure 7-6 is drawn with orderly particles forsi:rplicity. Comer molecules have the fewest neighbors, so they have the fewest.r:ermolecular forces. Because of the minimal forces, molecules evaporate most::aCi-lv from the comers.
i .:For pressure can be measured in either an open or closed system. In a closedi1il':tem, a partial pressure of vapor exists, because the rate of vapoization equals:e rate of condensation. This is most typically how vapor pressure isi*er"rnined. Flowever, the definition of vapor pressure does not apply only to a-,:sed system. In an open system, the vapor escapes, so it does not reach a staterr equilibrium between the rate of vaporization and the rate of condensation.lie r"apor pressure is a measure of the pressure just above the surface of thej,r'::d. It is difficult to measure, so we generally consider vapor pressure in anjr'-; system from a theoretical perspective, and apply values that have beenrer:rurined previously in a closed system at known temperature.
- :r",ight @ by The Berkeley Review lt Dxclusive MCAT Preparation
General Chemistry Phases and Phase Changes Liquid-Gas Dquilibrium
Vapor pressure is independent of the shape and volume of a container. Thevapor pressure above a dish of ethanol is the same as the vapor pressure above atest tube of ethanol. Despite the fact that the dish has substantially more liquidsurface area, vapor pressure is still measured as force per unit area. The sameamount of ethanol is vaporizing from each system per unit area of its surface, as
long as the temperature of both systems is the same. The rate of vaporization isgreater with the greater surface area in Container 2, but the vapor pressure is thesame in both containers. Figure 7-7 comparcs two containers that share the samevapor pressure, when they contain the same liquid at the same temperafure.
Container 2
FigureT-7
Because vapor pressure depends primarily on solution temperature (energy ofthe molecules) and AHtuporiralisn (the energy necessary to overcomeintermolecular forces), the atmospheric pressure does not significantly affect thevapor pressure. Boiling point depends on the atmospheric pressure, but do notconfuse boiling point and vapor pressure. The following relationships hold truewith regard to the vapor pressure of a system:
P.'apo, increases as temp increases; Prrupo'" decreases as AFluu'orirulior., increases
Example 7.10\Atrhich of the following statements is TRUE about vapor pressure?
A. The vapor pressure of a liquid always decreases with temperature.B. The vapor pressure of a liquid is always lower at higher elevations.C. The vapor pressure of a liquid is always higher at higher elevations.D. The vapor pressure of Liquid X is greater than the vapor pressure of Liquid
Z, if the normal boiling point of X is lower than the normal boiling pornt of Z.
SolutionVapor pressure is directly proportional to temperature, so vapor pressure alwaysincreases with temperature. This makes choice A incorrect. Elevation has no (ora negligible) effect on the vapor pressure, because the atmospheric pressure doesnot affect the vapor pressure (or rate of vaporization) to any appreciable extent.The temperature and AHrrupori"ali.n have the greatest effect on the vaporpressure. A lower atmospheric pressure lowers the boiling point, but it has noeffect on the vapor pressure. This eliminates choices B and C. As the boilingpoint increases, the ability for a compound to vaporize decreases. This meansthat as the boiling point increases, less of the vapor is formed and consequentlythe vapor pressure decreases. This makes choice D correct.
Container 1
Copyright @ by The Berkeley Review 7A The Berkeley Review
General Chemistry Phases and Phase Changes Liquid-Gas Equilibrium
Vapor Pressure GraphsVapor pressure increases as the temperature of the solution increases, but not in a
linear fashion. The graph in Figure 7-8 shows the relationship between vaporpressure and temperature. The graph cannot intersect the y-axis when the unitsof temperature are kelvins and the axis originates at zero.
Temperature
Figure 7-8
The relationship between vapor pressure and temperature is exponential. It ishighly unlikely that you have the Clausius-Clapeyron relationship memorized,so the only way to know this is using intuition. If you have ever boiled water forany reason, then you have observed the process that generates the graph inFigure 7-8. It take a while for the bubbles to form in water as it nears its boilingpoint. Once bubbles form, it is only a short period of time before the wateractually starts to boil. The graph shows a small increase in vapor pressure atiower temperatures (which fits with the lengthy duration before bubbles form),:ut a rapid ascent at higher temperatures (which fits with the short period oftime between bubble formation and boiling).
Equation 7.7, the Clausius-Clapeyron equation, can be used to explain the:elationship between vapor pressure and temperature. If T2 is greater than T1,
.hen the value of AHvaP /f -al is neeative, so the value of rr L is negative.
R \r' Tr /'" "-o"' P2
-',hen In h is negative, P2 is greater than P1. The conclusion from this is that asP2
:''': temperature of a liquid increases, the pressure of its aapor increases, but in an
"rponential manner. Vaporization is always an endothermic process/ soHr.aporizatior., is always a positive value.
tr
th6qJLO.tro
s
0.7\
:quation 7.7 canbe used to calculate the vapor pressure of a liquid at any given:.nperature, as long as the vapor pressure of the same liquid at another::mperature is known. The vapor pressure at the normal boiling point is 760 totr:re of a pair of values is usually vapor pressure at the normal boiling point). In
::dition to determining the vapor pressure of a liquid at a given temperature:--m knowing its boiling point, the boiling point for a compound can be::lculated at any given atmospheric pressure if any other vapor pressure for it isi:ronn. Calculations involving Equation 7.7 arc unlikely, but it should still be,:'Cerstood conceptually and graphically. It is a natural log function of pressure
=C temperature, but it is tricky to use because of the minus sign.
AH"^porization/1 - 1\lnPr - LnP2 = --fi:{t ,,1
- n,right @ by The Berkeley Review 79 Dxclusive MCAT Preparation
General Chemistry Phases and Phase Changes Liquid-Gas Equilibrium
Of more use than calculations is a conceptual understanding of the relationshipbetween boiling point, enthalpy of vaporization, volatility, and vapor pressure.Figure 7-9 demonstrates the relationship of these measurements.
As AHrru'orization t; Volatility J; Boiling point'l; Vapor pressure JFigweT-9
The reason the graph in Figure 7-8 reaches an endpoint is because the solutionreaches its boiling point, the highest temperature at which liquid may still exist.The boiling point is dependent on the atmospheric pressure, because the highestvapor pressure that the solution can reach is equal to atmospheric pressure. Theboiling point is defined as the temperature at which the vapor pressure is equalto the atmospheric pressure. The limits of the graph in Figure 7-8 are shownbelow in Figure 7-10.
Temperature ("C) Normal boiling point
Figure 7-10
Example 7.11lf Compound X evaporates more readilyfollowing statements must be FALSE?
than Compound Y, which of the
A. Compound X has a greater vapor pressure than Compound Y at roomtemperafure.
B. Compound X has a greater boiling point than Compound Y.C. Compound X has greater volatility than Compound Y.D. Compound X has a lower enthalpy of vapotization than Compound Y.
SolutionBecause Compound X is evaporating more readily than Compound Y,Compound X is more volatile than Compound Y, making choice C a validstatement. A more volatile compound has a greater vapor pressure (moreevaporated molecules) at a given temperature, so choice A is also a validstatement. With greater volatility, Compound X must have weakerintermolecular forces, and thus a lower AHvaporizatior", than Compound Y,making choice D a valid statement. The easier it is to vaporize a compound, theless energy that is required to vaporize it, so it has a lower boiling point. Thismeans that Compound X has a lower boiling point. Choice B is an invalidstatement, making it the correct answer.
l
m
fm
ml
&
ITlroqJH
Oaa,)Lo.k
o.d
Copyright O by The Berkeley Review 80 The Berkeley Reviex' ;
General Chemistry Phases and Phase Changes Liquid-Gas Equilibrium
Boiling PointThe boiling point of a compound is defined in two ways: one based on phasesand the other based on vapor pressure:
The boiling point is the tunperature abotse which a substance may not exist as a liquid.
The temperature at which the aapor pressure of a liquid is equal to the atmosphericpressure.
This means that at reduced atmospheric pressure (for example, in the mountains)the boiling point of a liquid is lower than normal. Under cases of increasedpressure (such as the typical pressure in a pressure cooker), the boiling point of aliquid is higher than normal. The normal boiling point of a compound ismeasured at L.00 atmospheres of pressure. The boiling process involvesvaporization of a compound in its liquid phase, which converts it to its gasphase. The boiling point is affected by intermolecular forces and atmosphericpressure.
Figure 7-1.1. compares the boiling points for the hydrogen-based compounds ofsixteen different elements. Notice that compounds containing elements alignedin the same column of the periodic table are connected in groups of four by solidor broken lines.
H2Te
NHa Hzo'
J/3Boiling point (K)
Figure 7-11
;i:s graph shows the effects of hydrogen bonding, molecular mass, and polarity:n the boiling point. The effect of hydrogen bonding on the boiling point of a::u'rpound is seen with the deviation of.H2O, HF, and NH3 from linear behavior.ae effect of molecular mass on the boiling point is seen in the gradual increasen boiling point as a periodic group is descended. The effect of polarity on the:'::,ling point is demonstrated by the higher boiling points associated with the::npounds having two lone pairs, rather than with the compounds of equalr;*.s having either one or three lone pairs. The molecules with iwo lone pairs;.re the most polar of any compounds having comparable molecular masses.
SnHa
! ;th
':!
| ^,a? l'*
"T
' PHg'
/c?rn HF
300100
l:nr,right @by The Berkeley Review al Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Liquid-Gas Equilibrium I
This graph (Figure 7-11) presents a large package of data, from which you mustbe able to extract useful information and understand it in terms of the definitionsand background knowledge that you already possess. For instance, the boilingpoints of HCl, HBr, and HI vary because of the increasing molecular mass of eachcompound. The greater the molecular mass of the compound, the harder it is tovaporize the compound. The harder it is to vaporize the compound, the greaterthe boiling point of that compound. This accounts for the increase in boilingpoint with increased mass for the haloacids. The same trend is observed with theother compounds in their periodic column (family).
Much can be observed when the boiling points of H2S and HCI are compared,because their molecular masses are roughly equivalent, and neither formshydrogen bonds (the hydrogen must be bonded to an N, O, or F for hydrogenbonding). The boiling point of H2S is greater than that of HCl, because the H2Smolecule is more polar than HCl, due to its bent geometry. To see this, thecompounds must be viewed in three dimensions. The slight deviation fromlinear behavior within the family that is observed with H2S and HCI is used bysome chemists to argue that the two compounds form very weak hydrogenbonds. This deviation is small compared to the ones observed in NH3, H2O, andHF, so any hydrogen bonding is minimal (if not negligible).
The boiling point of water is the greatest of the sixteen compounds listed inFigure 7-71,,because the oxygen has two lone pairs and two hydrogens that arecapable of forming hydrogen bonds. Therefore, all of the lone pairs andhydrogens have partners with which to form hydrogen bonds. In the case of HF,there are three lone pairs but only one hydrogen on fluorine, so only one H bondcan exist bond per HF molecule. This means that there are lone pairs on fluorinethat have no hydrogen with which to form a hydrogen bond. The lower amountof hydrogen bonding accounts for the lower boiling point of HF relative to H2O.The same holds true for the relative boiling point of ammonia (which has threehydrogens but only one lone pair present on the central nitrogen) relative towater.
Example 7.12\A/hich of the following compounds has the highest boiling point?
A. H3CNH2B. H3CCH2NH2C. H3COHD. H3CCH2OH
SolutionThe highest boiling point is associated with the compound with the greatestintermolecular forces and greatest molecular mass. Each compound can formhydrogen bonds, but the alcohols are more polar and form stronger hydrogenbonds than their amine equivalents. This eliminates choices A and B. Ethanol(H3CCH2OH) is heavier than methanol (H3COH), so the higher boiling point isfound with ethanol, choice D.
IIa
a
a
a
a
na
Ir
Tal
qT'
o1
*tai
tr*}i:uq
b'(
Pnbfr
fripruEm
mmr
Ilmffir@lhilt!qpfrn1
Copyright @ by The Berkeley Review a2 The Berkeley Review
General Chemistry Phases and Phase Changes Liquid-Gas Equitibrium
Boiling Point VariationsThe boiling point of a compound can be altered either by varying theatmospheric pressure or by varying the intermolecular forces (which in turnalters AHrru'orization). Boiling point decreases as the atmospheric pressure abovea liquid decreases (this is evident when you boil water in mountainous regionsand in vacuum distillation). Boiling point increases as the atmospheric pressureabove a liquid increases (this is evident in a pressure cooker). Boiling pointincreases as soluble impurities are added to solution (this is evident when youadd salt to boiling water).
Increasing Atmospheric Pressure (Example: A Pressure Cooker)
The boiling point of a liquid is increased by increasing the atmospheric pressureabove the liquid. This can be accomplished by vaporizing the liquid in a closedsystem. You should never heat a closed system, for it will eventually explode.This is why experiments that attempt to riise the boiling point of a iompoundoften use a partially closed system, one with some means of venting gas whenthe pressure gets too high. A pressure cooker is an example of a partially closedsvstem. As the water in the system is heated, it vaporizes. This increases theatmospheric pressure, and thus increases the boiling point. Eventually, when thepressure is great enough, it forces the pressure valve open (or lifts the lid) to ventthe atmospheric gases, thus reducing the pressure inside. This occurs in cycles,t'hich is why the lid (or valve) of a pressure cooker flutters as its contents heatr-rp. By increasing the mass of the lid or the tension in the pressure valve, theooiling point of the water can be raised. In essence, a pressure valve serves as aprimitive kind of thermostat. More sophisticated thermostats are usually a coiled:imetallic strip that expands or contracts as it is heated or cooled. A mercury:rip-switch is connected to the coil so that a bead of mercury balances at a certain:oint. When the temperature being maintained gets too high, the metal loop-rncoils slightly, tilting the glass tube with the mercury in it. This causes therlercury liquid to fall to one side thereby connecting the open ends of a wireiooked to a cooling fan, which turns it on.
Decreasing Atmospheric Pressure (Example: Vacuum Distillation)The boiling point of a liquid is decreased by decreasing the atmospheric pressure:rove the liquid. Reducing the atmospheric pressure makes it easier for a
:ompound to boil. This is why all liquids boil in a vacuum/ despite the extremely-,rrv temperature. A neat experiment to conduct is to fill a syringe partially with: r olatile organic liquid (acetone works well), making sure that no air is in the.','ringe. Plug the tip with your thumb, then pull up on the plunger. A vacuum,s generated in the syringe, so the organic liquid should begin to boil (small:ubbles can usually be seen). As it boils, the walls of the syringe actually get:rv cold. This experiment confirms that reducing atmospheric pressure reduces
-:e boiling point. The same principle is observed in vacuum distillation, where a::rture is distilled under reduced pressure, so that the boiling points are:=luced. Vacuum distillation is employed when a compound has an extremely-:*h boiling point or when it has a decomposition temperature lower than its
--,rrnal boiling point.
a5 Dxclusive MCAT Preparation
General Chemistry Phases and Phase Changes Liquid-Gas Equilibrium
Raoult's LawAccording to Raoult's law, the vapor pressure above a solution of two or moremiscible liquids depends on the mole fraction of each compound in solution. Theconcept behind this theory is that the mole fraction of a compound correspondsto the percentage of the surface of the liquid mixture due to that compound. Ifany component makes up only half of the surface, then only half as much as whatnormally evaporates from the pure liquid, vaporizes from the mixture. Raoult'slaw is shown below as Equation 7.2.
f) _ v_Drvapor = ,lir-vapor (pure) (7.21
The X1 in Raoult's equation is the mole fraction of the component ln solution, notin the vapor state. The mole fraction in vapor for the more volatile component isalways greater than the mole fraction in solution for the more volatilecomponent. As the mole fraction of a compound in solution decreases, the vaporpressure of that component above the solution decreases. This is attributed tothe reduced surface area of that component, and consequently less of thatcomponent evaporating away from the surface. P.,upo. (pure) is the vaporpressure of a pure sample of the component under those conditions, and P,rupo,is the measured vapor pressure of the component. Raoult's law is used tocalculate the vapor pressure of one component, which can be thought of as apartial vapor pressure.
Example 7.130.10 moles of NH3, PH3, and AsH3 are placed into a beaker at -100"C, where allthree exist in liquid phase. What can be said about their relative vaporpressures?
A. PAsHe < PpH: < PNH.B. PpHs < PA'H3 < PNH,C. PArH3 < PNH. < PpHs
P. PNHa < PAsHg < PpHs
SolutionAccording to Raoult's law, the vapor pressure depends on the mole fraction andthe pure vapor pressure of the component. The solution contains equal molarquantities of the three components, so the mole fraction is identical for all threecomponents. Consequently, the highest vapor pressure results from thecompound with the lowest boiling point (highest pure vapor pressure).Ammonia (NHe) has the highest boiling point, because it forms hydrogen bonds,so it has the lowest vapor pressure. Only choice D shows ammonia with thelowest vapor pressure, so choice D is the best answer. To verify the order inchoice D, the relative boiling points of PH3 and AsH3 should be determined.Both compounds are equally polar, so the most significant factor in determiningtheir relative boiling points is molecular mass, not intermolecular forces. Lightermolecules are easier to vaporize, resulting in greater vapor pressure for thecompound. PH3 is lighter than AsH3, so the vapor pressure of PH3 is greaterthan the vapor pressure of AsH3. This relationship is listed in choice D,confirming that it is a valid answer.
(
,]
:inye
pr]E
f'lli]
fum
- "1lr
iiu I
Plrll
:TL.-:I t
uit.L
:Tril1
ewul.r,
ilriles
glTIr.ll
llfiffir:
; m{-
-1.[l[aCopvright @ by The Berkeley Review a4 The Berkeley Review
General Chemistry Phases and Phase Changes Liquid-Gas F,quilibrium
Example 7.14Given a solution that is made by mixing 8.0 grams of methanol (CH3OH) with23.0 grams ethanol (c2H5oH) at a temperature where the vapor pressure of puremethanol is 150 torr and the vapor pressure of pure ethanol is 120 torr, what arethe vapor pressures of methanol and ethanol?
A. Methanol = 37.5 torr; ethanol = 50 torrB. Methanol = 37.5 torr; ethanol = 60 torrC. Methanol = 50.0 torr; ethanol = 60 torrD. Methanol = 50.0 torr; ethanol = 80 torr
SolutionEquation 7.2 can be used to determine the vapor pressure of methanol andethanol, as shown below:
P..upo, (CHeoH) = ":llnt x P'rrupo. (pure CHOH)' fltotal
P,rupo, (CHaCHzoHl = le$fPtr X P'.'upo. (pure CHfH2oH)ntotal
The first task at hand is to solve for the mole fraction of the two components inhe mixture. The molecular mass of CH3OH is 32 grams per mole, so 8 grams:esults in 0.25 moles CH3OH. The molecular mass of CH3CH2OH is 46 gramsper mole, so 23 grams results in 0.50 moles CH3CH2OH. It is important that you:emember to use the mole fraction, and not inadvertently use the moles of each:omponent. The mole fraction is found by dividing the moles of the component:,'' the total moles in solution. There are 0.75 moles total, so the mole fraction ofJH3OH is one-third, which is 0.33. The mole fraction of CH3CH2OH is two-rirds, which isO.67.
Prrupo, (CH3OH) = I x 150 torr = 50 torrJ
P,rupo. (CHgCHzOH, =?" 120 torr = 80 torr
lhe best answer is choice D. As a point of interest, the total vapor pressure of the.,'stem is L30 torr. The total vapor pressure of a solution must fall between the:':re vapor pressure of the least volatile component (L20 torr) and the pure vapor::essure of the most volatile component (150 torr). This is a good double-check:: use, as this would have eliminated choices A, B, and C.
llLe total vapor pressure above a solution consisting of a mixture of two or more*:uids is the sum of the individual vapor pressures of each component liquid.:: - is listed as Equation 7.3.
Prrupo. total = Prrupo. A * Pvapor B + PvaporC... (7.3)
:-: he boiling point of the solution, the total vapor pressure of the components:':-als the atmospheric pressure. However, because the cumulative vapor: r.ssure contains a mixture of compounds, there is not a single pure compoundj- --re vapor. This is a problem during distillation, because it is not possible to:::.erate an absolutely pure product. Because a mixfure of vapors can generate:: --,ugh vapor pressure to reach the boiling point, the addition of volatile;:.:onents to a mixture of liquids lowers the boiling point of a solution. This:=a is derived from the concept that all of the liquids in solution exert some:l rr pr€ssure of their own, independently of the other components.
- ::'.-right @ by The Berkeley Review a5 Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Liquid-Gas Equilibrium
It is ideally a linear relationship between the mole percent of a component andthe vapor pressure of that component, summarized in Figure 7-12.
ca
rr
100% Ao"hB Mole fraction
FigureT-12
0%A100% B
Figure 7-12 rcpresents an ideal mixture. However, for real compounds, there areother factors that come into play due to intermolecular forces between theliquids. Because the molecules interact in both the liquid and gas phases, thevapor pressure relationship is not as linear as Raoult's equation firstapproximates it to be. The intermolecular forces are greater in the liquid phasethan the gas phase, so vapor pressure is most affected in solution. If there is anoverall increase in attractive forces in solution when the components are mixed,then vaporization, and thus vapor pressure, decreases. This is a negativedeviation from linearity. If there is an overall decrease in attractive forces insolution when the components are mixed, then vaporization, and thus vaporpressure, increases. This is a positive deviation from linearity. These deviationsare applied to calculations using Raoult's law (used for ideal cases) so that a realvalue can be approximated.
DistillationDistillation is used to remove a liquid from a solution. To do so, the liquid mustfirst be converted into vapor, and then the vapor is allowed to flow up a distillingcolumn. Once at the top of the column, it can either return to solution, or take a
new pathway where it will be cooled and condensed back into liquid form. Theconcept seems easy enough at first glance, but there are some complications toconsider. Figure 7-13 shows the vapor pressure as a function of temperature fortwo hypothetical compounds, Compound X and Compound Y.
i0
fl
(il
il
0
td
tlfr
il1[
tc)L
ahtAOJhA.trg,(t
Prrapor x
Prrupo.X: PvaporY
Compound X Compound Y
at T6o (Compound X)
Prrapo, Y
Temperature
Copyright @ by The Berkeley Review a6
Figure 7-13
The Berkeley Review
General Chemistry Phases and Phase Changes Liquid-Gas Equilibrium
From FigureT-1'3, the relative vapor pressures of any two compound.s for anytemperature at which they are both liquids may be determined. The grapirshows the values for vapor pressure at the boiling point of Compound x. rheratio of vapor pressures is 3 : 1 in favor of Compout a x. rris means that for asolution where the mole fraction is one-half for each component in solution, themole fraction in the vapor phase is three-fourths for Compound X and one-fourthfor Compound Y. when the vapor is condensed, the solution becomes richer inCompound X, but it is not pure Compound X. After each evaporation andcondensation cycle, it gets richer, but never perfectly pure. Figure 7-14 shows thenumbers following a series of evaporation and condensation rycles.
I
\\.,
\1
X:
X: Y ratio (In
27:lvapor)
81 : 1 243: I3:1 9:
,tv 8t
il \- HlHl \$ll:l 3:
il\il\il\9:1 2'7:l 81 :1 243:1Y ratio (L:r solution)
FigweT-14
Figure 7-14 shows that the effectiveness of distillation does not become evident":ntil 3
series_of evaporation and condensation cycles have transpired. This is thegrinciple behind fractional distillation. In fractional distillatiin, the distilling;olumn has additional surface area, either from packing the column with an inerinaterial, such as glass beads, or by increasing the length of the distilling column.Froviding additional surface area results in more evaporation and condensationr-cles, and thus results in a more purified product.
')''er time, two things occur that reduce the effectiveness of disiillation. Theri'tillation pot becomes richer in the less volatile component, so the initial ratio is,tss than 1 : 1 in favor of Compound X. The second, ind more significant, factors that the glassware heats up, so condensation does not occur ai readily. with:':ss condensation, there are fewer evaporation and condensation cyclei, so the
apor becomes less rich in the more volatile component.
Erample 7.15,',:jch of the following distillations results in the purest product?4- \'Iethanol (b.p. = 56'C) from ethanol (b.p.= ZS"C)ffi Heptane (b.p. = 98"C) from hexane (b.p. = 69"C)
I Diethyl ether (b.p. = 35'C) from tetrahydrofuran (b.p.= eZ"C)D, Benzene (b.p.= 80"C) from 2,3-dimethylbutane (b.p. = 58"C)
i'nilutiona-e distillation that generates the purest product is the one with the greatestri:or pressure ratio of more volatile component to less volatile component. Ther::atest ratio is found in the pair of compounds that has the biggest difference inr':---::18 points. The difference in boiling points in choice A is zz'C, in choice B isli a in choice C is 32'C, and in choice D is22"C. The best answer is choice C.
a7 Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Colligative Properties
ffiriliffffiH;i|ffif$ffi #Concentration Effects
Colligative properties are properties of a solution that are affected by theconcentration of a soluble impurity. Colligative properties include boiling pointelevation (the same as vapor pressure reduction), freezing point depression, andosmotic pressure. The conductivity of a solution, although it is not formally acolligative property, shows a dependence on ion concentration.. Changes in thecolligative properties occur when solutes in solution bind the solvent molecules.The greater the number of impurities, the greater the effect. Figure 7-15 showsthe relative boiling point, fueezing point, osmotic pressure, and conductivity ofan aqueous salt solution as the concentration of salt is gradually increased.
Salt concentration
Figure 7-15
Figure 7-15 shows that as impurities are added to solution, the solvent has a
greater tendency to remain as a liquid. This is attributed to increasedintermolecular forces in the solution phase. Solute particles interact only with a
liquid. For instance, water tends to stay in the liquid phase with the increasedsolvation energy of impurities, rather than changing to another phase. This is
why the boiling point of an aqueous salt solution increases and the freezing pointof an aqueous salt solution decreases. The magnitude of the effect is not the
same for boiling point as for the fueezingpoint, as shown by the slope of the linesin Figure 7-15. There are different values for the boiling point elevation constantand the fueezingpoint depression constant for a given liquid. Both constants are
symbolized by the term k, but each has a different subscript character to identifyits purpose. The boiling point elevation constant is k6, while the freezing pointdepression constant is k1.
Figure 7-16 presents the heating curve for an aqueous salt solution overlappedagainst the heating curve for pure water (for comparison). Heating curves are
generated by adding heat uniformly at a constant rate to the solution of saltwater and monitoring the temperature of the solution. The freezing point of the
salt solution is lower than the normal freezing point of water, and the fueezingpoint of the salt solution is decreasing as more water freezes out from the
solution. The reason that the freezing point of the salt solution continues to
decrease as water freezes away from solution is that the molality of the impuritrin the solution is increasing as water is removed. As the molality of the solute insolution increases, the freezing point decreases, due to the greater relativenumber of impurities in the solution. The same is true for the boiling pointelevation of a solution. This is the explanation of why the last portion of water toboil away from salt water solution is so difficult to remove.
&-.q"
:n:n$"jj
uxr
S0t
wri
'wiiilm
m
tiotrUtrO.c.)
(6bo
U
Copyright @ by The Berkeley Review aa The Berkeley Review
General Chemistry Phases and Phase Changes Colligative Properties
ri5a6!
Standard heating curve
Heating curve of salt solution
Heat added
Figure 7-16
Boiling Point ElevationAs stated before, the boiling point of a solution increases with the addition ofrmpurities. This is why it is referred to as the boiling point eleaation. The increasein boiling point (and decrease in vapor pressure) is attributed to both a decreased.surface area from which the liquid can evaporate and an increase in theintermolecular forces binding the liquid in solution. In the case of a salt watersolution, water has stronger attraction to the ionic impurities than it does to other
"r-ater molecules. The intermolecular forces of water with water, and of water
',r'ith an ion are presented in Figure 7-17. The difference in strength is attributedto the larger positive charge on a cation than the partial positive charge present".n a protic hydrogen.
Weaker .- u. )';-'Interaction \H.
\..
)6-"H
Hydrogen Bonding
Figure 7-17
Jecause it requires more energy to break the ion-dipole interaction than the:'--drogen bond between water molecules, more heat energy is required to::lrove a water molecule from a solution with ions present than from a solution:: pure water. This is the molecular level explanation for the elevation of the:':rhng point for an aqueous solution containing ions. This idea is apptied to any:: -ute added to water, based on the notion that a solute is soluble in water when:: attractive force between the water molecule and the solute particle is greater:::r the attractive force between two water molecules. This is very similar to a;::rciple of solubility, that the most soluble species has the greatest solvation*5e:gy associated with it, while organic species do not readily dissolve into',:ier. The calculation portion of boiling point elevation is simplistic. Although::-:ulations are few on the MCAT, it is still essential that you understand the:r:rp of the equation, and the influence of variables on the boiling point of the1- -ition.
Na* , Stronger
"i-Zf Interactiontr... o_H
H
Ion-Dipole Interaction
a9 Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Colligative Properties (
Equation 7.4 is used to calculate the increase in boiling point when solute isadded to a solution.
AT6 = l6'i'* (7.4\
The k6 term is a constant for the solvent, i is the ionizability constant (essentially
the number of ions that form upon dissoltiog), and m is molality. The k5 valuefor water is 0.51 'C'k$l*o1". On the MCAT years ago, they introduced a forthterm called gamma. The gamma term represented the solute itself, based on theidea that not all solutes interact with the solvent in the same manner. This termmay be referred to as an actir.tity cofficient.
Example 7.16
\zVhich solution has the HIGHEST boiling point?
A. 0.10 moles magnesium chloride in 100 mL waterB. 0.15 moles lithium bromide in 150 mL waterC. 0.20 moles sodium iodide in L00 mL waterD. 0.25 moles potassium fluoride in 150 mL water
SolutionThe highest boiling point is associated with the solution that has the highestconcentration of impurities. The tricky part to this question is determining the ivalue. Magnesium chloride (MgCl2) has an i value of 3.0, while the other salts allhave i values of 2.0. Choice A has 0.3 moles of impurities in 100 mL water, choice
B has 0.3 moles of impurities in 150 mL water, choice C has 0.4 moles ofimpurities in 100 mL water, and choice D has 0.5 moles of impurities in 150 mLwater. The highest concentration of impurities is present in choice C. Whenlooking at questions like this, you should realize that there are several ways totest your understanding, The question could have asked for "GREATEST vaporpressure", in which case you would be hunting for the lowest boiling point.
Keep in mind that addition of a solute not only increases the boiling point of a
compound, but it also changes other colligative properties. For instance, b\-
adding salt to a pot of water when you cook pasta, you increase the boiling pointof the water slightly. However, contrary to what people think about adding salt
to the water when cooking pasta, it is not actually done for the purpose ofincreasing the boiling point. For a 4.0 liter pot of water (roughly a gallon), about
234 grams of sodium chloride are required to raise the boiling point by 1"C to
101"4. This is roughly half a pound of salt. The amount of salt needed to
effectively increase the boiling point is so great that the health risks are not worthit. The reason salt is added to water when cooking noodles is based on another
colligative property, osmotic Pressure. As the salt concentration increases, sc
does the osmotic pressure of the solution, preventing the noodles from gettingtoo soggy (water-logged). There are other interesting observations that can be
explained through colligative properties. When salt is added to hot water, itsuddenly froths (boils) for a moment, and then returns to a state of not boilingThis is because the dissociation of salt into water is exothermic, and the energ,.
released causes localized pockets of boiling water. Once the salt is dissociated
however, the boiling point rises, so the solution is no longer at its boiling point.
F;
]t\-;
P(beQt
SU
dt
ththpcfre
intso.
dtL
m(
Caboa5
rr-h
TlLl
LEL
ftE
:1 C,i
{_E
C,
D
5'r lu
-- l:
:--l J:
r- |-I1:
--t
--r-
Copyright @ by The Berkeley Review 90 The Berkeley Reviex
General Chemistry Phases and Phase Changes Colligative Properties
Freezing Point DepressionThe freezing point of a compound, like the boiling point, can be altered by" awing the intermolecular forces (which in essence varies AHfusion). Freezing
:oint decreases as soluble impurities are added to solution. -ih","uro.,irri
:ehmd the change in melting point is that impurities are found dissolved in a:ubstance in its liquid phase, not the solid phase. when salt is added to the'urface of an ice cube, the water moleculei on the surface of the lattice are:itracted to the salt, and thus can be removed from the lattice (which is melting:ie ice). This is an exothermic process, so additional energy is released to melt-:re ice further. This makes the melting process easier by Iowering the melting:oint. This is why salt is used to help melt the ice on roads and pievent futurl::eezing by lowering the {reezing po1nt. water molecules
""p"ri"r,.u stronger
:teractions in salt water than freshwater, so they do freeze orrifrorr, a salt water'rlution as_readily as they do from a pure water iolution. Solute particles exhibit:ltractive forces in solution. The effect that this has on melting is that solvent:,oiecules tends to stay in solution, so they melt at a lower tempeiature.l'lculating the freezing point depression for a solution is similar to calculating: riling point elevation, except that k1 replaces ku. The other terms are the same:s nr Equation7.4. Equation 7.5 is used to calculate the decrease in freezing point:"'-hen solute is added to a solution.
AT6 = |q;.i.m (7.s)
-: iind the correct temperature for the freezing point, subtract the change in-=:rperature (AT) from the normal freezing point (at 1 atm.). This new value is:::e lreezing point of the solution at that given concentration. Equation 7.6 shows:,:-,.,- this is done.
Tf = Tnf - AT, where Tnf = normal freezing point. (7.6\
-i i-ou are careless, it is easy to disregard the i on these problems. The i is the:i- -;]-number of ir:ns (partictes) from the given material. For instance, i for NaCls -' because NaCl forms two ions pet -ole.rrle as it fully dissociates in water.l--.: i value for sugar is 1, because when sugar dissolves into water, it does not:;:::-1 dny ions; it remains an intact sugar molecule.
;umple 7.17
' '; - al
1s the freezing point of a solution made by mixing 5.g4 grams of sodium
--,:ride into 200 grams of water? (kf = t.gO .C kglmole f6r watir)\- --t.+2'C \ '.\!'
\, ,.;;.; ^r.J-'\ rL':())( -7.,', - \iro- \(')-r
: -i.g6.c o.7.f -3.72"C
5'rlutionl-': moiecular mass of NaCl is 58.4 B/mole, so there are 0.10 moles of NaCl.l-'::e are 200 grams of water, which is 0.2 kg water. The molality is found byr':di''g the moles of solute (NaCl) by the Lilogru*s of solvenf (Hzo). Th;-: i":lion concentration in molality is 0.50 m NaCl. Each molecule of NaCl resultsr I nolecules of impurities, because NaCl fully dissociates into Na+ cations and** :lions when added to water. The total impurity concentration is 1.0 molal,r':r::: lowers the freezing point by 1.86 "C frorn the normal freezing point of 0"C.-*r :reezing point of the solution -r.B6"c,so choice C is the best answer.
?Nc)a
. :r:';6gh, @ by The Berkeley Review 9l Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Colligative Properties
Osmotic PressureWater has a natural tendency to flow from solutions of lower soluteconcentration to solutions of higher solute concentration to reach equalconcentrations. Pressure differences cause fluids to flow, and the driving forcecausing water to flow is known as osmotic pressure. Osmotic pressure is the forceper unit area exerted by a solution through osmosis across a semipermeablemembrane. Osmotic pressure is calculated using Equation7.7.
n = MiRT (7.71
where zr is osmotic pressure, M = molarity, i is the ionizability constant (numberof ions upon dissociation), R is the energy constant, and T is temperature inkelvins. Osmotic pressure is used to measure the mass of polymers, includingnatural polymers like proteins, in a U-tube apparatus. The two sides of the U-tube are separated by a semipermeable membrane, which segregates moleculesaccording to size. Water molecules can pass through the membrane, becausethey are small. However, proteins cannot pass through the membrane. A knownmass of protein is added to one side of the U-tube, creating a difference inosmotic pressure between the two sides. Water flows from the side withoutprotein to the side with protein, so the water levels become uneven. The proteincannot migrate across the barrier, so the concentration is always greater on oneside than the other. Water stops flowing when the osmotic pressure equals thehydrostatic pressure. Equation 7.8 is used to calculate the hydrostatic pressure.
Phydrostatic = pgAh (7.8\
where p is the density, g is the gravitational force constant, and Ah is the heightdifference between the solution in the left side of the U-tube and the solution inthe right side of the U-tube. The experiment is shown in Figure 7-18.
lonro.o,,u,,. = PgAh
t l-
J
;rr
:
fte
Txrr
id,tr lt'
.,&.
fit3l
Before solute has been added,each side of the U-tube containsthe same amount of H2O at 25"C.
Figure 7-18
As shown in Figure 7-78, the osmotic pressure and hydrostatic pressure opposeone another. Once the water levels reach a fixed height, then Equation7.T is se:.
equal to Equation 7.8. From that, molarity can be determined, which can be
converted to moles. A given mass of protein was used, so the molecular mass isfound by dividing this number by moles.
A height difference existsafter solute has been addedto the left side of the U-tube.
$mru
ilruffi!
rlms
pflfiu
:uutnr:l
C-um
Frrunt
F',lrxlis'
fftr*r*
$iltlrl
ffi,s:;m [r:
Semi-permeablemembrane
Copyright @ by The Berkeley Review 92 The Berkeley Reviex,
General Chemistry Phases and Phase Changes Coltigative Properties
Example 7.L8
How many grams of a 12,000 Itu*/*ol" polymer must be added to water tomake a 10-mL solution with an osmotic pressure of 0.0246 atm. at ZT'C? (R =9.932 L'atm./mole.K)
A. 0.012 grams polymerB. 0.120 grams polymerC. 1.200 grams polymerD. 12.00 grams polymer
Solutionusing Equation 7.7, n = MiRT, we can determine how many grams are required.The polymer does not dissociate, so for a polymer (or protein-- a biologicalpolymer) the value of i is 1. Substituting into the equation yields:
zg/712,000g'mole-1 x 1x 0.0g2L.atm.mole-1.K-1 x 300K0.0246 atm - '
0.01 L
_0.0246 x 120 _24.6 x 0.12 = 0.1291x8.2x3 24.6
0.12 grams of polymer are needed, so the best answer is choice B. Not very muchpolymer is necessary to generate enough pressure difference to be measurable.The units are tricky, because Equation 7.8 expresses pressure in terms of pascals,rvhile EquationT.T is in terms of atm. To do a problem where osmotic pressure isequated with hydrostatic pressure, a conversion between pascals and atm. isnecessary. The likelihood of seeing such a calculation on the MCAT is minimal.
Example 7.19The GREATEST osmotic pressure is associated with which of the followingsolutions?
A. 10 grams polymer with molecular mass 10,000 g/mole in 10 mL water.B. 100 grams polymer with molecular mass 10,000 g/mole in 100 mL water.C. 10 grams polymer with molecular mass 10,000 g/mole in 100 mL water.D. 100 grams poll.rner with molecular mass 10,000 g/mole in 10 mL water.
SolutionThe greatest osmotic pressure results from the solution with the greatest polymerconcentration. The polymer is the same in each answer choice, so they each havethe same molecular mass. This question reduces to a hunt for the most grams ofpolymer in the least amount of solvent (which is the highest concentratedsolution). The most grams of polymer in the least water is found in choice D.
ConductancePure water does not conduct an electrical current, because there are no ionspresent in solution to transfer electrons. In order for water to conduct electricity,there must be ions present in solution. The specific conductance of an aqueoussalt solution is directly proportional to the concentration of salt in solution.Electrical conductance is observed only with ionic solutes. All specificconductance is measured relative to aqueous sodium chloride solutions.
0.0246- z xIx8.2x3+z720
Copyright @by The Berkeley Review 93 Exclusive MCAT Preparation
General Chemistry Phases and Phase Changes Colligative Properties
There is also a term for electrical conductance in an aqueous salt solution knownas condosity. The condosity of a solution is defined as the molar concentration ofan aqueous sodium chloride solution that has the same specific conductance as
the aqueous salt solution. Salts that form aqueous solutions capable ofconducting electricity better than sodium chloride solutions have a condositygreater than their molarity. The greater the ratio of condosity to molarity, the
better the salt at conducting electricity in an aqueous environment. For instance,
if a salt were twice as good as NaCl at conducting current in an aqueous solution,then 2.0 M NaCl(aq) solution would have the same conductivity as a 1.0 Msalt(aq) solution. In that case, the 1.0 M salt(aq) solution has a condosity of 2.0.
Example 7.20If an ice cube floats in a glass of water in such a way that the surface of the wateris flush with the brim, and the ice cube sticks out above the level of the water,what will occur as the ice cube melts?
A. The water level will drop below the top of the glass.B. The water will overflow the top of the glass.
C. The water level will remain flush with the top of the glass.
D. The water level will rise above the top of the glass, but will not overflow it.
SolutionFor this question, it is important to remember that when ice melts, it becomes
water. The mass of the water displaced by the floating ice cube is equal to the
mass of the water generated by melting the ice cube. When the ice cube melts,
the water that is formed has exactly the same mass and density as the waterdisplaced. This means that it also has the same volume, so it fills the volumeoccupied by the submerged portion of the ice cube. The net result is that the
level of the water remains constant at the top of the glass. As more and more ice
melts, the ice cube drops lower into the water, but the top of the waters remainthe same. Choice C is the best answer.
Copyright @ by The Berkeley Review 94 The Berkeley Review
I specializLngin MGAT preparation-ill
I
PhasesPassages
l5 Passages
I OO Questions
Suggested phases passage Schedule:I: After reading this section and attending lecture: passages r, rv, v, & IXQrade passages immediat ery after.o-pt.tion and log your mistakes.II: Following Task I: passages II, III, vr, vrl, & x (56 questions in 47 minutes)Time yourself accurately, grade your answers, and review mistakes.III: Review: passages VIII, XI , XIII & euestions 9b - I OOFocus on reviewing the concep[s. Do
"Lt."".ry anout timing.
B
u!
mMlW
|offiint&i
riilud'
dfifiiro
6ilmmftihimI. trreeze-Drying and Cloud-Seeding
II. Phase Diagrams
III. Heating Curve
IV. Bizarre Phase Diagrams
V. Methanol-Ethanol Vapor Pressure
VI. Mixed Vapor of Methanol and Acetophenone
VII. Raoult's Law
VIII. Desalination and Osmotic Pressure
IX. Freezing Point Depression Comparison
X. Freezing Point Depression Experiment
XI. Dumas Experiment
XII. Physical Properties of Liquids
XIII. Allotropes
Questions Not Based on a Descriptive Passage
Phases and Phase Changes Scoring Scale
Kaw Score MCAT Score
85 - 100 15-156A-a4 to-t248-67 7 -9
55-47 4-6r-34 l-5
(r -7)
(8 - 14)
(r5 - 2r)
(22 - 28)
(2e - 3s)
(36 - 45)
(44 - 50)
(51 - 56)
(57 - 65)
(64 - 70)
(7r - 76)
(77 - 84)
(85 - e2)
(e5 - lOO)
ffi,c
ilt*
h
I
Passage I (Questions 1 - 7)
The changing of phases for matter has a commercial use
-n heating and cooling, and a use in the creation of many:roducts that are a part of our daily lives. Phase changes are:ither endothermic (heat absorbing) or exothermic (heat.eleasing). The phase changes that are endothermic aretblimation (solid to gas), melting (solid to liquid), and
...aporation (liquid to gas). The phase changes that arer\othermic are condensation (gas to liquid), deposition (gas to
'rlid), and/reezing (liqluid to solid). Artificial snow-making:r'olves the deposition of water vapor.
Commercial freeze-drying processes employed to make:rny of the coffees marketed today, involve exactly what the- 'ne implies. A batch of freshly brewed coffee is first-rzen, and then dried by means of sublimation. In the first:r of the process, the coffee is cooled until it has frozen:jch occurs at a temperature below 0"C, because impurities
- -re water lower the freezing point). In the second step, the. -rospheric pressure is reduced so the ice can be sublimed. , rv to form the dehydrated coffee. As the ice sublimes. " ";, . the temperature of the frozen coffee slurry lowered'::ause sublimation is endothermic). This results in the::ration of freeze-dried coffee. The sublimation of ice is the
'-. :3ss shown in Reaction 1 below.
H2O(s) + heat -+ HZO(g)
Reaction 1
l-he condensation process occurs naturally when rain-:s form. To enhance the chances of rainfall in draught-
-:r,-n areas, techniques for cloud-seeding have been: : rped over the years. Clouds are masses of condensed. :: ','apor. Rain is formed when the droplets of this vapor
..;::-:Jate into rain drops as the water mist in the clouds:::s on the surface of fine ice crystals, which one might
:,,; . to form at 0"C. In actuality, however, the cloud must-:ercooled to -10'C in order for the ice crystals to form.
lr induce the formation of ice crystals, tiny dry ice::' of solid carbon dioxide can be dropped from an
,i ": ::.e. These pellets freeze the water mist in the clouds.'' , i:-id dry ice is maintained at its sublimation pointr : r.imately -78"C). Once some ice crystals have begun
-.i in a cloud, the water mist can aggregate on the''..:i. to form many raindrops, and the result can the
,.-:r of a small shower.
l:e drying procedure involved in freeze-drying of coffee, .n example of what physical process?
r , MeltingB. Boilingi- , Sublimationf|, Condensation
An alternative to the freeze-drying procedure that wouldstill yield dehydrated coffee is which of the following?
'2d R""ryrtallization using a water-ether mixed solvent-$i omimdon of the water to collect the residue
C . Distillation and collection of the solvent
,p . Flltration of the solute with sieves
2. The optimum conditions forwhich of the following?
F. 25"C and 1.0 atm.
,n. ZS"Cand l.0torrC. -10'C and 1.0 atm.
,. d. -f O"C and 1.0 torr
A. VaporizationB. MeltingC. FusionD. Sublimation
.,).',,}
freeze-drying coffee are
3.
i.
5.
{-.'
To lower the temperature of a cloud, what can bedropped into its core? {'::1 ' ''":::;':'"'A . lce crystals at 0'CB. Water at 0"C -1' ':
C. Dry ice at its sublimation pointr D) Sodium chloride at25'Ct-
a, * i.- :,..''':Supercooled water would be which of the following?
,4. Water at 0'C +iB. Ice at 10'CC. IceatO'C
,'D?Water at -10'C il.---; \
Cloud-seeding tdform rain exemplifies which kind ofphase change involving H2O?
I c r'- '(t 'r' \" 1\
\ . r ,\n Y\' \-{,i'.-
. ,l '
\
-),,,.-1 E-c->*-'
A
v-rt l-\: , '-:J ': ' (
7 . Each locale in the following answer choices has a
different standard boiling point for water. Whichsequence of places reflects the correct relationship ofthese boiling points in DESCENDING order?
A. A seaport > the mountains > a midwestern town:H'The mountains > a midwestern town > a seaport
-€. A midwestern town > the mountains > a seaport
(:") A seaport > a midwestern town > the mountains
ln' '-. l*-o1J
t"1,
i. "' .. r. .,'
GO ON TO THE NEXT PAGE'i
i..r.:. .-<- i -, .' t.J..,i- : . 1 ' ". :
:-:rt O by The Berkeley Review@ 97
Passage ll (Questions 8 - 14)
Standard phase diagrams graph the phase of a material orcompound against ambient pressure (abscissa) andtemperature (ordinate). The lines of the graph represent phase
changes for the compound. The two phase diagrams drawn inFigure I and Figure 2 below show that at low pressure and
temperature, it is possible to convert directly from the solidphase to the gas phase (referred to as sublimation). Phase
diagrams do not intersect with either axis, because it is
physically impossible to reach conditions of 0 kelvins and 0
torr pressure. The two phase diagrams drawn below forstandard compounds are typical. The phase diagram in Figure1 is for an unknown substance, Compound A, while the
phase diagram in Figure 2 is for Compound B. The dashed
lines are drawn to aid your seeing the point on the graphwhere the indicated temperatures and pressures intersect.
Figure 1
Figure 2
Both phase diagrams show a similar slope and have onlythree phases associated with them. There are some phase
diagrams that show more than three phases. For instance,
sulfur can assume two different crystalline structures, so ithas two unique solid phases. As a consequence, the phase
diagram for sulfur shows four phases.
Temperature (K) 273 298
Temperature (K) zzg zsg
Copyright O by The Berkeley Review@ GO ON TO THE NEXT P
8. Which labeled point in the phase diagram belowcritical point?
Temperature
4-pointx, where all three phases coexist
28. Point x, above which gas and liquid do not exist
'e. Point y, where a1l three phases coexist
{!}tt"t y, above which gas and liquid do not exist
If you were to reduce the pressure on Compound A w298 K from 1.3 atm. to 0.9 atm., the compound would:
"#. undergo melting (from solid to liquid).
d. lrndergo vaporization (from liquid to gas).'C. undergo condensation (from gas to liquid).
D. undergo deposition (from solid to gas).
10. At which of the following temperatures isNOT possible for Compound B?
..A. 0'c
...B'. 10'C
c. 25"C\,' D.\ loo"c -' /t-,
1 1 . Which of the compounds would be a gas at
10'CJ '2t'i;'K. Cornpound A onty
78. Compound B only
(Both compounds A and B
(P- rNeither compound A nor B
12. ln what phase does Compound B existtemperature and pressure? -- .fi.'sotia
^z'7 T
,8. Liquid
-C. Gas
D. Lambda
(-)!
Il]i
9.
9a
1 3. Which arrow represents an isothermal evaporation?
!
()!
Lg
15 . According to Figure 1, the heat capacity of Compound
W in the solid phase is:
!. eryal to the heat capacity in the liquid phase.
.8. ,,greater than the heat capacity in the liquid phase.
,.41tess than the heat capacity in the liquid phase.\-/ -/ exactlV twice the heat capacity in the liquid phase.
tJ"f , \'L--t r*"*;L'
L 6. According to Figure 1, the enthalpy of fusion is:
1K- equalto the enthalpy of vaporization.
*f greater than the enthalpy of vaporization.
fj""t than the enthalpy of vaporization.
t' equal to the enthalpy of formation.
17 . If 10.0 grams of an unknown substance, Compound Q(l) heats up to a higher temperature than 10.0 grams ofCompound W when exposed to the same amount ofheat, which of the following conclusions is valid?
;yd*Compound Q has a larger heat capacity than
Compound W.
,,,p1" Compound Q has a larger enthalpy of fusion than
.^ Compound W.
CrJ Compound Q has a smaller heat capacity than\-/ Compound W.
B{ Compound Q has a smaller enthalpy of fusion than
Compound W.
18. How much heat would be required for 10.0 grams ofCompound W to go from a liquid at ten degrees below
the boiling point to a gas at ten degrees above the
boiling point?
A. 3,015 calories
,-ll) 3,150 calories
V. zo,otscalories
,16 . zo,tsocalories
19. Which of the following conclusions can be made about
Cgmpound W relative to water?
/. Co-pound W has a lower vapor pressure than/
waler at room temPerature.
B. Compound W has a lower AHfusion than water'
T4.jonpound W has a lower AHvaporizati6n than(---' water.
D. Compound W has a lower heat capacity in the
liquid phase than water.
Copyright @ by The Berkeley Review@ GO ON TO THE NEXT P
2 0. Which of the following statements about phase change
in Compound W is NOT true?
-*; During phase changes, temperature remains
constant.
Km" process of going from the solid phase to the
. liquid phase is an endothermic process.
{d The process of sublimation is exothermic.
-fr fn" heat capacity for the compound varies for each
phase.
21 . Which statement accurately relates the heat capacity ofCompound W and the slope of the ascending line in its
heating curve?
-K1he slope is directly proportional to heat capacity.
{._tr-}ne slope is inversely proportional to heat\----z
capacrty.
*C. tire slope is independent of heat capacity.
af ne slope and heat capacity differ by a constaol
amount.
J oo .-l
fd.-.i
^-* ACt:*^\-
r t,j
!\.." d- ..,-'. 'l a.
r,l * \-"
r 11 l': J"b* "LJ b's {' \/t
'$.t' l"*
d:r-#i "4{ i'
_
s'> {' St\ "*1
Passage lV (Questions 22 - 28)
Phase diagrams relate ambient temperature and pressureIo the most abundant phase for a material to exist in under.hose conditions. Under conditions where the phase diagramshows that a liquid is present, a small amount of vapor mayalso be present, but not in a high concentration. Fornstance, water at 0"C and 1.00 atm. is present in all three
:hases, although the phase diagram shows that only the solid.nd liquid may coexist. The amount of water vapor isrinimal. The triple point of water, where any of the three:hases may be present in abundance, is 0.01'C and 4.57 torr.
Complications arise in materials that have more thanrree phases. In the phase diagrams shown in Figure I andl:eure 2, each material has two solid phases, both of which-:r'e different lattice structures and physical properties.
(!()!
q6)
q
298
Figure 1
Temperature (K)
298 Temperature (K)
Figure 2
it 1.0 atm and25"C, both materials exist as Solidg.
For which material can both solids exist simultaneouslv',', ith the liquid phase?
{ . Both Material T and Material CB. Only Material TC. Only MaterialCD . Neither Material T nor Material C
Material T
oa2
: " right @ by The Berkeley Review@ lol GO ON TO THE NEXT PAGE
b Which of the following statements is NOT true?
A. It is possible to sublime the solid4 form ofMaterial C at temperatures below the first triplepoint by changing the pressure.
p'. It is possible for gas of Material C to undergodeposition to solidg at temperatures between the
,- -first and second triple points.(-g.'Fiotl^, Material T and Marerial C have rwo rriple
points each.
f. Xstandard temperature and pressure (0"C and 1.00atm.), Material T exists as solid4. \
'.' ''', o,"\',
.. i..\'"t''" ui'
"''2 4 . Which direct conversion is NOT possible?
f . Material T: Sotidg into liquid.
__.8. Material C: Liquid inro solidg.C. Material C: Gas inro solid4.
- D. Marerial T: Solid4 inro gas.
2 5. For both Material T and Material C, which phase cancoexist with the other three at some point?
A. Solida- - -' - - "-A
i..fl.:" ''
,'R.'sotidg ] ,,,-''
C. Liquid
D. Gas,*',1
1.r' ii .., .r . . " ,
16. How can the two different solids be explained for both' of the comoounds?Nt' A . The two solids have different lattice structures.
B. The two solids have different connectivity ofatoms.
C . The two solids have different molecular masses.
D. The two solids have different chiral centers.
27 . At a temperature of 25"C, which of these statementsmust be true?
I. For Material T, solid4 is less dense than solidg.I
. j II. For Material C, conversion from solid4 into solidg
,' ir an endothermic process.
J ,O. For Marerial T, it is possible to undergo depositiononly to form solidg. bur nor solid4.
A. I only
3. II only
C . I and II only
O. U and III only
J- ''r,
28. Which solid deforms upon heatingpressure?
A. Both Material T and Material C
B. Only Material T
C. Only Material C
D . Neither Material T or Material C
at atmospheric
Copyright @ by The Berkeley Review@ t02 GO ON TO THE NEXT
Passage V (Questions 29 - 35)
Three beakers were prepared with a mixture of methanol
and ethanol. In Beaker I 50 grams of methanol and 50 gramof ethanol were mixed. In Beaker II 50 milliliters ofmethanol and 50 milliliters of ethanol were mixed. In Beaker
m 1.00 moles of methanol and 1'00 mole of ethanol werc
mixed. The three beakers were separated and placed into
individual l0liter sealed containers. The total vapor
and vapor pressure due to methanol were recorded for the
separate beakers. This was done by analyzing the gas
each beaker in the closed system. Listed in Table 1 below
data for both methanol and ethanol:
Table 1
According to Raoult's law, the vapor pressure ofcomponent of a mixture can be determined by multiplthe mole fraction of that component in solution by the
vapor pressure of that component at the same tempe
The mole fraction is defined as the moles of a c
divided by the total moles of the solution. Raoult's law
for an ideal mixture and holds true as long as the
of the solution do not interact.
If there are attractive forces between the components
the solution mixture, then the vapor pressure is lower
would be calculated by Raoult's law. If there are repu
forces between the components in the solution mixture,
the vapor pressure is higher than would be calculated
Raoult's law. These are referred to as negative and posi
deviations from Raoult's law, respectively.
2 9. Above which beaker is the vapor pressure of mthe GREATEST?
A. Beakerl
B. Beakertr
C. Beakerltr
D . The vapor pressure of methanol is the same
all three beakers.
30. As the temperature of the solution in Beaker I iwhat is observed for the vapor pressure of methanol'
the total vapor pressure above the beaker?
A. Ptotal vapor increases; Pmethanol increases.
B. Ptotul vapor increases; Pmslhmsl decreases.
C. Ptotul vapor decreases; P,,'s16unol increases.
D . Ptotul vapor decreases; Pmg1h661 decreases'
Compound Formula p at 20"c MW b.p.
Methanol CH3OH 0.7914 glmL 32.04 56"C
Ethanol C2H5OH 0.1893 glmL 46.0',7 '79"C
31. During the first minute of evaporation, before
equilibrium is established and the temperature changes
to any significant degree, what is observed for the
solution in Beaker III?
A. The ratio of moles methanol to moles ethanol
increases, and the rate of vaporization from the
beaker also increases.
B. The ratio of moles methanol to moles ethanol
increases, while the rate of vaporization from the
beaker decreases.
C. The ratio of moles methanol to moles ethanol
decreases, while the rate of vaporization from the
beaker increases'
D. The ratio of moles methanol to moles ethanol
decreases, and the rate of vaporization from the
beaker also decreases.
,r 2. The same experiment is carried out with propanol and
methanol in one case, and with ethanol and propanol in
another case, and the following data are collected:
Mixture Pvapor at 35'C
10 mL methanol w/ 10 mL propanol 42.7 ton10 mL methanol wl 20 mI' propanol 37.1 torr10 mL ethanol w/ 10 mL propanol 38'4 torr10 mL ethanol wl 20 mL propanol 33'5 ton
What can be said about the vapor pressure of pure
propanol compared to both ethanol and methanol?
A. Pmethanol > Ppropanell Pethanol ) Ppropanol
B. Ppropanol ) Pmethanoll Pethanol ) Ppropanol
C. Pmethanol > Ppropanol; Ppropanol ) Pethanol
D. Ppropanol ) Pmethan6li Ppropanol ) Pethanol
What is the total pressure above Beaker I at27"C, if the
vapor pressure of pure methanol is 40.0 torr and the
,'upo, pt"tture of pure ethanol is 30'0 torr at that
emperature?
-{ . Greater than 40.0 ton
B. Greater than 35.0 ton, but less than 40.0 torr
C . Greater than 30.0 torr, but less than 35.0 torr
D . Less than 30.0 torr
. ::-,ight @ by The BerkeleY Review@ r05 GO ON TO THE NEXT PAGE
3 4. If the vapor pressure of pure methanol is greater than
the vapor pressure of pure ethanol, then what must be
true about the relative temperatures of each solution in
order to have equal total vapor pressures above all three
beakers?
A. Tt > TII > TUt
B.TUr>Tt>TnC. TI>Tu>TnD. TU>Tl>TtU
3 5 . If the vapor above Beaker I were collected and condensed
into another beaker, what would be true about the vapor
pressures of ethanol, methanol, and the total vapor
pressure above the new beaker, relative to what was
originally observed above Beaker I?
A. The total vapor pressure and the vapor pressures ofmethanol and ethanol would all be greater above the
new beaker.
B. The total vapor pressure and the vapor pressure ofmethanol would be greater above the new beaker'
The vapor pressure of ethanol would be less'
C. The total vapor pressure and the vapor pressure ofmethanol would be less above the new beaker' The
vapor pressure of ethanol would be greater'
D. The total vapor pressure would be the same, while
the vapor pressure of methanol would be greater and
the vapor pressure ofethanol would be less'
Passage Vl (Questions 36 - 43)
To reduce the temperature necessary to distill a liquid,distillation can be conducted under vacuum conditions.Appropriately enough, this procedure is refened to asvacuumdistillation. In vacuum distillation, the solution is heated ina closed container with reduced atmospheric pressure. Thesystem is not closed, but the flow of gas into the system ishindered by one-way valves. By definition, the boiling pointof a liquid is the point (temperature) at which the vaporpressure equals the atmospheric pressure. By reducing theatmospheric pressure, the vapor pressure required for boilingis lowered and thus the energy needed to boil the liquid is
reduced.
A researcher carries out vacuum distillation on 120.0grams of a mixture that is 50Vo by mass 2-propanol with50Vo by mass acetophenone. This solution is made by adding60.0 grams of 2-propanol to 60.0 grams acetophenone in a
250-mL flask. The boiling points, molecular masses, and
room temperature vapor pressures of the two compounds are
listed below:
Compound Boiling Point MW P'""po,Acetophenone 203"C 1 20.1 5 12 torr
2-Propanol 82'C 60.1 0 48 torr
Table 1
Table 1 lists the normal boiling point, molecularweight, and room temperature vapor pressure. The structuresof acetophenone and 2-propanol are shown in Figure l.
CH:
Figure L
The researcher reduces the pressure above the mixture and
collects the vapor in successive aliquots of 5.0 mL. A totalof twenty-five 5.0-mL aliquots are collected before the
amount of solution remaining in the flask becomes too smallto measure. The twenty-five samples are labeled I through25, in the order in which they were collected.
3 6. At room temperature, what is TRUE about the vaporpressure of 2-propanol above the mixture?
A. The vapor pressure of 2-propanol equals the vaporpressure due to acetophenone.
B. The vapor pressure of 2-propanol is double the
vapor pressure due to acetophenone.
C . The vapor pressure of 2-propanol is eight times the
vapor pressure due to acetophenone.
D. The vapor pressure of 2-propanol is equal to the
total vapor pressure.
odAcetophenone
HOH1l
u.cAcH,2-Propanol
Copyright O by The Berkeley Review@ to4 GO ON TO THE NEXT P
3 7. How many moles of acetophenone (C6H5COCH3) are
contained in the mixture?
A. 0.25 moles
B. 0.50 moles
C. 1.00 moles
D. 2.00 moles
38. What is TRUE for the boiling point for pureacetophenone at 500 torr?
A. It is greater than 203'C, because decreasing the
atmospheric pressure makes it easier to boil.
B. It is less than 203'C, because decreasing the
atmospheric pressure makes it easier to boil.
C. It is greater than 203'C, because decreasing the
atmospheric pressure makes it harder to boil.
D. It is less than 203"C, because decreasing the
atmospheric pressure makes it harder to boil.
3 9. The mole fraction of acetophenone in the origindsolution is which of the following values?
A. 0.25
B. 0.33
c. 0.50
D. 0.67
4 0. What is observed when comparing Aliquot 5 wiinAliquot l0?
A. Aliquot 10 has a greater total vapor pressure and
larger mole fraction of acetophenone than Aliqucn
5.
B. Aliquot 10 has a smaller total vapor pressure and
lower mole fraction of acetophenone than Aliqud5.
C . The total vapor pressure is greater above Aliquot "
than Aliquot 5, while the mole fractionacetophenone is greater above Aliquot 5.
D. The total vapor pressure is greater above Aliquotthan Aliquot 10, while the mole fractionacetophenone is greater above Aliquot 10.
41 . After time, the mole fraction due to acetophenshould do which of the following?
A. It should increase, due to evaporation.
B. It should decrease, due to evaporation.
C. It should stay constant, due to evaporation.
D. It should stay constant, regardless ofevaporatioc.
42. Addition of another 60.0 grams of 2-propanol to the
original mixture would affect the vapor pressure ofacetophenone in what way?
A . It would reduce the vapor pressure of acetophenone.
B. It would have no effect on the vapor pressure ofacetophenone.
C. It would increase the vapor pressure ofacetophenone.
D. It would increase by exactly one-half the vaporpressure of acetophenone.
{ 3 . Aliquot 1 from the mixture has which of the followingmole distribution percentages?
A . 50.OVo acetophenone and 5O.0Vo 2-propanol
B. 33.3Vo acetophenone and66.'7Vo 2-propanol
C . 2O.0Vo acetophenone and 80.0% 2-propanol
D. ll.l%o acetophenone and88.9Vo 2-propanol
r'-right @ by The Berkeley Review@ l05 GO ON TO THE NEXT PAGE
Passage Vll (Questions 44 - 50)
According to Raoult's law, the vapor pressure above asolution consisting of more than one liquid is equal to the
sum of the individual vapor pressures. The individual vaporpressures for each component can therefore be calculated by
multiplying the mole fraction (Ia) of a given component by
the vapor pressure (Pa') of a pure sample of that same
component. The equation for the total vapor pressure due to
all of the components is expressed in Equation 1.
Ptotal = IePa" + IgPn" + I6Pg" + "'Equation I
This equation is based on ideal behavior for solutions. Ifthe attractive force between the component liquids in solution
is strong, a decrease in the vapor pressure above the mixtureis observed. This decrease in vapor pressure is referred to as
negative deviation from Raoult's law. The strong attractiveforces are associated with a large negative heat of dilution.The opposite holds true for positive deviations from ideal
behavior. Drawn in Figure 1 are graphs showing positive
and negative deviations for hypothetical solution mixtures.
Xc= 7
Xo=0Mole fraction Xc= o
Xo= |
Figure 1
The first graph shows a positive deviation from ideal
behavior, because the observed vapor pressure is greater than
the ideal vapor pressure. The second graph shows a negative
deviation from ideal behavior, because the observed vapor
pressure is less than the ideal vapor pressure.
44. What is TRUE about the dilution energies for a mixturewith positive deviation from Raoult's ideal behavior?
A. AG>0;AH>0B. AG<0;AH>0C. AG>0;AH<0D. AG<0;AH<0
LH
; 100!
o0*
F
g 100
aI
lfr
Observed behavior
I Ia"l trrav;or
otr."rr/"a behavior
4 5. Which of the following pairs of solvents would beexpected to have the LARGEST negative deviation fromideal behavior?
A. Ethanol and carbon tetrachloride
B. Ethanol and acetone
C. Ethanol and hexane
D. Ethanol andcyclohexene
46. If the vapor above a 507o mixture of methanol withbutanol at 22"C were collected and placed into a flaskwhere it condenses, what is the mole percent ofmethanol in vapor above the new (second) flask? [Puremethanol has a vapor pressure of 87 torr and purebutanol has a vapor pressure of 29 torr at22'C.l
A. Less than25%o
B. Between 25Vo and5j%o
C. Between 50Vo andTS%o
D. Greater thanT1%o
47. Positive deviations from ideal behaviorpressure above a mixed solution can bewhich of the following?
A. Attraction between liquids, AH > 0
B. Attraction between liquids, AH < 0
C. Repulsion between liquids, AH > 0
D. Repulsion between liquids, AH < 0
in the vaporattributed to
48. Using the information in Figure 1, what is the vaporpressure due to Substance B (l) in a 40Vo (by mole)mixture of Substance B in Substance A, if the purevapor pressures of A and B are 150 ton and 75 torrrespectively?
A. 30 torrB.60torrC. 90 torrD. 105 ton
Copyright @ by The Berkeley Review@ l06 GO ON TO THE NEXT
4 9. Referring to the liquids in the lower graph in Figure I,what can be said about the mole fractions of each liquidsolution over time, if the flask is open to theenvironment and the mixture is allowed to evaporate?
A. C and D both increase.
B. C and D both decrease.
C. C increases and D decreases.
D. C decreases and D increases.
50. Two liquids exhibiting a drastic negative deviation fronideal behavior can attribute the behavior to which offollowing?
A. A bond-forming reaction between the tcompounds in solution
B. A bond-breaking reaction between the tcompounds in solution
C . An increase in one compound's polarityD. A decrease in one compound's polarity
Passage Vlll (Questions 51 - 56)
Reverse osmosis can be employed to extract fresh waterfrom salt water. The process involves applying a force to thesalt water that is greater than the osmotic pressure of aqueoussalt solution, consequently forcing the flow of fresh water outfrom the aqueous salt solution. The apparatus used toaccomplished this consists of a container for the salt watersolution and a cellophane filter through which no ions canpass, so that only water may flow through it. Thecellophane filter is referred to as a semipermeable membrane.It distinguishes and segregates molecules by charge. Thereare other semipermeable membranes which segregateaccording to particle size. The apparatus is shown below inFigure 1.
00
Purified water
Figure ITo convert sea water to water that is suitable for
- :nsumption, an applied pressure of I100 p.s.i. (-70 atm.) is::lployed. This process of water purification is referred to as
-rsalination" The osmotic pressure (n) of the solution can be: ";ulated using the following formula:
n = MiRT(where M = molarity and i = ionizability)
Equation 1
A solution with 5.84 grams NaCl dissolved into 100:*:ms water (with a final volume that is just over 100 mL) is; .:ed into a desalination tube. The desalination tube is then- ::rnected to a plunger used to vary the applied pressure. The;:ralination tube employs a cellophane membrane for': -egation of the impurities from the water.
i i. An external pressure of 10 atm has what effect on anaqueous solution of MgCl2 with an osmotic pressure of5 atm?
A. External pressure forces water out from the saltsolution, which increases [MgCl2] over time.
B. External pressure forces water out from the saltsolution, which decreases [MgC12] over time.
C. External pressure forces water into the saltsolution, which increases [MgC12] over time.
External pressure forces water into the saltsolution, which decreases [MgCl2] over time.
D.
:rr-right O by The Berkeley Review@ t07 GO ON TO THE NEXT PAGE
5 2. Which of the following solutions has the GREATESTosmotic pressure at 20' C?
A. 0.20 moles NaCl(s) with one liter water (whereKsO for NaCl is greater than 1)
B. 0.30 moles KCI(s) with one liter water (where Krnfor KCI is greater than 1)
C. 0.25 moles MgCl2(s) with one liter water (whereKrO for MgCl2 is greater than 1)
D. 0.30 moles PbCl2(s) with one liter water (whereKro for PbCl2 is equal to 1.6 x 10-5 M3)
53. In order to reduce the osmotic pressure of a 100-mLsample of brine solution from25 atm. to 10 atm., whatamount of pure water must be added?
A. 100 mL
B. 125 mL
C. 150 mL
D. 250 mL
5 4. What is the external pressure necessary to stop osmosisof a 1.0 M NaCl(aq) solution at 31"C where, R = .0821L atm. mole-l K-l?
A. Between 2.5 and 5 atm.
B. Between 5 and 10 atm.
C. Between 10 and 35 atm.
D. Between 35 and 70 atm.
55. In desalination, once an external pressure source isremoved, the flow of water:
A. continues from higher salt concentration to lowersalt concentration.
B. from higher to lower salt concentration reverses tolower to higher salt concentration.
C. continues from lower salt concentration to highersalt concentration.
D. from lower to higher salt concentration reverses tohigher to lower salt concentration.
5 6. Which of the following methods could also be used forpurifying salt water?
A. Filtering through glass beads to remove the salt
B. Freezing the solution and removing the relativelypure ice that forms
C. Centrifuging the solution into cotton
D. Increasing the pressure to force the salt toprecipitate out of solution
Passage lX (Questions 57 - 63)
A researcher wanted to study the effects of saltconcentration on the physical properties of a solution. To dothis, she set up six flasks that varied in content andconcentration (see Table 1 for contents). She used distilledwater that had been recently heated to remove any dissolvedcarbon dioxide. In each flask, the solution was made by firstadding the solid to the flask, followed by adding the liquid tothe flask. In each case, the salt fully dissolved into thesolution at the ambient temperature of the solvent. Theresearcher concluded that the solutes are fully soluble in waterat room temperature.
Flask Grams salt Grams water added
# I 2.0 grams NaCl 100 grams H20# 2 2.0 grams NaCl 200 grams H20# 3 4.0 grams NaCl 100 grams H20# 4 2.0 grams KCI 100 grams H20# 5 2.0 grams KCI 200 grams H20# 6 4.0 grams KCI 100 grams H20
MW (NaCl) = 58.4 grams/moleMW (KCD ='74.5 grams/mole
Table IProperties of the solution that are affected by the
concentration of solute are referred to as colligativeproperties. The colligative properties include the freezingpoint, the boiling point, osmotic pressure, and conductance.As more non-volatile soluble impurities are added to asolution, the freezing point decreases and the boiling pointand osmotic pressure increase. Conductance increases, ifthesolute is ionic. For covalently bound solutes, conductance isnot detectably affected.
As an aqueous salt solution boils, the kg of water(solvent) decrease, so that the molality of the solutionincreases. This means that as the solution boils away, theboiling point is increasing. A similar effect is observed withfreezing point depression. An interesting observationinvolves ice cubes. Ice cubes freeze from the outside inward,so as they freeze, the exterior is pure, but solute impuritiesget trapped in the core. This explains why the center of icecubes are cloudy. If it is a gas particle that is trapped in theice cube, it makes a loud noise as it escapes, once the icecube is broken open.
57. When considering only solutions in Table 1, what canbe said about the melting of the solution in Flask #3?
A. The freezing point is the greatest of all thesolutions, and it increases as the solution freezes.
B. The freezing point is the lowest of all solutions,and it increases as the solution freezes.
C. The freezing point is the greatest of all solutions,and it decreases as the solution freezes.
D. The freezing point is the lowest of all solutions,and it decreases as the solution freezes.
Copyright @ by The Berkeley Review@ roa GO ON TO THE NEXT
58. The HIGHEST boiling point would be associated withwhich of these solutions in Table 1?
A. Solution B (Flask # 2)B. Solution C (Flask # 3)
C. Solution E (Flask # 5)
D. Solution F (Flask # 6)
59. Upon comparing solutions in Table 1, which offollowing is NOT true?
C.
A.
D.
The solution in Flask # t has a higherconductivity than the solution in Flask # 4.
The solution in Flask # 2 has a higher vapressure than the solution in Flask # 5.
The solution in Flask # 3 has a higherpressure than the solution in Flask # 6.
The solution in Flask # 4 has a higher freepoint than the solution in Flask # 6.
60. Which of the following aqueous solutions hasLOWEST boiling point?
A. 0.20 m BeCl2(aq)
B. 0.20 m NaCl(aq)
C. 0.30 m KCI(aq)D. 0.30 m MgCl2(aq)
61. If K1 for water is 1.86 "C/molal, then the freezingof Solution C (Flask # 3) is which of the following?
A. -0.12 "C
B. -0.24'Cc. -1.20 "c
D . -2.40'C
62. As more solute is added to solution, which offollowing does NOT increase?
A. The freezing point
B. The boiling point
C. The osmotic pressure
D. The density
63. The LOWEST freezing point for the solutassociated with which of the following solutions?
A. Solution A (Flask # l)B. Solution B (Flask # 2)
C. Solution D (Flask # 4)
D. Solution E (Flask # 5)
B.
Passage X (Questions 64 - 70)
A student set out to determine k1, the freezing point
depression constant, for cyclohexanol' The freezing point
depression constant is a property of a solvent, so solutes
must be added to cyclohexanol to study its colligative
properties. In two separate experiments, the student added
known quantities of menthol and camphor to separate test
tubes containing 10.000 grams of cyclohexanol liquid at
)'7'C. The solutions were thoroughly mixed and then
rmmersed in an ice bath until they were frozen' Then the test
rubes were removed from the ice bath and placed into a water
rath held constant at 35'C. The temperature of the mixture
:n the test tube was recorded every twenty seconds' until the
solution was completely liquefied again' The time and
.3mperature data for Experiment 1, using camphor as the
,r]ute, are listed in Table 1 below:
Camphor
Mass = 1.0102 grams MW = 1524Time (seconds) Temperature ("C)
0 3.9
20 4.3
40 t4.'7
60 5.1
80 5.6
100 6.0
120 6.4
140 6.1
160 7.1
180 r7.8
200 18.7
220 19.6
240 20.5
260 21.4
280 22.4
300 z3.J
320 aA a
340 25.1
360 26.1
Table I
For Experiment 2, using menthol as the solute' the
'-;Jure was carried out in exactly the same manner as in,:.rment 1. The formula for camphor is C16H160' and
, 'rrmula for menthol is C16H2gO' This time the data
:': :ecorded by graphing temperature as a function of time'
',: than listing the data in table form' The graph for the
:" roi experiment is shown in Figure 1' The freezing-: of the solution is determined by finding the inflection-.. A horizontal line is extrapolated back from the
' -:rtion point to the temperature axis to determine the
::-:r1g point of the solution. The horizontal line used to
,, '-nine the freezing point is also shown in Figure 1'
t,{G[ :'. right @ by The BerkeleY Review@ GO ON TO THE NEXT P.{GE
1.0102 grams menthol rn10.000 grams cYclohexanol
80 120 I60 200 240
Time (seconds)
280 320 360
Figure 1
The normal freezing point of cyclohexanol is found to be
22.6"C, according to the thermometer used in Experiment 2'
64. Why must the normal freezing point be determined
using the thermometer, rather than using the standard
value from the literature?
A. The standard value may not hold during the
exPerimental conditions'
B. The thermometer may not be calibrated' so the
normal fteezing point must be determined to
determine the AT accuratelY'
C . The normal fteezing point is not the same for all
samples of cYclohexanol'
D. The freezing point varies with the shape of the
container.
65. Menthol and camphor do not dissociate once in
solution. If an impurity were chosen that can dissociate
into two particles in solution, how would the freezine
point be affected?
A . The freezing point would decrease by twice as much
as expected, if the impurity did not dissociate'
B. The freezing point would decrease by as much
expected, if the impurity did not dissociate'
C . The freezingpoint would decrease by half as muJi
as expected, if the impurity did not dissociate'
D. The freezingpoint would remain constant'
66. The freezing point for the camphor-cyclohe''ar'--i
solution in Experiment 1 was reached:
A . during the first minute of observation'
B . during the second minute of observation'
C . during the third minute of observation'
D . during the sixth minute of observation'
22
21'
20
l918
t'7
16
U
I
6)o.6-)F
40
5 7. f"w an experiment in which the compound chosen was a$Dlid at room temperature rather than a liquid, thenn-i-rture must be heated from room temperature to itsrnelting point instead of being cooled from roomEmperature to its freezing point. This would be thecase with naphthalene. Which of the following graphs
rpresents what should be observed, if a naphthalene-camphor mixture were heated to its liquid state and then
'lbserved while it was allowed to cool to its solid form?
A.
2a
E
<.)
Ec)F
I
dHO
.c)H
c)gdH
8.6)t-i
Time
Time
Time
Time
Copyright @ by The Berkeley Review@ lto
point.
GO ON TO THE NEXT P
6 8. The freezing point for which solution is lower, the one
used in Experiment I or the one used in Experiment 2?
A. The camphor solution has a lower freezing point,because the camphor solution has a higher molalitythan the menthol solution.
B. The camphor solution has a lower freezing poingbecause the camphor solution has a lower molalitythan the menthol solution.
C. The menthol solution has a lower freezing poing
because the menthol solution has a higher molalitythan the camphor solution.
D. The menthol solution has a lower freezing point,
because the menthol solution has a lower molalitythan the camphor solution.
6 9. What would be expected for the freezing pointsolution made by mixing 20.000 grams of cyclohewith 1.0102 grams of menthol?
A. The freezing point would be22.6"C.
B. The freezing point would be20.2'C.
C. The freezing point would be 17.8"C.
D. The freezingpoint would be 13.0"C.
70. Which of the following statements BEST explainsreason for the observed depression in the freezingas an impurity is added to the solution?
A . The molecules can form a lattice structureeasily with impurities present, because
impurities repel the solvent molecules to formlattice-like structure. The solution thusmore easily, which results in an increase infreezing point.
B. The molecules can form a lattice structureeasily with impurities present, becauseimpurities repel the solvent molecules to formlattice-like structure. The solution thus freezes
easily, which results in a decrease in the
point.
C. The molecules cannot form a lattice structure
easily with impurities present, becauseimpurities attract the solvent molecules and
them in solution. The solution thus freezes
easily, which results in an increase in the
point.
D. The molecules cannot form a lattice structurceasily with impurities present, because
impurities attract the solvent molecules and
them in solution. The solution thus freezes
easily, which results in a decrease in the fi
Passage Xl (Questions 71 - 76)
The Dumas experiment involves filling a flask of knownnass and volume with an organic liquid. The organic liquid.s heated, so that it begins to vaporize. Assuming that the,:r in the flask is lighter than these organic vapors, the air isJisplaced out of the flask by the organic vapors through a
inall hole at the top of the flask. The ideal flask is a
,:herical flask with a stopcock attached, but a good substitute
'r ould be any flask with a foil cap that has a small pinhole in.:.e foil cap. Foil is chosen in lieu of plastic or rubber,:3cause organic solvents will dissolve both plastic andr*lber. The solution is heated by flame until the last trace of,qurd residue has evaporated.
Once the liquid has completely evaporated, it is assumed-.rt the flask is filled entirely with organic vapors. As the'=sk cools, the vapors condense back into a liquid, and air':,.'r's back into the flask through the pinhole. Once the flask, rack to its initial temperature (ambient temperature), the
" , :rlbined mass of the flask and condensed organic liquid is-:asured. The increase in mass over the initial weighing of.- i 3mpty flask can be attributed to the organic liquid in the".sk. The small amount of air displaced by the organic: "rd is assumed to be negligible.
-\n example of the flask is drawn in Figure I below:
Organic liquid
Figure 1
I:e mass of the organic liquid can be used to determine- :,ecular mass of the unknown organic liquid. Because:::ssure, temperature, and volume of the gas in the flask,,: ,-,rlr &t the boiling point of the liquid, the moles of the: .n the flask can be determined using PV = nRT.
-J,h)' should this experiment NOT be carried out with a
:la:dc cap on the flask?
{ , The plastic cap can expand during the experiment.
B . The pore size in the plastic cap would be too small.
C . The plastic cap does not allow air to flow back intothe flask as it cools.
D, The plastic cap can dissolve in the organic vapors.
'' :l
't:
t'
Pinhole
r:ht @ by The Berkeley Review@ GO ON TO THE NEXT PAGE
7 2. Which of these equations provides the correct way tocalculate the molecular weight of the liquid in the flask?Note: m"on6ensed vapor = mass of the organic liquidcondensed in the flask.
A. MW = fitcondensed uroo, * U.PV
B. MW = illcondensed uuoo, * W.RT
C. MW- I xRTmcondensed vapor PV
D. MW- 1 xPVmcondensed vapor RT
7 3. What would be the result if the heating source wereremoved before all of the organic liquid had completelyvaporized?
A. The mass measured at the end of the experimentwould be too high, and the molecular masscalculated from the experiment would be too high.
B. The mass measured at the end of the experimentwould be too high, and the molecular masscalculated from the experiment would be too low.
C. The mass measured at the end of the experimentwould be too low, and the molecular masscalculated from the experiment would be too high.
D. The mass measured at the end of the experimentwould be too low, and the molecular masscalculated from the experiment would be too low.
7 4. Which of the following changes to the flask wouldNOT increase the accuracy of the experiment?
A. Rather than using a foil cap, use a permanentattachment with a small valve that can be openedand shut manually.
B. Use an electric heating mantle to encompass the
flask, rather than a flame from a gas burner.
C. Increase the size of the pore in the aluminum foilcap.
D . Use a spherical flask with less internal surface area.
7 5. Which type of organic liquid could be measured MOSTaccurately using the Dumas technique?
A. A less volatile organic liquid with a high molecularMASS.
B. A highly volatile organic liquid with a highmolecular mass.
C. A less volatile organic liquid with a low molecularrnASS.
D. A highly volatile organic liquid with a lowmolecular mass.
7 6. Which of the following liquids would form theDENSEST vapor upon heating?
A. CH3CH2CH2OH
B. CH3(CH)3CH2CIC. CH3(CH)aCH2BrD. (CH3)3CCH2OH
Copyright O by The Berkeley Review@ tt2 GO ON TO THE NEXT P,
Passage Xll (QuestionsTT - 84)
Liquids have certain properties that provide an indirecrmeasure of their intermolecular forces. These properties areviscosity, surface tension, and vapor pressure. Table I liststhe values for these properties for various selected solvents.
Table 1
The definition of each physical property is listed belo*.The SI units for each value are indicated in Table I above.
Viscosity: A resistance to flow exhibited by al0
liquids.
Surface tension: The amount of energy required to increasathe surface area of a liquid by a specifiedunit amount.
Vaporpressure: The pressure exerted bythrough evaporation abovethe liquid.
a gas f,
7 7. Which of the following does NOT affectpressure above a liquid?
A. Hydrogen bonding
B. PolarityC . Molecular mass
D . Surface area
78. For non-hydrogen-bonding liquids, as the vincreases, what is observed with the vapor pressure
surface tension?
A. The vapor pressure increases, while thetension shows no distinct trend.
B. The surface tension increases, while the rpressure shows no distinct trend.
C. Both the surface tension and vapor pressurelncreases.
D. Neither the surface tension nor the vaporshows a distinct trend.
the surface
the
Viscosity SurfaceTension
Pt"po.(@ 20'c)
CHB13 2.0 x 1o-3 kg
m.s4.2 x l0-2 J
m23.9 ton
CHCI3 5.8 x 10-4 kg
m.s2.5 x l0-2 J
^2173 torr
CCla 9.7 x lo-4 kg
m.s2.5 x l0-2 J'
m286.8 torr
C3HsO3 l.s7lgm.s
6.3 x 10-3 Ja
tt1 L.00018 ton
Hzo 1.0 x lo-3 kg
m.s7.3 x lO-2 J
m218.2 ton
CoHr a 2.6 x rc-4 kE
m.s1.6 x 10-2 J
n244 ton
7 9. What explains the higher vapor pressure for chloroformthan for carbon tetrachloride?
A . Carbon tetrachloride is more polar than chloroform.
B. Carbon tetrachloride has more hydrogen bondingthan chloroform.
C. Carbon tetrachloride has greater surface tension thanchloroform.
D. Carbon tetrachloride has a greater molecular massthan chloroform.
8 0. A piece of paper when laid flat to maximize its surface
area would remain floating on the surface of whichsolvent for the longest time?
A . Carbon tetrachloride
B. ChloroformC. Hexane
D. Water
E 1 . Through a column of which solution would a metal ballfall the fastest, if the only external force exerted on the
falling ball were gravity?
A. BromoformB. Carbon tetrachloride
C. GlycerolD. Water
$ l. Which compound is MOST likely to form beads on the
surface of the indicated solid?
A. A polar liquid with lowsurface of a polar solid
B. A polar liquid with highsurface of a polar solid
C. A polar liquid with lowsurface of a non-polar solid
D. A polar liquid with highsurface of a non-polar solid
I 3 - \\hat is the more critical factor when determining theP'upo, and viscosity of CCl4 compared to CHCI3?
--L. For both Puunor and viscosity: the molecular mass
B. For both Puu'or and viscosity: the polarity
C. For Puapor: polarity; and for viscosity: the
molecular mass
D . For Pvapor: the molecular mass; and for viscosity::polarity
surface tension on
surface tension on
surface tension on
surface tension on
the
the
the
the
rr",ight @ by The Berkeley Review@ I r5 GO ON TO THE NEXT PAGE
84. Which set of values is the BEST estimate for thesurface tension, vapor pressure at20'C, and viscosity ofCH2C12?
A . Surface tension = 2.5 x IO-2 -Jm2
Vapor pressure = 223 torr
Viscosity = 4.7 x 1g-a kg
m.s
B. Surface tension = 4.9 x rc-2 J-m2
Vapor pressurc = 67.4 ton
ViscositY =4.J x 10-4 kg
m.s
C . Surface tension = 4.9 x IO-2 J
m2Vapor pressure = 223 torr
Viscosity = 4.7 x 1g-Z kB
m.s
D . Surface tension = 2.5 x rc-z -Im2
Vapor pressure = 67.4 torr
Viscosity =4.7 x l0-2lem.s
Passage Xlll (Questions 85 - 92)
Allotropes are made of the same element, but havedifferent physical properties. The difference in physicalproperties can be attributed to the type of molecular bonding,as is seen with diamond and graphite, both of which are madeof carbon. In diamond, all carbons have sp3-hybridizationand the lattice is held together completely by sigma bonds.In graphite, all carbons have sp2-hybridization and their aresheets of conjugated carbon atoms in a network of six-membered rings. The structure of a sheet of graphite isshown in Figure 1.
Figure 1
Buckminster fullerene, buckyball, is also an allotrope ofcarbon. One example is C6g, which is a spheroidal structuremade entirely of carbon atoms. Buckyball can be separatedfrom other allotropes, such as C79 and ClZ, by sublimation.These allotropes have different physical properties than C6gdue to different packing in their respective lattice structures.Because the molecules are spheroidal in nature, it canfunction like microscopic ball bearings, so buckyball is oftenused as a lubricant. The molecular mass is the mostsignificant factor in the energy required to undergo a phasechange.
8 5. How can it be explained that diamond is more densethan graphite?
A. Double bonds are longer than single bonds.
B. Layering in graphite allows atoms to get closer,because they are in direct contact.
C . Carbon atoms in diamond are smaller than carbonatoms in graphite.
D. Packing is closer in diamond's tetrahedral latticethan graphite sheets.
86. How can the difference in melting point betweenallotropes BEST be explained?
A . The two have different polarities.
B. The two have different molecular weights.
C. The two have different interatomic bonding.
D . The two have different electronegativities.
Copyright @ by The Berkeley Review@
8 7 . How can the black color associated with graphite beexplained in terms of structure?
A. Graphite has conjugation, so it absorbs all light.B. Graphite has conjugation, so it reflects all light.C. Graphite has no conjugation, so it absorbs all
light.D. Graphite has no conjugation, so it reflects all lighr
8 8. Which of the following materialsconductor?
A. COO Buckminster fullerene
B, ClZBuckminster fullerene
C. Diamond
D. Graphite
is the best electrical
8 9. What are the bond angles found in graphite?
A. 90"
B. 109.5'
c. r20'D. 144"
9 0. Which of the followingform an allotrope?
A. Argon
B. Carbon
C. Phosphorus
D. Sulfur
materials is LEAST likely
')'
P
c.
91. Which of the following three dimensional stMOST accurately depicts P4?
A. p-p B. ptlAD.p
il
,/t\n\
9 2. Which of the following allotropes of carbonGREATEST sublimation point at 0.010 atm.?
A. cooB' co+c. czoD. Ctz
has
tt4 GO ON TO THE NEXT P
Questions 93 through 100 are NOT based on a
descriptive passage.
93. If a solution made by adding 0.10 moles KCI to 1.00liters of water conducts electricity better than a solutionmade by adding 0.10 moles AgCl to 1.00 liters ofwater, then:
A . 0.10 M KCl(aq) has
0.10 M AgCl(aq).
B. 0.10 M KCl(aq) has
0.10 M AgCl(aq).
C. 0.10 M KCI(aq) has
0.10 M AgCl(aq).
D. 0.12 M KCI(aq) has
0.10 M AgCl(aq).
I -l . Which of the following phase change processes absorbsthe GREATEST amount of energy?
A. Evaporation
B. Condensation
C. SublimationD. Deposition
Given that the normal boiling point for Compound A is121.0"C, and the enthalpy of vaporization is 33.26kJ/mole, what would the boiling point of Compound Abe in Running Springs, California (elevation 9,870 ft.and P316 = 603 torr)?
A . 115.0"C
B. 133.3'C
c . 131.4"C
D. 151.0'C
greater osmotic pressure than
the same osmotic pressure as
lower osmotic pressure than
lower osmotic pressure than
-., right @ by The Berkeley Review@ I 15 GO ON TO THE NEXT PAGE
9 6 . CHa has a lower boiling point than SiH4, due to:
A. its lower molecular mass.
B. its stronger ionic interactions.
C. its lower affinity entropy.
D. its increased hydrogen bonding.
9 7. Which of the following is a property of a solid, underisobaric and isothermal conditions?
A. The ability to flowB. An amorphous structure
C . Molecules that undergo displacement
D. Static dimensions
98. Water has a vapor pressure of 149.4 torr at 60"C. Anunknown liquid is found to have a vapor pressure of130.6 torr at 60"C. If the vapor pressure of water at
room temperature is 22.65 ton, then at roomtemperature, the vapor pressure of the unknown is:
A. 7.61 ton.
B. 20.00 torr.
C. 22.65 torc.
D. 45.35 torr.
99. A liquid is defined as:
A . having a definite shape and definite volume.
B. having a definite shape and indefinite volume.
C . having an indefinite shape and definite volume.
D. having an indefinite shape and indefinite volume.
10 0. Chloroform is a liquid at room temperature (roughly20'C). Which set of physical constants is possible forHCCI3?
A. Boiling point: 88"C; melting point: 3l"CB. Boiling point: 18"C; melting point: -31"C
C. Boiling point: 61'C; melting point: -64'CD. Boiling point: 32'C; melting point: 31'C
1.C 2.D 3.B6.C 7.D 8.D
11. D 12. A 13. B16. C 17. C 18. B21. B 22. B 23. D26. A 27. D 28. C31. D 32. A 33. B36. C 37. B 38. B41. A 42. A 43. D46. D 4t. C 48. A51. A 52. C 53. C56. B 57. D 58. B61. D 62. A 63. A66. C 61. B 68. A71. D 72. A 73. A76. C 77. D 78. B81. B 82. D 83. A86. B 87. A 88. D91. B 92. D 93. A96. A 97. D 98. B
4.C 5.D9.8 10.D
14. D 15. C19. C 20. C24. D 25. B29. B 30. A34. B 35. B39. B 40. D44. B 45. B49. C 50. A54. D 55. B59. B 60. B64. B 65. A69. B 70. D74. C 75. A19. D 80. D84. A 85. D89. C 90. A94..C 95. A99. C 100. c
Copyright @ by The Berkeley Review@ 116 YOU'RE THROUGH WITH PHA
1.
Phases Passages Answers
Choice C is correct. In freeze-drying, the ice in the frozen coffee is exposed to a reduced pressure environmentwhere the water molecules in ihe ice sublime into a gas. The process of converting from a solid directly into agas is known as sublimation, making choice C the best answer.
Choice D is correct. Because water is in the form the ice when it sublimes away from its frozen state, thetemperature must be set so that the ice does not melt (into liquid). This means that the temperature must be lessthan 0"C. This eliminates choice A and B. The atmospheric pressure must be low enough to allow forsublimation, so 1 torr is better than 1 atm. The best answer is thus choice D, a temperature well below thenormal freezing point of water and a pressure well below atmospheric pressure.
Choice B is correct. The goal of freeze-drying is to remove the water from the coffee solution. Any altemativeto freeze-drying would lead to the same result. In the second choice, the water is distilled (as opposed tosublimed) away from the solution, so the residue left behind should be similar. The high temperaturesrequired for distillation destroy the flavor of the coffee, however. So although distillation works well fordehydrating coffee in theory, freeze-drying works better in practice. The best answer is choice B.
Choice C is correct. The temperature of a cloud can range from 0 to -10"C, according to the passage. To cool thecloud to a lower temperature, a substance must be introduced into it that is either cooler than that temperatureor that reacts in an endothermic fashion. The only choice that meets either of these requirements is c-hoice C.It is stated in paragraph four of the passage that the sublimation point of CO2 is -78"C.
Choice D is correct. Supercooled water is water that is cooled below its normal freezing point, but it remains aliquid. Water normally freezes at 0'C, so pure water at any temperature below this value is considered to besupercooled. Water at -10"C meets this description, making choice D the best choice. Choices B and C shouldhave been eliminated, because they are in the wrong phase.
Choice C is correct. Cloud-seeding involves turning pockets of water moisture in the air into ice crystals. Theconversion of a liquid to a solid is referred to as fusion, making choice C the best answer. This answer can bedetermined by elimination of the wrong answers. The dry ice is sublimed in the process, but dry ice is carbondioxide, not water.
Choice D is correct. The boiling point of a substance is defined as the temperature at which the vapor pressureis equal to the atmospheric pressure. This means that the boiling point of a solution is the greatest where theatmospheric pressure is the greatest. The greatest atmospheric pressure of these at the three locations is at sealevel (an ocean port), which eliminates choices B and C. The mountains have the lowest atmospheric pressureof the three locations, so the boiling point is lowest in the mountains, making choice D the correct answer.
ft. Choice D is correct. The critical point is point y, where the liquid-gas interface line terminates. Above thistemperature and pressure, the material may not exist as a true liquid or true gas, and it is referred to as a super-uitical fluid. It is no longer possible to distinguish between a liquid and a gas above the critical point. Asuper-critical fluid exhibits properties of both a liquid and a gas, including the ability to flow and anamorphous nature. This makes choice D the best choice. Point x is referred to as the triple point, the point atrvhich all three phases may coexist.
n Choice B is correct. Compound A is represented by the phase diagram in Figure 1. Figure 1 shows that at 298 Kand 1.3 atm. (just over 1000 torr), Compound A exists as a liquid. After reducing the pressure from 1.3 atm. to 0.9atm. (ust less than 700 torr), Compound A crosses into the gas phase (following the vertical line downward forthe decrease in pressure). Therefore, the correct answer is B. Even if you are not certain from drawing lines onthe diagram that it actually crosses into the gas phase, there is no answer choice saying that it undergoes nophase change and remains as a liquid. When reading graphs, it is often a good idea to use your answer sheet forfold the pages of the test to generate a reliable straight edge.
i ;rr, right @ by The Berkeley Review@ tt7 Section VII Detailed Explanations
10. Choice D is correct. Sublimation is the process by which a substance converts from the solid phase directly intothe gaseous phase without passing through the liquid phase. At an appropriate pressure, any temperaturcbelow the triple point (in this case, 298 K) is adequate for sublimation of Compound B to occur, since under these
conditions the compound has an interface between solid phase and the gas phase on the phase diagram. From
the choices given, the only temperature above 298 K (25'C) is 100'C (373 K), which is choice D. At this
temperature iublimation cannot occur, since the compound cannot exist in the solid state for any of the possible
pressures.
11. Choice D is correct. At 1.0 atm. (760 torr) and 10"C (283 K), Compound A (Diagram I) is in the liquid phase-
Looking at Figure 2 under the same conditions places Compound B in the solid phase. The question asks which,
of the compounds is a gas (under the given conditions); and neither one is, so choice D is the best answer.
CompoundBexists as a solid .a,--"'(y
Temperature (K) zss
12. Choice A is correct. STP is 0"C and 1 atm. (273K and760 torr). Looking at Figure 2, at a temperaturc of 273
and a pressure of 760 totr (the perpendicular lines are already drawn in the passage), Compound Bexists as a solid, so choice A is the best answer.
13. Choice B is correct. The key to answering this question is to know that the lower right quadrant of th" $rq(at low pressure and high temperature) indicates the conditions under which the material exists as a gEvaporaiion is defined aJ the conversion from a liquid to a gas, so a vertical arrow down from the liquid regi
to the gas region represents the isothermal liquid-to-gas phase change (evaporation). The correct Arrowarrow b, so the answer is choice B.
1"4. ChoiceDiscorrect. CompoundAdoessublime al25'C(298K),if atapressurebelowT60tor;CompoundAisolid at 250 K and,760 tori; and Compound A does have a triple point roughly equal to water (the triple poirtwater is at 0"C). The only uncertainty is the relative densities of the solid and liquid phases. There is no p
at which an increase in pressure at constant temperature converts the solid to a liquid. An increase in
pressure increases the density of the material. This makes choice D the best choice.
15. Choice C is correct. The graph plots temperature as a function of heat, while heat capacity describes heat
temperature change. fhis-meani that the relative heat capacities for the three phases can be determined fthe graph. From*Figure 1, we know that the solid phase heat capacity is proportional to 1 divided by
slop! oi hne a. The liquid phase heat capacity is proportional to 1 divided by the slope of line c. The gt
shows that slope u r itop" c, so the heat capacity of the solid phase is less than the heat capacity ofliquid phase.
'Choi"" C is the best answer. it is given numerically in Table L that the heat capacity of
liquid pnur" is twice the heat capacity of the gas phase. For the heat capacity of the liquid phase to be
th; he;t capacity of the solid phase, the slopes of lines a and e would have to be equal. The slopes of I
and e are not equal.
16. Choice C is correct. Enthalpy is a measure of energy in the form of heat, so the heat added to the system is
increase in enthalpy. fne enihalpy of fusion is represented by line segment b in Figure 1, and the enthalpl
vaporization is repiesented by line segment d. The length of segment b is less than the length of segment
choice C, which states that the AH1,rsi6r., is less than the AHruporiration/ is correct.
Temperature (K) 283
Copyright @ by The Berkeley Review@ r la Section VII Detailed ExPl
17. Choice C is correct. Heat capacity is the amount of energy required to raise the temperature of a substance by1'C (or 1 K)' It is observed that Compound Q heats to a higher temperature than Compound W when both areexposed to the same amount of heat; this means that Compound Q requires a smaller input of heat energy toreach the same increased temperature that Compound W reaches. Becairse less heat energy is required to raisethe temperature of Compound Q by one increment, the heat capacity of Compound e is smaller than the heatcapacity of Compound W, so choose C. Because no-phase change transpired for either Compor:nd e or CompoundW, no conclusion can be drawn about their enthalpies of firsion. For this reason, choices B and D are notnecessarily wrong, but they are not valid for the question.
Choice B is correct. The heat required for this transformation is the sum of the heats for the three steps:IAHsteps = AHliqrrid heating (from.10'C below the boiling point to the boiling point) + AHrupo.ization *
AHgas heating (frbm the boiling point to 10"C above the boiling point).The MW of Compound W is given in the passage as 100 grams per mole, so 10 grams of Compound W is equal to0.1 mole of Compound W. The heat capacities are given as Cfquid = I.0 call f..gand Cgas = 0.5 cal7r."6.
AHhq,rid = m.Ctquid.tr1= (10 S)(1.0 calT*."gX10"C) = 100 calories.
AHvaporization = 30 kcalT-o1" (0.1 mole) = 3 kcal = 3000 calories.
AHgas = m.Cgas.AT = (10 g)(0.5 cal7r."g;119'C) = 5o calories
AHtotul = 100 + 3000 + 50 = 3150 calories, so choice B is the best answer,
Choice C is correct. Compound W has a lower boiling point than water (64"C for Compound W versus 100 .C forwater), so it has a higJrer vaPor pressure than wateial all temperatures at which both compounds are liquids(temperatures lower than 64"C). This eliminates choice A. Theie is nothing mentioned about the melting pointof Compound W or its heat of fusion, so choice B can be eliminated. Compoind W has a lower boiling poiif tnur,water, so it must be easier to vaporize than water, resulting in a lower heat of vaporization for CoLpound Wthan for water. Choice C is the best answer. Both Comporlnd W and water have a liquid phase heai capacityof 1.0. Choice D is therefore eliminated.
Choice C is correct. From the graph in Figure 1 of the passage, we see that the phase changes are representedby segments b and d. It can be seen that the temperature t"maitrs constant during ihe phase .iung"r, ,o A i, tr,r".Melting (the process of going from solid phase to liquid phase) requires the #dition of heat, which makes itendothermic, so B is true also. After looking at the data given-in the passage for the heat capacities atdifferent phases, we find that D is true. \A/hen going from the"solid phase to *r"!ur phase, heat is udd"d ut ,constant rate, so the process must be endothermic, not exothermic. The false statem"ent is choice C.
Choice B is correct. ]he slope of the line is defined as change in temperature over heat added. The heatcapacity of a material is the heat added
_per unit change in temperature. The two values are inverselyproportional to one another, which can verified by comparing their ntritr. This makes the best answer choice B.
18.
19"
l,"l
point (roughly 1.0 atm' and greater than 298 K). The two solid phases cannot coexist with the liquid ph"r" ?o,Material C at either triple point. The lower triple point is the intersection of the two solid phases ani the gasphase, while the uPper triple point is the intersection of solidg, liquid, and gas. Only Material T can have thetwo solid phases coexisting with the liquid phase, so choice B is the best ansi'er.
:3" Choice D is correct. For Material C, the first triple point is the intersection of the gas, solid4, and solidgphases. This means that at temperatures below this point, the only possible forms in which Material C canexist are either the solidg or gas form. It is a true statement that solida may be sublimed into gas, so choice Ais valid. Between the two triple points of Material C, the line segmenl-is the interface of sohdfi and gas. Thismeans that at temperatures in this range, it is possible to convert gas into solidg, which is referred to asdeposition. This makes statement B valid. Both of the phase diagrami shown have two triple points each, sostatement C is valid. At the intersection of 1.0 atm. and273 K on the upper phase diagram 1tf," pnur" diagramof Compound T), the phase present is solid6, not solid4. This makes choice D^an untrue statement.
-::',-nght O by The Berkeley Review@ ll9 Section VII Detailed Explanations
24. Choice D is correct. Direct conversion refers to the direct interfacing of the two phases, so that the phasesinterconvert without first passing through an intermediate phase. For Material T, solidg may be converteddirectly into the liquid form at temperatures between the two triple points. This makes choice A valid.Material C, liquid may be converted directly into solidg at temperatures above the second triple point,solid4 may be converted directly into the gas form at temperatures below the lower triple point. This machoices B and C valid. To convert solid4 into gas for Compound T, the conversion must pass through ei
liquid or solidg. Direct conversion is nof possible; choice D is the best answer.
25.
26.
Choice B is correct. For Material T, solid4 and gas cannot coexist; and for Material C, solid4 and liquidcoexist. The only phase that can coexist with the other three phases in both Material T and Material C
solidg. Pick choice B for optimal satisfaction.
Choice A is correct. The two solids, in order to belong to the same compound, must have the same a
composition. Therefore, they must both have the same molecular mass. This eliminates choice C.
connectivity implies different sigma bonds, which identifies structural isomers, not the same comDifferent chiral centers describes stereoisomers, not different solid phases. The best way to explain d
solids is to say that they form different lattice structures. Examples are calcium carbonate and sulfur, weach have different solid forms that are stable under different conditions. The best answer is choice A.
27. Choice D is correct. For Material T, when the pressure is increased at298K, the phase converts from solidgsolid4. The increase in pressure compacts the solid, so the denser solid phase is solid4. Statement Itherefore invalid. For Material C, heating solid4 converts it into solidg, so the conversion from solid4 isolidg must be endothermic (heat-consuming). Statement II is therefore valid. For Material T, there isinterconversion between solid4 and the gas phase, so deposition of gas into solid4 is not possible. Thisstatement III valid. Only statements II and III are valid, making choice D the best answer.
28. Choice C is correct. The deformation of a solid can be associated with the change of the solid phasesolid4 into solidg. At 1.0 atm., the solid4 and solidg forms of Material T do not intersect, while they doMaterial C. Only Material C shows deformation of the solid phase with temperature change at 1.0 atm.makes choice C the best answer.
29. Choice B is correct. The solution with the greatest mole fraction of methanol has the highest vapor pressuremethanol. Methanol has the lower molecular mass between ethanol and methanol, so Beaker I havemoles of methanol than ethanol. This means that Beaker I will exhibit a higher vapor pressure duemethanol than Beaker III. Because the density of methanol is greater than the density of ethanol, Beakerwith equal volume quaniities of methanol and ethanol has more methanol by mass than ethanol by mass.
mole fraction of methanol in Beaker II is even greater than the mole fraction of ethanol in Beaker I.highest vapor pressure of methanol is therefore above Beaker II. Pick choice B.
30. Choice A is correct. As the temperature of the solution increases, the energy of the system increases, so thatamount of evaporating liquid increases. This means that both the vapor pressure of ethanol and ofincrease, resulting in an increase in the total vapor pressure as well. The best answer is choice A.
31. Choice D is correct. Over time, both ethanol and methanol evaporate from the solution. Because methanola higher pure vapor pressure than ethanol (this can be concluded from the lower boiling point associated
methanol), it is evapbrating faster than ethanol. This means that the solution is losing methanol faster
it is losing ethanol. The remaining solution is thus growing rich in ethanol. This eliminates choice A and
Because ethanol has a lower vapor pressure than methanol, the rate of vaporization decreases as thefraction of methanol decreases. Choice D is the best answer.
32. Choice A is correct. As more propanol is added to the mixtures of both methanol and propanol, and ofand propanol, the vapor pressure decreases, as seen with the reduced vapor pressure associated withaddition of 20 mL propanol. The reduced mixed vapor pressure shows that propanol has a lower purepressure than both methanol and ethanol. This makes choice A the best answer. The relative vaPorof pure samples of the three alcohols is Pmslhanol ) Pethanol ) Ppropanot, which can be deduced frompassage and the information in this question.
Copyright @ by The Berkeley Review@ l2O Section VII Detailed
JJ. Choice B is correct. Beaker I has equal amounts of methanol and ethanol by mass. Because methanol has alower molecular mass than ethanol, the number of moles of methanol is greiter than the number of moles ofethanol. If the moles of ethanol were equal to the moles of methanol, then ihe total vapor pressure would be anaverage of the two pure vapor pressures (30.0 torr and 40.0 torr), which is 35.0 torr. Due to the excess methanolin the mixture, the vapor pressure is greater than 35.0 torr. It can be no higher than pure methanol, with avapor pressure of 40'0 torr. This means that the vapor pressure above Beaker I is between 35.0 and 40.0 torr.The best answer is choice B.
Choice B is correct. The solution richest in methanol has the greatest vapor pressure. The greatest molefraction of methanol is found in Beaker II, so at equal temperatures, the vapor piessnr" above Beaker II is thegreatest. The mole fraction of methanol is greater in Beaker I than in Beaker III, so at equal temperatures, thevaPor pressure is given by, Prrupo. Beaker II > Pvapor Beaker I > Pvapor Beaker III. To increase the vapor pressureabove the solution, the temperiture must be incieased, so for the'Beaker III solution to exert the same vaporpressure as the Beaker II solution, the temperature of the solution in Beaker III must be increased. This meansthat the temperature of the solution in Beaker III must be greatest, if the vapor pressures above all threebeakers were equal. This eliminates all of the choices except B.
Choice B is correct. Because the methanol has a higher vapor pressure than ethanol, the vapor above Beaker Iis richer in methanol than the solution in Beaker I. If this vapor were collected and added to a new beaker,then the solution in the new beaker would have a higher mole fraction of methanol than the Beaker I solution.This means that the vapor pressure of methanol above the new beaker is even greater than above Beaker I,while the vapor Pressure of ethanol above the new beaker is less than above Beaker I. The total vaporpressure above the new beaker is greater than the total vapor pressure above Beaker I, because the solution inthe new beaker is richer in methanol, which vaporizes more easily than ethanol. The best answer is choice B.
Choice C is correct. The molecular mass of 2-propanol is 60 g/mole. The molecular mass of acetophenone is 120g/mole. The mixture is made with equal parts of the two components by mass. Because the mass ofacetophenone is twice that of 2-propanol, there are twice as many moles of 2-propanol as acetophenone. Thepure vapor Pressure of 2-propanol is normally four times that of pure acetophenone (as given in the chart), sothe mixture should show both effects in that the vapor pressure of 2-propanol iJ eight times that ofacetophenone. The correct answer is therefore choice C. To solve for the vapor piess.tte exactly, you must firstsolve for the mole fraction of 2-ptopanol, because the vapor pretr.tie of Z-ptopanol'depends on it.PcHscH(oFI)CHs = (XcHscH(oH)CH)(Ppure CH3CH(oH)CH3). The mole fraction of 2-propinol is:
l mole 2-propanol 7 =21 mole 2-propanol + 0.5 mole acetophenone 1.5 3
This means that the Pr_upo. of 2-propanolis2/g (48 torr) =32torc, and the p.,upo, of acetophenone is 1/3eztorr) = 4 torr. You should pick C for best results.
Choice B is correct. Acetophenone is 50%by mass of the 120 grams. The mass of acetophenone is (0.5)(120 g) =60 g. The MW of acetophenone can be calculated by adding up the atomic weights of the formula given in theproblem:
C=128/rnol, H=18lmo;and O=L68/rnol'8(128/moi=968/molandS(18lmoD=88/mol
96 + 8 + 16 =120i/r.,.ol... moles of acetophenone = 60 grams 11mole; = 0.50 molesI20 g
Choose B for best results. Of course, the 60 grams is given in the passage, and the 120 is given in Table 1.
Choice B is correct. At reduced pressure (500 torr is less than 1 atm.), the boiling point is reduced to a value lessthan the normal boiling point. The normal boiling point of acetophenone is-liited in the chart as 203"C, sochoices A and C are eliminated. Because boiling point is defined as the temperature at which vapor pressureequals atmospheric pressure, a reduced atmospheric temperature makes it easier to reach the Ubiting point.The best answer is choice B.
-rtlri.
i r:;right O by The Berkeley Review@ t2t Section VII Detailed Explanations
39. Choice B is correct. The equation for the mole fraction of acetophenone is: XacetoPhenor," = TTH{S'molestotai
The total number of moles in the mixture is the sum of the moles of acetophenone and 2-propanol. The mole
fraction of acetophenone is:
0.5 mole acetophenone 0.5_1
40.
1 mole 2-propanol + 0.5 mole acetophenone 1"5 3
Choice B is the best answer.
Choice D is correct. The vapor pressure of pure acetophenone is less than the vapor pressure of 2-propanol at
all temperatures below the boiling point of acetophenone. This means that 2-propanol evaporates faster than
acetophenone. With each succesiive aliquot, there i"s a larger percentage of acetophenone. This means that
the mole fraction of acetophenone is gr"ui", above Aliquot io tnut above Aliquot 5. This eliminates choices B
and C. Because Aliquot 5 contains a greater percentage of 2-propanol, there is a greater vapor pressure abole
Aliquot 5 than above Aliquot 10. The correct choice is answer D.
Choice A is correct. Because 2-propanol has a lower boiling point than acetophenone, it evaporates more
quickly than acetophenone, and thusihe mole fraction of 2-propanol in the original solution reduces over time-
As the mole fraction of 2-propanol lessens, the mole fraction of acetophenone increases, and thus the vapor
pressure of acetopheno.," io.."uses. The sum of the two mole fractions is equal to 1.0 (a constant), so if one mole
iraction decrease-s, the other must increase. As a point of interest, as the vapor Pressure due to acetophenone
increases, the vapor pressure due to 2-propanol decreases. Pick A to reach correctness nirvana.
Choice A is correct. Because the mole fraction of 2-propanol increases with the addition of 60 grams of 1-
propanol, the mole fraction of acetophenone is reduced. The sum of the mole fractions of the components in the
^i*trr" must equal L.0 (i.e., the sum of the parts is the whole). A lower mole fraction of acetophenone results in
a reduction in ttre vapor pressure of acetophenone. The correct choice is A.
Choice D is correct. Because the vapor pressure above the original mixture is initially 32 torr for 2-propancfl
and 4 torr for acetophenone, the amount of 2-propanol in the ,ripor phase (and thus Aliquot 1) relative to tlrc
totalvapor i"g2/zi.This fractionred.uces p8/g=L-7/g = 1-.111 =.889. Thepercentage of 2-propanoli
therefore 88.9"/o, making the percentage of acetophenone 11.1%. Given this, choose D. I:1: choice A,ori
because the mole fractioir of 2-propanol is greaterihan the mole fraction of acetophenone. Choice C would
true if the mole fractions *"r" "qnut,
beciause the pure vapor pressure of 2-propanol is four times that
acetophenone. This eliminates choices A, B, and C'
Choice B is correct. A positive deviation from Raoult's ideal behavior is due to unfavorable interactions in
solution phase. Unfavtrable interactions are repulsive interactions that have a positive AH value' B"::
the material dissolves into solution, the AG for solvation must be negative. The favorability of the dissolvi
process is thus due to entropy. The correct combination of signs for the thermodynamic values is choice B' Fn
ihe second paragraph of the passage, a positive AH value can be inferred.
Choice B is correct. The largest negative deviation would involve the greatest increase in the intermo
forces. Acetone in solution 6y itsef cannot exhibit hydrogen bonding (it does not. have an active hyd
Once ethanol has been added to the acetone, the acetonl .utt fot* hydrogen bonds using one of its lone pair fthe carbonyl oxygen. The increased attraction due to hydrogen bonding results in stronger Tl"]i:::i:,iliquid phase ani-thrls a reduction in the vapor pr"rr.rtu.
- Choose B for blissful feedback. Carbon tetrachl
n&u.t", and cyclohexene are essentially non-polar and exhibit no hydrogen bonding'
Choice D is correct. From aS}"hby moles mixture of the two liquids, the vapor due to methanol above the
flask is 87 /@Z + 29). When the vapor is condensed, collected, and placed into the second flask, the conde
solution js 7s%methanol by moles. The vapor pressure of methanol above the second flask is greater than Ibecause methanol u,ruporui", more readily tnu.t butanol evaporates. T b" the star you want to.O:,
!-1:*This question present's the principle behind fractional disiillation. By continuously condensing and
distiling, the distillate becornes richer with regard to the component with the lower boiling point'
41.
L'
43.
44.
45.
46.
Copyright @ by The Berkeley Review@ t22 Section VII Detailed
1/,
48.
Choice C is correct. Positive deviations are associated with a mixture that generates a vapor pressure greaterthan expected according to Raoult's law. A positive deviation is due toleduced intermollcular foices insolution, which can be correlated with repulsion. Because mixing the liquids results in repulsion (which can bethought of as bond-breaking), the dilution endothermic, so AH is positive. This is statedin the passage, so nobackground knowledge is needed. The best answer is choice C.
Choice A is correct. The vapor pressure is calculated by multiplying the mole fraction of Component B by itspure vapor Pressure. The mole fraction of Component B is 0.40 and its pure vapor pressure is 75 torr. The math isas follows:
0.40 x75 torc =2 / Sx 75 torr = 2 xlitorr = 30 torr
The best answer is choice A. The answer must be less than 75 (the vapor pressure of pure Component B), sochoices C and D are eliminated.
Choice C is correct. Compound D has a higher pure vapor pressure than Compound C (as shown in the graph bythe higher vapor pressure for pure Compound D than pure Compound C), so Compound D evapotui"s -otureadily. This must be read from the second graph. The vapor pressure when the moleJraction of Compound D isequal to 1.0 is greater than the vapor pressure when the mole fraction of Compound C is 1.0. Therefore, asCompound D evaporates, Compound C does not evaporate as readily, and so the solution loses Compound D morereadily than it loses Compound C. With time, the solution becomes enriched in Compound C from a molepercent perspective. The mole fraction of Compound C increases while the mole fraction of Compound Ddecreases. This is choice C. With evaporation, the moles of both Compound C and Compound D deciease, butthe question asks for the mole fraction, not moles. The sum of the mole fractions must alwlys be 1.0.
Choice A is correct. A drastic negative deviation results in a sharp drop in the vapor pressure. This impliesthat more than just an increase in intermolecular forces has transpired. A sharp drop ilvapor pressure can beattributed to dimerization between the molecules (which is caused by a bond-forming reaition;. This is bestexplained by choice A. Changes in polarity would cause only minor deviations in vapor pressure, not majorones/ so choices C and D should have been eliminated. The product has a greater molecular mass than theoriginal molecules, making it harder to vaporize than the original molecules.
Choice A is correct. Because the Ps11s1.61 exceeds the Pes6e1ig, the applied pressure forces the water to flowfrom the salt solution (the more concentrated solution) across the semipermeable membrane (to the lessconcenttated solution). This is because the net force is from the solution to the pure water cell. This makes bothchoice C and choice D invalid. As water leaves the aqueous magnesium chloride solution, the MgCl2concentration increases, because the volume of water is decreasing, while the moles of salt are remainingconstant. This makes choice A the best answer.
Choice C is correct. The solution with the greatest osmotic pressure is the solution with the greatest total ionconcentration. The greatest total ion concentration results from the highest combination of molarity (saltconcentration) and ionizability (i). Choice A yields a total ion concentration of 0.4 molar, because the saltsolution is 0.20 molar, and there are 2 ions per NaCl salt molecule. Choice B yields a total ion concentration of0.6 molar, because the salt solution is 0.30 molar, and there are 2 ions per KCI salt molecule. Choice C is thebest choice, because the salt solution is 0.25 molar and there are 3 ions per salt molecule. This results in 0.75molar solution in total ion impurities, assuming the salt is fully soluble. The solubility products are listed toshow that the salts are either fulty soluble or partially soluble. Choices A, B, and C all dissolve completely,because their solubility product, and therefore their molar solubility, are greater than 1.0. Answer choice D istempting, because if you blindly considered the concentration listed, you would get the largest value. But foriead chloride D (PbCl2), the solubility product is 1.6 x 10-5 M3, so the molar solubility is 1.6 i 10-2 M.
Molar solubility =
This means that not all of the lead chloride solid added to the water dissolves. According to the molarsolubility, the molarity of the ions in fully saturated aqueous lead chloride is 0.048 M. For this reason, leadchloride is eliminated from the answer choices. Choice C is the best answer choice.
49.
:1.
5eP4
:",-right @ by The Berkeley Review@ 123 Section VII Detailed Dxplanations
53. Choice C is correct. This question is asking about dilution, for which one normally uses the equation M1V1 =MZYZ. Because the osmotic pressure of the solution decreases as the volume of solvent (and thus solution)increases, the equation can be rewritten as: n1V1 = IE2V2. The osmotic pressure of the solution decreases, becausethe salt concentration decreases (and osmotic pressure depends on the molarity of the solution). If themathematical approach does not make sense, it should intuitively make sense that decreasing the solutionconcentration would reduce the osmotic pressure, because osmotic pressure is directly proportional to molarity (r= MiRT). To reduce the osmotic pressure to 0.4 times its original value, the concentration must be reduced to 0.4
times its original value, and thus the volume must increase by a factor of 1.0 /0.+ = 2.5. The final volume ofsolution should be 250 mL. For the final volume to equal 250 mL, 150 mL of water must be added to the inirial100 mL of solution. The calculation is as follows:
n1V1 = n2V2,makingYz = ut{g) ThusV2 = lmm1-(25-at40=) = 250mr.' '\"zl - \10 atm. /
Again, 250 mL is the final volume, so 150 mL water must be added to the original 100 mL of brine solution to readra total volume of 250 mL. This is choice C.
5b.
54.
55.
Choice D is correct. To stop osmosis, the extemal pressure must equal the osmotic pressure. The osmotic
Choice B is correct. Water can be extracted from salt water by exploiting the other colligative properties
Choice B is correct. In desalination, applied extemal pressure forces the flow of water from the cell of hi
(n) can be calculated using the equation n = MiRT (as presented in the passage). The osmotic pressure is thus:
m = 1.0 mole L-1 x 2 ions x 0.0821 L atm. mole-l K-1 x 304 K = 2 x25 atm. = 50 atm, choice D
concentration to the cell of lower concentration. Once the applied pressure is reduced, nature takes over andflow of water reverses so that it flows from the cell of lower concentration to the cell of higher concentraThis is best stated in answer choice B.
the solution. Water could be distilled away (boiled off and then recondensed). It can also be frozen awalthough the freezing point of the salt water is lower than the freezing point of pure water. This makesB the best answer.
3/. Choice D is correct. The aqueous salt solution in Flask #3 has the highest concentration of all of thebecause NaCl has a lower molecular mass than KCl, and Flask #3 has the greatest mass of solute per leastof solvent. Because the freezing point is depressed by the addition of solute, the freezing point is the lowestthe most concentrated solution, which is the one in Flask #3. As the ice freezes out of the solution, it frelatively pure (salt-free); thus, as ice freezes from the solution, the solution becomes more concentrated.the solution becomes more concentrated, the freezing point is lowered even further, so choice D is correct.
58.
59.
Choice B is correct. The presence of a solute, in addition to lowering the freezing point, also elevatesboiling point of a solvent. This means that the most concentrated salt solution is associated with the hiboiling point. AT = kbim, so the salt of choice once again is NaCl (since it has the smallest mass), andsolution we want is the one with the largest molality. The most concentrated NaCl solution is the one in#3 (most grams and least volume). Choice B is the correct answer.
Choice B is correct. The solutions in Flask #1 and Flask #4 look comparable, but because KCI has amolecular mass, there are more moles of salt in Flask #1 than Flask #4. This means that the solution in#1 has more ions than the solution in Flask # 4, so the solution in Flask #L has a higher electrical conductiChoice A is a valid statement. The solution in Flask #2 has more solute particles than the solution in Flaskso the solution in Flask #2 has less water vapor escaping. This results in a lower vapor pressure, so choicean invalid statement, and thus the best answer. The solution in Flask #3 has more solute particles thansolution in Flask # 6. Osmotic pressure (n) is calculated using n = MiRT, so the solution in Flask #3 has a hiosmotic pressure than the solution in Flask # 6. Choice C is a valid statement. The solution in Flask # 6higher concentration of KCI than the solution in Flask # 4, so the solution in Flask # 6 has a more dep(lower) freezing point. The solution in Flask # 4 has a higher fueezing point, so choice D is a valid state
Copyright @ by The Berkeley Review@ t24 Section Vtr Detailed Dx
50' Choice B is correct' The lowest boiling point is associated with the sorution that has the smallest number ofsolute ions in solution' All of the salts"d^issociate "o-pt"tury ; fi;"r*'"ltrrrror,, so the smalrest product ofmolality and ionizability (m x i.value) yields tn"-r-uil"r;
"or.,";;t*ir" "r solute ions in solution. Thisultimately results in the iowest uortrng p"i"t riti* ,orrrrior,r. Th" lr;"rt-";ncentration of the choices is 0,2 m:n3,.'lj;fill",TJffi':;::,?,:'*aT};:1,:fl:i:11,,H"*;i#:#ation and ,-urru,ir va,ue or the
b1' Choice D is correct' To determt"",jl,"^:ij t-::"i^F.!rint of a solution, the morarity of rhe solution, the i valueof the salt' '"11:'*g noll a -enression
constant of the sorvent must be knowil:Tfl"IT:'fft?:',:Xil':i;:::f;tLX*,,il:11;',::\#:;r#i$?X"?'ft:*:ff*:i;:il1"JnTil?
(4 g solute)(l mole solute'1
molality = _ 59.4 g solute ,
= 40 mole solute = 40 mole solute 0.67 mn^r ,.
.1 kg solvent 58.4 k;;;;"t = 6 "*
solvent
."i::1}ilT:"*:f: -rH. point requires first calculating the tueezingpoinr depression value, which can be
AT = krim = (1.86 'C / molat)2(0.62 molal) = 1.86 x 1.34"C > 1.86.C
ffinggsfil'":1j::.::::LT,""l^,:#fr"ii:iilg,',".#sli:il"",T:lilj1s case 0.c) minus the rreezing
Tbp = Tnbp - AT = 0 - i.g6 - _1*g6,so T6O < _1.g6 .CThe freezing point of the solution is lower than -1.86 "C and only choice D is ress than -1.g6.C, so choice D is thebest answer.
6:' Choice A is correct' As more solute is added to the solution, the concentration increases. The freezing pointdecreases with increasing concentrutio.,, -ur.*jJoi"" e ,r,u best answer.- rr," a"r",rity is assumed to increase,because the addition of Jolute it't;;";t", a solutiJn's mass without significantly affecting its volume.choice A is correct' The Presence of a solute depresses (lowers) the freezing point of a sorvent, so the mostconcentrated solution of silt water must have ui" r"*"" r.e:zlng point. inii*, B and D are eliminated,because the volume of waLel- is ;;;;;;"" in choices A and c, .gjt"l equar mass sart quantiries. This reavesanswers A and C as possibilities' AT = kfim, hence il* ,"it -ith the lirgeri *J"riry (m) has the greatesttemperature depression' The greatest temperature depression results in the lowest freezing point. NaCl has aIarger number of mt'les trtu'' rtr-i""u"r" N"cr has a'lower molecurar mass than Kil, and both NaCr and KCI;::"i::T'l,Tf":::',ffi'":r1ff:#:,l'#";i';,"';;x":.xltr*irtl'i,llil',o,r +r e", ,n"l,,gh*,
ffidetermining the normal fr"";;-;;;";. 't'h" i*;";;; "i1y" ," this experiment is the difference inffiff;ilH? _T::lr,:.:i:"r:1", *"n"tn", *," *,"r_oirr"t", is otr oy a few degrees or nor. Rv rrcino +1o ..*^thermomerer throushout the experim;;;;# ":::il:T;#J;tj,l :":,T, *fl:il,&ji."uil ;XT"*JX",.,,*"
Choice A is correct' If the impurity dissociates into two particles, then there are twice as many impurities aswould normally be expected, so the freezing point is J"pi*r"a by twice ,r* "*r"^, that wouid normally beexpected' This makes choice A correct, so chJice A ir t";;?oj"u ro,. a brighter tomorrow.Choice C is correct'
-ItT 0 seconds up to 160 seconds, the temperature changes by an average of 0.4 "C per 20seconds' From 180 seconds up until thi experiment is terminatei,-th^e ,"*p"r""*r""'.hur,g", by an average of 0.9
"C per 20 seconds' It was betieen the tZ0-second mark ur,J *,u 1g0-second -urt ilu, the rate of temperaturechange was not one these two 'oatt'e"'--*".urrr" the rate of increase changed, r, "ur",
u" assumed that the srope ofthe graph corresponding to tn" J"t" *."rd show "r,
i"ir""tirr^, point bJtween the r}o- ur.,a lg'-second marks.This means that ihe meliing p"i.t;"r;;Jd rorr,"*h".";;;;;; these two marks, which means thar it occurredduring the third minute or ihl er.peri*""i. The best answer is choice C.
'right @ by The Berkeley Review@ t25 Section VII Detaited Explanations
67. Choice B is correct. In the proposed experiment, the mixture is observed as it cools. Because the temperature
decreases as it is observed on"i ti*", choices A and C (which show an increase in temperature over time) are
eliminated. The curve for cooling should be the inverse of the curve observed with cyclohexanol in the menthol
experiment, so the best choice is answer B. Choice D is incorrect, because it shows a relatively flat line to starlfoliowed by a rapid temperature change during the phase change process. As seen with heating curves for pure
substances, the tlmperaiure change during a phase change process should be zero. The curves representing the
cooling of a pure compound 'oursns one with solute impurities is drawn below. Notice the similarities betwem
the two graphs in terms of slope changes.
68. Choice A is correct. The same mass of camphor was added to Test tube #1 as the mass of menthol that w
added to Test tube #2. Because menthol has a larger molecular mass, the moles of camphor added are grea
than the moles of menthol added. This means that the molality of camphor is greater than the molalitymenthol, thus eliminating choices B and C. The freezingpoint for the camphor solution is somewhere betr
17.L"C and17.8"C, accoiding to the data in Table 1. The freezing point of the menthol solution is 17
according to the graph, so the freezing point for the camphor solution is less than the freezing point for
menthol solution. This makes choice A correct.
Choice B is correct. Using twice as much solvent as was used in the menthol experiment would result in
solution with half the molality of the menthol solution in the experiment. This results in a AT value (decrea
in freezing point) that is half as large as the AT from the experiment. The freezing point still drops from
normal frle"itg point of 22.6'C (eliminating choice A), but not by as much as what was observed inexperiment. TIiu ft""ring point is greater than 17.8 'C, thus eliminating choices C and D. The only choice
isinswer B. The observed AT inihe experiment was22.6 - 77.8 = 4.8. The observed AT, if 20.000 gramsr
cyclohexanol were used instead of 10.000 gtams, would be 2.4. The freezing point would therefore be 22.6 - 2-4
20.2'C. Either way, pick B.
70. Choice D is correct. The solution shows a decrease in freezing point as impurities are added to solutiorL
choices A and C are eliminated. With impurities present, the lattice cannot form as easily, because
impurities can interfere with the lattice structure as it grows, and the impurities attract the mretiining them in the solution phase. The best answer is choice D.
71. Choice D is correct. As is stated in the first paragraph of the passage, organic vapors dissolve rubber
plastic, so the cap carurot be made of these substances. The best answer is choice D. The plastic cap may exl
durlng the expeiiment, but that would not necessarily harm the results. Choice A is good, but r] is n9J the
urrr*Jr. The pore size is not too small, as long as vapor can pass through the pore, so choice B is elimi
Air flow is in both directions, so choice C is eliminated.
72. Choice A is correct. The molecular mass is measured in terms of grams per mole. This means that mass (m)
be in the numerator of the calculation. Because the mass (m) is in their denominator, rather than n
a.)tr
trq)A.
ti
69.
choices C and D are eliminated. From the equation PV = nRT, the moles (n) equal PV divided by RT'
me€u:rs that the inverse (1/r.,) equals RT/pV. The molecular mass is m x 1/r', = rn * RT7py, choice A'
/5. Choice A is correct. If not all of the organic liquid had vaporized, then there would be excess liquid ren
in the flask at the end of the condensin! step, so the mass of liquid in the flask would be too high. This
result in a calculated mass that was also too high. The correct answer is choice A.
Graph of solvent with imPuritY
qJti
(gti0.)
O.
0lFTime
Graph of solvent without imPuritY
Time
Copyright @ by The Berkeley Review@ 126 Section VII Detailed
74. Choice C is correct. For choice A, a permanent cap that can be opened and closed would allow for the flask to bes_ealed, so that no vapor could escape while the flask cools. A closed system is more stable than an open system.Choice A would make the experiment more accurate. For choice B, the electric heating mantle1rorld disiributethe heat more uniformly around the flask than a flame. This would allow for even heiting and a homogeneousmixture, making the measurement more accurate. For choice C, an increased pore size would make the"systemmore open and thus allow more vapor to escape as the system cools back down. This would result in a loss in themoles of gas and consequently, the measurement would notbe more accurate. Pick choice C for happiness andcorrectness on this particular question. For choice D, the more spherical the shape of the flask, the less heatthat is lost to the environment. A spherical container offers the smallest amount of surface area per unitvolume (of any container shape), and therefore the least area through which the heat is lost. The less heatthat can be lost, the more accurate the measurement of temperature is, and thus the more accurate theexperiment.
Choice A is correct. The organic liquid that has the smallest measurement error is the one that doesn't readilyevaporate from the flask during the final massing. The organic liquid should condense completely back to itsliquid form at room temperature. This is a characteristic associated with a less volatile liquid (a liquid witha high boiling point). This eliminates choices B and D. To minimize error, the molecular miss should be high,so that any measurement error in the experiment would be small relative to the actual mass of the organiccompound. Choice A is the best answer.
Choice C is correct. -The
densest vapor forms from the liquid with the greatest molecular mass. All gases atstandard conditions have roughly the same molar volume, so the greatest mass per voiume is founJ in thecompound with the greatest molar mass. The molecular mass of propanol (choice A) is 50 grams per mole, themolecular mass of chloropentane (choice B) is 106.5 grams per mole, the molecular mass of biornohexane (choiceC) is 165 grams per mole, and the molecular mass of 2,2-dirnethylpropanol (choice D) is 88 grams per mole.Choice C has the greatest molecular mass.
T. Choice D is correct-. The vapor pressure above a liquid depends on intermolecular forces within the liquid andthe mass of the molecule. Hydrogen bonding increases the intermolecular forces, and thus lowers the amountthat vaporizes. As hydrogen bonding increases, the vapor pressure decreases, so choice A is eliminated.Polarity increases the intermolecular forces, thereby lowering the amount that vaporizes. As the polarity ofthe molecule increases, the vapor pressure decreases, so choice B is eliminated. As the molecular mass increasesfor a compound, the energ'y required to convert the molecule into its gaseous form increases. The vapor pressurethus decreases as the molecular mass of the compound increases, so choice C is eliminated. As the surface areaincreases, the amount of vapor formed is greater, but the area over which it is distributed is also greater. Theresult is that the amount of vapor striking a unit area (pressure) does not change. The surface arei of a liquidaffects the rate at which it evaporates away, but it does not affect the vapor pressure above the liquid, sochoice D is the best answer.
*!' Choice B is correct. For the non-hydrogen bonding liquids in Table 1 (comprising all of the liquids exceptglycerol and water) the sequence of relative viscosities is: CHBr3 > CCl4 > CHCI3 > COlts.1+. For ihe non-hydrogen bonding liquids in Table 1 the sequence of relative surface tensions is: CHBr3 > CCl4 = CHCI3 >C6H1.4, but the trend in the vapor pressure is: CHCI3 > CCla > CoHr+ > CHBr3. Surface tension followsroughly the same trend as viscosity, but the vapor pressure deviates from the trend. It is an inverse trend,except for the hexane, which means that it shows no distinct trend. Staying within the parameters of theanswer choices, the best answer is choice B.
T Choice B is correct. Gravity is the same in all of the trials, so the ball that falls fastest, is the one in thesolution where there is the least resistance to flow. This is observed in the solution with the lowest viscosity.According to Table 1, carbon tetrachloride has the lowest viscosity of all the choices. The correct answer ischoice B.
Itl|u Choice D is correct. The piece of paper stays atop the solvent with the greatest surface tension (even after ithas absorbed solvent). This is assuming that the density of the paper is greater than that of all of the solventchoices. Of those choices, the solvent with the greatest surface tension is water. The best answer is choice D.
75.
76.
tlrr",right @ by The Berkeley Review@ t27 Section VII Detailed Explanations
81. Choice D is correct. The lower vapor pressure of carbon tetrachloride compared to chloroform results from less
carbon tetrachloride vaporizing than chloroform. Carbon tetrachloride is non-polar, so it cannot be more polarthan chloroform (which is a polar solvent), eliminating choice A. Neither carbon tetrachloride nor chloroformform any hydrogen bonds, so choice B is eliminated. According to Table 1, carbon tetrachloride and chloroformhave the same surface tension, so choice C is eliminated. Carbon tetrachloride does have a greater molecularmass than chloroform, so it requires more energy to vaporize than chloroform. This results in a lower vaPorpressure for carbon tetrachloride. The best answer is choice D.
Choice D is correct. Beads form when a compound has a high surface tensiory because its molecules are highlyattracted to one another. This eliminates choices A and C. Beading is most exaggerated when the surface on
which the bead sits has different properties than the beading liquid, so there is little or no attractive force
between the liquid (beading solvent) and the surface. The best answer is choice D, where the solvent is polarand the surface is non-polar. It may help to think of water beads that form when water is dropped on waxpaper.
Choice A is correct. According to thb data in Table 1, the vapor pressure is less for carbon tetrachloride than forchloroform, while the viscosity is greater for carbon tetrachloride than for chloroform. Chloroform is polar,and carbon tetrachloride is non-polar, so the non-polar compound has both the lower vapor pressure and greateviscosity, both of which are normally associated with stronger intermolecular forces. This deviation can be
attributed to the greater molecular mass associated with carbon tetrachloride than with chloroform. In tlncase of carbon tetrachloride compared to chloroform, the molecular mass is more important than polarity indetermining both the vapor pressure and viscosity. The best answer is choice A. Picking A should give ymmuch happiness, extreme joy and deep satisfaction in your soul. If not, it should at least give you a point on
82.
83.
exam.
84. Choice A is correct. To determine the physical properties of dichloromethane (CH2C|), it must be com
to chloroform and carbon tetrachloride (the other chlorohydrocarbons in Table 1). The molecular mass
carbon tetrachloride is greater than the molecular mass of chloroform, which is greater than the mass
dichloromethane. Because dichloromethane is lightest, it has the highest vapor Pressure (and undervaporization to the greatest extent), which makes the vapor pressure greater than 173 torr. This eliminachoices B and D, which are invalid. The surface tension should be so*e rriln" aronttd 2.5 x 16-2 l per m2, makichoice A the best answer. To confirm that choice A is best, the viscosity of dichloromethane should be a vjust trnder 5.8 x 10-4 kg per m.s.
85. Choice D is correct. The densily of a material depends on the closeness of the particles of which it isThis makes choice D the best answer. Choice A is eliminated, because double bonds are shorter thanbonds. Because diamond is denser than graphite, atoms are closer in diamond, so despite the fact thatatoms touch and there exists layering in graphite, choice B does not explain why diamond is denser thgraphite. Carbon atoms are the same size, regardless of the molecular packing and bonding. Choice C
eliminated.
packing. This makes choice B the best answer.
Choice A is correct. A species that appears black in the presence of white light is one that absorbs a1l I
that strikes its surface. This eliminates choices B and D. As seen in Figure 1 in the passage, graphitesignificant amount of conjugation. The best answer is choice A.
88. Choice D is correct. Electrons travel through delocalized molecular orbitals. In the case of ca
allotropes, conductivity would occur through conjugated n-bonds. Graphite has the most conjugation (from
end of the sheet to the other) of any carbon allotrope. The best answer is choice D.
86.
87.
Copyright @ by The Berkeley Review@ t2a Section Vtr Detailed
89.
90.
91.
92.
93.
94.
Choice C is correct. As stated in the passage, all carbons in graphite have sp2-hybridization. This means thatall carbons have trigonal planar geometry and all of the bond angles are 120'. The best, and correct, answer ischoice C.
Choice A is correct. Allotropes are most commonly solids, so it is unlikely that a compound that existspredominantly in the gas phase has allotropes. Argon is an inert gas, so it is the leist likely to formallotropes. Choice B is eliminated based on the information about carbon allotropes listed in the passage.Phosphorus and sulfur can form isotopes.
Choice B is correct. Phosphorus, P, is directly below nitrogen in the periodic table. Like nitrogen, phosphorusdesires to make three bonds. In choice A, each phosphorus makes only two bonds, so choice A is eliminaied. krchoice C, two phosphorus atoms make three bonds and two phosphorus atoms make two bonds, so choice C iseliminated. In choice D, one phosphorus atom makes four bonds and three phosphorus atoms make two bonds, sochoice D is eliminated. In choice B, each phosphorus makes exactly three bonds, so choice B is the best answer.
Choice D is correct. A stated in the last sentence of the passage, molecular mass is the most significant factor inthe energy required to undergo a phase change. This means that the greatest number of carbons in the allotroperesults in the greatest sublimation point. The best answer is C72, choice D.
Choice A is correct. Conductivity of electricity through water is dependent on the number of ions present insolution. A greater number of ions results in better conductivity. Because the conductivity of the KCI iolution isgreater than the conductivity of AgCl solution, the 0.10 M KCI(aq) solution has more ions than the 0.10 MAgCl(aq) solution. This means that KCl is more soluble than AgCl in water. This also means that the 0.10 MKCl(aq) solution has a greater osmotic pressure than a 0.10 M AgCl(aq) solution. This makes choice A the correctanswer.
Choice C is correct. Energy is absorbed by the system when the compound goes from a lower energy state to ahigher energy state. Choices B and D can be eliminated, because in each case the compound is going from ahigher energy state to a lower energy state. The conversion from a solid into a gas requires a greater amount ofenergy than the conversion from a liquid to a gas. This means that more energy is absorbed during sublimationthan during evaporation. This makes choice C correct.
Choice A is correct. This is an easy question hidden in extraneous and intimidating facts. All you need to knowis that the boiling point is defined as the temperature at which the Prru'o. equals Patmosphere. Reducedatmospheric pressure (which occurs at higher elevations, like the mountaini) means that less vapor pressure isnecessary to reach the boiling point. Consequently, less energy is needed to achieve boiling. This lowers theboiling point from the value of the normal boiling pont (1.27'C for Compound A). Only choice A is less than727"C; therefore, choice A is the best answer.
Choice A is correct. Ionic interactions, affinity, and H-bonding are negligible when dealing with CH4 andSiH4, because both compounds are symmetric and thus non-polar. The fact that CH4 has a lower boiting pointthan SiH4 is because heavier molecules have more mass to put into motion when the molecules vaporize, andthus heavier molecules require more kinetic heat energy to escape from the liquid phase into the gas phase. Inconclusion, more heat is required for heavier molecules to vaporize and reach their boiling point. The boilingpoint is greater for SiH4 than CH4, because the silicon compound is heavier. The best answer is molecularmass, choice A.
Choice D is correct. A solid is defined as having its component atoms (or molecules) in fixed positions,undergoing no translational motion (displacement). This means that a solid has a definite shape and adefinite volume. A solid is rigid unless subjected to extreme pressures, and it maintains a lattice structure. Assuch, solids do not flow, they have a shape (and thus are not amorphous), and their molecules undergo nodisplacement. This eliminates choices A, B, and C. Because the shape of a solid remains constant at a fixedtemperature and pressure, it maintains set dimensions. Choice D is the best answer.
pvright @ by The Berkeley Review@ 129 Section VII Detailed Explanations
98. Choice B is correct. Because the unknown liquid has a lower vapor pressure than water at 60"C, the unknownliquid should also have a lower vapor pressure than water has at room temperature. Because the vaporpressure is determined as a natural log function, the vapor pressure of the unknown should be only slighth-lower at the lower temperature than the vapor pressure of water. This makes choice B the intuitively correctchoice. The graph below shows the vapor pressure as a function of temperature for the unknown liquiocompared to water.
99. Choice C is correct. This question requires that you spew back the definition of a liquid. A liquid has n'-'
definite shape, but it does have a definite volume. This is reflected in choice C.
100. Choice C is correct. For a compound to be a liquid at room temperature, its boiling point must be greater tharn.
room temperature (otherwise, it would be a gas), and its melting point must be lower than room temperatu:e(otherwise it would be a solid). The only choice with a melting point less than 20"C and a boiling point greaHthan 20'C is answer C. This in itself is not necessarily an MCAT-style question, but it is necessary to be able:oread charts on the exam. At times, the phase of a substance must be determined, making it necessan- :ro'
evaluate the phase based on the physical properties of the substance. For instance , if you are asked to dei,:ewhether a compound can be distilled, you must first determine whether it is a liquid at the given temperature
lrtro
o)g
U)oCJlro.l-ioo.(!
w
'Qtt-{t!)
-:9
Sp
-Water-----'Unknown liquid
Temperature ('C)
- ::-,:rght @ by The Berkeley Review@ r30 Section VII Detailed Expl
o?
Thermochemistry Section GoalsKnow the deft4ltions of free energy, enthalpy, and entropv.Th"t"",@i,*yo'tohar'eaconceptualdefinitionof-thebasic.termsusedinthermodvna mics. Free enersv fC) ls tt-re accessible energy oT the system , enthalpr1 (H) is the heat energyof the syitem, and entropf(Sl is the disorder of the *stem. The three are related to each other bythe equltion: AG = ll-t -'flS.
Thermodvnamics is the study of energy distribution in a chemical reaction or equilibrium system.There are'six fundamental equations th-at you must know' These are:
Be able to determine the enthalpy change for a reaction from other information.The chanee in enthalpv associated with a reaction can be determined in one of three ways. The firstmethod ilcalorimetiy tdirect measurement of the heat change during the reaction). This involvessuriounding the sysiem with a solution capable_of absorbiig or relEasing a large supply of heatenersv. Bv fireasuiing the change in the surioundings, the change in enthalpy of the system can beinfe#6d. fhe second ilrethod iniolves Hess's law, which states that the enthalpy change for a reactionit th" trr* of the enthalpv changes for any component set of reactions. The fhird m-ethod involvesin" fr-o"a *"**i"" and ii'u"rv .6*-on in'organic chemisbry. Energy must be added to a system tobreak bonds, ind energy is ieleased from a Iystem when bonds are formed. The result is that theheat of the reaction"iun be determined'from the change in bonds and bond energies.
Be able to determine the eguilibrium copstant frg{n thqrqodYnamic.data. - -
The esuilibrium constant for a reaction can be determined from the free-energy change associated;ith i[; ,"uition, starting with 1.0 M reactants and 1.0 M products (referri:d to as AG"). Theequilibrium constant can b"e used also to find AC'.
There are stand"ard energy diagrams. that show the heat of a teaction as it proceeds along the reactionpathwav. There is alsoTlre heiting curye, which shows a phase-change process and the associated[h;;; i" temperature over a purifid of time. You should.6e familiar with both types of graphs andunde?stand thb fundamental information that each presents. Energy diagrams are critical in organicchemistry.
A.AG =AH-TAS B. AG=AG'+RTlnQrxD. AC' = -RT L:r Keq E. AC = RT ln Qrx/Keq
C. AH1s6q1iot1 = AHprodusl - AHlgsslanls
F. AH = bonds broken - bonds formed$
l{,l
0
iT
fl
;fr
M
The heat absorbed by a material can be determined from the "bulg"
in tqmperature. Each materialhas an associated heit capacity (known more precisely as specific hVat). Theheat cqpqcity i-s definedas the heat required to raise the temperature of 1.00 gram of a material .b_y
1.00_'C. Specific heat isthis same valr-ie relative to water. Witer is assigned a}eat capacity of 1.00 calories pei gram'Kelvin
Understand heat ca and its a n to calori
by definition. Heat capacity is used to deteririne the heat-charige in a calorimetry experiment.
Know the basic of an ne and a
Be able to work with energy diagrams of all types.
fft" Cuitrot cvcle is the theoretical mechanics linking the interconversion of work energy and heat
"1"tgy \A,rhefi heat energy is converted into work energy,;q typjcalaqph.catlgl"{i. t:,,?.ilj T.:i?i
io"?1a"r the steam engYi" and the combustion engine. The iteam bngine js a closed system, whilethe combustion eneine Ir utt ot
"n svstem. When wdrk energy is conver-ted into heat energy, a typical
aoolication of it ii to run a iefrigerator. The heat engine and heat pumP use evaporation and;5fr4;;;"tion to store and release Heat. Associated with*the phase change in matter in these devicesis a drastic volume change, which can perform work,
l'?
General Chemistry Thermochemistry Introduction
Thermochemistry is the study of the energy associated with a chemical reaction.Thermochemistry answers the fundamental question "How much is involved ina reaction?" By using thermochemistry principles, we can determine thequantity of products formed, the quantity of energy released (or absorbed), andthe quantity of reactants left unreacted for a given reaction. Energy diagramsshow thermochemistry values. The net change in an energy diagram shows theenthalpy and free energy changes associated with the overall reaction.
Thermochemistry tells us whether a reaction is favorable or unfavorable andwhether it is exothermic or endothermic. Exothermic reactions yield heat, so thetemperature of the system rises as the reaction proceeds. Endothermic reactionsabsorb heat, so the temperature decreases as the reaction proceeds. In thissection, we shall focus on the heat and energy associated with a reaction.
Once a reaction has released or absorbed heat, the suroundings change.Thermodynamics is the study of the energy associated with a system andsurroundings. There are three fundamental laws of thermodynamics. Eachaddresses a different aspect of energy as it relates to all systems. We accept theselaws at the general chemistry level, although some physicists theorize that thefirst law is incorrect, given that energy and matter can be interchanged.
First Law of Thermodymamics: The energy of the unioerse is constant,
Second Law of Thermodynamics: In any spontaneous process, there is always anincrease in the entropy of the unirserse,
Tliird Law of Thermodymamics: The entropy of a perfect crystal at 0 keluins is zero.
The first law of thermodynamics is applied to the internal energy of a system.Kinetic energy can assume the form of work or heat. The Carnot cycle takesrdvantage of the first law of thermodynamics by addressing the interconversionrf work energy and heat energy. The heat engine and the heat pump are thefreoretical models of this conversion process. In the heat engine, because energys conserved, the heat that is absorbed by a system is converted into work. The:,et energy change of the system is zero. In the heat pump, because energy istelrs€rv€dr the work that is applied to a system is converted into heat that is:errLoved from the system. Again, the net energy change of the system is zero. In:lemical reactions, the heat energy associated with a reaction is a result of the-lterconversion of kinetic energy and potential energy in the form of bonds.
lLe second law of thermodlmamics is applied to explain the universe's tendencyr reach a natursl state. This is to say, that nature proceeds to the most probablexate, which is one of disorder. So while the energy of the universe is constant, its heading towards disorder. In a chemical process, the hope is that the entropy:r the universe remains constant. The change in the entropy of the universe is asurr of all changes in entropy for a process. Equation 8.1 shows this relationship.
ASuniverse = ASsyslsm + ASsurroundings (8.1)
Ibe third law of thermodynamics is the standard against which other entropyr:lues can be measured. A perfect crystal at zeto kelvins has its atoms arrangedrn an organized lattice, in which the motion of all atoms has ceased. Disorder:annot exist under these conditions. This is a theoretical situation, because zero*elr,ins is unattainable. This law also sets a standard for measurement, in this
=se ensuring that entropy is always a positive term.
-opyright @by The Berkeley Review t53 Exclusive MCAT Preparation
General Chemistry Thermochemistry Terminology and GraPhs
H i'ffir i!i
Energy Diagrams
Energy diagrams show the relationship between the energetics of a system and
the rEactioi steps (listed along the reiction coordinate). The diagrams can be
used for kinetic purposes (by looking at the activation energy) or thermod;rnamic
purposes (by looking at the energy difference between the reactants and the
products). Two enerly diagrams are given in Figure 8-1below. The first graph
,ho-r tlre energy of i reaciion as function of the reaction pathway for either a
spontaneous (exergonic) or exothermic (heat releasing) reaction. We cannot
identify whet|er ii is exergonic or exothermic unless the type of energy is
specified. If it is free energy (G), we use the term exergonic' lf it is heat energ,Y
1it;, we use the lerm exothetinlc. The second graph shows the energy of a reaction
as a function of reaction pathway for either a non-spontaneous (endergonic) or
endothermic (heat absorbing) reaction. If it is free energy (G), we use the term
endergonic. If it is heat enerly (H), we use the tetm endothermic. T}te first graph
(1eft)"distinguishes between the kinetic region and the thermodynamic region-
ihe kineticiegion is where activation energy can be read. The thermodynamic
region is wheie the free energy change can be read. The second graph Gigf-,t)
sh"ows the same features, Uut wittL the regions labeled more specifically'
Although it may not seem so, the graph are labeled in a similar fashion.
Xbotr0)
I!
Exergonic reaction (AG < 0)
orExothemic reaction (AH < 0)
Reaction coordinate
Endergonic reaction (AG > 0)
orEndothemic reaction (AH > 0)
Figure 8-L
The energy diagrams shown in Figure 8-1 represent one-step reactions. Multi-
step reaJtio.r, i-rurr" graphs with multiple apexes. The reaction coordinate
represents the chronollogical steps that the molecules take during the reaction
pathway from reactants-to prodlcts. The kinetic region defines the activation'enurgy,'and thus defines the rate of the reactions. The thermochemistry region
a"fiti"r the energy change for the reaction. These two regions are generally the
most significantispects-of utt "tt"tgy
diagram. Activation energy is read from a
free enJrgy diagram, not from atr energy diagram based on enthalpy. The free
energy uii "tttf,utpy
are related uccordit g to Equation 8.2, which requires that
temperature be given in kelvins'
Reaction coordinate
Copyright @ br The BerkeleY Review 134
AG=AH-TAS
The BerkeleY
(8.21
General Chemistry Thermochemistry Terminology and Graphs
Free Energy
Gibb's free energy, associated with a closed system, is the accessible energy in asystem that allows chemical reactions to proceed to a state of equilibrium. Thefree energy (G) of the system accounts for both the entropy (S) and the enthalpy(H) of the system. The free energy change for a chemical reaction or physicalprocess is a sum of the enthalpy change and the negative of the entropy changetimes temperature. Equation 8.2 shows that these three thermodynamicvariables, along with the temperature, are all related and thus can be used topredict the value of one another. The free energy change of a reaction is usedprimarily to determine the favorability of the reaction. If a chemical reaction hasa negative value for AG, then the reaction is defined as being favorable in theforward direction. If a chemical reaction has a positive value for AG, then theieaction is defined as being unfavorable in the forward direction. Table 8.1shows different combinations of thermodynamic values as they relate toEquation 8.2.
Case ResultAS positive, AH nesative Favorable at all temperaturesAS positive, AH positive Favorable at hieh temperaturesAS negative, AH negative Favorable at low temperaturesAS negative, AH positive Unfavorable at all temperatures
Table 8.1
-:rle 8.1 summarizes the four combinations of values for AH and AS. In two of:: cases, the temperature must be known in order to determine the sign of the::e energy change (AG). This means that to determine the favorability of a:=::tion, temperature must be known in many cases. A reaction that is favorable:' -ow temperatures may not necessarily be favorable at higher temperatures..-,'. is the mathematical reasoning behind the dependence of the equilibrium: .:,,.tant (Kun) of a reaction on the temperature.
:'.ample 8.1' :eaction that is favorable at25'C but unfavorable at 500"C has what signs for* - :rd AS?
ioth AH and AS are positive.ioth AH and AS are negative.fi is positive, and AS is negative.-J{ is negative, and AS is positive.
..r:r.on";'.':rable reaction has a negative AG, while an unfavorable reaction has a' --,. e AG. AG is negative at lower temperatures (25"C) and positive at higher:-:-ratures (100'C), so AG increases with T. Temperature is measured in
-:,*i, so T is always a positive value. Using Equation 8.2, LG = AH - TAS, the, -: rf AS must be negative, (negative number), because only the TAS term"" :.. rvith temperature. As T increases, AG increases, so AG = AH - T(negative::-rr. AH must be negative for AG to be negative at lower temperatures,, -.. inegative number) = AH - T(negative small number). Both AH and AS
'' - : iative, so choice B is the best answer. This can be confirmed by Table 8.1.
,: :l:ht O by The Berkeley Review 155 Exclusive MCAT Preparation
General Chemistry Thermochemistry Terminology and Graphs
Reactions that are favorable in the forward direction are said tobe spontaneous.Spontaneity refers to the favorability of a reaction, not to the rate of a reaction. Ifthe reaction is favorable in the forward direction (proceeds in the forwarddirection to reach equilibrium), then the reaction is spontaneous in the forwarddirection. This is a thermodynamic perception of the reaction, which means it isbased solely on the free energy change (AG) for the reaction. \Atrhether a reactionis observed may also depend on kinetic factors. If there is not enough activationenergy present in the system (that is, if the temperature is too low), then thereaction cannot proceed in the forward direction even though it is spontaneous.AG.* must be negative in order for the reaction to be spontaneous in the forwarddirection.
There is also the standard free energy change (AG') to consider, which is the freeenergy change when a reaction goes from standard conditions to equilibrium-Standard conditions are defined as 1 atm. of pressure, and a temperature of 25'C,and all solute species (reactants and products) present at L.00 M concentration-Under conditions where all solute concentrations are equal to L.00 M, the value ofQ.* is L.0, so ln Qr* is zero. In such a case, only K"O needs to be considered whmdetermining the free energy change, which leads us to Equation 8.3.
AG' = - RT ln Keq (8.3)
When the AG" value for a reaction is close to zero, the reaction has an equilibriumconstant of roughly 1.0. Reactions with K"n roughly equal to 1.0 are driven bythe addition of excess reactant or the removal of products, which changes ttevalue of AG, but not the value of AG'. These techniques exploit Le ChAtelierbprinciple to shift the equilibrium to the product side of the reaction. Thetechnique is illustrated mathematically using Equation 8.4.
Example 8.2If L:r Keq = 5.0 and the temperature is 25'C, what is the value of AG'?
A. +12.5 kJB. +8.3 kJc. -8.3 kJD. -t2.skJ
SolutionTo solve for fhe numerical value, use Equation 8.3, AG" = -RT ln Keq.is in kJ, so R is 8.31.4J/mole K. We are given ln K"n, not I("0, soeasier.
AG' = - (8.314)(2e8xs) I = - (10)(250Xs) J = -(2.s)(5) kJ = - 12.5 kJ
The best answer is choice D. Some of you may have approximated the valuefollows:
AG" = - (8.314X2e8X5) I = - (8X300X5) J = -(2.4)(5) kJ = - 12 kI
The answer choices are spaced far enoughapproximation will lead to the best answer. Mathit is also important to understand relative AGcrunching is not directly involved.
apart that any reais simplified on the MCAT,
The ansthe math i
calculations where n
Copyright @by The Berkeley Review 136 The Berkeley
General Chemistry Thermochemistry Tenninology and Graphs
Any equation used to determine the free energy change of a reaction mustaccount for the free energy change from the i"itiii conditlions of the reaction tothe equilibrium conditions of ihe reaction. A reaction going from initialconditions-to equilibrium conditions can be broken into tw6 partial reactions:initial conditions to standard conditions, and standard conditions to equilibriumconditions. The derivation below summarizes this reasoning in Equation g.4:
AGrx - AGlinitial conditions -+ equilibrium conditions)AGry = AGliniual -+ standard conditions) + AGlstandard -+ equilibrium conditions)
AGlinltiat conditions -+ standard conditions) = RT ln Qo - RT ln 1 = RT ln er*AGlstandard -+ equilibrium conditions) = RT In 1 - RT ln &q = - RT ln I("q
AGrx - RT ln Qrx - RT In &q = - RT ln Gq + RT ln e."Substituting in AG' for - RT 1r1IGt yields Equation g.4
AGrx = AG" + RT ln Qrx (g.4)
This same derivation can be altered slightly to generate Equation g.5:
AGrx = RT L:r Qrx - RT ln Keq = RT (h e* - In &q) = p11r.,9ftKeg
AG= (8.s)nr r'€ry-Ke9
You may recall from the equilibrium section that Qr* defines the reaction state,3d 5"q defines equilibrium. \A/hen e.* is less than ieo, the e."_to_Ko., ratio is-ess than 1.0, and the log of a number less than 1.0 is negative. This miies AGr*:'regative, according to Equation 8.5. when the value of"er* is less than K.o, the:eaction h.1g too many reactants and too few products, ro it pro.""ds forw"ald to:each equilibrium, reaffirming that the reaction is favorable as written.
Example 8.3--r'. reaction that is spontaneous in the forward direction correspond.s with which:'t ihe following features?{. It has an equilibrium constant that is greater than 1.0.ts. The reaction is spontaneous in the ,"rrerru direction.C. It has a AG11 that is positive for the reaction as written.D. it has a ratio of Qrx 16 GC that is less than 1.
S'olutionlpontaneous in the forward direction" means that the reaction is favorable as'u.itten. Favorability in the forward direction implies that there is an excess of:=actants and a shortage of products relative to equilibrium. This tells us nothingt:'out the value of the equilibrium constant. The ialue of IGq may or may not bireater than 1.0. This eliminates choice A. A reaction that iJspontaneous in thei::-,rard direction is non-spontaneous in the reverse direction. This eliminates:: lice B. A reaction is spontaneous in the forward direction when AG is:e'gative. This eliminates choice C. \Atrhen the ratio of erx to Kuo is less than 1,r€r€ is an excess of reactants, so the reaction proceeds forwardtb equilibrium.l* Qr* to Kuo is less than L, the log of the ritio in Equation g.5 is nlgative, soJ:1 rs negative. A reaction is spontaneous in the forward direction when AG.r.
s :.egative, so choice D is the best answer.
l:r";right @ by The Berkeley Review 137 Dxclusive MCAT Preparation
General Chemistry Thermochemistry Terminology and Graphs
EnthalpyEnthalpy (H) is defined as the quantity of heat present in a system. Enthalpychange (AH) is the heat that is either lost from or added to a system (a reaction)during a chemical reaction or physical process. There is an enthalpy changeassociated with every chemical reaction or physical process. The value of AH ispositive when heat is absorbed by the reaction (as is the case in an endothermicreaction) and negative when heat is given off by the reaction (as is the case in anexothermic reaction). The AH value of a reaction is derived from the system'sheat change, not the heat change of the surroundings. The difference betweenthe energy change of the system and the surroundings is just a sign, because thetotal quantity of energy released by the system is the same as the amount ofenergy absorbed by the surroundings. The sign convention is based on the ideathat the AH value describes the change in heat for the chemical reaction.Enthalpy change alone does not describe the favorability of a reaction, just theheat. However, the value of AH plays a significant role in determining AG.
Example 8.4\tVhich of the observations below is NOT associated with an exothermic reaction?
A. An increase in the temperature of the solutionB. The formation of products that are more stable than the reactantsC. A net loss in the heat energy of the surroundingsD. A value for the change in enthalpy that is negative
SolutionAn exothermic reaction is defined as a reaction that releases heat energy from the
system to the surroundings. This results in an increase in the heat energy of the
surroundings, so choice A is valid. Energy is released by the reaction, becausethe products are at a lower energy level (are more stable) than the reactants. Thismakes choice B valid. An exothermic reaction has a negative value for AH, so
choice D is valid. The false statement of the answer choices is choice C, becauseheat energy is gained by the surroundings, not lost from it. This means that the
correct answer (the statement that is NOT true) is choice C.
Example 8.5If the solvation of NH4Cl(s) by water solvent results in the cooling of the solutiorLthen the sign of the AH value for the solvation reaction is:
A. positive at all temperatures.B. negative at all temperatures.C. positive at low temperatures and negative at high temperatures.D. positive at high temperatures and negative at low temperatures'
SolutionBecause the solution becomes cooler after the reaction, heat must have been
absorbed by the system from the surroundings. The aqueous solution isconsidered to be the surroundings, as that is where the heat is absorbed from-When heat is absorbed, the enthalpy change is positive, making choice A correctThe reaction is thus endothermic. This means that the driving force behind ttndissociation of ammonium chloride into water must be linked to the favorablcchange in entropy, not the unfavorable change in enthalpy. The sign of thcenthalpy change does not change with temperature, so choices C and D shorrldbe eliminated immediately.
Copyright @by The Berkeley Review l3a The Berkeley
General Chemistry Thermochemistry Terminology and Graphs
.:.9 xdu!
rFi tilXu d
I#E II]
|.e -PldUIU.dJE8
>\o.
l]
Energy diagrams can refer to heat energy, in which case the heat of reaction canbe extracted from the graph. When heat is absorbed, the reaction is endothermic.It is uphill from the reactants to products in the energy diagram. When heat isreleased, the reaction is exothermic. It is downhill from the reactants to productsin the energy diagram. Figure B-2 shows the enthalpy diagrams for anendothermic reaction and for an exothermic reaction.
Endothemic reaction (AH is positive)
heat+A+B * P+ZExothemic reaction (AH is negative)
Reaction coordinateFigure 8-2
Example 8.6Combustion of a hydrocarbon in the presence of excess oxygen gas is BEST:lassified as which of the following?{. An exothermic reactionts. An endothermic reactionC. An exoconductive processD. An endoconductive process
Solutioni-icause the temperature of the surroundings increases during a combustion:=action (combustion is the complete oxidation of either a hydrocarbon or-::bohydrate), heat is given off by the system to the surroundings. When heat::.ergy is released, the enthalpy change for the chemical reaction is negative. A-rgative enthalpy change is associated with an exothermic reaction, which makes
":.,rice A correct. The terms "exoconductive" and "endoconductive" ate contrived::ls that have no useful meaning in chemistry. The oxidation of a hydrocarbon-. : reaction, not a process, so for several good reasons, choices C and D should: - -|L be eliminated.
*: - heat change for a reaction may be determined in many ways. It can be found
'::r the AH values for component reactions that sum to the overall reaction.--*. is Hess's law. The change in heat energy may also be determined from:",::e:ences in the bond-energy values for the bonds that are broken and bondsl:":: are formed. Heat energy is needed to break bonds, while heat energy is",1:'=:-.ed when bonds are formed. The last method to consider is an experimental-.
"r'here the heat change for a chemical reaction is measured by calorimetry.
r- ::iorimetry, the change in temperature for a known quantity of surrounding:.:.:ial is measured. This temperature change can be converted into a heat--::.ge for the reaction.
Reactant
A+B -* P+Z +heat
Reaction coordinate
c
cE0Jd
lH3f:FIr€JE6
:;r-ight @ by The Berkeley Review 159 Exclusive MCAT Preparation
General Chemistry Thermochemistry Terminology and Graphs
EntropyEntropy (s) is the measure of randomness (disorder) within a system. Entropytakes into account molecular randomness (degrees of rotational freedom for eachatom within a molecule) which reflects the abitity of a molecule to rotate freelyabout its bonds. Cyclic compounds have less entropy than linear compounds,because they are more rigid and cannot rotate as freely. Cyclic structures arethus less favorable entropically than linear strucfures. Alkanes are more randomthan alkenes, because alkenes have a double bond (n-bond) in particular, aboutwhich the molecule may not rotate. Entropy also accounts for the overallrandomness of a system (freedom for the molecules to move in a translationalmanner). Gases have the most entropy of the three common phases, becausetheir molecules are freest to move throughout the container. the gas phase isentropically the most favorable. Entropy (in a more the advanced definitionpresented in physical chemistry) is the reversible heat change of the system(ereversible) divided by its temperature.
An increase in entropy (and disorder) is defined as a positive change in entropy(AS). We are typically more interested in the entropy change for a chemicalreaction or physical process than we are in the absolute entropy of anycomponent. Entropy can be thought of as organizational potential energy. \zVhenyou organize something, it has the potential to become disordered, and releasethat organizational energy. This organizational energy becomes part of the freeenergy of the system.
Example 8.7\tvhich of the following processes is Nor associated with its correct sign for AS?
A. Assys for a liquid-to-gas phase change is positive.B. ASsys for a reaction in which the volume increases is positive.C. ASsys for a reaction in which a n-bond is formed is negative.D. Assys for sublimation is negative.
SolutionIncreases in the randomness of a system (molecular chaos) have a positive ASsy5value. Conversely, decreases in the randomness of a system (molecular chao"s)have a negative ASr^ value. Because the molecules are becoming more randomin a liquid-to-gas phase change, the entropy change for the process (Assvs) ispositive. Choice A is a valid statement, so it is eliminated. When the voluire ofthe system increases, the molecules are capable of occupying a greater number ofpoints in space, making the molecules more random. The entropy change for theprocess (ASrvr) is positive. Choice B is a valid statement, so it is eliminated-When a n-b<jnd is broken, the molecule has more ability to rotate freely, whichmakes the system more random. Thus, when a n-bond is formed, the moleculeloses some ability to rotate freely, which makes the system less random. Theentropy change for the reaction (ASryr) is negative. Choice C is a valid statement,so it is eliminated. Sublimation is the physical process whereby a solid isconverted into a gas. Because the molecules are becoming more random in asolid-to-gas phase change, the entropy change for the reaction (ASryr) is positivgnot negative. This makes choice D an incorrect statement and th6 best answer-Positive entropy changes are often associated with increases in the number ofmolecules and phase changes to phases of lower density (greater volume).
Copyright @by The Berkeley Review 140 The Berkeley Review
General Chemistry Thermochemistry Terminology and Graphs
Example 8.8What can be said about the entropy of cyclohexane(l) as compared tocyclohexane(s)?
A. The entropy of cyclohexane(s) is greater than cyclohexane(l).B. The entropy of cyclohexane(l) is greater than cyclohexane(s).C. The entropy of cyclohexane(l) is equal to that of cyclohexane(s).D. Neither cyclohexane(l) nor cyclohexane(s) has any entropy, because entropy
applies only to non-cyclic molecules.
SolutionWhile cyclic molecules have lower entropy than their linear counterparts, theystill have entropy. Choice D is eliminated. Cyclohexane is the same molecule, nomatter what phase it is in. The randomness of the molecule itself is unchanged,so there is no change in the degrees of freedom (ability to rotate about bonds andchange angles.) The difference in entropy between the two systems is due to thephase difference. The solid phase is more structured than the liquid phase, so inthe solid phase the molecules are more ordered (less random). Because themolecules are more random in the liquid phase, choice B is correct. You shouldbe aware that the sequence of relative entropy in the three phases is as follows:Sgas > Sliquid > Srotta' Every compound ii aefined as hiving some positiveentropy, according to the third law of thermodynamics.
There is an entertaining and simple example of entropy in action that you can dowith a rubber band. It involves the stretching and relaxing of a rubber band.When a rubber band is stretched and held in the stretched state for a fewmoments, its temperature equilibrates with the environment. Using your lip as athermometer, place the stretched rubber band against your lip to gauge itstemperature. When you are convinced that it is at room temperafure, remove therubber band from your lip and let it return to its natural state of relaxation.Then, immediately place it against the same lip. The rubber band will feel cold.The reason for this is rooted in the thermodynamics of the process. The rubberband naturally tends to go from a stretched state to relaxed one, makingAGstretched-to-recoiled a negative number. The rubber band becomes cold whenit goes from stretched to relaxed, making AHstretched-to-relaxed a positive number(endothermic processes absorb heat and therefore feel cold). Because AG isnegative and AH is positive, AS must be a positive number. This means that as
the rubber band goes from stretched to relaxed, it becomes more disordered.This may seem odd at first, but when you consider any net (like a tennis net or avolleyball net), you should note that when it is stretched, it is orderly. Whenbundled up, it is disordered, In essence, a rubber band is a microscopic net. Thedriving force for relaxing from a stretched state is entropic in nature.
Copyright @by The Berkeley Review t4l Bxclusive MCAT Preparation
General Chemistry Thennochemistry Terminology and Graphs
Hessts Law
Hess's law states that the AG, AH, or AS for a particular reaction can bedetermined by summing the AG, AH, or S values for any cumulative series ofsubreactions for that particular reaction (or in the case of AS, you can sum theentropy values for each reactant and product in the reaction). Typical MCATquestions about this include summing the AG'661p, AH'form, or S" for thereactants and products (reversing the reactant values is necessary to addcorrectly). The overall AG, AH, or AS can be found by subtracting the AG"fot11u
AH"form or S' for lhe reactants from the AG'form, AH"form, or S' for the products.
Equations 8.6,8.7, and 8.8 are the application of Hess's law using formationvalues to calculate free energy, enthalpy, and entropy, respectively.
AG"reaction = ! AG'1616ation products - IAc"go.-ation reactants (8.5)
AH"reac'fi'n;_T:::H;,':::.T ;ilT::"nreac'fan'is lll
The free energy change of formation (AGformatior-,) is the change in free energythat transpires when a compound is formed from its component elements in theirmost stable state at standard temperature (25"C) and pressure (1.00 atmosphere).The enthalpy of formation (AH1o.-atiod is the heat change that transpires whena compound is formed from its component elements in their most stable state.There is no entropy of formation for compounds, as they are defined bythemselves as having a set amount of disorder (entropy). In most generalchemistry textbooks, there are exhaustive tables of free energies and heats offormation, so the values are common information. The AGioltoxli6n andAH6s1661is1 for an element in its most stable form is 0. This applies to oxygmgas, which is found in combustion reactions. Combustion reactions are amongthe common MCAT examples using Hess's law with the heats of formation.
Example 8.9\A/hat is the formation reaction for C2H5Br(g)?
A. 2 C(sr) ** ,rftlB. 2C(gr) *\ n frlc. 2C(gr) +z!H21g1
D. 2C(gr) *\^rtrl
SolutionBromine is found in nature as a diatomic liquid under standard conditions. Youmay recall this from an organic chemistry lab, in which you added Brz in CCI+an unknown organic compound to test for the presence of an alkene n-bond-Bromine (Br2) adds anti across a carbon-carbon n-bond that is present inmolecule, which results in the disappearance of the bromine liquid. Because *rebromine liquid is brown, the presence of an alkene is supported by thedisappearance of the brown color from the solution. Based just on the phasemolecular state of bromine in the answer choices, the best choice is answerThere are certain miscellaneous facts that you should know for some question the MCAT. Although the test writers give you a great deal of informatithere is other required information (background knowledge) that you must
+!Brz(g) -.--> C2H5Br(g)
+ ! Br21l1 -* C2H5Br(g)2-'+ Br(g) * C2H5Br(g)
+ Br(l) * C2H5Br(g)
Copyright @ by The Berkeley Review 142 The Berkeley
General Chemistry Thermochemistry Tenninology and Graphs
To apply Equations 8.6,8.7, and 8.8 correctly, the overall reaction must bebalanced first. Stoichiometric coefficients are critical to Hess's law, as eachcomponent is multiplied by its stoichiometric coefficient prior to summing thevalues. Example 8.10 shows the application of Equation 8.7 in calculating theenthalpy for the combustion of methane. The solution is a step-by-step processthat serves as a case-based derivation of Equation8.7.
Example 8.10\Ahat is the standard enthalpy of combustion for methane?
A. AH'6e11nation CO2 + AH'161p3tion H2O - AH'formation CHaB. AH"formation CO2 + 2 AH"Sslmation H2O - AH"formation CHaC. AH"formation CH4 - AH'formation Co2 - AH"formation H2oD. AH"6s16ation CHa - AH'formation CO2 - 2 AH"lelpation H2O
SolutionThe first step is to balance the overall reaction:
Balancedreaction: lCHa(g) + 2O2G) --+ 1CO2(g) + 2H2O(g)
The next step is to view the formation reactions and standard enthalpy offormation for each compound in the balanced equation:
AH"formation CHa(g) = AH'rxn for: 1C(gr) +2IbG) + CHa(g)
AH"formation Oz(g) is not applicable, because OZ(g) is in its most stable state
ihe final step is to sum up the formation reactions in a way that equals thet.,'erall reaction. This requires multiplying reactions by their stoichiometric:oefficients, reversing the reaction for compounds on the reactant side, and:rossing out molecules that appear on both sides of the subreactions:
AH"formation CO2(g) = AH"rxn for:
AH"formation H2O(g) = AH"rxn for:
- CHa(g) --+ 1€(Br) +2112@
re€d + 1o2(g) --+ 1Co2(g)
-l2fd + 1O2(s) -+ 2 H2O(g)
C(gr)+oz(s) + COz(s)
Hz(g) + torftl -+ Hzo(s)
-1 (AH"i6snation CHa(g))
1 (AH"16.*ution CO2G))
2 (AH'i6rllntion HzO(g))
- CHa(g) + 2 O2@) --+ 1 CO2(g) 2H2O(g) AH"r Coz + 2 LH" ,1g,ro - AH"f CH4
-re best answer is choice B. Because the formation reactions for reactants are:er-ersed, the equation for calculating any thermodynamic value from the: -,rmation values involves formation values for products minus formation values::,r reactants. Equation 8.7 shows this for determining the enthalpy of reaction::om the enthalpy of formation values.
-:,pvright @ by The Berkeley Review 143 Exclusive MCAT Preparation
General Chemistry Themrochemistry Terminology and Graphs
For examples 8.11 through8.1.4, use the values in Table 8.2.
Compound AH'61kl7-o1 S" (I/mot.r) Compound AH'61kJ7-o,; S'(I/mot.r)
co(g) -1.1L.2 198 CozG) -393.5 21.4
Hzo(g) -242 189 H2O(l) -286 70
C2H5OH(l) -278 161 C5H6O2(s) -1011 196
C6H1206(s) -1278 212 ozG) 205
Table 8.2
Example 8.L1\A/hat is the standard enthalpy change for the water shift reaction?
Coz(g) + Hz(g) --+ CQg) + H2o(g)
A. +40.3 kJlmole
B. +3.2 kll-o1"c. ajkl/moIeD. -+0.3 kllmole
SolutionThis question is solved using Equation 8.2 AH"reaction = IAH"form products - IAH'form reactants. The reaction is balanced as given, so we can procbed directlyto the insertion of numerical values:
AH'1sx6ti61 = AH'form qg + AH'661m H2O -AH'fo.* CO2 - AH'6o.r1 p,
AH"reaction = (-111.2) + (-242)- (-393.5) - 0 = -353.2 + 393.5 = + 40.3 kJlmol"
The correct €urswer is choice A.
Example 8.12What is the standard enthalpy change for the combustion of ethanol?
C2H5OH(I) + Oz(g) -+ CQG) + H2O(l)
A. -Is6zkJlmoleB. -t235 kJlmole
c. -zqikl /moleD. -401.5 kJlmole
SolutionThis question is solved using Equation 8.7, AH'reaction = I AH'form products - IAH"for* reactants. The reaction must first be balanced:
CzHsOH(l) + 3 O2(g) --+ ZCOzG) + 3 H2O(l)
From here, we can proceed to the insertion of numerical values:
AH'reaction - 2 AH'6611 CO2 + 3 AH'66r- HrO - AH"form C2H5OH - 3 AH"5or* 61
AH'reaction = 2 C393'5) + 3 (-286) - (-278) - 3 (0)
= 2 (-400 + 6.5) + 3 (-300 + 14) + 300 - Zz
= -800 + 13 -900 + 42 + 900 -22 = -1400 +33 = -1867 kJlmote
The correct answer is choice A.
Copyright @ by The Berkeley Review 144 The Berkeley
General Chemistry Thermochemistry Tenninology and Graphs
Example 8.13\A/hat is the standard enthalpy change for the combustion of glucose?
C6H12O6(s) + Oz(g) -+ CQ(g) + H2O(g)
A. +598.5 kIlmoleB. -tS23kI/moteC. -Z799kJ /moteD. -5355 kllmole
SolutionThis question is solved using Equation8.7, AH"reaction = LAH'form products - IAH"forrr, reactants. The reaction must first be balanced:
C6H12O6(s) +6 O2G) *-+ 6COzG) + 6H2O(g)From here, we can proceed to the insertion of numerical values. oxygen is moststable as 02 under standard conditions, so it is omitted from the expression:
AH"reaction = 6 AH"form CO2+ 6 AH"fo.* HrO -AH"form C6H12O5,
AH'reaction = 6 (-393.5) + 6 (-286) - (-1278) = 6 (-679.5) + 1278
= 6 (-700 + 20.5) + 1300 - 22 = - 4200 + 723 +1300 - 22 = - 2900 + 101
= _2799 kJlmole
The correct answer is choice C.
Example 8.14\\hat is the change in entropy for the combustion of pentyl lactone (C5Hgo2(s))?
C5HsO2(s) + 6 O2@) --+ 5 CO2(g) + 4 H2O(l)
A. +3701l-o1".6B. +2 J/-61".14C. -741 /
^oL.yD. -1.17Jlmole.f
SolutionThis question is solved using Equation 8.8, AS"reaction = L S"form products - IS'fo.11 reactants. The reaction is balanced as given, so we can proc""l directly to]re insertion of numerical values:
AS"reaction - 5 S"COz + 4 S"gr9 -S"CsHsOz - 6 S'Oz
AS'reaction = 5 (214) + 4 (70) - (196) - 6 (205) = !070 + 280 - 196 - 1230
= 1350 -7424=-74The correct answer is choice C. Be sure to pay attention to the phase, especially:or water, because inadvertently using the value for H2o(g) instead of u2o1iy',.,-ould lead you to pick choice A.
-opyright O by The Berkeley Review t45. Exclusive MCAT Preparation
General Chemistry Thermochemistry Terminology and Graphs
Bond Energies
Heat of reactions can be calculated using the heat added to the system to breakthe bonds and the heat released when new bonds are formed. It takes energy tobreak a bond, and energy is given off when a bond is formed. This informationcan be applied to an energy diagram, as shown in Figure 8-3 below:
Reaction coordinate
Figure 8-3
Figure 8-3 shows an exothermic reaction. Enthalpy data helps to determinerelative bond strengths. In an exothermic reaction, net energy is released, so thebonds that are formed are stronger than the bonds that are broken. Equation 8.9can be used to calculate the enthalpy change for a reaction from bond energies.
AHreaction = Energlbonds broken - Energy6er.rds formed (8.e)
A great example of bond energies is found in biochemistry. Energy is storedwhen adenosine diphosphate (ADP) is converted into adenosine triphosphate(ATP). This energy can be released when ATP is converted back to ADP, shownas Reaction 8.1 below:
ATP + H2O --+ ADP + Pi + heat
Reaction 8.1
The release of energy is predictable from the relative energies of the bondsbroken and the bonds formed. Reaction 8.1 is shown in more detail in Figure B-4.
rli
(d
€t!
oo.llll^li-TP-o" + t-oTH +60 loo
I0.t0",", brfk"r,Bonds broken: O-P and O-H
ooll-"T" + '-ot {t-."60 I loo
r.,l-"a rol-"aBonds formed: O-H and G-P
Figure 8-4
By general chemistry conventions, because the bonds broken are the same as thebonds formed, we predict that the AH for the reaction is zero. However, thisreaction is used to provide cellular energy, so we know it is actually heatreleasing. The discrepancy is explained by examining the bonds in more detailBecause the reaction is exothermic, the bonds broken are weaker than the bondsformed. The unusually weak bond in the ATP molecule is the O-P bond-Because of electrostatic repulsion by the oxygen atoms, the O-P bond iselongated. Longer bonds are weaker, so the O-P bond of ATP is a weak bond.
Bonds are broken earlyin a chemical reaction.Energy mustbe addedto break the bonds.
Bonds are formed later in a
chemicalreaction. Energyis released when bonds areformed.
Copyright @by The Berkeley Review 146 The Berkeley Revier
General Chemistry Thermochemistry Terminology and Graphs
The O-P bond in ATP is referred to as a high-energy bond tn biochemistry. Thispresents a slight naming problem, because in chemistry, a high bond-energydescribes a bond that is strong. A "high-energy bond" (note the inversion of then'ord sequence) is not a bond with high dissociation energy. The high-energybond in the phosphate linkage is actually a weak chemical bond. Perhaps it isbest to think of a high-energy bond as a bond that is high on the energy diagram,making it unstable (reactive). It is that very weakness that leads to the overallrelease of energy when this bond is cleaved in the hydrolysis of ATP. Example3.15 shows how the bond energies are used to quantify the change in enthalpy.
Example 8.15tr\hat is AH for the following reaction, given that the bond energy for C=O is 795kJlmole, for H-H is 428 kJlmole, for C-O is 361 kJlmole, for C-H is 411k]/mole, and for O-H is 468 kjlmole?
H3CCHO(g) + Hz(g) -+ H3CCUzOH(g)-{. -90 k}/moleB. -77 kl /rnoleC. +17kJ/moLeD. +90 k]/mole
SolutionThe first step is to determine the bonds that are broken and the bonds that areiormed. The diagram below shows the bonds that are broken and formed:
o
-t+HIC H
Broken: C=O, H- H
II
H*H +I
HHt, I)..
HrC' 'o--- H
Formed: C- O, C- H,O- H
trn reality, only the n-bond of the carbonyl is broken, but we are given numbers interms of single and double bonds, so we consider the entire C=O double bond to're broken and the single C-O bond to be reformed. Using Equation 8.9:
AHreaction = BEC=o+ BEg-g - BEc-o - BEc-u - BEo-H
AFlreaction =795 + 428 - 361' - 41'1' - 468
AHreaction = 800 - 5 + 400 +28 - 400 + 39 - 400 - 11- 500 + 32
1200 - 1300 + 99 - 1,6= -100 + 83 = -17 kI / moleThe correct answer is choice B. The reaction is exothermic, which means that theronds broken are weaker than the bonds formed. This implies that n-bond are',veaker than o-bonds.
The values given in Example 8.15 are for heterolytic bond dissociation in the gasphase. There are some assumptions in general chemistry that are not true. Ingeneral chemistry, all C-H bonds are treated as equal in strength, although weknow from organic chemistry that they are not. For instance, a 3" (tertiary)carbon forms a weaker bond with hydrogen than a 2' (secondary) or 1' (primary)carbon forms with hydrogen. The bond energy also varies with hybridization,ivhere bonds involving sp carbons are stronger than bon* involving spzcatbons, which are in turn stronger than bonds involving spr carbons. Bondenergies are also considered in organic chemistry, but in more detail.
Copyright @ by The Berkeley Review t47 Dxclusive MCAT Preparation
General Chemistry Thermochemistry Calorimetry
Heat Energy
Calorimetry is the measurement of the heat released from or absorbed by achemical reaction or physical process. The heat change can be measured bycarrying out a reaction in a sealed reaction vessel that is surrounded by somematerial (a liquid is best) that is capable of absorbing the heat given off by anexothermic reaction, or capable of supplying heat to an endothermic reaction.Once the heat has been transferred, the liquid (known as the heat sink) eitherincreases in temperature due to the gain of heat energy or decreases due to theloss of heat energy. A decrease in temperature is observed if the reaction isendothermic. The heat of the reaction can be calculated from the change intemperature for the system using the heat capacity and mass of the solution.Equation 8.10 shows the relationship between heat energy (q) and the change intemperature (AT). The mass of the solution is represented by m, and the heatcapacity of the solution is represented by C.
9 = MCAT (8.10)
By using measured values of reactants, the enthalpy change per mole for thereaction can be calculated, assuming that the enthalpy change for the reaction isequal to the heat energy of the reaction (q). Th" stereotypical lab experimentfrom general chemistry, lab involves the running of a reaction in a styrofoam cupfilled with an aqueous solution and a thermometer. A stirring rod is alsoprovided to ensure that there is homogeneous distribution of heat throughoutthe system. The temperature is read in consistent intervals and then plotted as afunction of time. The true final temperature after the complete mixing of thereactants and the end of the reaction cannot be read from the thermometerdirectly, because by the time that the thermometer has actually reached thetemperature of the solution, heat has already been exchanged with theenvironment. This means that to determine the final temperafure after reaction"a line must be extrapolated from the data points on the graph back to thetemperature axis at the t = 0 mark. This technique is common in laboratoryprocedures. The plotting of temperature as a function of time must be carriedout in uniform increments (intervals) in order to get useful data.
Example 8.16How many calories does your body use to heat up 100 g of water from 0'C to37'C?
A. 7400caloriesB. 3700 caloriesC. 740 caloriesD. 37]calories
SolutionTo determine the heat needed to raise the temperature of 100 g of wEquation 8.10 is used. The mass is L00 g, AT is 37'C, heat capacity for water iscaI/ g.K, so q is found by basic multiplication. Plugging known values into qmCAT yields: (100 gX1.0 callg.K)(37K) = 3700 cal. Choice B is the best answer.you drink all of your beverages at icy temperatures, you will expend caloriheating them to body temperature. If you were to drink two liters of ice waterday, that would result in the expenditure of 74 nutritional calories to heat wa
Copyright @ by The Berkeley Review t4a The Berkeley
General Chemistry Thermochemistry Calorimetry
lnample 8.17r,rl;L;r i5 the final temperature of a solution made by mixing 30 mL of H2O at:il :'C n'ith 60 mL of H2O at 60.0"C?
.4. rl.5'Ct| ;1.0'c:. {_r.O'c
-. ;.) L
ir, utionT,r::lse the larger volume has the greater temperature, the final temperafure iss:::-:er than the mean temperature (45"C). This eliminates choices C and D. To:*-le between choices A and B, we must rely on a more rigorous calculation.lie answer can be found by one of two methods. The first method is to use ar":::l-!ted average of the two volumes and their temperatures. The final volume
:r ,,,-ater = 90 mL, "o
30 mL (one-third) of the solution starts at 30"C and 60 mL90 mL 90 mL
1'';-ihirds) of the solution starts at 60"C. The weighted average is calculated as'r : "
l1\-S:
I rso"cl + 2 reo'c) = 10"C + 40'C = 50 'CJJ
],:,rce B is the correct answer. The second method is based on Equation 8.10. Itr,:ates the heat lost by the warmer solution to the heat gained by the cooler;:.rfion. If no heat is lost to the environment, then E"oo1ir.g + Eheating = 0. This::.:. be manipulated to read: Eheating = - Ecooling. Substituting Equation 8.10 for::-:erg.!'on each side of the equation yields the following:
mCAThs61l1g = - mCAT66eling
mAThs6tilg = - mAT66eling
30 g (T1- 30) = - 60 g (T1- 60) = 60 g (60 -Tf)
30 (T6- 30) = 69 (60 - Tf) .'. T1 - 30 = 2 (60 - Tt)
T1- 30 = 120 - 2T1 .'. 3 Tf = 150 .'. Tf = 50
:::1r'smokes Batman, it's choice B yet again.
Heat Capacity
:{eat capacities are just what the name implies; they are a measurement of thei:rrage capacity of absorbed heat for a given material. The unit for heat capacity< energy per mole.Kelvin. Energy can be measured in either joules or calories:ld the temperature change is the same whether measured in kelvins or:erbigrade (Celsius). Heat energy is kinetic energy, which for our considerations;:rall include only translational energy and vibrational energy. For the most part,;,-.lids have lower heat capacities than liquids, because they have only vibrationalr:netic energy.
:{eat capacities are used in the calculation of transferred heat, which is usually:ansferred within a material via conduction. Conduction involves the transfer of:eat by way of collision when molecules are in direct contact with one another.-'*.q two atoms within a molecule vibrate about a bond, the atoms may collide-,','ith a neighboring atom and transfer some of the vibrational kinetic energy torat neighboring atom. That is the essence of conductive heat transfer on the:ricroscopic level. The greater the amount of energy required to increase the'-ibrational energy of a substance, the greater its heat capacity.
Copyright @ by The Berkeley Review 149 Exclusive MCAT Preparation
General Chemistry Thermochemistry Calorimetry
The rate of temperature change for a material depends on its heat capacity. Amaterial with a high heat capacity requires more time to reach a targettemperafure than a material with a lower heat capacity, when heat is applied at auniform rate. Because of this, materials with high heat capacities generallyexperience small temperature increases. This is why the temperature near a bodyof water is relatively consistent compared to temperature fluctuations in morearid regions. Water has a high heat capacity, so it absorbs a great deal of heat onhot days, preventing the air temperature increasing that much. On cold days,water can release heat to the environment, preventing the air temperature fromdecreasing that much, making it warmer than it would be in an aridenvironment.
Example 8.18\A/hat is the heat capacity of a material that requires 3.0 kcal to increase a 15.0 gsample from 25'C to75'C?
A. 1.9 calTr.g
B.2.gcal1f.yC. 3.6 calT*.6
D. 4.0 callf.y
SolutionTo calculate the heat capacity (C) from heating data, Equation 8.10 is used. Themass is 15.0 g, AT is 50'C, and q is 3000 cal, so C is found following basicmultiplication. Plugging known values into q - mCAT yields:
3000 cal = (15 g)(CX50 K) .'. C = 3000 ,ul / 1tsgx50 K) = 3000 "uI / rSo *x
After manipulating the numbers to isolate C, the value is found to be 4.0 callf.tq-Choice D is the best answer.
Example 8.19\Alhich of the following observations is valid when comparing two materialsequal mass starting at the same temperature?
A. The material with the greater heat capacity has a higher melting point.B. The material with the greater heat capacity has a higher boiling point.C. The material with the greater heat capacity reaches a higher tem
when they are both exposed to the same amount of heat energy.D. The material with the greater heat capacity reaches a lower
when they are both exposed to the same amount of heat energy.
SolutionPhase change processes have little to do with heat capacity. They both relateintermolecular forces to an extent, but they are not related to one another.such, heat capacity cannot be used to estimate melting point or boiling point,choices A and B are eliminated. If the two materials are exposed to theamount of heat energy (q) and both have equal masses (m), then thewith the greater heat capacity (C) experiences the smaller temperature(AT), according to Equation 8.10. A smaller temperature change resultsIower final temperature, so choice D is the best answer.
Copyright @by The Berkeley Review 150 The Berkeley
General Chemistry Thermochemistry Calorimetry
Heat of Phase Changes
In addition to heat being absorbed or released as the temperature of a material,mixture, or solution changes, heat can also be exchanged when a materialundergoes a phase change. However, during a phase change, the temperafure ofthe system does not change. Phase changes are reversible physical processes. Ifa phase-change process is endothermic, then the reverse process is exothermic,yielding the same absolute value of energy. For instance, the melting of a solidinto a liquid is endothermic, with an enthalpy change of AH;rrr1or.,. The freezingof that same liquid into a solid is exothermic, and the enthalpy change forfreezinghas the same absolute value as AH6rrriorr, but it is a negative number.
The heat exchanged during a phase change can explain some seemingly bizarrephenomena. As strange as it may seem, when water heezes, the surroundingenvironment becomes warmer. An interesting application involves frozen lakes.As water freezes downward from the bottom of the ice layer at the surface of thelake, heat is released to the surrounding water. This warms the water just belowthe surface of the ice. Water is densest at 4'C, so as the water warms from 0'C to4'C, it becomes denser. Warm water (if you can call water between 0"C and 4'C"warm") sinks to the bottom of the lake. This is responsible for the circulation ofwater under frozen lakes.
The heat of phase changes can also be applied to the concept of refrigeration.The enthalpy of vapofization is significantly greater than the heat of fusion, so aphase change between liquid and gas can be used to absorb or release a greatdeal of heat. For instance, steam bums are far more severe than hot water burns,even when the steam and water are both at 100"C. The greatest amount of heat isassociated with condensation. This is also the idea behind perspiration. Whenwe perspire, the moisture on our skin evaporates, absorbing a large amount ofheat from the surface of our body. This works only in urrrriror,*ents that arearid. L:r a humid environment, moisture from the air can condense on your skin,rvhich negates the perspiration process. This is why a "dry heat" feels preferableto a "hot and muggy" climate for most people. Our bodies are physiologicallyequipped to deal with heat in a dry climate.
Example 8.20How much energy is required to melt 30 grams of ice at 0'C? (AH6.rsion = 6.0kIlmole)
A. 5.4 kJB. 5.0 kJc. 9.0 kJD. 10.0 kJ
Solutionln this question, AHfusion is expressed in units of energy per mole, so the amountlf water in the ice should also be expressed in terms of moles, rather than grams.',\'e must start by converting grams of ice into moles of ice. Upon dividing the 30
-ams by water's molecular mass (18 grams per mole), we get 1.667 moles of ice.Ihe math is simply: 1.667 moles x 6.0 kilojoules per mole = 10 kJ. The correcte:$wer is choice D. When you consider the amount of calories needed to melt:;e, it becomes evident that snacking on ice is a great diet strategy from a caloricstandpoint. Ffowever, from a nutritional standpoint, a diet of nothing but ice:nips would probably not be the best for anyone in the long run.
-opyright @ by The Berkeley Review l5l Exclusive MCAT Preparation
General Chemistry Thennochemistry Calorimetry
Calorimetry Experiments
The enthalpy change values we see in general chemistry textbooks are the resultsof calorimetry studies. A calorimetry study involves carrying out a reaction in acontainer surrounded by a heat sink (usually water) to absorb or provide heatenergy. One passage on the MCAT several years ago presented a typicalcalorimetry experiment using a styrofoam cup, an experiment that is conductedin most general chemistry laboratory classes. The reaction is an exothermicreaction between a strong acid and strong base. A styrofoam container is chosenbecause it is a good thermal insulator, and it fits within most laboratory budgets.The styrofoam calorimeter experiment is shown in Figure 8-5 below:
Thermometer
Initially filled with100 mL 1.0 M HCl(aq)
Figure 8-5
A baseline temperatures for the hydrochloric acid solution in the styrofoam cupand for the L00 mL L.0 M KOH(aq) in a second container are determined. Once a
steady baseline temperature for each solution is established, the potassiumhydroxide solution is added to the hydrochloric acid solution in the styrofoarncup. The temperature is then recorded at regular twenty-second intervals.Because the reaction is exothermic, the temperature of the system increases. Adouble styrofoam cup system is used to increase insulation. Figure 8-6 showsdata that were collected and graphed at regular time intervals.
80 100 120 140 150
Time (seconds)
Figure 8-6
34.0
32.0
Cr.- 30.0o.)k
fr za.oti0)
t zo.o0)li
24.0
60
Double styrofoam cup
Copyright @ by The Berkeley Review
20 40
152 The Berkeley
General Chemistry Thermochemistry Calorimetry
Analysis of the data has the goal of determining a value for AT for the mixture.The reaction is not instantaneous, because the heat is not transferred instantlp sothe thermometer does not register the final temperature right away. Heat is lostfrom the system to the environment by the time the thermometer catches up andreads the actual solution temperature. To compensate for this loss of heat, thedata points are used to extrapolate a line back to the time of mixing (the z0-second line in this experiment). This leads to a value for AT, which can then beplugged into Equation 8.10, q = nCAT. The value used for AT is the change intemperature from initial temperature (at the time when the solutions are firstmixed) to the projected temperature (extrapolated back to the time of mixingfrom the data), as shown in Figure 8-7 below:
34.0
32.0
30.0
28.0
26.0
24.0
22.0
20 40 507080 100 120 r40 160 180 200 220 240 260Time (seconds)
Figure 8-7
The styrofoam cup is not a perfect insulator, so some heat is lost to theenvironment, which accounts for the drop in solution temperature over time.with better insulation, the slope of the exirapolated line is .lor", to zero (a flatline). There are several conceptual and error-analysis questions that canaccompany this experiment, which makes it a prime topic for the MCAT.
Example 8.21tVhy is the solution stirred after mixing?
A. To generate the activation energy needed for the reactionB. To prevent the mixture from forming a biphasic systemC. To increase the effectiveness of insulationD. To distribute the heat evenly throughout the system
Solutionstirring does not introduce much energy into the system, so choice A iseliminated. Two aqueous solutions do not form a biphasic layer, so choice B iseliminated. Insulation depends on the material, not the motion of the solution, so;hoice C is eliminated. Solutions are stirred to generate homogeneity of solutemolecules and heat. Heat must be evenly distributed through the solution. If the>olution were not stirred, then the temperature of the solution at the point where*te thermometer rests might not be accurate, because heat pockets in the solutioncould exist. Choice D is the best answer.
U@L
!
o.Ec)F AT=34"C -2TC=12"C
Copyright @ by The Berkeley Review Exclusive MCAT Preparation
General Chemistry Thermochemistry Calorimetry
Example 8.22What would be observed if the concentrations of HCI and KOH were cut in half?
A. The change in temperature would be the same.B. The change in temperature would increase by a factor of two.C. The change in temperature would decrease by a factor of two.D. The change in temperature would decrease by a factor of four.
SolutionIf the concentrations of both solutions were cut in half, assuming that theirvolumes stayed the same, then the moles of reactants would be cut by half. Witha reduction to half the quantity of reaction, the amount of heat released is cut byhalf. This would reduce AT by half, and makes choice C the best answer. Ifeither reactant is decreased, then the amount of heat released must also decrease.In this case, both reactants are decreased by the same amount, so neither onebecomes the limiting reagent.
Example 8.23\A/hy is styrofoam chosen for the experiment?
A. For its high heat capacityB. For its isothermal compressibilityC. For its thermal insulating capabilityD. For its ability to conduct heat
SolutionThe goal in a calorimetry experiment is to measure the heat associated with a
reaction. Paramount in this task is the ability to retain all energy within thecontainer, minimizing the interaction with the surrounding medium. Styrofoamis chosen, because it is a good insulator. This makes choice C the best answer.You do not want the container to absorb heat, so choice A is eliminated. The
experiment entails monitoring a temperature change, so no "isothermal" propertris pertinent. This eliminates choice B. Styrofoam is not very dense, so it isassumed that its molecules are relatively far apart from one another. Moleculesthat are far apart (not in direct contact with one another) do not conduct heatvery well. Thus, styrofoam reduces heat loss through conduction. This
eliminates choice D. A thermos design is used in more accurate calorimetn-experiments. The design of a thermos includes an evacuated chamber betweenthe inner container and the outer wall. An evacuated space prevents heattransfer either through convection or conduction.
Example 8.24VVhy must a lid used be used in the calorimetry experiment?
A. To prevent heat loss via conductionB. To prevent heat loss via convectionC. To prevent heat loss via evaporationD. To prevent heat loss via radiation
SolutionThe lid serves as a barrier to prevent gaseous molecules from escaping. This
reduces both convective heat loss and evaporative heat loss. The greateslamount of heat is lost through evaporation, so the more significant role of a lid isto prevent evaporative heat loss. Choice C is the best answer.
Copyright @ by The Berkeley Review 15'4 The Berkeley Relic
General Chemistry Thermochemistry Calorimetry
Heat TransferHeat is naturally transferred from a warmer object to a cooler object. The greaterthe difference in temperature between the objects, the more heat that transfersper unit time. Heat is transferred in four ways: convection, conduction,radiation, and evaporation-condensation. These terms are defined below, alongwith examples of each type of heat transfer.
Conaection (aia fluid medium): Convection is the movement of heat through a fluidmedium. The particles of the medium collide with the surface of the hotobject, resulting in the transfer of heat from the hot object to the medium.The medium can flow, allowing it to travel through the space between thehot and cold objects. When a warm medium strikes the surface of a coldobject, heat is transferred through collision to the cold object. The net resultis that heat is transferred from the warmer surface to the cooler surfacethrough the fluid medium. Thb heat is said to travel via thermal currents.
convection example: Convection ovens function by creating thermal currents,which heat food. The heat source (electric coil or flame) is placed in thebottom of the oven where the cooler (denser) air is found. The cooler aircollides with the heat source, increasing its thermal energy. As the airbecomes warmer, it becomes less dense, and thus rises. The food is typicallyplaced near the top, above the heat source. The warmer air strikes thecontainer holding the food and transfers heat to the container. This cools theair, which then becomes denser, and sinks back to the bottom. The flow ofthe warm and cool air creates thermal currents. Convection ovens are theconventional ovens found in most kitchens. Some convection ovens augmentthe air flow with fans that help circulate the warm and cool air.
Jonduction (uin direct contact): Conduction is the transfer of heat through directcontact (vibrational energy is transferred via collisions between neighboringmolecules). The particles of a solid vibrate faster as the object becomeswarmer. Vibrating molecules can transfer heat energy to neighboringmolecules when they collide. This allows neighboring molecules toequilibrate. Heat is transferred from the warmer point to the cooler point inthe object. Because of the continual heat transfer and equilibration ofvibrational kinetic energy between neighbors, heat gradients are formed.
londuction example: When heating a metal rod (as when roastingmarshmallows over an open flame), the end of the rod in the flame becomeshot. However, although one end is placed in a heat source energetic enoughto cook food, the other end of the metal rod is not too hot to hold. This isbecause the temperature is not uniform throughout the rod. There is agradient of temperatures between the hot end and the cold end of the rod.
:,:.Jiation (absorption and emission of IR photons): Radiation is the emission ofphoton energy in the form of electromagnetic waves. The energy of thephoton varies with the temperature of the object that emits it, as expressed inphysics by Wien's displacement law. An object becomes cooler and itsmolecules move less energetically as it continues to radiate infrared energy.
r :.diation examples: Heat-seeking devices are designed to detect regions of highinfrared emission. Heat is responsible for making molecules vibrate, and IRphotons, when absorbed, stimulate a molecule to reach an excited vibrationalstate. This means that hot objects emit infrared photons. Microwave ovensfunction by irradiating food with EM waves of a specific frequency thatrotates the water molecules inside the food. By heating the water moleculesselectively, water molecules can transfer heat to neighboring food moleculesthrough conduction.
, :r'right @ by The Berkeley li.eview 155 Exclusive MCAT Preparation
General Qhemistry Themrochemistry Calorimetry
Eaaporation-Condensation: Evaporation is an endothermic process in which a
liquid absorbs heat and is converted a liquid into a gas. Condensation is anexothermic process whereby heat is released from a gas as it is converted intoa liquid. A hot body can be cooled when a liquid evaporates from its surface.A cold body can be warmed when a liquid condenses on its surface.
Evaporation-condensation example: Perspiration involves the evaporation of athin film of liquid on the surface of the skin to cool the body. To maximizethe effects, surface area must be maximized. In very humid environments,where the rate of condensation equals the rate of evaporation, evaporation is
ineffective in cooling the body. For instance, when you first enter a gym, itfeels hot and sticky. This is the perspiration from the other inhabitantscondensing on the surface of your skin (you unsuspecting cold body you.)
\Atrhen designing many kinds of machinery, engineers must consider all forms ofheat transfer. A good example is the Miracle Meat ThawerrM, as seen on some ofthe finer infomercials television has to offer. The Miracle Meat ThawerrM takesadvantage of several principles of heat transfer. It is a metal tray with a blacktop, and ridges. The dark surface absorbs light energy more effectively than a
light surface would. The ridges increase the surface area, optimizing theabsorption of photons. Being made of metal allows for easier conduction of heatenergy. The metal should have a low heat capacity, so that minimal heat isretained by the metal itself. A frozen piece of meat placed on the surface of thethawer acts as a heat sink for all of the energy absorbed by the metal tray. Awell-designed tray has a slight slope and a drainage hole leading to a basin. Ifwater collected on the surface of the plate, then heat energy would be wasted inevaporating the water. The apparatus does not work as well if the object beingthawed covers the entire surface, because fewer photons can be absorbed.
An engine radiator is another example of a device designed to transfer heat. Itworks by moving a fluid through the core of a hot engine. The fluid absorbsheat, provided that the coils through which it moves have enough contact withthe hot engine. The fluid then passes through more coils in the front of the car,which are covered with thin fins. The purpose of the fins is to maximize the
surface area from which heat may be lost via convection. As air passes over thefins, heat from the engine is released to the air.
Example 8.25\A/hich of the following is NOT a reason why the space surrounding the resistivecoil in an incandescent light bulb is evacuated?
A. To maximize energy loss via radiationB. To prevent pressure buildup at higher temperaturesC. To prevent oxidation of the filamentD. To maximize energy loss via convection
SolutionA light bulb is designed to convert heat (generated through resistance in the coilinto light. By evacuating any gas from the bulb, we reduce the amount of hea;energy transferred from the coil through convection. The result is that energy is
dissipated as light, and not heat. This maximizes radiation and minimizesconvection, making choice A valid and choice D invalid. Choice D is the bes:answer. If the bulb contained gas, the gas could expand upon heating, and the
bulb might blow up. This makes choice B a valid statement. In an evacuateibulb, there is no oxygen, so oxidation is eliminated. This makes choice C valid'
1
il
{
Copyright @ by The Berkeley Review r56 The Berkeley Revieu
General Chemistry Thermochemistry lleat and Ulorh
fleat and Wofl1 : ,: , ,, ,,,,,,, :,,; ,,:, " ,, ,,,',
EnergeticsEnergy is a topic conunon to both chemistry and physics. In a sense, chemistry isthe study of energy production while physics is the study of energy application.In chemistry, we think of energy in terms of heat, which is defined by the calorieunit. Heat is absorbed or released during chemical reactions. In physics, wethink of energy in terms of work, which is defined by the joule unit. Work isdone by either the system or the surroundings in a physical process.
Energy can be converted between heat and work. Heat may be absorbed by asystem, expanding the gas contained within, which in turn does work on thesurroundings as it expands. This is the essence of any machine that converts heatdirectly into work. In this case, heat is defined as positive, because heat energy isgained by the system. Work, however, is defined as negative, because work isdone by the system on the surroundings. The total energy of a system is a sum of theheat and work. Equation 8.11 defines the change in energy for the system.
4f, = q-PAV (8.11)
E is the energy of the system, q is heat, P is pressure, and V is volume. Work canbe described as either FAd or PAV, depending on the system. For a pistonsystem, the volume changes, so work is PAV. As a piston expands (positive AV),energy is released by the system to the surroundings, so work is defined as -PAV. Values are defined from the system's perspective. When q is positive, heatf-lows from the surroundings into the system. \Alhen q is negative, heat flowsfrom the system out to the surroundings. When w is positive, work is done onthe system by the surroundings. \A/hen w is negative, wotk is done by the systemon the surroundings. \rVhen a piston retums to its original position, the energy ofa system does not change (AE = 0). This is to say that energy is neither absorbednor released, but converted between work and energy (q = plV).
Some terms that describe the conditions involved in energetics are listed below:
Adiabatic: A process where there is no change in heat (q = O). Heat is neithergained nor lost. The system is perfectly insulated thermally.
Colorie: A unit of energy defined as the heat energy required to raise 1.000gram of water by 1.000 "C (specifically from 14.50"C to 15.50 'C).
Energy (E): The capacity to do work or to produce heat. There are two typesof energy with which we shall concern ourselves: kinetic (energy of motion)and potential (position of an object or chemical composition).
Heat (q): The form of energy involving the motion of molecules. Motionincludes translational displacement, and vibrational and rotationalmovement. The heat associated with a process depends on the pathway.
.{otile: A unit of energy defined as the work energy exerted when oneNewton of force is applied to an object to move it a distance of one meter.
SLtrroundings: The surroundings are defined as the environment neighboringa chemical reaction or physical process. The surroundings can either donateenergy to the system or absorb energy from the system.
System: The system is defined as the contents of a chemical reaction orphysical process that produce or absorb energy. We often study the system,ithen we look at a chemical reaction.",\'ork (w)'. The ability to move an object over a given distance by applying a
iorce. This form of energy is harnessed to move mass. Like heat, workSepends on the pathway. w = F.d = - P.AV
--.ovright @ by The Berkeley Review 157 Exclusive MCAT Preparation
General Chemistry Thenrnochemistry lleat and lllork
Carnot Cycle
The Carnot cycle is a cyclic process carried out in a piston, where heat isconverted into work or work is converted into heat. Practical examples of theCamot cycle include engines and refrigerators. We inherently know the conceptbehind the Carnot cycle. I4trhen we blow on a hot liquid, it is done so withpursed lips. If you exhale through your mouth with a normal relaxed degree ofaperture, your breath comes out at body temperature. But if you exhale throughyour mouth with a small opening, the air feels cooler as it passes across yourskin. That is due to the compression of the gas (exothermic) as it passes throughyour lips, and the expansion of the gas (endothermic) once it leaves your mouth.The air feels cooler, because it is expanding as it passes across the surface of yourskin. When a gas expands, the molecules increase their intermolecular distance,which breaks intermolecular forces. Just as bond breaking is endothermic, so isthe expansion of a gas.
The process of blowing air on your skin through pursed lips results in heattransfer from a cold body (your skin) to a hot body (your mouth). This isunnatural heat flow, so it is similar to the function of the Camot heat pump. Weshall look at a heat pump (refrigerator) in more detail. The Carnot cycle, in apractical sense, can be applied in the form of a heat pump or applied in reverse inthe form of a heat engine. It involves phase changes, so the process is oftendrawn overlaid onto a phase diagram, but the traditional lines of the phasediagram are sometimes removed. Figure 8-8 shows a typical phase diagram onthe left and then the Carnot cycle for a heat pump overlaid onto that same phasediagram on the right.
Figure 8-8
The heat pump follows a counterclockwise path about the phase change plocess
cycle. The following lists the phase details of the four-step cycle shown in Figure
8-8:
1,. A material that is normally a gas at room temPerature and pressure iscompressed into a liquid. Condensation is an exothermic process, so thematerial heats up and finishes as a liquid at a higher temperature andpressure than it originally had.
The material is maintained at a high pressure, so it remains a liquid. It isallowed to cool back to room temperature by dissipating heat to theenvironment. It finishes at a lower temperature, but the same high pressure-
J. The pressure on the system is returned to normal, so the material ex
from a liquid into a gas. Evaporation is an endothermic process, so
material cools down. It finishes as a gas at a lower temperature andthan it originally had.
(,q)tr0)dga
H
q)t{(^C,
lipi
2.
Temperature (kelvins) Temperature (kelvins)
Copyright @ by The Berkeley Review l5a The Berkeley
General Chemistry Thermochemistry Heat and lllork
4. The material is maintained at normal pressure, so it remains a gas. It isallowed to warm back to room temperature by absorbing heat-from theenvironment. It finishes at a higher temperature, but the same pressure.
After the four-step process, the material returns back to its original state. Theinternal energy does not change. Figure 8-9 shows the energetics of the heatpump cycle shown in Figure 8-8.
Figure 8-9
As you follow the heat pump around in its counterclockwise path, you see thatwork is done so that heat energy can be released. The following is a list of theenergy details of the four-step cycle shown in Figure 8-9:
1. As the material is compressed into a liquid, work is done on the system (w1).Heat is released from the molecules as they form stronger intermolecularforces. The heat remains a part of the system in this step, so we cannotconsider it transferred yet.
2. Heat is released from the system as the material cools to ambienttemperature (q*). As it cools at constant pressure, the volume decreasesslightly (liquids do not exhibit significant volume changes), so work is doneby the surroundings on the system (w2).
3' As the material expands back into a gas, work is done by the system (w3).Heat is absorbed by the molecules as they form weaker intermolecular forces.The heat is not replenished into the system in this step.
1. Heat is absorbed by the system as the material warms to ambienttemperafure (q+). As it heats at constant pressure, the volume increases, sowork is done by the system on the suroundings (w+).
-{-fter the four-step process, the material returns to its original state. Overall,rnore heat is released than absorbed and more work is done on the system thanb)' the system. The net result is that work goes into the system and heat flowsaut from the system. Another common graph associated with Carnot heatriunps shows pressure on the y-axis and volume on the x-axis. The process is-Jre same, but a phase diagram cannot be understood easily with such axes.
(,1
l{oq(,
ot{(,/)oCJlr
Temperature (kelvins)
Copyright @ by The Berkeley Review Exclusive MCAT Preparation
General Chemistry Thermochemistry Ileat and Work
Refrigerator (Applied Heat Pump)
The refrigerator, in its simplest form, uses work to absorb heat as a fluid passesthrough the four stages in a closed system. In a refrigerator, a change in volumeis ultimately a change in heat. The basic idea behind the refrigerator is to putwork energy into the system to compress a gas and condense it into a liquid.Compression of a gas and condensation are both exothermic, so heat is released.The system is then allowed to thermally equilibrate with the environment byreleasing heat to the surroundings. Following thermal equilibration, the liquidevaporates and the gas expands back to its natural state. Evaporation and theexpanding of a gas are both endothermic, so heat is absorbed.
Condensation and evaporation are carried out at different pressures, so it takesmore work energy to compress and condense the gas than the work energ'yreleased when the liquid evaporates and the gas expands. Ideally, the differencein work energy equals the heat energy released, assuming that energy isconserved in the overall process. Figure 8-10 shows a simplistic schematicdiagram for the process, where some material flows repeatedly counterclockwisethrough the system.
Translating from the theoretical schematic in Figure 8-10 into the actualrefrigerator requires a piston and a network of coils. A piston is used tocompress the gas into a liquid. This is represented by the gas-to-liquidconversion shown in the exothermic oval of Figure 8-10. This region equates tothe compressor in a refrigerator. The compressor converts the refrigerant gas toits liquid form, which is an exothermic process. After compressing therefrigerant, the liquid refrigerant flows into a high-pressure reservoir coil, whidtis represented by the shaded tube from which heat is released in Figure 8-10-
Because the reservoir coil is hot, the outer surface is fitted with fins to enhance
convection. In order to keep the material a liquid, the warm coil must be
pressurized. This is accomplished using a pressure valve at the end of the coil.
Once the liquid refrigerant has cooled to ambient room temperature (RT), itpasses through the pressure valve, and then expands into a gas within a low-pressure sealed aluminum coil. As the liquid refrigerant expands within the low-pressure coil to form a gas, heat is absorbed from the environment during thisendothermic process. Heat is absorbed from the coils, so the coils become coldIn turn, the coils absorb heat from the surrounding air. Any gas, liquid, or solidin contact with the coils loses heat by transferring it to the walls of the coils. The
cold coils are placed in an insulated, sealed container, so the core of the contaibecomes cool. This is the interior of the refrigerator. The compressor andpressure coils must be kept as far away from the core of the container as
because of the heat they release. Figure 8-11 is a more detailed pirepresentation of the refrigeration process shown in Figure 8-10.
l;r,qr:eirrt Oby The Berkeley Review r60 The Berkeley
General Chemistry Thermochemistry Ileat and lllorlr
Inlet valve
Gas is compressed intoOutlet valve a liquid (exothermic process)
Figure 8-11
The four stages of the refrigerator shown in Figure 8-11 are listed below:
1. Freon gas flows from the low-pressure coils into the piston of the compressorthrough an inlet valve at the top of the piston. The piston is thencompressed. The increased pressure exerted on the freon gas converts thefreon gas into its liquid phase. As the piston descends further, the hot freonliquid is forced through an outlet valve. The liquid is hot due to thecondensation process. (Check a compressor if you wish to verify this).
2. When the piston reaches the bottom of its cycle, the outlet valve is forcedopen, and hot freon liquid escapes into the high pressure coils. Because freonliquid is hot (condensation is a heat-releasing process), the high-pressurecoils are hot. They are found on the back (exterior) of the refrigerator, oftenprotected by a metal grid or thin plate. The freon liquid cools andequilibrates with the air by dissipating the heat to the external environmentthrough the walls of the coils. Coils are chosen to maximize surface area.
3. Freon liquid at ambient temperature is forced from the high-pressure coilsthrough a pressurized outlet valve into the low-pressure coils. The freonliquid is forced into the low-pressure coils by the increased hydraulicpressure generated when hot freon liquid is forced into the high-pressurecoils through the inlet valve by the compressor.
1. Freon liquid evaporates to form freon gas in the low-pressure coils. Freonevaporates by the absorption of heat from the core of the container in whichthe low-pressure coils reside. During this step, heat is absorbed from theinside of the refrigerator. The freon gas fills the low-pressure coil once it hasevaporated. The cycle repeats.
The net result is that work is applied at the compressor, and heat flows out fromthe high-pressure coils, so heat is removed from the core of the refrigerator to theenvironment. The compressor acts as a pump, and freon flow is in one directionthrough the system. Heat is absorbed and released at different points.
-\ir conditioners and freezers have the same mechanics as a refrigerator. TheCifference between freezers, air conditioners, and refrigerators is the temperature
'rf the internal coils. Because a phase change from liquid to gas absorbs the heat,lhe temperature of the cold coils equals the boiling point of the refrigerant. A:efrigerant is selected on the basis of its boiling point and the target temperature.rt the system. The temperature in a freezer is less than the temperature of a anelr conditioner or refrigerator, so the boiling point of the refrigerant used in a:eezer is lower than the refrigerant used in an air conditioner or a refrigerator.
fopyright @ by The Berkeley Review 16l Exclusive MCAT Preparation
General Chemistry Thermochemistry tleat and lJllork
wc-a & 9z
Heat Engine
The Carnot heat engine is the reverse of a Carnot heat pump. This is to say thatwhen a heat pump is run backwards (heat is added in and work is released), a
heat engine is created. The diagram for a heat engine, Figure 8-12, shows arrowsin a clockwise loop, as opposed to the heat pump, which has its arrows in a
counterclockwise loop, like Figure 8-9. The heat engine operates by convertingheat into work. In its simplest definition (Carnot definition), heat is used toexpand a gas in a closed system within a piston. When the gas expands, thewalls of the container move. Because the container is a piston, only one wallmoves. Work is associated with force times distance, so the motion of the wall is
harnessable work. The basic concept behind a heat engine is rooted in the idealgas equation PV = nRT, where altering one variable changes another variable. lnan engine, a change in heat is ultimately a change in volume. This means that nochange in intemal energy transpires, only the conversion from heat into work.The following is the operation of a Carnot heat engine:
1. The first step starting from the upper left comer of the diagram in Figure 8-12below is the reversible isothermal expansion of the gas from state a to state b.During this step, work (defined as wu-6 in the diagram) is done by the pistonon the surroundings (defined as negative for the system); and heat (qf ) tsabsorbed by the gas. In a combustion engine, it is the explosion of thegasoline that generates this heat.
2. The second step is the reversible adiabatic expansion of the gas from state bto state c. As implied by the term "adiabatic," no heat is gained or lost by thesystem. As a consequence of no heat transfer, the temperature drops as heatenergy is absorbed by the expanding gas. In this step, the piston does work(wU-.) on the surroundings. In a combustion engine, this is that small periodof time after the gasoline has exploded where the piston is still rising.
3. The third step is the reversible isothermal compression of the gas from state cto state d. During this compression step, the surroundings do work (w6-4on the gas (defined as positive for the system), and heat (q2) flows out of theengine. In a combustion engine, this is the venting of the piston allowing thehot exhaust to escape and cool air and gas to flow in.
4. The fourth step (which completes the cycle) is the reversible adiabaticcompression of the gas from state d back to state a, During this step, thesurroundings do work (wa-u) on the gas as the piston descends, but heat G
not gained or lost. In a combustion engine, this is the equilibrating of the
piston chamber back to its resting state just before the spark ignites the gas.
Volume
Figure 8-12
The graph may also be diagrammed as pressure as a function of volume. I:r*volume term is important, because work is defined by the equation: w = -Pl'i-As of 2000, the heat engine has yet to appear on the new version of the MCAT.
:tlt:
J"
n
Copyright @ by The Berkeley Review 162 The Berkeley Revieu
General Chemistry Thermochemistry Ileat and Work
A good example of the application of the principles behind the heat engine is thesingle-piston steam engine. In the steam engine, we start by considering areservoir filled with hot water (water at 100"C). Water is taken through a four-step process that changes its phase and pressure and then refums it to its originalstate. The following steps outline the operation of a steam engine:
1. Water evaporates to form steam as heat is stoked in the boiler. During thisstep, heat is added from the surroundings (defined as heat energy into thesystem). The steam builds up pressure in the reservoir.
2. Steam flor,t's from the reservoir into the piston when the inlet valve to thepiston is opened. The pressure exerted by the steam that enters causes thepiston to rise from its lowest point to its highest point (apex). As the pistonrises, work is done by the system on the surroundings (defined as energy outfrom the system).
3. \zVhen the piston has reached the top of its cycle, the outlet valve is opened,and the steam escapes into the exhaust column and collects in the condenser.The piston returns to its lowest point through inertia and the force of acounterweight (in a one-piston engine). As the piston falls, work is done bythe surroundings on the system (defined as work energy into the system).
4. The steam condenses in the condenser and is pumped back into the boiler.During this step, heat is released to the surroundings (defined as heat energyout from the system) as the steam condenses back into water. The cycle thenrepeats step one again.
The condenser reduces the pressure so that when the exhaust valve is opened,:he steam within the piston can flow from a region of higher pressure (the piston):o a region of lower pressure (the condenser). Heat is released in the condenser,so an efficient engine finds a way to recycle that emitted heat. Figure 8-13 showsl pictorial diagram of a steam engine.
Boiler Water pump Condenser
Figure 8-13
: steam engine is a closed system through which water circulates. A. ::'Lbustion engine differs from the Camot heat engine and steam engine in that' --. an open system. Petroleum mist and air are injected into the piston. The::-rfure is ignited, and the piston rapidly expands. The power stroke of the:.::!rn does work on the crankshaft, tuming the crank 180 degrees. Exhaust is-'::.oved from the piston and released to the atmosphere. The conversion of, :-osive linear work into circular motion requires mechanisms that have been,:=rully engineered with an understanding of the basic physics involved.
Exhaust
- . :', right @ by The Berkeley Review t63 Exclusive MCAT Preparation
General Chemistry Thermochemistry Ileat and Work
one of the more important physics-based features of an engine is expansion andvalve timing. Because the motion generated by an engine is periodic, there mustbe periods (timing) for each operation of the machine. The inlet valve must openand close (allowing fuel and air to enter the chamber) before the spark can ignitethe mixture, followed by the opening of the exhaust valve to vent the productsfrom the chamber. This is what is referred to as the timing of the engine. Modemengines have each piston fitted with four valves (two intake and two exhaust), sothat the gas diffuses more evenly in the chamber. Because the up-and-downmotion of the piston ultimately turns a crank, torque must be invoked. Tomaximize torque, a lever arm system is used. This allows the rod to be attachedat the edge of the crank (maximizing moment arm) and allows the rod to move insame direction as the piston, for the most part. This maximizes the applied forceof the system shown in Figure 8-13.
Engine EfficiencyThe efficiency of an engine is simply the output over the input. Work energy isproduced from heat energy, so engine efficiency is work out divided by heat in.The most efficient engines have low exhaust temperature (minimal heat Gwasted), few moving parts (minimizes friction), and low weight (work energl-produced is not wasted in moving the engine). Efficient engines are multi-pistonsystems rather than a single piston connected to a counterweight. A goodanalogy involves a bicycle crank. The pedals are placed out of phase for greatestefficiency. A bicycle could be designed with two pedals in phase attached to a
crank housing a counterweight that is out of phase with the two pedals. Thecounterweight system works well in theory, but this would not be an enjoyablebike to ride uphill because of its extreme mass.
Copyright O by The Berkeley Review 16,4 The Berkeley Revier
ThermochemistryPassages
l5 Passages
I OO Questions
Suggested Thermochemistry passage Schedule:I: After reading this section and attending lecture: Passages I , lll, VI, & Xl
Qrade passages immediately after completion and log"your mistakes.II: Following Task I: Passages IV, v, IX, & x (28 questions in 3i6 minutes)
Time yourself accurately, grade your answers, and review mistakes.III: Review: Passages VII, VIII, XII - XIII & euestions 9J - I OO
Focus on reviewing the concepts. Do not worry about timing.
I2EffiL)R.E.V.r."E,.wt
, '!t,t,..,..
ffi
Specialtzrng in MCAT Preparation
l!
F:lili
iiflti
1.I
1e
I.
II.
III.
IV.
V.
VI.
VII.
VIII.
IX.
X.
XI.
XII.
XIII.
Energy Diagrams
Bond Energies
Born-flaber Cycle
Free-Energy Calculation
Rocket Fuels
Enthalpies and Sugars
Salt-Solution Calorimetry Experiment
tleat Pack
Bomb Calorimeter
Hot and Cold Compresses
Calorimetry Experiment
Single'Piston Engine
Carnot Cycle
Questions not Based on a Descriptive Passage
Thermochemistry Scoring Scale
Raw Score MCAT Score
84 - loo l3 l566-85 lo l247 -65 7 -934-46 4-6t-33 L-3
(r -7)
(8 - 14)
(t5 - 2r)
(22 - 28)
(2e - 35)
(36 - 45)
(44 - 4e)
(5O - 55)
(56 - 62)
(65 - 6e)
(7O - 78)
(7e - 85)
(86 - s2)
(95 - 1OO)
:l.er
:rO,it, Llfr, :
tai rrineL
1.
Passage I (Questions 1 - 7)
Drawn below are two energy diagrams for two separateprocesses. Figure 1 shows the energy diagram for Reaction1, and Figure 2 shows the energy diagram for Reaction 2.
For Reaction 1, there are two pathways drawn leading to twoseparate products, X and Y. Both pathways represent two-step reaction mechanisms. In Reaction 2, therc are two steps
to the reaction, as represented by the two peaks, but only one
reaction pathway is shown on the energy diagram.
Figure 1
Figure 2
In Reaction 1, either pathway can occur, depending on.r: conditions. The pathways are referred as the kinetic and':.ermodynamic pathways. Lower activation energy is.,sociated with the kinetic pathway, while a more stable, rduct is associated with the thermodynamic pathway. At
'.', temperatures, the reaction is forced to take the pathway^..t requires the least activation energy. This means that the
' ,,-tic product is observed at lower temperatures. At elevated;:rperatures, the thermodynamic product is observed.
- . In Reaction 1, what is true of the rate-determining step?
A ".) The first step is rate-determining, because it has thehighest activation energy.
B. The first step is rate-determining, because it has theIowest activation energy.The second step is rate-determining, because it has
the highest activation energy.D. The second step is rate-determining, because
the lowest activation energy.
Reaction Co-ordinate
. :r'right O by The Berkeley Review@ 167 GO ON TO THE NEXT PAGE
2 . If AS is positive for Reaction 2, then as the temperatureis increased, what is observed?
A. Both the IQO-to-Q ratio and AG increase.B.
_ The K.q-to-q ratio decreases, while AG increases.
,_'.C .)The &q-to-q ratio increases, while AG decreases.D. Both the IQO{o-Q ratio and AG decrease.
3 . For Reaction l, what can be said about the enthalpies ofboth pathways?
--.A. AH for Pathway X is positive; more energy is
released from Pathway X than Pathway Y..-8. AH for Pathway X is positive; less energy is
released from Pathway X than Pathway Y.. C. AH for Pathway X is negative; more energy is
_ -_released from Pathway X than Pathway Y.
a 9_. -AH for Pathway X is negative; less energy isreleased from Pathway X than Pathway Y.
f:>.k. At higher temperatures, what is true about Pathway X
and Pathway Y of Reaction I ?
A.i Pathway X is the kinetic pathway and is more
, probable than Pathway Y.
.f. Pathway X is the thermodynamic pathway and ismore probable than Pathway Y.Pathway Y is the kinetic pathway and is moreprobable than Pathway X.Pathway Y is the thermodynamic pathway and ismore probable than Pathway Y.
e.
D.
\ :''.
i
-4.'. What is observed in Reaction 2?
{m" intermediate builds up to a detectable level,because the rate-determining step is the first step.
B. The intermediate builds up to a detectable level,because the rate-determining step is the second step.
.d. fn" intermediate cannot build up to a detectablelevel, because the first step is rate-determining.
D. The intermediate cannot build up to a detectablelevel, because the second step is rate-determining.
,[- *ntrh of the graphs represents the percentages ofproducts X and Y as the temperature is increased?
A.
---....--.>
Temperature ---------->
Temperature -------------->
Temperature +
P.,1
g
Temperature
Copyright O by The Berkeley Review@ l6a GO ON TO THB NEXT
T)A' ,tReaction I is favorable at lower temperatures but
unfavorable at higher temperatures, then what is trueabout the change in enthalpy and entropy?
7/1,. nu> o; AS > o
-), LH>0;AS<0(clu<o;AS>o)41 aH<o;AS<o
xG- x'l{ - 1 FoI
ctdiGI
g!
/ --\ - \I L-:' \-'( * \-i'\
)I
!!!!!!n
Passage ll (Questions 8 - 14)
The enthalpy of a reaction can be found using the bondenergies. The change in enthalpy can be attributed to thedifference between the energies of the bonds broken and theenergies of the bonds formed. Equation 1 shows howenthalpy change is obtained from bond energies.
AH = B.E. (bonds broken) - B.E. (bonds formed)
Equation ITable 1 lists the bond energies of some of the more
common bonds within organic molecules.
Table 1
Table 2 lists the bond energies of some of the less
Jommon bonds within organic molecules. Both tables fail tojccount for hybridization or the effects of neighboring,ubstituents on the stability of a bond.
Table 2
Values were derived from bomb calorimetry experiments
-,rng the gas phase reactions of several molecules. The kJ
- it is converted into kcal by dividing by 4.18.
\ . Which bond is the WEAKEST in diethyl peroxide?
A. H-c ) t1'\,'')
B. C-O -: I .r, r-'. 1..
c. G-o. :
D. C-C
What is the bond energy of thenitrogen gas?
A. 258.0 kJ per moleB. 390.5 kJ per moleC, 313.1 kJ per moleD . 523.0 kJ per mole ,.1 'i ,
i, l, i ,- rovright O by The Berkeley p"rri"*@.*-l--J;
r- 'l
second n-bond in
- ^,'ltNi - .'\- t
Bond B.E.(KJ/moie)
Bond B.E.(KJ/mole)
Bond B.E.(KJ/mole)
H-C 4t3 C-C 34',1 N-C 305
H-H 432 C-H 413 N-H 391H-N 391 C-N 305 N-N 160
H-O 461 c-o 358 N-O 201H-F 565 C-F 485 N-F 212H-CI 42'7 C-CI 339 N-CI 200H-Br 363 C-Br 276 N-Br 243
H-I 295 C-I 240 N-I 111
B.E.(KJ/mole)
l.b.
l\t
Why does an tp2^C-tp3C bond have a higher bond
"n"igy than an sP3C-sP3C bond?
A. An sp2 orbital is shorter than an sp3 orbital, hencethe sP'C-sP'C bond is shorter and thus stronger.
E n" sp"2 orbital is shorter than an sp3 orbital, hencethe sP"C-sP'C bond is longer and thus weaker.
.O. An sp2 orbital is les^s electronegative than an sp3
orbital, hence the sp'C-sprc bond is more polarrand thus stronger.
D. An sp2 orbital is more electronegative than un tp3orbital, hence the sp'C-spJC bond is more polarand thus stronger.
\'( 1-t"' '
t' t' < \ "rr
I 1. The breaking of an H-H and an F-F bond to formtwo H-F bonds is: :. ....
-A. highly endothermic. rl' 1'-'
-B. slightlyendothermic. ., ., r \-r.i{ -r \ - 2 a
€. -slightly exothermic.Ar{ ' " :, 'D. hishlv exothermic.I /
I 2. How can the bond energy trend of the sulfur-halide bond
be explained?
. k1"'As the halide gets larger, the bond gets stronger.B. As the halide gets more electronegative, the bond
gets stronger.The strength of the bond depends on both the srze
and polarizability oithe halide.As the electron affinity of the halide increases, the
bond gets stronger.
13 . What is the approximate bond energy of a Br-F bond?
9iD.
A. zil kImole 'i ' il i
B. 242 kImole ' .t )
c.206 kJmole
D. 161 kJmole
ft
14. Why is the F-F bond weaker than the Cl-Cl bond?
X(fn" F-F bond is shorter, and shorter bonds are
- always weaker.
B. Chlorine is more electronegative than fluorine.
lQ-;Fluorine atoms repel when close together.
-D-. Chlorine is smaller than fluorine.
r69 GO ON TO THE NEXT PAGE
Passage lll (Questions 15 - 21)
The Born-Haber cycle refers to the energy released when asalt is formed from its most stable elements in their naturalform. The enthalpy change for the overall reaction can befound by summing the individual steps of the overall process.The formation of lithium fluoride from lithium metal andfluorine gas is shown in a stepwise fashion below;
Step I: Sublimation of lithium metal:Li(s) -+ Li(e) AHsublimation = +153 kJ
Step II: Ionization oflithium gas:
Li(g) -+ Li+(g) Ionization energy = +513 kJ
Step III: Dissociation of F-F Bond:
I fz6l -+ F(g) Half bond energy F2= +77 H2 --'Step IV: Electron affinity of atomic fluorine:
F(g) + 1 e- -+ F-(g) Electron affinity = -324H
Step V: Formation of lattice from gaseous ions:
Li+(g) + F-(g) -+ LiF(s) Lauice energy = -1045 kJ
Overall:
Li(s) + ! pzgl -; LiF(s) AHreaction = -626k12'-The total energy of the reaction is found by summing the
five steps of the reaction. The lattice energy can be foundusing Equation l, where Q is charge and r is the distancebetween ions in the lattice:
g = p QcationQanionr
Equation I
15. If sodium metal were used instead of lithium metal,what values would be affected?
L The sublimation energy would increase. ., ^ t.
tr. The ionization energy would increase..-^,-.{ "'., :,':m. The lattice energy would increase. -) ' v"'
@ I only -4 fu^ -{ L*, -}r.i*J
--4._ IIonlyz{. land II onlyD. I and III only
: | "rin| .-,,.' /'
.f&. *n, is only half of the bond dissociarion energy usedin Step III?
. { . The F-F bond is only partially broken.B. The table value for the bond dissociation energy is
r.. -' double the actual value for the bond dissociitionenergy.
!. Because the bond is broken in a heterolytic fashion,' the bond dissociation value is only halfofthe valuefor the homolytic breaking.
-D. The breaking of the bond produces two fluorineatoms, and only one is needed.
Copyright @ by The Berkeley Review@ 170
i.,,-'--u'
GO ON TO THE NEXT PAGM
I
#. Ho* do the lattice energies of NaCl, LiF, and MgO
'* compare to one another?
l/\' .--A. Err,lgO > ELir.> ENaCt
I. ErureO > ENuCt > ELipt C)E11p>EN'CI>EVgO. --D. ENaCt > ELip > EUgO
n1* ?., '' r. t '-' .' "r " "-',i t''
I ,l ,-,4t
-ll[- If calcium metal is used instead of lithium metal, hosc4n-the value in Step II be determined?
'\') Cn ,,/Onlv the value fbr the first ionization should be\-/
ur"i.B. Only the value for the second ionization should be
used.
C. An average ofthe first and second ionization value:should be used.
D. The sum of the first and second ionization valuesshould be used.
A:.)onlyX I andll onlv
-4.. I and III onlyD ) t, II, and III
t9. The electron affinity can BEST be described as:
. the energy associated with the excitation of arelectron.
"8. the energy associated with the de-excitation of an
electron.
-El'tlte energy associated with the loss of an electron.t
\ ;..the energy associated with the gain of an electron.
2 0. Which change results in LESS energy being released inthe overall process?
A. Using a metal that is easier to sublime thamj r * lithium
I 1 B. Using a halogen that forms a stronger bond a-s *- diatomic molecule than fluorine.C. Using a halogen atom with a greater electrom
affinity than fluorineD . Using a cation that can easily lose a second elecrcm
21. What changes would reduce the lattice energy?
L Substituting a larger cation for a smaller cation
II. Substituting an anion of less charge for the anion
m. Reducing the cation charge, anion charge, anddistance between ions each by half
Passage lV (Questions 22 - 28)
The free energy of a reaction depends on the favorabilityof the reaction and the position of the equilibrium. Reactionsthat spontaneously proceeds in the forward direction are saidto have negative free energy changes. This is defined as afavorable reaction. The equation below shows therelationship ofthe free energy change (AGrJ and the standard
equilibrium shift from equal parts products and reactants.
AGobserved = AG" + RT lnQr*
Equation 1
The value for AG' can be found by measuring the AGqwhen one mole of each product and reactant are mixedtogether in a one liter container. The value for AG" is givenby the following equation:
AG.=_RTlnKsq
Equation 2
The free-energy change can also be found using the.o11owing equation:
AG11 =AHrx-TASfxEquation 3
The AH.* represents the enthalpy change for the reaction,i hich is measured as the heat change (either gained or lost):uring the reaction. The ASr* represents the entropy change
. rr the reaction, which is measured as the change in-.ndomness of motion during the reaction.
I L Which process is NOT entropically favoratle?
I /'oirtittutioirtoietrranol from 6ct'# -i' 11'f i ""
Ii'. Solvation of a salt by water --\ -\
C . Sublimation of, iodine solid
D . Diffusion of a pure gas into the air - I -
I3 , Which reaction has a AS value > 0?
A. NaCl(s) + HZO(I) --- Na+(aq) + Cl-(aq)
J. PCt:(r) + Cl2(g) : PCl5(s)
C . 2 HzG) + Oz(g) --- 2HzO(1)
E.Nz(e)+Oz(e)
-
2NO(e)
\ddition of a catalyst MOST changes:
A . the free energy ofthe reaction.
B . the enthalpy of the reaction.
C. the entropy ofthe reaction.
D . the rate of the reaction.
: . right @ by The Berkeley Review@ t7t
25. If a reaction that is spontaneous as written lowers thesolution temperature as it proceeds, then the AS for thereaction is:
.A. negative at all temperatures.
,}i negative at low temperatures and positive at high.temperatures.
,C. positive at low temperatures and negative at hightemperatures.
D. positive at all temperatures.
5.:
26. The AGobserved for a reaction can be found by which ofthe following equations?
A. AGe6sslved = RT ln
B. AGe6sslved = RT ln
ar*K"q
IsAt
C. AG66sglued = =Qrx eRT
K"q
K-D. AGob.erug6 =:E sRTa* .,.-,o... ,. . ''' t i I ''
t' *1-: ':' "
i'.' i)', , -'' t: ' '-"-
?7 . For a reaction where the reaction quotient (Qr") is
",.. greater than the equilibrium constant (Keq), what is true
: ' about the reaction and AG66ss1yg6?
_..A. AGe6ssrved > 0; the reaction shifts to the reactant
side to reach equilibrium.B. AG66selved < 0; the reaction shifts to the reactant
side to reach equilibrium.
,,C . AG66.erved ) 0; the reaction shifts to the product
side to reach equilibrium.
D. AG655g1ved < 0; the reaction shifts to the product
side to reach equilibrium.
28. The AH for a given reaction is 12.5 kJ/mole. The AS
for the reaction is 25 J/mole'K. At what temperature
does the equilibrium constant equal 1.0?
A. O'C
8,.. 227'C
c. 327"C
D. 500'c
I-
.1 I
,:
GO ON TO THE NEXT PAGE
Passage V (Questions 29 - 35)
Over the past few decades, several advances have beenmade in rocket propulsion systems. These center on theengineering of the rocket equipment, as well as on advancesin the fuels used to propel rockets. The ideal fuel system islightweight and capable ofproviding large sustained bursts ofenergy. A perfect fuel system is liquid hydrogen (b.p. 18 K)mixed with liquid oxygen (b.p. 91 K). Reaction I below isthe gas phase combustion reaction of hydrogen and oxygen.
HZ(g) + ! Oz(el -+ HzO(g); LH' = -242 kJ2 mole
Reaction 1., oxidation of hydrogen gas
The oxidation ofhydrogen gas has been used in the laterstages of several launches oforbiting space craft, but it is notused for the initial launch from earth, where greater energy isrequired. Reactions 2, 3, and 4 have also been used togenerate the energy required to move massive projectiles.
C12H26(t) + 181Oz(g) ) 12 COz(e) + 13 H2O(g)2
AH" = -75t3 kJmole
Reaction 2, oxidation of kerosene
2 N2Ha(l) + NzO+(e) -+ 3 Nz(e) + 4HzO(e)
AH" = -t049 kJmole
Reaction 3, hydrazine oxidation via dinitrogen tetraoxide
2Al(s)+6NHaClOa(s)
-+ I NO2(g) + 6 HCl(s) + I Al2O3(g) +9 H2O(s)
AH" = -1216 kJmole
Reaction 4, AU(C2Ha)1 oxidation via NH4CIO4
The reactions all generate a great amount of heat by wayof oxidation. By rapidly oxidizing the fuel, a concentratedstream of energy is produced to lift the rocket and project itin the determined pathway.
2 9. The AHformation of H2O(l) is -286 kJ per mole. Howcan the difference from the value listed in the passage be
explained?
A. The condensation of water is endothermic, so theAHformation for H2O(1) is more negative thanAHformation of H2O(g).
B. The condensation of water is exothermic, so theAHformation for H2O(l) is more negative thanAHformation of HZO(g).
C. The condensation of water is endothermic, so theAHformarion for H2O(l) is less negative thanAHformation of H2O(g).
D. The condensation of water is exothermic, so theAHformation for H2O(l) is less negative thanAHformation of HzO(g).
Copyright @ by The Berkeley Review@ 172 GO ON TO THE NEXT
3 0. What makes kerosene a better fuel for early rocketstages than hydrogen gas?
A . At standard temperature, kerosene is far denser thmhydrogen, and thus more convenient to store.
B. Kerosene provides more energy per gram thanhydrogen.
C . Kerosene burns at a higher temperature.D . Oxidation of kerosene produces fewer products.
31. Which of the following describes the ideal mixture dhydrazine and dinitrogen tetraoxide?
I by mass in favor of hydrazineI by mass of hydrazine with N2O42 by mass in favor of N2O4I by mass in favor of N2O4
3 2. Which of the following reactions CANNOT be usedpropel rockets?
C6H1a(l) + gL OzG) -+ 6 CO2(g) + 7 H2O(g)
N2H4+lKC1O4 + N2 +2H2O+lKClC. Fe + (NH4)2CrO4 -+
L Fe'rOr + HrO + .L CrrOr + 22'2'
D. H2SO4 + 2 KOH -+ K2SO4 + 2H2O
33. Which of the following is NOT a requirement ofrocket fuel?
A. Highly exothermic combustion reactionB. A large heat-to-gram fuel ratioC . Gas phase at room temperatureD. Highly reactive as an oxidant or reductant
34. In the N2H4(1) + NZO+(g) reaction, what is true?
A. N2Oa(g) is a reducing agent.
B. N2Ha(l) loses four electrons per nitrogen.C . Nitrogen-nitrogen bonds are broken.D . Hydrazine is being oxidized.
3 5. What is the heat per gram value for kerosene?
A. Greater than 75.0 kJ per gramB. Greater than 50.0, but less than 75.0 kJ per
C . Greater than 40.0, but less than 50.0 kJ perD. Less than 40.0 kJ per gram
L.2B. 1
c. 3
D.2
A.
B.
Passage Vl (Questions 36 - 43)
Thermodynamics is the study of energy and itsdistribution within a physical or chemical system. One ofthe common mathematical practices in thermodynamics is thecoupling of several processes to sum up to an overallprocess. For example, the overall enthalpy is the sum of theindividual enthalpies for each reaction. This is known as
Hess's law. In addition to enthalpy, entropy and free energy:an also be determined using Hess's law. The entropy andtiee energy change for a reaction is equal to the sum of theentropy and free energy changes for any set of reactions thatsum to the same reaction.
A common application of this procedure is the summingrf the heats of formation for each reactant and product toobtain the heat for the overall reaction, such as the
--ombustion of a hydrocarbon. By definition, the heat oflormation for an element in its most natural state at 25'C ist kJ/mole. This is to say that converting an element such as
O2(g) into O2(g) (the most stable form of oxygen at room:emperature) requires 0 kJ/mole. A small sampling from a:hart of enthalpies of formation is listed below. The chart:'elow can be used in conjunction with Hess's law (which isrot one of the laws of thermodynamics) to calculate the.' alues common to thermodynamics.
Compound^H'{
kJ I\mole/ ^c'l
k.r I\mole/
Cor(e) -393.5 -394.3
CzHzG) +226.8 +209.1
CrHa(e) +52.4 +68.3
CzHsOH(l) -21',t.1 -114.9
Hzo(e) -241.8 -228.6HrO(l) -285.9 -237.2
Table 1
The phase of the compound is specified, because the
=iergies are different for each phase of the compound. The::fference between the thermodynamic values associated with:.ch phase is the thermodynamic value for the phase change::ocess. The enthalpy of fusion (associated with melting)-.n be found by subtracting the enthalpy of formation for thei,:lid from the enthalpy of formation for the liquid. The
=:thalpy for a reaction is found by subtracting the enthalpy of::,rmation for all of the reactants from the enthalpy of. :mation for all of the products.
-r 6. What is the free energy change for the vaporization ofwater under standard conditions?
k":
,d,
D.
-44.r kJmole
-8.6 kJmole
+8.6 kJmole
+44.1 kJmole
(. rt)'/
-.L " ) ';
\)-r.o !'; -'''.-'''
_z-7t.t-\:{
-opyright @ by The Berkeley Review@ t73
(- ":. \'(* '"'
GO ON TO THE NEXT PAGE
il,,1
3V. The difference in the enthalpies of combustion for\ xylose (an aldopentose) and ribose (an aldopentose).canBEST be attributed tor ,^ ] e -' " , "''"1 '-r'".*'
d thedifference in the number of O-H bonds.-D . the difference in the number of C=O bonds.
/\^r-I
3)Q. From the values in Table I, what can be said about the' hydrogenation of acetylene (CZH2;' to ethylene (CZHq)?
.{Uyarogenation of acetylene is favorable.
, - p. HyOrogenation of acetylene is unfavorable.(J-
-Ei The free energy change for the reaction cannot be
calculated without knowing the enthalpy offormation for H2(g).
_.p';- The free energy change for the reaction cannot be' calculated without knowing the enthalpy of
combustion for HZ(g).
\ C2L{2:.i{, ---$ C'.*ILV
(ot,i ''2-o1.( -'{:a'}3 9. Which of the following is the change in enthalpy for
the formation of chloroethene under standard conditions?
'A i AH for the reaction:
C.
AH for the reaction:
2 C(er) + tl Hz(e) + I Ctz(t) + C2H3C1(g)22AH for the reaction:
2 C(d) + tI Hz(e) + I Ctz(e) -+ C2H3Cl(g)22AH for the reaction:
2 C(d) + lI H2(c) + 1 Cl(g) -+ C2H3Cl(g)2
4 0. Which of the following is the change in enthalpy forthe combustion ofpropene under standard conditions?/
rAo3 AHTHTO +2 AH1CO2 - AHgC3H6.( ) '/' Br' 3 AHiH2O + 3 AHpCO2 - AH1C3H6vre, 4 AH1H2o + 3 AH1co2 - AHpc3H6
-D: AHlC3Ho - 3 AH6CO2 - 3 AH1H2O
ll'J , *.:
B.
At-_O-- --itjCo->-L\
Given that AHformation for H3C-CH: is -87.4 --kJ-,mole
, how can the AHr* of -271.4 kJ fo. the followingI,h mole( hydrogenation reaction be explained?
Hz9=CFI-CH=CH2 + 2H2 + H3CCH2CH2CH3
..K'. Thevalue for AH.* is more negative than expected,
because the reactant is stabilized by conjugation.
,"'{ fn" value for AHr* is less negative than expected,
. ,because the reactant is stabilized by conjugation.
,' C . .The value for AHo is more negative than expected,-'/ because the product is stabilized by conjugation.
'D. tne value for AH.* is less negative than expected,it because the product is stabilized by conjugation.
- t (','*
(-t t-\ 6'ot& t n:' . .-, .*(
\ , - tl-, '+ -r -i\."t"t \ t
fl
nO,; *nr.n of the following statements is true about the
, '/ heat produced from burning one gram of ethyne and one
'!! ' gram of ethene in the presence of excess oxygen gas
under standard conditions?,/,/ Ai 1.0 g C2H2(g) yields more heat than 1.0 g
CzHq(e).
..,-B-."-.1.0 gCzHzG) yields less heat than 1.0 g C2Ha(g).,.
l,.(. t.O g C2H2(g) yields as much heat as 1.0 g
czHq(9.D. The amount of heat cannot be calculated without' first knowing the AH1s1rn31i61 of oxygen gas.
43. Which of the following is the free energy ofcombustion for ethanol (CZHSOH) under standard
conditions?
A:'-1325.3 kI'. mole
B. -1368.3 kJmole
c. -1675.1 kJmole
D. -1920.1 kJmole
-;.
. -. i' , ,-r' -''J I ! '>'
+r- /-
13r-'1'': (J
\t2't i \\
.r\
3Cl t< te ', \) ( - ?
t *1: l"
- =-Go - '? t'- ' flf'r
Copyright O by The Berkeley Review@
_1\
t
Passage Vll (Questions 44 - 49)
The enthalpy of a reaction can be calculated empiricallyusing data from calorimetry experiments. In the standariprocedure, referred to as bomb calorimetry, a known quanriry
of a compound is burned in a containment vessel surroundoiby a liquid with a known heat capacity (usually water). Tlie
heat absorbed by the surroundings is assumed to equal the
heat released by the reaction. This indirectly gives the AH forthe reaction. An error that arises using this method is the
loss of heat to the environment. To avoid this, the liquid riencased in an insulated container.
The results of calorimetry experiments can be combina:to determine the AH for other reactions using Hess's lan
Hess's law states that the AH for a reaction is equal to the
sum ofthe AH values for any series ofreactions that sum ryto the overall reaction.
A researcher conducts the following two experimenrc rrdetermine the molar heat of formation for magnesium oxide:
Experiment IA researcher places into an insulated beaker 1.00 kg of0.ll-M HCI solution (C = 4.34 J/g'K) at 22'C. This beaker is
labeled Beaker l. 2.43 grams of magnesium metal (24.-:
grams/mole) are then added to the solution. A thermometer
calibrated in 0.10"C increments is placed into the beaker'
The temperature at uniform time intervals is recorded irFigure 1. Reaction 1 is shown below:
Mg(s) + 2 H+(aq) -+ Mg2+1aq) + H2(c)
Reaction L
20 40 60 80 100 120 140 160
Time after addition to Beaker I (seconds)
Figure 1
Experiment 2
A second beaker, identical to that in Experiment 1, is ssup and labeled Beaker 2. 4.03 grams magnesium orie(40.3 grams/mole) are then added to the solution. A
thermometer calibrated in 0.10"C increments is placed i-nar
the beaker. The temperature at uniform time interval: urecorded in Figure 2. Reaction 2 is shown below:
MgO(s) + 2 H+(aq) -+ Mg2+1aq) + H2O0)
Reaction 2
GO ON TO THE NEXT PA
9rsI=32d!
8. :o
8zt
174
P7)oE:o&zs&ze
20 40 60 80 100 120 140 160Time after addition to Beaker 2 (seconds)
Figure 2
The same researcher conducts the following experimentio study an endothermic reaction:
Experiment 3
A third beaker is filled with 1.00 kg pure water also at22'C and labeled Beaker 3. 5.35 grams ammoniumchioride (53.5 grams/mole) are then added to the water. Athermometer calibrated in 0.10'C increments is placed intothe beaker. The temperature at uniform time intervals isrecorded in Figure 3. Reaction 3 is shown below:
NHaCI(s) + H2O(l) -+ NH4+(aq) + Cl-(aq)
Reaction 3
20 40 60 80 100 120 140 160Time after addition to Beaker 3 (seconds)
Figure 3
Each graph shows that the thermometer lags for aboutrrrtv seconds, before equilibrating with the solution. Tortain the highest solution temperature, a line must be
' itrapolated through the data points to the y-axis. This is
-.ed to determine the change in temperature, which-..imately is used to calculate the AH for the reaction.
i L Which of the following statements is a valid conclusionfrom the observations ofBeaker 3?
A . Solvation of NH4CI(s) is exothermic.
B . Solvation of NH4CI(s) is endothermic.
C . Solvation of NH4CI(s) is endergonic.
D . Solvation of NH4CI(s) is unfavorable.
9zrC)
=20Cdr&ts818
O by The Berkeley Review@ t75 GO ON TO THE NEXT PAGE
45. Adding 1.21 grams of Mg(s) into Beaker 1 instead of2.43 grams would have led to a temperature increase upto what temperature?
A. 26"C
B. 30'C
c. 38'C
D. 54"C
4 6. With respect to Experiment 2, whatenthalpy for the following reaction?
MgO(s) + 2 H+(aq) -+ Mg2+1aq; +
,r 1.0 x 4.34 x 7.5 x 4.03 kJ40.3 mole
p 1.0 x 4.34 x 9.0 x 4.03 kJ40.3 mole
(- 1.0 x 4.34 x 7.5 x 40.3 kJ4.03 mole
n 1.0 x 4.34 x 9.0 x 40.3 kJ4.03 mole
is the molar
HzO(l)
the gas given off in thethe magnesium metal was
4 7. Given that the enthalpy of formation for H2O(l) is -286
kJ/mole, what is the enthalpy of formation of MgO(s)?
A. AH@eaker 1) + AHlBeaker D - 286 kJ / mole
B. AHlneaker 1) - AHlneaker 2) - 286 kJ / mole
C. AHpeaker 1) + AHlBeaker D + 286 kJ / mole
D. AHgeaker l; - AHlBsaker 2) + 286 kJ / mole
4 8. Which of the following isreaction of Beaker I afteradded?
A . Carbon dioxide gas
B. Chlorine gas
C . Water vapor
D. Hydrogen gas
4 9. In calculations, it can be assumed that the heat capacityof 0.1 M HCI is equal to that of water. If the heat
capacity of 0.1 M HCI is greater than the heat capacityof water, how is the calculated value affected?
A. The calculated AH is too small.
B. The calculated AH is too large.
C . The difference is insignificant compared to heat lostto the environment.
D . The reaction rate increases, so the thermometer doesnot accurately record the temperature.
Passage Vlll (Questions 50 - 55)
Heat packs are designed for commercial use to produceheat rapidly. One of the reactions employed to generate the
heat is the oxidation of iron. The reaction is exothermic and,
if controlled, it can produce uniform heat for several hours.
This reaction is shown below as Reaction 1:
4 Fe(s) + 3 O2(g) + 2 Fe2O3(s) AH" = -1652 kJ/mole
Reaction L
The oxidation-reduction reaction employed is the same
reaction observed when iron rusts in moist air. In dry air,iron does not rust as rapidly. The greater rate ofoxidation inthe presence of water is attributed to the dissolving of oxygeninto water, and the consequential increase in interactionbetween oxygen and iron. The presence ofions in the waterincreases conductivity, which also escalates the rate ofoxidation.
To aid in the transfer of electrons within the heat pack,
the reaction is carried out in aqueous sodium chloride withsawdust, activated charcoal, and zeolite. To increase the rate
of the reaction, and consequently the production of heat, the
iron is finely powdered. The lifetime of the packet is
indirectly proportional to the degree of powdering of the iron.
5 0. Roughly how many grams of iron should be added to
the pack to produce 1000 kJ ofenergy?
A. Less than 50 grams
B. More than 50 grams but less than 100 grams
C . More than 100 grams but less than 150 grams
D. More than 150 grams
51. How does the weight of the pack compare before and
after opening?
A. It weighs less after opening, because of the loss ofheat.
B . It weighs more after opening, because of the loss ofheat.
C . It weighs less after opening, because of the gain ofoxygen.
D. It weighs more after opening, because of the gain
of oxygen.
5 2. What can be concluded about the change in entropy and
change in free energy for this reaction at roomtemperature?
A. AS">0andAG">0B. AS'>0andAG"<0C. AS'< 0 and AG" > 0
D. AS'< 0 and AG'< 0
Copyright @ by The Berkeley Review@ 176 GO ON TO THE NEXT PA
Time
5 3 . What is the role of the minimum amount of salt waterpresent in the heat pack?
A. The salt water absorbs the heat produced by the
reaction, so that the process is adiabatic.
B. The salt water regulates the rate of heat productionas the reaction proceeds, so that the pack does no,t
raise to a temperature that is too high.
C. The salt water dissolves the iron metal, providingmore surface area.
D. The salt water facilitates the transfer of electronswithin the reaction.
54. Which of the following graphs represents ihe
temperature of the heat pack versus time for both finelypowdered iron and less finely powdered iron filing:lAssume that equal masses of iron are present in botrheat packs.
Finely powdered iron
- kon filings semi-powdered
B.
5 5. What is the purpose of placing the packet into a papupacket inside of a plastic wrapper?
A . The paper allows for the flow of air to carry out rht
reaction. It is kept in plastic to prevent oxidatimfrom the air before it is opened.
B. The paper is impermeable to air, preventingoxidation of the activated charcoal.
C . The paper allows for the flow of air to carry out ftG
reaction. The plastic prevents iron from vaporizing
out of the packet.
D. The paper is semipermeable to air, allowing onlythe activated charcoal to be oxidized.
{
{u
itI
t
A.(-)rraCd
6)gEoF
r@
6nilftili0
@
illIoilil
!d
Time
Time Time
Passage lX (Questions 56 - 62)
The heat released from an exothermic reaction can be
determined accurately by carrying the reaction out in a bomb
calorimeter. The bomb calorimeter is comprised of a sealed
lead reaction vessel encased in a water bath. A thermometeris used to determine the change in temperature for the
surrounding circulating water. The heat released from the
reaction is assumed to equal the heat absorbed by the
calorimeter. A typical steel-walled bomb calorimeter is
shown in Figure 1.
Thermometer
Excess02 gas
Clamp
Leadcasing
Sample andIgnition coil
Figure 1
The sample is placed in a cup which contains an ignitioncoil connected to the ignition wires of a circuit. Possible
errors in the experiment are introduced by igniting the
sample, because the amount of heat generated by resistance in
rhe ignition coil is not measured directly. Excess oxygen gas
is used in combustion reactions to ensure that oxygen is not
rhe limiting reagent. The temperature of the water is recorded
against time from the start of the reaction until the
:alorimeter has cooled back to room temperature.
The exact time at which the reaction starts must be
estimated. The final temperature of all of the components inthe calorimeter are assumed to reach the same temperature as
$ater. Because the reaction is to be carried out underadiabatic conditions, the experiment is designed in way where
:he water temperature rises only a small amount, thus the
effects of convection and conduction are minimized.
5 6. What is the role of the stirrer in the experiment?
A. To help to form more stable products so that the
reaction reaches an even more favorable equilibrium
B. To provide activation energy to initiate the reaction
C . To lower the pressure and thus push the reactionforward
D. To homogenize the heat distribution throughout the
solution
Isnition
/,
Copyright @ by The Berkeley Review@ 177 GO ON TO THE NEXT PAGE
57. What liquid could be substituted for water in thisexperiment?
A. Decane b.p. 174.1"C, heat capacity 1.44 cilg."C
B. Diethyl ether b.p. 34.6"C, heat capacity 0.70 cl!- g''C
C. Isopropanol b.p. 82.3"C, heat capacity 0.96 i+g''C
D. Pentanol b.p. 138.2'C, heat capacity t.:: -Sal-
5 8. Why are the walls of the bomb made of steel?
A. Steel has a high heat capacity
B. To prevent any change in pressure
C . Rigid walls maintain the volume of the system
D. Reactive with water
5 9. Why is excess oxygen gas added to the container?
A . To ensure a complete reaction
B . To increase the amount of convection
C . To store the heat energy
D . To prevent the formation of CO2
6 0. What ignites the sample?
A . The oxidation of the wire in the ignition coil
B. The heat generated by resistance in thc wire of the
ignition coil
The friction generated by spinning the wire of the
ignition coil
The reaction of wire vapor with graphite after the
ignition coil sublimes
61. To study a reaction that generates very little heat, what
change in the apparatus will NOT help?
A . Using a liquid with a lower heat capacity than water
B. Using more reactant in the bomb
C . Increasing the stir rate
D. Using a thermometer with more calibrations
62. What reaction, assuming all reactions are favorable, is
most likely to result in a decrease in temperature?
A. Oxidation of a metal by oxygen
B. Combustion of a hydrocarbon
C . Precipitation of solute from water
D . Dissolving of solute into aqueous solution
c.
D.
Passage X (Questions 63 - 69)
Hot and cold packs are commercially available for bothmedical treatment and food storage. These packets rely on theheat released or absorbed by a chemical reaction. The changein heat energy is referred to as the enthalpy change for thereaction. A typical pouch contains an anhydrous salt in onemembrane and water in the other membrane. Rupturing thewall between the two membranes allows a solution to form,which results in either heat being absorbed (if solvation isendothermic) or heat being released (if solvation isexothermic). It is important that the salts be inert and non-toxic. Two typical reactions are given below:
CaCl2(s) + H2O(l) -+ Ca2+1aq; + 2 Cl-(aq)
AHsolvation = - 82.4 kJ/mole
Reaction 1
NHaNO3(s) + H2O(l) + NH4+(aq) + NO3-(aq)
AHsolvation = +26.9 kJ/mole
Reaction 2
The increase or drop in temperature can be estimated bymultiplying the moles of salt by the enthalpy (as measured incalories) to determine the energy change, and then dividingthis value by the mL of water (which is approximately themass). This estimate is close, because the heat capacity ofwater is defined as 1.00 calorie per degree per gram water. Asthe salt dissolves into solution, however, the heat capacity ofthe aqueous salt solution increases. For this reason, alongwith the decreasing reaction rate over time, the packets showthe greatest temperature change initially.
Ten grams of calcium chloride can raise the temperatureof 100 mL of water approximately eighteen degrees. Tengrams of ammonium nitrate can lower the temperature of 100mL of water approximately seven degrees. Once used, thepackets cannot easily be recycled, because the anhydrous saltsare difficult to regenerate. Alternative salts for heat packsinclude magnesium sulfate and sodium sulfate, both of which(when anhydrous) are used as dehydrating agents for organicsolvents. Alternative salts for cold packs include ammoniumchloride and ammonium iodide.
63. Which of the following fluids would be BEST for a
generic radiator for a power plant?
A. Distilled water
B. Salt water
C. Nitrogen gas
D. Dehumidified air
64. For a heat pack designed to reach 100"C, the solvationreaction must be:
A. exothermic and rapid.
B. exothermic and slow.
C. endothermic and rapid.
D. endothermic and slow.
Copyright O by The Berkeley Review@ t7a GO ON TO THE NEXT P
65. Which of the following solvation reactions could beused for a heat compress?
A . A reaction breaking weak lattice forces and formingstrong solvent-to-ion interactions
B. A reaction breaking strong lattice forces adforming strong solvent-to-ion interactions
C . A reaction breaking weak lattice forces and formingweak solvent-to-ion interactions
D. A reaction breaking strong lattice forces adforming weak solvent-to-ion interactions
6 6. What is the highest temperature reached after 40 gramrof CaCl2 are added to 200 mL of water at20'C?
A. 36"C
B. 56'C
c.72'CD.92'C
67. How can a solution temperature of 107"C beafter sodium sulfate is added to water at 50'C?
A. The solvation reaction is highly exothermic,the boiling point of the aqueous solution isthan 100'C because of boiling point elevation.
The solvation reaction is highly exothermic,the boiling point of the aqueous solution isthan 100'C because ofboiling point depression.
The solvation reaction is highly endothermic,the boiling point of the aqueous solution isthan 100"C because ofboiling point elevation.
The solvation reaction is highly endothermic,the boiling point of the aqueous solution isthan 100'C because ofboiling point depression.
6 8. Which of the following anhydrous salts wouldlikely be found in a cold pack?
A. MgSOa
B. CaSO4
C. MgCl2
D. NHaCI
6 9. What is true about the salts used in cold packs?
A. Lattice forces are strong; solvation is e
and driven by an increased entropy.
B. Lattice forces are weak; solvation is end
and driven by an increased entropy.
C. Lattice forces are strong; solvation is ex
and driven by an increased enthalpy.
D. Lattice forces are weak; solvation is exothermicdriven by an increased enthalpy.
B.
C.
D.
Passage Xl (Questions 70 - 78)
A two-part experiment is carried out involving fourmetals and four liquids. In Experiment I, 1O.O-gram samplesof each of the four metals are placed into an insulated heatingchamber with uniform heat distribution. The initialtemperature for each metal was recorded before it was placedin the heating chamber. After ten minutes in the heating:hamber, the metals were removed and the temperature foreach metal was immediately measured. Table 1 below showsthe initial and final temperature for each metal:
Solid Initial Temperature Final TemperatureMetal I 22.0 "C 51.0 "C
Metal II 22.1 "C 81.2 "C
Metal III 22.O'C 73.4'CMetal IV 22.0'C 46.9'C
Table 1
In Experiment II a 1O.0-gram cylinder of Metal IV isreated to exactly 50"C from its initial temperature of 25.0'C.The metal is added to a 40.0-gram sample of unknown liquid,rt 25'C in an insulated container. The metal is fullysubmerged into the liquid, and the system is allowed to:quilibrate until the two temperatures are equal. The final:emperature of the solution is recorded when the liquid and\letal IV reach the same temperature. The temperature of therutside wall of the insulated container did not increase duringhe experiment. The experiment is repeated three more times:sing a different liquid each time. The results from the four.rials are recorded in Table 2 below:
Solution FinalTemoerature
Liquid I 28.3'CLiquid tr 29.2"CLiquid Itr 28.1'CLiquid IV 30.7'C
Table 2
The two experiments were repeated an additional four:.mes each, to minimize any errors that may have arisen:uring the procedure. The numbers listed in both Table I andlable 2 are the data collected for the first trial only. No,,gnificant errors were detected in the subsequent trials, so the:ata are assumed to be valid.
-0. To determine the heat capacity for any of the metals,what else must be known precisely in the firstexperiment?
A . The initial temperature of the heating chamber
B. The molecular mass of the metal and the initialtemperature of the heating chamber
C . The initial temperature of the heating chamber andthe total energy absorbed by the metal
D. The total energy absorbed by the metal
-opyright @ by The Berkeley Review@ t79 GO ON TO THE NEXT PAGE
71. The metal with the GREATEST heat capacity is:
A. Metal I.B. Metal ILe . Metal IIL
,, g. MetattV.
\'//
If a metal and a liquid of identical heat capacities wereused in the second experiment, what would thetemperature be once the system reached equilibrium afterthe energy was fully transferred? Assume that no heatis lost to the environment.
..A. Less than 25.0'C"S. Less than 37.5"C, but greater than 25.0"C
.'- C-n' Exactlv -37.-5"C
5 . c."u,.r,nun 31.5"C
7 3 . The heat capacity of Metal IV is less than the heatcapacity of any liquid EXCEPT:
A. Liquidl
"B. LiquidtrC , Liquid Itr
- D. LiquidIV
7ai. Had MetalMetal IV,have been:
III been used in Experiment II instead ofthe final temperatures for the liquids would
F,\)
M7 greater than the temperatures found using Metal IV,because more heat would have been absorbed by the
solution using Metal III.,-Ii. less than the temperatures found using Metal IV,
because more heat would have been absorbed by the
. solution using Metal IIL1 C .l greater than the temperatures tbund using Metal IV,t---"''
b".uure the heat capacity for Metal III is less thanthe heat capacity for Metal IV.
D. less than the temperatures found using Metal IV,because the heat capacity for Metal III is less thanthe heat capacity for Metal IV.
The GREATEST amount of transferable kinetic energyis found in which of the following?
A. Metal I at 60'C
B-. Metal II at 60'C
C. Metal III at 80'C
D. Metal IV at 80"C
7 6. The relative heat capacities of the liquids are arranged inwhich of the following sequences?/'^:\)(a-."ffi>t>II>IV"8.. IV>II>I>III{. w >III>II>I{. trII > III > IV
7 7 . Had the temperature of the outside wall of the containerincreased, how would the results have bee affected?
A . The observed final temperature would be lower thanexpected, and the calculated heat capacity of theliquid would be too high.
B. The observed final temperature would be lower thanexpected, and the calculated heat capacity of thesolid would be too high.
vel me observed final temperature would be higherthan expected, and the calculated heat capacity ofthe liquid would be too high.
,{. tt't" observed final temperature would be higherthan expected, and the calculated heat capacity ofthe solid would be too high.
7 8. Which of the following relationships represents theratio of the heat capacity of the liquid to the heatcapacity of the solid in Experiment II?
O. Ctiquia=4ATliquid
Csolid ATsotirl
". Clquid
= I ATliquid
Csolid 4 ATs616
". Clquid
=4 ATrotto
Csolid ATtquiO
O. CLquiO - 1 ATsolid
Csolid 4 ATLqu;6
Copyright @ by The Berkeley Review@ rao GO
Passage Xll (Questions 79 - 85)
The focus of the basic design of the one-cylinder engineare a piston and a counterweight, placed on a cyclic crank insuch a way that the maximum potential energy of thecounterweight occurs when the piston is fully extended. Theextension of the piston turns the crank 180', and the descen:of the counterweight turns the crank the second 180' of thecycle.
Drawn below is a diagram of a basic one-cylinder engineThe boiler, when rapidly heated, creates a flow of steam thari
forces the piston upward. Once the piston is fully extendedthe intake valve is closed and the exhaust valve is opened so
that the counterweight can fall with less resistance. Thesteam from the piston is cooled and condensed in thecondenser. The water is then pumped from the condenser [cthe boiler. A pump must be used because of the pressu-edifference between the boiler and condenser. Figure I shovsa basic single-piston engine.
Crankshaft Counterweight
Exhaust valve
Intake valve
CondenserBoiler
Water pump
Figure 1
The one-piston engine is perhaps best recognized as the
steam engine that has moved large trains and electricaLgenerators for years. The point at which the rod connects &ctop of the piston to the crank is linked to the top of thcpiston by a hinge. The rod must be able to rock back and
forth as the piston ascends and descends. The counterweig{nis not exactly out of phase with the piston but is sligh{offset to ensure that the crank turns in only one direction.
7 9 . Which of the following valves is a one-way valve thm.
allows flow from left to right?
B.A.
reiFD._r,la-l\__:iJ' I
ON TO THE NEXT PAGM
8 0. If a cooling fan were added to the system' where would
be the BBST Place for it?
A. Adjacent to the boiler
B. Adjacent to the crankshaft
C . Adjacent to the condenser
D . Inside of the Piston
81 . To raise the piston, what must be true about the two
valves in the engine?
A . Both the intake and exhaust valves must be open'
B. Both the intake and exhaust valves must be closed'
C. The intake valve must be open' and the exhaust
valve must be closed'
D. The intake valve must be closed' and the exhaust
valve must be oPen'
B 2. When the counterweight is at its lowest point on the
crankshaft, the piston is not completely compressed'
What is the reason for this?
A. Not all of the steam can escape rapidly enough to
compress the piston completely' -The residual
steam offers too much resistance for full descent'
B. All of the gas escapes' so to compress the piston
completely, steam must be pumped back into the
chamber.
C. The exhaust valve is situated too high on the
chamber wall.
D . The intake valve is situated too low on the chamber
wall.
Which of the following statements BEST describes the
flow of heat in the sYstem?
A. Heat is added at the condenser and removed at the
piston.
B. Heat is added
piston'
C. Heat is added
condenser.
D. Heat is added
condenser.
at the boiler and removed at
at the Piston and removed at
at the boiler and removed at
the
the
the
-rpyright @ by The Berkeley Review@ l8l GO ON TO TIIE NEXT PAGE
8 4 . Water is chosen for which of the following reasons?
A . Water has a high heat capaciry in the liquid phase'
B. Water has a high heat capacity in the gas phase'
C . Water can sublime and condense at the temperatures
of the engine'
D . Water can vaporize and condense at the temperatures
of the engine.
8 5. What would make this engine MOST efficient?
A . If no heat were emitted from it
B. Ifheat radiated evenly throughout the engine
C . If heat radiated more from the condenser than from
the Piston
D . If heat radiated more from the boiler than the piston
Passage Xlll (Questions 86 - 92)
The diagram shown in Figure 1 below is a simplifiedrepresentation of a Carnot engine. The diagram shown inFigure 2 below is a simplified representation of a Carnot heatpump. By definition, the change in energy for the systemcan be calculated as the sum of work and heat, as shown inEquation 1.
AF-q+wEquation 1
where q is heat and w is work (found by w = -PAV for the
system). The actual operations of an engine and refrigerator(heat pump) are not accurately represented by the diagrams;
but because the overall pathway between two points can be
calculated by any combination of reversible pathways thatsum to the overall pathway, it is possible to calculate the
values for the Carnot engine and heat pump from these
diagrams.
Figure 1
Figure 2
The arrow between any two adjacent letters refers to a
step in the overall process. Each step can be treated as
independent or in combination with adjacent steps forcalculation purposes.
8 6. Which diagram from the passage represents an engine?
A. The diagram in Figure 1, because work flows intothe system and heat flows out of the system.
B. The diagram in Figure 1, because work flows outof the system and heat flows into the system.
C. The diagram in Figure 2, because work flows intothe system and heat flows out of the system.
D. The diagram in Figure 2, because work flows outof the system and heat flows into the system.
Copyright @ by The Berkeley Review@
Volume€
Volume-------*
8 7. For the diagram in Figure 2, the full cycle shows:
A. AE > 0; work in > work outB. AE < 0; work in < work outC. AE = 0; work in > work outD. AE = 0; work in < work out
8 8. If nF = 0 for the system, and the work for the system rs
positive, then which of the following is true?
A. Heat (q) is negative for both the system and thtsurroundings.
B. Heat (q) is positive for both the system and &esurroundings.
C . Heat (q) is positive for the system and negative fcrthe surroundings.
D. Heat (q) is negative for the system and positive icrthe surroundings.
8 9. After a hand pump has been operating, the shaft xwarm and the tip of the needle is cool. This is becau-*a
A. air is compressed in the shaft of the pump, and n
expands as it enters the needle tip.
B. air is compressed in the shaft of the pump, anC m
expands as it leaves the needle tip.
C . air expands in the shaft of the pump, and it rs
compressed as it enters the needle tip.
D. air expands in the shaft of the pump, and it ls
compressed as it leaves the needle tip.
9 0. From point a to point c in Figure 1, what is true?
A. The gas expands, and wsy5lsm > 0.
B. The gas expands, and wsy5lgm < 0.
C . The gas compresses, and wsyslgm ) 0.
D. The gas compresses, and w5yslsrn ( 0.
91. An efficient engine would have which of the follou"ingT
A. A high exhaust temperature and many points rr
energy is transferred
B. A low exhaust temperature and many points v-
energy is transferred
C . A high exhaust temperature and few points uenergy is transfened
D. A low exhaust temperature and few points uenergy is transl-erred
92. The arrow from point b to point c in Figumreprgsents:
A . a pure work step.
B. a pure heat step.
C. a step with both heat and work, where AE > 0
D . a step with both heat and work, where AE < 0.
la2 GO ON TO THE NEXT P
Questions 93 - 100 are NOT basedon a descriptive passage
9 3. The fact that, as order is gained by a system, itssurroundings are losing order is a consequence of whichof the following?
A. The First Law of ThermodynamicsB. The Second Law of Thermodynamics
C . The Third Law of Thermodynamics
D. The Rutherford rules
94. What is the final temperature after 32 mL H2O at36.0'C is mixed with 96 mL HrO at 64.0'C?
A. 57.0'CB. 54.1"C
c . 51.3'CD. 50.0"c
9 5. How much work is done on a system when an ideal gasis expanded adiabatically from 0.4 liters to 2.4 litercunder a constant pressure of 950.0 torr?
A. 5.00 liter.atmospheres
B. 2.50 liter.atmospheres
C, -2.50 liter'atmospheresD. -5.00 liter.atmospheres
96. If PAV = +24.4kIandq=22.2which of the following values?
A. -46.6 k.r
B. -2.2kJc. 2.2 kJ
D. 46.6 k.r
kJ, then AE is equal to
If AH is positive and AS is positive, then AG is whichof the following?
A. Positive at all temperatures
B. Negative at all temperatures
C . 7,ero at all temperatures
D. Can be zero, positive, or negative, depending onthe temperature.
AS for a gas-to-solid phase change is always:
A. positive at all temperatures and pressures.
B. negative at all temperatures and pressures.
C . zero at all temperatures and pressures.
D. can be zero, positive, or negative, depending on thetemperature.
't1
i8.
- rpyright @ by The Berkeley Review@ NOW THAT WASN'T SO BAD!
9 9. Expansion of a gas is:
A . exothermic and entropically favorable.B. endothermic and entropically favorable.C. exothermic and entropically unfavorable.D . endothermic and entropically unfavorable.
10 0. Which of the following relarionships may be TRUE?
I. Vaporization absorbs more heat than fusion.
II. More heat is released during sublimation thancondensation.
m. When a liquid freezes, the environment getswarmer.
A. I onlyB. II onlyC . I and III onlyD. II and III only
1.A 2.C 3.D6.A 1.D 8.C
11. D 12. C 13. B16. D t7. A 18. D21. D 22. A 23. A26. A 21. A 28. B3r. c 32. D 33. C36. C 37. B 38. A41. B 42. B 43. A46. D 47. B 48. D51. D 52. D 53. D56. D 51. C 58. C61. C 62. D 63. B66. B 67. A 68. D7t. D 12. B 13. D16. A 17. A 78. D81. C 82. A 83. D86. B 87. C 88. D91. D 92. B 93. B96. B 97. D 98. B
4.D .5.89.D 10.A
14. C 15. A19. D 20. B24. D 25. D29. B 30. A34. D 3s. C39. A 40. B44. B 45. B49. A 50. C54. C 55. A59. A 60. B64. A 65. A69. A 70. D14. D 75. D19. A 80. C84. D 85. A89. B 90. B94. A 95. C99. B 100. c
J.
\-
,
6.
7.
Thermochemistry Passages Answers
Choice A is correct. In Reaction 1, for both the thermodynamic and kinetic pathway, the larger activationenergy (higher energy transition state) is associated with the first step of the two-step reaction. The greaterthe activation energy, the slower the step. The slowest step is the rate-determining step, so the transitionstate of highest energy corresponds to the rate-determining step. The first step is rate-determining in Reaction1. The correct answer is choice A.
Choice C is correct. The value of the free energy change (AG) is found from the equation AG = AH - TAS. Thequestion states that AS is a positive value, and given that T is measured in kelvins, it too must be a positivenumber. This means that as the temperature increases, the value of AG decreases (because TAS is increasing)-This eliminates choices A and B. Using a second equation, AG = RT ln (Qr*/Keq), it can be seen that as AGdecreases, the Q.*-to-Keq ratio decteases, so the K"O-to-Q.* ratio increases. The best answer is choice C.
If you didn't recall the second equation above, you could have also used the equation AG = AG" + RT ln Qo,where AG' = - RT ln &q. The standard free energy change (AG') and the equilibrium constant (K"q) are relatedaccording to AG' = - RT 11 K"q. As AG' decreases, the value of AG decreases and the value of K"n
-increases, so
consequently the K"O-to-Q." ratio increases.
Choice D is correct. Both Pathway X and Pathway Y begin at the same point (implying that the reactants arethe same). This means that only the endpoint of the two pathways must be compared to answer this questicn-The endpoint of Pathway Y is lower than the endpoint of Pathway X, so more energy is released from Pathv"a;nY than Pathway X. This eliminates choices A and C. Because the endpoint of Pathway X is lower than thestarting point, the enthalpy change is negative (it is an exothermic reaction). This makes choice D the bestanswer choice.
4. Choice D is correct. At higher temperatures, the thermodynamic pathway is preferable, because it yields tfugreatest amount of heat. At lower temperatures, the reaction is forced to go by the lower energy transitimrstate. Pathway Y is the thermodynamic pathway, because it produces more energy than Pathway X. Athigher temperatures, Pathway Y is preferred, so choice D is the best answer.
5. Choice B is correct. Because the second transition state (apex in the graph) is of higher energy than thetransition state, the second step of the reaction is the rate-determining step in the reaction. This elchoices A and C. The intermediate occurs at the nadir (low point) between the two transition states (bethe first and second steps). If the second step is the rate-determining step, then the intermediate is allowedbuild up its concentration. This is referred to as bottle-necking of the intermediate. The best answer is choice
Choice A is correct. As the temperature is increased, more of Product Y is formed, and therefore less of Productis formed. This eliminates choices B and D, which both show increases in Product X and decreases in ProductThe sum of Product X and Product Y must always be 1,00"/o, so choice C is not possible (the sum is less than 1
at the middle temperatures). The best answer is choice A.
Choice D is correct. Because Reaction t has the endpoint of the energy curve lower than the starting point, itexothermic. This means that the value of the enthalpy change (AH) is negative. This eliminates choicesand B. Because the reaction is less favorable as the temperature increases, AG must be increasing withtemperature. The equation for the free energy change (AG) is AG = AH - TAS. In order for AG to increasetemperature, the TAS term must be negative. The temperature is measured on the Kelvin scale, so it ispositive. Only the entropy change (AS) can be negative. This makes choice D the best choice.
8. Choice C is correct. This question is a "read the chart" question. At 146 kJ per mole, the O-O bond isweakest. Even if you didn't remember what diethyl peroxide was, you could have matched the answerto the values in the table. Pick C.
Copyright @ by The Berkeley Review@ ta4 Section VIII Detailed
9.
10.
11.
12.
13.
1{.
Choice D is correct. The energy of the second n-bond in N=N can be approximated by subtracting the bondenergy of a nitrogen double bond from the bond energy of a nitrogen triple bond. Although the eflects of thecloser nuclei (associated with the shorter triple bond) may include strengthening of the rlg*u bond, for thisquestion it is best to assume that the change in energy between the double and triple bond is due entirely to thenew (second) n-bond. The energy difference is 947 - 418 = 523. The best answer is choice D.
Choice A is correct. Because of the reduced p-character in the sp2 orbital compared to the sp3 orbital, an sp2orbital is shorter and more electronegative than the sp3 orbital. This eliminaies choices B ind C. The bestanswer is choice A, because the bond strength is more associated with bond length than the bond polarity.Choice D is a true statement, but not the best answer.
Choice D is correct. The bonds broken are both relatively weak (154 and 432), while the bonds formed are bothrelatively strong (565 each). The heat released is greater than 500 k] per mole (586 - 1130), which makes thereaction highly exothermic. The best answer is choice D.
Choice C is correct. The energy trend of the sulfur-halide bonds is not a clear trend. The strongest halide bondto sulfur is with iodine (the largest halide), but because of the weak bond between sulfur and bromine, a trend insize cannot be drawn. This eliminates choice A. The most electronegative halide bond to sulfur listed is withchlorine, but because this is the second strongest bond to sulfur and the weak bond between sulfur and bromine, atrend in size cannot be drawn. This eliminates choice B. The halide with the greatest electron affinity ischlorine, so as with electronegativity, a trend in size cannot be drawn. This eliminates choice D. The bestconclusion that can be drawn is that the bond strength depends on two opposing factors, which would explainwhy bromine has the lowest bond energy. Based on halogen size, the relalive strength order would be: S-Cl >9-Br > S-I. Based on polarizability of the halogen, the relative strength order would be: S-I > S-Br > g-Cl. The actual order is S-I = Hl > S-Br, so both effects must be involved. The best answer is choice C.
Choice B is correct. A Br-F bond would be stronger than a Br-Cl bond, so the bond energy is greater than 218kjlmole. This eliminates choices C and D. A Br-F bond would be weaker than a Cl-F bond, so the bondenergy is less than 253 kJlmole. This eliminates choice A. The only choice that fits in the energy range is B.
Choice C is correct. As a general rule, shorter bonds are stronger than longer bonds, because the overlap oforbitals allows for the electrons to be shared between nuclei-most readily. This makes choice A a falsestatement. According to periodic trends, chlorine is both less electronegative than fluorine and larger thanfluorine. This eliminates choices B and D. The weakness of the F-F bond can be attributed to the repilsion ofthe nuclei of the two fluorine atoms. As the fluorine atoms get close enough to form a bond, the two +9 nucleibegin to repel. The repulsion results in the overall weakening of the bond. The best answer is choice C.
Choice A is correct. Because sodium metal is heavier than lithium metal, the energy required to sublimesodium is greater than the energy required to sublime lithium. This makes statement I true. As a column in theperiodic table is descended, the ionization energy of the element decreases, so the first ionization energy oflithium is greater than the first ionization energy of sodium. This makes statement II a false statement. Theatomic radius of sodium cation is larger than the atomic radius of lithium cation, thus the lattice energy of thesodium salt involves a larger r value in the denominator. The lattice energy is decreased when lithium isreplaced by sodium, so statement III is a false statement. The correct answer is choice A.
Choice D is correct. In the reaction, the F-F bond is completely broken (partial breaks are not possible), sochoice A is eliminated. The value for the bond energy for F2 is 154 k]/mole, although the value ii not from atable. Half of the value is used, because only one fluorine atom is needed. The bond is broken in a homolyticfashion (with each atom receiving one electron), so choices A, B, and C are eliminated and choice D is the bestanswer.
Choice A is correct. The lattice energy increases with increasing charge (q) or decreasing radius (r). Magnesiumcarries a +2 charge, and oxygen carries a -2 charge, so the greatest lattice energy is associated with MgO. Thiseliminates choices C and D. Because Li+ is smaller than Na+ and F- is smaller than C1-, the r""ot d highestlattice energy is associated with LiF. This makes choice A the best answer.
t85 Section VIII Detailed Explanations
18.
19.
20.
Choice D is correct. If calcium metal is used in lieu of lithium metal, the ionization energy changes, becausecalcium forms a dication, which requires two ionizations. The total ionization energy for calcium is a sum ofthe first and second ionization energies. As a note, the extra energy invested in the ionization is recovered bythe doubling of the lattice energy. This is because the ionization energy doubles when the cation chargedoubles from +1 to +2. The best answer is choice D.
Choice D is correct. Electron affinity involves the gaining of a free electron by a neutral atom. The excitationand relaxation (de-excitation) involve the absorption and emission of a photon when an electron moves betweenthe ground state and an excited state. The charge of the atom does not change in either process. The ionizationenergy is associated with the loss of an electron, while the electron affinity is associated with the gain of anelectron. The best answer is choice D.
Choice B is correct. Using a metal that is easier to sublime than lithium requires the input of less energy, so theoverall process yields more energy. Choice A is consequently eliminated. Using a halogen that forms a strongercovalent bond than fluorine requires the input of more energy, so the overall process yields less energy. Thismakes choice B the best answer. Using a halogen with a greater electron affinity than fluorine (which is notphysically possible, given that fluorine has the highest electron affinity of the halogens) releases moreenergy, so the overall process yields more energy. Choice C is consequently eliminated. Using a metal that caneasily lose a second electron requires more ionization energy, but double the amount of energy is released in thelattice formation, so the overall process yields more energy. Choice D is also eliminated.
Choice D is correct. Substituting a larger cation for a smaller cation increases the distance between ions. Thfureduces the lattice energy, because the denominator has been increased. This makes statement I true-Substituting an anion of lesser charge for the current anion in the salt reduces the lattice energy, because thenumerator has been decreased. This makes statement II true. Cutting the cation and anion charges by one-haffieach reduces the lattice energy to one-fourth its original value, because the numerator has been decreased br nfactor of four. Cutting the internuclear distance by one-half doubles the lattice energy, becausedenominator has been decreased by a factor of two. Overall, the effect of cutting all three values in halJ ithat the lattice energy decreases by a factor of two, making statement III true. The best answer is choice D.
7' Choice A is correct. In distillation, a mixture is converted into pure components. Distillation results inorder for the system, which is entropically unfavorable. This makes choice A a correct answer. The confrom a solid salt into two ions in solution is entropically favorable, because the system becomes more rThis eliminates choices B. In sublimation, a solid is converted into a gas, which increases the randomness,the process is entropically favorable. This eliminates choice C. Diffusion of a gas into the air results inloss of a pure component into a mixture, which increases the randomness. This is entropically favorable,choice D is eliminated.
)? Choice A is correct. A value of AS > 0 results from an increase in disorder. In choices B and C, the numbermolecules decreases from reactant to product, so both can be eliminated. In choice D, the number ofremains the same, so choice D can be eliminated. In choice A, the salt becomes more random as itinto ions. Pick A if you desire the jubilation and enjolnnent associated with correct answers.
24. Choice D is correct. Addition of a catalyst does not affect the thermodynamic values of a reaction, such as
energy (AG), enthalpy (AH), and entropy (AS). A catalyst stabilizes the transition state, loweringactivation energy for a reaction. This affects only the reaction rate and makes the best answer choice D.
25. Choice D is correct. Because the reaction is spontaneous as written, the change in free energy (AG)reaction is negative. Because the solution temperature lowers as the reaction proceeds, theendothermic. This makes the value of AH positive. Rearranging Equation 3 yields: TAS = AH - AG; so ifpositive, and AG is negative, then the value of AS must be positive at all temperatures, making choicebest answer.
27.
Copyright @ by The Berkeley Review@ ra6 Section VIII Detailed
26. Choice A is correct. This question is answered by combining Equations 1 and 2 into one composite equation asfollows:
AGobserved = AG" + RT kI QrxAGobserved = -RT In K"O + RT ln Qr* = RT ln Qr* - RT * Kee
AGobserved = RT (ln Q.* - h K"o) = o, 1r', Qt"
, Ke9
The best answer is choice A.
Choice A is correct. \A/hen the reaction quotient is greater than the equilibrium constant, too many products arepresent, so the reaction must shift to the reactant side in order to reach equilibrium. This etiminates choices Cand D. When the reaction spontaneously shifts in the reverse direction (to the reactants), it is said to bepositive, so the best answer is choice A.
Choice B is correct. According to Equation 2, LG = -RT ln KuO, the value of AG must be zero for the equilibriumconstant to be equal to 1,.0, because the natural log of 1 is 0. Eqriation 3, AG = AH - TAS, must be used to determinethe temperature at which AG is zero.
AG = AH - T AS = 0 .'. by adding TAS to both sides of the equation, AH = T AS
Dividing both sides of the equation by AS leads to the relationship: T = 4I{
aH =
12,500-+ * 12,500 K = 25oo 6 = SooKAS5I 255
mole.KDo not blindly choose answer choice D, however, because the answers are in Celsius. The temperature must beconverted from Kelvin into Celsius. Subtracting2T3 from the Kelvin value yields 500 K = 227'C, choice B.
Choice B is correct. The value for the water reaction in the passage involves the formation of water vapor. Inthe question, the value is listed for the formation of water liquid. Conversion from vapor to liquid isexothermic, so the formation of water liquid yields more heat energy than the formation of witer ,rapoi. Th"best answer is choice B.
Choice A is correct. The first stage of a rocket requires that the greatest total amount of heat be given off,because the first stage is responsible for liftoff and the acceleration away from the earth's gravitational pull.The fuel must provide a great amount of heat, while not occupying too much volume (the storage space m.tit beminimized). It is ideal if the fuel provides a large amount of heat per gram as it burns, making the fuel mostefficient without significantly increasing the mass of the rocket. Hydrogen would be ideal, because it is solight (except for the fact that it is a gas at nearly all temperatures). At room temperature, kerosene is a liquid,while hydrogen is a gas. Because liquids ate denser, they are far easier to store conveniently and efficiently.This makes choice A a good choice. Choice B should be eliminated, because kerosene provides less energy pergram than hydrogen. This is to say that 242 divided by 2 is greater than 7513 divided by 170. Whetherkerosene bums hotter than hydrogen or not should have no bearing on its use as a rocket fuel. The hotter itburns, the more heat that is wasted by loss to the environment. Choice C can be eliminated. Oxidation ofkerosene produces both water and carbon dioxide, which is more products than hydrogen gas (which producesonly water). The answer choice D is a wrong statement.
Choice C is correct. From the balanced equation, two moles of hydrazine are required for every one mole ofdinitrogen tetraoxide. Hydrazine weighs 32 grams per mole, while dinitrogen tetraoxide has a molecular massof 92 grams per mole. This means that for 64 grams of hydrazine, 92 grams of dinitrogen tetraoxide are required.This means that roughly one and one-half times as many grams of dinitrogen tetraoxide are needed ashydrazine. This is answer choice C.
27.
28.
,:rvright @ by The Berkeley Review@ ta7 Section VIII Detailed Explanations
JZ.
JJ.
34.
35.
36.
37.
Choice D is correct. The first three reactions are all oxidation-reduction reactions that mimic the sample
reactions from the passage. The last reaction is an acid-base reaction, and it does not produce sufficient energl-
to propel a rocket. - The iorrect answer choice is answer D. Although the passage does not blatantly state that
anbxidation-reduction reaction should be used, it can be inferred from the sample reactions.
Choice C is correct. From the sample reactions in the passage, it can be seen that rocket fuels engage in highir
exothermic combustion reactions, provide a large amount of heat Per gram, and are highly reactive illoxidation-reduction reactions. This makes statements A, B, and D valid. A rocket fuel is better if it is a solid or
liquid at room temperafure, rather than a gas, so it can be stored (packed) more efficiently. This means that
choice Cis not a requirement of a rocket fuel. You must pick C, you must, you must, you must!
Choice D is correct. Dinitrogen tetraoxide gets reduced in the reaction, so it is the oxidizing agent in the
reaction. Statement A can therefore be eliminated. Hydrazine is oxidized into nitrogen gas. The oxidation
state of nitrogen changes from -2 to 0. This implies that the entire compound loses four electrons (two per
nitrogen), not"that eact nitrogen loses four electrons. Choice B can therefore be eliminated. Because the fine-
prodict iras a nitrogen-nitrog6n triple bond, it can be concluded that nitrogen-nitrogen bonds are formed durins
ihu "o.rrr"
of the reaction, ut d t ot broken. This eliminates choice C. Hydrazine is losing hydrogens, so it isbeing oxidized. This makes choice D the best answer.
Choice C is correct. The heat per gram for kerosene can be found by dividing 7573 kJ per mole by 170 Srams pe:
mole. The value falls in the range of 40 to 50 k] per gram. The math is set up as follows:
50 = 8500 > 7513 ;. 6800 = 49170 170 170
38.
Because the range is between 40 and 50 k| per mole, the best answer is choice C.
Choice C is correct. Yaporizatron is a change in the physical state of matter, whereby a liquid evaporates :;form a gas. Writing out ihe reaction for the vaporization of water and using values Table 1" yields:
HzO(l) -+ H2O(s) AGformation H2O(l) = -237.2 AGformation Hzo(g) = -228.6
AGformation of products - AGformation of reactants = AGrx + -228.6 - (-237.2) = + 8.6 k] per mole, choice C' Enerp- ut"qnir"a to vipbrize water, so we expect a positive number. A value of 44.1' results from using AH numbers'
Choice B is correct. Both xylose and ribose form five-membered (furanose rings) rings, according to t.re
information in the question. The five-membered rings are not subject to much ring strain (if they were, fir-*'
membered rings would not be so common), so choice A is eliminated. Both xylose and ribose are aldopentoses rIthey each havJ one C=o (in the form of an aldehyde) and four O-H bonds (in the form of hydroxyls on carbonr'
two through five). This eliminates choices C and D. The only choice left is answer choice B. The repul':;on
experier,cei by the eclipsed hydroxyl groups in the five-membered ring appears in the enthalpy of reactrc'::
Th" *o." ,up.rlrion, the less siable the molecule, and thus the greater the enthalpy of reaction (more heat r.M
is released when the ring steric hindrance is relieved). Using the enthalpy of reaction to determine 5estability of structural features is common in organic chemistry. Using Hess's law with the general cheml':u
values does not show the other features that affect a molecule's stability. Many experiments in orga:,,rr
chemistry focus differences in the enthalpy changes between reactions with similar bond changes.
Choice A is correct. From Table 1, the AG" values of + 209.1. for acetylene (the reactant) and + 68.3 for ethvlo'nr'
(the product) can be read from the free energy change column._ Be sure that you use the values for AG", and rutr
AH'ty mistake. The free energy of formation for H2(g) is 0, because hydrogen gas is a diatomic gds at rorrnm
temperature in its natural elemental state. This eliminates choices C and D' For the hydrogenation r€df:-r4r
the change in free energy is found by AGlsmrtion of products - AGformation of reactantr' The value AG'1a ef,jath
68.3 - ZO9.I, a negative number. A negative numbler for AG' makes the reaction favorable in the fon"
direction, so the best answer is choice A. The hydrogenation reaction is shown below:
1.CzHz(6,) + 1H2(g) -+ 1C2Ha(g)
Copyright @ by The Berkeley Review@ l8a Section VIII Detailed Ex
39.
40.
Choice A is correct. At standard conditions, carbon exists as graphite, not as diamond, eliminating choice C.Cl2 and H2 are gases at 25"C in their most stable elemental form, so this rules out choice B. Al;, chlorineexists as a diatomic molecule in its most stable form at 25'C, so the answer is not choice D. The only answerchoice left is A, which is a good thing considering it is the correct answer. Be sure that the equation baiances.
Choice B is correct. The first thing to do is to balance the combustion reaction:
1C3H6 + 4.502 + 3H2O + 3CO2The phases are ignored, because they are not listed in the answer choices. Normally, phase is significant,because water can be listed as a liquid or a gas.
AHo - AH16.rrlulion of products - AHformation of reactants = 3 AH1611 H2O + 3 AHpe11 CO2- AH6or- qr"uThis is choice B. Oxygen gas (O2) is ignored, because its most stable form at 25"C is the diatomic molecule 02,so its enthalpy of formation is 0.
Choice B is correct. From the values in Table 1, the hydrogenation of one n-bond (from an alkene to an alkane)produces a change in enthalpy (AH"nyarogenation) of -139.8 kj per mole. This value is determined from thefollowing reaction and thermodynamii datl:
CzHsG)+HzG) -+ C2H6(g) AHformC2H41r1=+52.4 AHformC2HulgS=-87.4 MformH2lg;=0
AHhydtogenation = AHformation of products - AHformation of reactants = (87.4) - 52.4 = -f gg.S -lI-mole
The reaction in the question involves the hydrogenation of two n-bonds in butadiene, so a good approximation isthat the change in enthalpy for the complete hydrogenation of butadiene has a value of Zl-t3e.d i1 p". mole) =-279.6 k] per mole. The value is given in the question as -271,.4 kf per mole, which is less negative thin expected.This eliminates choices A and C. The logic behind the reduction in heat energy released is rooted in either aincreased stability in the reactant or decreased stability in the product. In this case, it is the stability of thereactant (conjugation) that causes the reduction in heat released. The best answer is choice B.
Choice B is correct. The heat produced for the reaction is calculated using the equation: AHy* = AHproducts-AFlreactants. Using values from Table 1 (values are in kJ per mole), the two AH are calculated as fo[o#s:
CzHz -=+ HzO + 2CO2
AH: +226.8 -285.9 -393.5 AFII* = t(-285.9) + 2(-393.5)l - 226.8 = -1299.7 kImole
CzHs -->
2H2O + 2CO2AH: +52.4 -285.9 -393.5 AH1* =[2(-285.9) + 2(-393.5)] - 52.4 = -141,12 kI
moleThe problem asks for heat per grarn, not the heat per mole, so the values must be divided by molecular massbefore comparison to one another.
C2H2 yields 1299'7 W, which is just less 11',ut'r 1300 \[ which is 50 K26 g' ' 269' g
C2H4 yields 1477'2 kL, which is iust over 1400 kL which is 50 kI28 g 289 I
Thus, C2H2 produces just under 50 kJ per gram of heat, while C2H4 produces a little more than 50 kJ per gram ofheat, making answer B the correct choice.
Choice A is correct. The free energy produced for the reaction is calculated using the equation: AG.* -AGproducts - AGreactantr. Using values from Table 1 (values are in kJ per mole), AG is calculated as follows:
C2H5OH --+ 3 H2O + 2CO2
AG: -774.9 -237.2 -394.3 AG1, = tg (-237.2) + 2 (49aQl - (-174.9) - kl-
= (-711.6) + (-788.6) + 174.9 = -1325.3 kImole
Choice A is the best answer.
l::.,-right @ by The Berkeley Review@
47-
!)
Section VIII Detailed Explanations
44. Choice B is correct. The initial temperature of the water in Beaker 3 before the NHaCI salt L. added to th;water is 22"C. In Figure 3, the temperature decreases to a value less than 2TC (exfiapolation shows that thttemperature is aboui17.S 'C). Once the temperature reaches its lowest point, it slowly climbs up until it leveL
off it ambient temperaturc (22'C). This means that the reaction is absorbing heat from the solution during the
dissociation of NHaCl. Because heat is being absorbed during the solvation reaction, the solvation of Nt!C-must be an endothermic reaction. The passage also states that the reaction is endothermic. To score big, volshould pick B.
Choice B is correct. Adding 1.21 grams of Mg(s), instead of 2.43 grams as were added in Experiment 2, resuits r.only half of the heat being releasid that was released in Experiment 2. The temperature increase in Beaker l(using 2.43 grams) was 1d"C, so the temperature increase using 1.21 grams should be about 8'C. This, ivhsadded to the starting temperature of 21'C, would yield a final temperature of 30"C. Choice B is the bes;
answer, for those of you interested in best answers.
Choice D is correct. To solve for the molar enthalpy of this reaction, you must first solve for the heat m
reaction for MgO in terms of k] per gram (using the data from Beaker 2), and then convert from kJ per gram NIeC
to kJ per mole*MgO. To determine the heat released during the reaction, we use the equation E = mCAT, n'he--t
m equals 1.0 kg, e = +.5+J per g.K, and AT is extrapolated from the graph. To extrapolate, you draw a-strai53m
line ihrough tn"" dutu points to ihe time = 0 axis, in order to estimate the highest temperature of the solution
20 40 60 80 100 120 1,40 1.60
45.
46.
Time after addition to Beaker 2 (seconds)
As extrapolated from the graph in Figure 2 (shown above), the apex temperature is roughly 31'C, so the r a-:rc
of AT is ioughly 9"C. The -heat r"l"ased when 4.03 grams MgO was added to Beaker 2 is calculated as follort -'r
E = 1.0ks,x 4.34 I x 9.0K = 4'34x9kJ = 43'40 -4'34k1 = 39'06kJ" g.K
The molecular mass of MgO is given in the passage as 40.3 grams/mole, so 4'03 grams is exactly 0.10 mo-les tr{$The molar enthalpy is cJlculaled by dividing the heat of the reaction by the moles of reactant (MgO, in 5m
case). The calculation is set up as follows:
1.0 x 4.34 x 9.0 kJ x 40.3 Srarns =1.0 x 4.34 x 9.0 x 40.3 kJ
4.03 mole
Choice D is correct. Because the temperature increase during the reaction, heat is given off by the reaction
the reaction is an exothermic reaction. Because the reaction is exothermic, the sign of the enthalpy cha
(AH) is negative. All of the answer choices are negative, so this doesn't help in eliminating wrong answers
4.03 Sam mole
AT = +9.0"C
Copyright @ by The Berkeley Review@ 190 Section VIII Detailed
47. Choice B is correct. Using Hess's law, we know that the molar heat of reaction is obtained from the heats ofreaction for any series of reactions that sum to the overall reaction. For this experiment, the two reactionsalong with the formation of water are as follows:
Mg(s)+2H+(aq) -->
Mg2*(uq)+Hz(g) AH=heatreleasedfromBeakerlxl0=-66g.g k]mole
Mgo(t)+2H+(aq) ----+ Mg2*(uq)+H2o(1) AH=heatreleasedfromBeaker2xL0=-826.2 kImole
Hz(g) + lozfzl _* H2o(t) AHformation = -286 kJ
mole
The overall reaction is: Mg(s) * l}rtfl ---l> VfgOtrl
We must reverse the second reaction and then sum the AH values for the three component reactions above, inorder to obtain the overall reaction. This means that the AHformatio_n (MgO) = AHlBeaker f; - AHpeak er z1 - 2g6kJ per mole. The best answer is choice B. The actual value is -669 + Wt'- fgi'= -sGA:"-
Choice D is correct. The reaction of a metal with hydronium results in the oxidation of the metal by the H+ions. The reaction is as follows: Mg(s) + 2 H+(aq) -+ Mg2+{aq) + Hz(g). H2 gas is given off, so choice D is best.
Choice A is correct. If the heat capacity (C) is greater for the HCI solution than water, then the value for C(4.18 for water) plugged into E = mCAT is too small. The calculated value is therefore too small. This makeschoice A correct. An increased heat capacity does not allow the temperature to increase as much, which lowersthe container temperature and actually decreases the reaction rate. This makes choice D invalid.
Choice C is correct- According to the information for Reaction 1, four moles of iron produce 1652kI of heat. Thismeans that 2 moles of iron produce 826 kI, and 3 moles of iron produce 7239 kI. In order to produce 1000 kJ ofenergy/ just over two but less than three moles of iron are needed. Two and one-half moles of iron produce 826kJ+ 207 kJ, which is greater than 1000 kJ. This means that the amount of iron needed is less than 2.5 moles of iron.Iron weighs 55.85 grams per mole, so the mass needed to produce 1000 kJ of heat is greater than 111.2 grams (themass of two moles) and less than 739.62 grams (the mass of two and one-half moles-of iron). The best"answer ischoice C.
Choice D is correct. The pack initially contains iron and the solution in which the reaction transpires. Onceexposed to the air, oxygen can enter the pores of the paper container and oxidize the iron. Because'the product(Fe2O3) has gained mass from the environment, the mass after reaction is greater than the initial mais. Theexact amount of mass increase is the mass of the oxygen that has reacted. The best choice is answer D.
Choice D is correct. Because the reaction takes place at room temperature and the iron is oxidized tocompletion, it is assumed that the reaction is favorable. This results ln a AG' value less than zero, whicheliminates choices A and C. The reactants are four solids and three gases that go on to form two solids. This isa. loss in entropy by the system, which carries a negative AS'. The value of A6' is also less than zero, makingchoice D the correct choice.
Choice D is correct. The role of salt water, as mentioned in the passage, is to increase the interaction of oxygenwith iron and to conduct electricity. None of the answer choiceJ addresses the interaction of iron with oxygen.This means that the role of the salt water of interest to us is to facilitate the flow of electrons. Salt waterprovides the medium through which the electrons may transfer. The best answer to this question is choice D.Heat is released by the reaction, so the reaction cannot be adiabatic. Choice A is thus eliminated. The saltwater has no role in thermal regulation, because it does not undergo any chemical or physical changes duringthe reaction' Choice B is thus eliminated. Iron metal does not dissolve into water, as ytu"perhaps hJve noticedwhen iron structures are able to stand through the rain. Choice C is thus eliminated.
18.
19.
:,,-right @ by The Berkeley Review@ l9r Section VIII Detailed Explanations
54.
55.
55.
being produced, so the reaction must have stopped'
Choice A is correct. Answer choice A best explains the logic behind the packaging. Because oxygen gas l
enter the packet to r"u.i*itr'' the iron, t1e *"'ipJ."1e mustil-t1?1ll';" tj3:::'^1||o3,1"Yilt"#'?''n"*il:H;;;.i:"il pr*"", the packet from oxidizing prematurery, it must not be.exposed to air. The pl
wrapper is impermeJfu lo uit, so- the pack is stable-in the anaerobic environment' The plastic wraPper
nelps to prevent the loss of water due to evaporatlon'
choice D the best answer.
Choice D is correct. Stirring the solution does not increase the thermodynamic values (equilibrium consft
free energy, and enthalpy), but it does-"f;;rhe reaction t",PT:":* 1t^1^,t1tj*#1"^-SY::"* :]r;;|1ffi;T:[l'fi#,il;'f;i#;;;;;."'d;;;;"sh ""ersy to aciivate the reaction. rhe energv used to acti
the reaction is provided by the svstem's heat energy' as meas-urt Pl,lYi:ff"T:1"^,":lT""rf-,it";:ilffiil:'f,J;;;r;;-riir'" '"r,i,l'"-h"ur r,o effectin the pressure of ihe solution, so choice C is elimi
The stirring of the sol.rTion allows the heat to be distributed uniformly throughout the solution' which
3/.
58.
Choice C is correct. The liquid substituted must have a heat capacity similar.to rlater and not have a bo
point that is too row. t b; boiling p'i";;'+9 iYtl1 -1n,1:*l"i::"5c:1":,t:,t11* ir"Hl'ffit"$iipolnT tnal rs LU() ruw.
vaporizes. Isopropanol has a boilfuig^point and heat "upu"ity
close to those of water' If the heat capacity
too high, the AT is too small, and thui ih" u."rrru"y of the temperature change is decreased' Choice C is besL
Choice c is correct. whether steel has a high heat capacity or low heat capacity' the effects of both can b
calculated for. The lower the heat.upu"ity, tfie better, b".u*" water can nse- to ".I"]q:iT*":1til?::l,l:a trivial point. By having steel walls, the container does not expand during the reaction' so the vo
remains constant white the pressure .nungui. -rrtir *"t* choice.B rifl,ana :h"fi :,lti:'^ T^1T,Yt'*:
is not reactive -ith ;'#;,";*;j;*tn? nuat change cannot be attributed sorery to the reaction. The
answer (so that the heat capacity at constant volume cin be used in calculation) is choice C'
Choice A is correct. The excess oxygen is added to ensure a complete reaction. ]h1 amount of heat generated
measured relative to the mass of the sample (the limiting reagent i1 tirg reaction)' 3"-.."1ff::q1.t-T::IILgdbLrIEu lclqLrv ! Lv
convection, but that is not the reason for aiding excess oryg:n.. This is a true, but irreleva*.:t?::::t; "#5gas stores little to no heat energy, because guJ"s ha'oe.t"H*:ly low heat capacities relative to water'
D is eliminated, because the presence of o*yf"tt results in CO2 formation' The best €rnswer is choice A'
Choice B is correct. The sample is ignited by heat -emitted from the ignition coil' The question is'
generates the heat *,r,".'rn 'nl" 'Jru T:;lT:i*,:?^1:":l1l:"-*'",T il'lflT:ffi: *ilYfi?lfffi"#:il:U'?'# :ffi ,
.i,"'",'' ;""* il ;;; FI :t{:' :T .T'";* * ":I"".:ffj1 3 ff;LS"
*'''that the sample is ignited when th-e ,"rirtoit"*peratu'e is high enough' The best answer is choice B'
Choice C is correct. In determining the temperature change from a reaction that generates very little heat'
problem is that a small Ar results i''.u" ""oiln it' a1111i:l::^:v :ilF":
titl'i T'iri."l:ffii:f ffii;ffTH,";i:,""ffi greater and thus more accuratery measured. choice A is a valid statement'
sample (reactant) ir tr&, more heat i' g""e*tudi.to q" lllt-tfq:l:111:tr",T,*,?"ff:T:"f.:l:;ffiii: $::'r;1,,: H,H#:'i#;il;** 'ii' '"1" naps tn!.svstem ," dt:llll":..*:1"i::'_t?^1*Y
more quickly. rhis does not herp the '"uitio";f: * 11T"'*-T:?:::ffi::h *:liJgilfi::I'fi;:ffiT;:il*Ji;Lil"'ilH;;: p'".t'" *""surement or Ar. rhis makes choiceb a valid statement' rhe
answer is choice C.
59.
60.
61.
Copyright @ by The Berkeley Review@ 192 Section VIII Detailed E
62' Choice D is correct. As a rule, both the oxidation-reduction reaction of a metal with oxygen and the combustionreaction of a hydrocarbon (which is an oxidation-reduction reaction) are exoth".-i"- reactions. They bothresult in an increase in temperature of the solution, not a temperature decrease. If the precipitation of a soluteis favorable, then the AG that reaction is negative. In a precipitation reaction, the system becomes moreordered, so AS is {etinrlgty negative. The signlf AH must be negative to make the sign fr ac 1.r"guti,ou, basedon the equation AG = AH -TAS (negative nuriber - T(negative "ytil9| = negative number). Dissolving a soluteinto solution is the only one of the choices given that cin possibly be end#hermic, if all of the answer choicesrepresent favorable reactions. An endothermic reaction reiults in'a temperature drop. The best answer is thuschoice D.
63. Choice B is correct. Whether the solution is for the radiator of a power plant, or any system designed to absorbgreat amounts of heat, the best solution is the one with the greatest heat capaciiy ind greateJt contact areawith the heat source. Choices C and D should not be considerld, because they are toth ga"ses. The addition ofsalt to water increases the he-at capacity, allowing salt water to absorb more heat per degree increase, and toreach a higher temperature before boillng. This can be inferred from the passage where it states that thetemperature increase begins to slow, because the heat capacity changes u, ,r,o." salt dissolves. This is whymany power plants use salt water in their circulating radialor system. Cars use a mixture of ethylene glycol indistilled water' using an organic compound tike eth"ylene glycoi prevents against salt buildup thut *o.r'id o"".r,if salt water were used, due to the evaporation of *ui"r. rf,e Uest answer is choice B.
Choice A is correct. For a heat pack to reach a temperature greater than room temperature, it must exploit areaction that releases heat energy' This describei an exolhermic reaction. Because the heat is neededimmediately, tapid reactions are more favorable. Slow reactions allow more heat to dissipate to theenvironment. This combination makes choice A the best answer.
Choice A is correct- To be to use in a heat compress, the reaction must be exothermic. To be exothermic, thereaction must require minimal energy to break the bonds and obtain a great deal of energy when the new bondsare formed. hr the case of a salt dissolving into water, this would resui from weak lattice forces (bonds broken)and strong solvent to ion interactions (solvation). The best answer is choice A.
Choice B is correct. The passage states that 10 grams of calcium chloride raise 100 mL of water approximatelyeighteen degrees from room temperature. This -ear,, that forty (40) grams of calcium chloride should increase109^ $ water by 36"C. The final temperature is the initial temperaiure (20"C) plus the temperature increase(36'C), resulting in a final temperature of 56'C. The actual value is a little t""s, due to the reaction time beinglonger (allowing heat to dissipate to the environment) and the changing heat capacity of the aqueous solutionas the reaction proceeds. The best answer is choice B.
Choice A is correct. For the temperature to increase, the reaction must be exothermic, so choices C and D areeliminated. The boiling point is greater than that of pure water (100'C), so the boiling point must have beenelevated. The best answer is choice A.
Choice D is correct. Sulfate salts are exothermic when they dissolve, so they cannot be used in cold compresses.This eliminates choices A and B. Ammonium salts are endothermic when they dissolve, so heat is absorbed andthe
-so-lution begins to cool. The best answer for an endothermic solvation involves ammonium chloride(NHaCl). The best answer is choice D. The answer is provided in the last sentence of the passage.
Choice A is correct. Cold packs absorb heat because of a chemical reaction, so the solvation reaction must beendothermic for cold packs. This causes the temperature to decrease. The favorability of a reaction depends onthe value of AG. If the value of AG is negative, then the reaction is said to be favorable. An endothermicreaction has a positive value for AH, so the only way that the value of AG can be negative is if the value of ASis positive (AG = AH - TAS). This automatically *"uttr that the salt dissociates beciuse of increased entropy,not. favorable enthalpy' This eliminates choices C and D. The lattice forces are most likely strong, requiringhigh energy to break. The more e_nergy required to break the lattice bonds, the more likelyihat thJ'reac'tion isendothermic. The best answer is choice A.
64.
65.
66.
68.
f,q
Copyright @ by The Berkeley Review@ 193 Section VIII Detailed Explanations
70.
7'1..
72.
Choice D is correct. The heat capacity depends on three values: energy absorbed (E), mass of substance (m), andthe change in temperature (AT). The mass of each metal is known and the change in temperature is known, soonly the heat absorbed, while inside the heating chamber must be known to calculate the heat capacity foreach metal. This makes choice D correct.
Choice D is correct. Heat capacity has the units calories per gram.Kelvin in the standard case. Using theformula E = mCAT, the heat capacity (C) for each metal sample can be found by dividing the energy absorbeiby the mass of the sample and temperature change for the process. Each metal absorbed the same amount c:heat energy and had an identical mass. The only difference between the metal samples was the change irtemperature observed for the process. Because AT is in the denominator when calculating the heat capacity fe:each metal (C = E7-.O1), the greatest heat capacity is associated with the metal that showed the smalles;increase in temperature. According to Table 1, the metal with the smallest final temperature (and thereforesmallest AT) is Metal IV, so pick choice D with great pride and satisfaction.
Choice B is correct. If the heat loss for the liquid and heat gain for the metal were identical (which is lruewhen the process is adiabatic), then the following mathematical equality holds true:
Eheating of liquid = -Ecooling of metal
mliquid Cliquid (Tiinat - Tnit Giquid)) = - mmetal Cmetal (Tful - Tinit (metal)) = mmetal Cmetal (Tinit (metal) - Tritlrr
Given that the heat capacities are equal, they can be canceled from each side of the equality:
mliquid €uqoia (Trinat - Tinlt (liquid)) = mmetal €*.tut (Tinit (metal) - Ttnat)
mliquid (Trinat - 25) = rnmetal (50 - T1621) .'. 40 (T6i61 - 25) = 10 (50 - Ttnail + 4 (T1in21 - 25) = (50 - T6"u1)
4 Tfinal - 100 = 50 - T6tt61 ;. 5 T6n21 = 150 .'. T6r.,4 = 30
The best answer is choice B. This could have been solved intuitively by saying that if the heat capacities ::mthe liquid and solid are equal, then mixing equal mass quantities of the metal and liquid would lead tr otemperature exactly between the two starting temperatures (the average of 25 and 50 is 37.5). Because thmwas excess liquid, the final temperature would be closer to the initial liquid temperature than the inirnlflmetal temperature, which makes it less than 37.5"C. The temperature must increase somewhat from seliquid's initial temperature, so the final temperature is between 25"C and 37.5"C. The best answer is choice B
73. Choice D is correct. If the heat capacity of the liquid and metal are the same, then the final temperature30'C. Considering that the mass of the liquid is four times the mass of the solid, the difference in thetemperature (and thus temperature change) can be atiributed to different heat capacities. If the itemperature is greater than 30"C, then the heat capacity of the liquid is less than the heat capacity ofsolid. The math is shown below:
mliquid Cliquid (Trinat - Tmlt (llquld)) = mmetal Cmetal (Tinit (metal) - Ttnat)
Given that m1'n14 = 4 x m11s161, the following substitution and subsequent cancellation can be made:
4 xm5p1g Chquid (Tmd - Tmit gtquid)) = mmetal Cmetal (Ti''1t (metai) - Tmrat)
4 Ctquid (Ttnat - 25) = Cmetal (50 - T6"u1)
By cross-dividing (if there is such a mathematical term), we find that:
Timd - 25 = Cmetal ... n * Tflnal - 25 = Cmetal
50 - T66u1 4 x C1iqs14 50 - Tiitrul Ctiquid
If T66u1 = 30'C, 1',"r', Cmetal = 1
CIquidIf T1i121 > 30"C, thsn 9metal t 1
CliquidIf Tiinal < 30"C, thsn Cmet:- s
Cliqrn;
This means that any liquid that shows a final temperature greater than 30"C has a heat capacity less t},.roheat capacity for the solid. According to Table II, this is true only for Liquid IV. If this wasn't vou:thought, then your test-taking skills should have pulled you through. Only one choice is correct, so dliquids, you should pick the one with either the highest or lowest final temperature, never one in the mjThe lower the heat capacity, the more the temperature increases for a given quantity of heat, so the ;choice is the liquid with the highest final temperature, Liquid IV. The best answer is choice D.
Copyright O by The Berkeley Review@ t94 Section VIII Detaited Ex
74. Choice D is correct. Because Metal IV reached a lower temperature in Experiment I than did Metal III, theheat capacity of Metal IV must be greater than the heat capacity of Metai III. This means that when bothMetal III and Metal IV are at 50"C, Metal IV has more kinetic energy than Metal III. This means that whenMeial III is added to the liquids, it has less energy to transfer to thE"hquids, so that the final temperature ofeach liquid must be lower than it would have been had Metal IV been used. More kinetic energy iir the metalresults in more kinetic energy being transferred to the liquid and thus a higher final temperature for the liquid.This is_best explained in answer choice D. Choices e ana B are eliminatj, because the heat transferred fromMetal III is less than the heat transferred from Metal IV.
Choice D is correct. The greatest amount of transferable kinetic energy is found with the metal with thegreatest combination of temperature and heat capacity. Metal IV has Iire greatest heat capacity, based oninformation from Experiment I. of the choices piesenied, Metal IV also has" the greatest temperature. Thismakes Metal IV at 80'C the correct choice, so chooie answer choice D.
Choice A is correct. Using your test-taking skills, you see that the correct answer involves the sequentialordering of
119 liquids in Table II, according to their iit al te*peratures. Because equal amounts of heat were
applied to all four liquids and the amount of liquid was identical in each trial, the lower AT is due to a largerheat capacity' This_-means that the greatest heat capacity is associated with the liquid that showed thesmallest AT (Liquid iII). The correct choice is III > I > II > iV, ur,r*u, choice A. The answer choices shouldhave been narrowed down to either A or B based strictly on test-taking skills, using the sequence oftemperafures from Table 2. For optimum satisfaction, choose A.
Choice A is correct. If the outer wall of the calorimeter increase in temperature during the experiment, heatmust be lost during the process. Any heat that is lost to the environ*".,i i, not absorbeld by the liquid, so thefinal temperature of the liquid is less than is expected. The final temperature of the solid is equal io the finaltemperature of the liquid at equilibrium, so the final temperature ii lower than expected. this eliminateschoices C and D. Because the final temperature is too low, the change in temperature for the liquid is too|mall, and the change in temperature for the solid is too high. Because the calculation for the energy transferinvolves equating the two temperature changes, the inaccuiate temperature changes affect the heatLpacities.Because the temperature of the liquid does not increase as high as itihould, the loiver temperature change maybe mistaken for a larger heat capacity for the liquid. Becauie the energy (n; equats mCAT, a AT value that istoo small will be balanced out by a C that is too lirge. The best answer is choice A.
Choice D is correct. The ratio of the heat capacities can be determined by equating the energy changes for boththe liquid heating up and the solid cooling dbwn.
Eheating of liquid = -Ecooling of solidmliquid Chquia (Trinat - Tinit (liquid)) = - msolid Csolid (Trinat - Tinir (solid)) = msohd Csohd (Ti.rit (rota) - Ttnat)
mliquid Cliquid (Trinat - Tmit Giquid)) = msolid Csolid (Thit (solid) - Trinat)
40 C11qqi4 AThquid = 10 Cs6p4 ATsotid
...Chquld-lATsolidCsohd 4 ATli'rri4
The best answer is choice D.
75.
76.
77.
-8.
-9' Choice A is correct. A one-way valve allows fluid to flow in only one direction. Because the flow isasymmetric, the valve must be asymmetric. Choices C and D have symmetry, so they can be eliminated. Forfluid from left to right, the valve must seal on the left and remain open on"the righi. Choose A. The valvemust be viewed in three dimensions to appreciate its operation. If pt"irrrr" is apphJd (due to the flow of fluid)from right to left, the ball seals with the port and no fluid can now. f press.rre-is applied (due to the flow offluid) from left to right, the ball does not form a seal with the port, so fluia continues to flow. This means thatfluid flows in only one direction (left to right) through the valve.
,i:lvright @ by The Berkeley Review@ 195 Section VIII Detaited Explanations
80.
81.
82.
83.
84.
85.
86.
Choice C is correct. A cooling fan, by definition, is designed to cool by convection. In the one-piston steamengine, cooling is required to condense the steam into liquid water. Condensation takes place in the condenser.so the fan should be set up near the condenser (so the condenser can be cooled). The best answer is choice C. Acondenser would have many thin fins on its surface to increase the surface are and thus increase heat transfer br-way of convection.
Choice C is correct. For the piston to rise, the intemal pressure must exceed the external pressure. To build upthe internal pressure, steam is added to the core of the piston through the intake valve. Pressure builds up a-.
the moles of gas increase. This is because according to the ideal gas equation, when volume and temperature are
constant, pressure increases as moles increase. The temperafure is considered to be constant, because steam in thepresence of water is at 100"C. Steam is added to the core of the piston by allowing steam to flow into the piston.but not flow out of the piston. This means that the intake valve should be open and the exhaust valve shouldbe closed. The best answer is choice C.
Choice A is correct. When the counterweight is at its lowest point, the cycle of the piston is complete. The
counterweight is lifted again as steam is added to the core of the piston. It would be ideal if the piston werecompletely compressed; but because there will always be some gas present in the core of the piston, fuJ-
compression is not possible. This is best explained in choice A. In modern engines, the gas is evacuated from thechamber to help the piston descend with less resistance. Modern engines also employ multiple pistons that are
out of phase, rather than use a counterweight.
Choice D is correct. Because the liquid is converted into a gas in the boiler, heat must be added to the boi-ler t;induce evaporation. This eliminates choices A and C. The gas is then converted back into a liquid in thecondenser, so heat must be removed from the condenser. A well-designed automobile has a passenge:compartment heater that takes advantage of the heat released from the condenser to help heat the passengecompartment. To carry this out, a fan blows across the outside surface of the condenser, and the heat flows intcthe passenger compartment with the air flow. The correct answer for this question is thus choice D.
Choice D is correct. The engine is operated through the interconversion between liquid water and steam. Th:,iis known as vaporization, so the correct choice is answer D. Choice A is a good explanation for why lr-a:*would be used in a radiator.
Choice A is correct. An engine to converts heat into work, so an efficient engine is an engine that converts a-l-1 rthe heat into work. If all of the heat is converted into work, then no heat is emitted. A perfect engine is n;*possible. An ideal engine would run at ambient temperature, so that no heat is lost to the environment. Pr.[choice A if you know what is best for you.
Choice B is correct. A Camot engine is designed to convert heat into work. An engine takes heat in to give rffiwork energy overall. The diagram that represents an engine is the diagram that represents a work- releas:ryprocess (one with work equal to a negative number). The vertical steps in the diagrams (b to c, d to a, e to f. ardg to h) represent steps in which the volume does not change, so no work can be done in these steps (if A\i = I["
then -PAV = 0). The trick to this question is to pick one diagram and solve it. The correct answer is eiLlem
choice B or D, because "work flows out of the system" must be true in the statement for an engine. In the diagrm,in Figure 1, the step from a to b represents the expansion of the gas. When a gas expands, the proc#endothermic and work energy is released. This implies that work is a negative number, because work energ"r ri
given off by the system. The change in volume (AV) is positive for the process, so -PAV must be negative. Indiagram in Figure 1, the step from c to d represents the compression of the gas. When a gas compressesprocess is exothermic and work energy is absorbed. This implies that work is a positive number, because rtenergy is absorbed by the system. The change in volume (AV) is negative for the process/ so -PAV mus;:positive. The pressure is greater from a to b than it is from c to d. This means that the magnitude of the ufrom a to b is greater than the magnitude of work from c to d. The absolute value of the wu-6 > the absc:value of w"-4. The overall work is the sum of both work steps. Because the negative value has a la:magnitude than the positive value, the process must be negative overall. This makes the diagram in Figu:ean engine. This also makes choice B correct.
Copyright @ by The Berkeley Review@ 196 Section VIII Detailed
87. Choice C is correct. In a full cycle on the diagram for an ideal system (like the Carnot refrigerator in thediagram in Figure 2), the change in internal energy is zero. This eliminates choices A and B. For a refrigerator,work is added into the system, as heat is removed from the system. The "work in" must therefore be greaterthan the "work out" for the overall process to be heat absorbing. From the graph in Figure 2, going from f to ginvolves a positive change in volume, so a negative value for work (w = - PAV) is calculated. From point f topoint g, work energy is released, so the process is defined as "work out." From the graph, going from h to einvolves a negative change in volume, so a positive value for work (- = - PAV) is calculated. From point h topoint e, work energy is absorbed, so the process is defined as "work in." The pressure is greater moving from h toe than it is moving from f to g. The magnitude for the "work in" step (h to e) is therefore greater than themagnitude for the "work out" step (f to g). This makes choice C the best choice.
Choice D is correct. If work for the system is positive and AE (the change in internal energy) is zero, then q (theheat of the system) must be negative (given that AE = q + w). If q is negative for the system, heat was releasedfrom the system to the surroundings, meaning that q must be positive for the surroundings. The best answer istherefore choice D.
Choice B is correct. Because the shaft of the pump is becoming hot, it must be that air (and heat) is beingcompressed in the column of the shaft (assuming the heat is not due to friction). If the heat is greatest at thebottom of the shaft (where the pressure is greatest), then the heat is due to the exothermic nature ofcompressing gas, not friction. If the heat were due to friction, it would be uniform throughout the column of theshaft. Because the needle tip is cool, the gas must be expanding at that point. The compressed gas expands as itexits from the needle point. Because air is compressed in the shaft, choices C and D are eliminated. Becausegas expands as it leaves the needle tip, choice B is the best answer.
Choice B is correct. From point a to point c in the diagram in Figure 1., the volume of the system has increased,so choices C and D are invalid (and thus eliminated), because the gas has expanded. As a gas expands, heatmust be absorbed (expansion of a gas is endothermic). When the gas expands, work energy is given off from thesystem to the surroundings (when AV is positive, work (which equals - PAV) must be negative). This makeschoice B correct.
Choice D is correct. The purpose of an engine is to convert heat energy into work energy. The warmer theexhaust temperature, the more heat that is dissipated to the environment rather than being converted intouseful work energy. If heat energy is wasted, the engine is not being efficient. This implies that both choice Aand choice C are invalid. The more points at which energy is transferred, the more energy that is dissipated inthe way of friction. If energy is dissipated as friction, it is not being used efficiently. This means that manypoints of energy transfer would result in decreased engine efficiency. The best answer is therefore choice D.
Choice B is correct. From point b to c in the diagram in Figure 1, the volume does not change, so no work can be
done. Work is defined as -PAV, so a change in volume (AV) of zero means that PAV is zero and thus no work isdone. The best answer is therefore choice B. In choices A, C, and D, there is work being done, and that is notpossible at constant volume.
t3. Choice B is correct. The First Law of Therrnodynamics deals with the conservation of energy. Energy can be
neither created nor destroyed, and as such, energy is neither lost nor gained in any process. The Second Law ofThermodynamics states that in any spontaneous process there is an increase in the entropy of the universe. Themagnitude of the entropy change of the surroundings is always greater than or equal to the magnitude of theentropy change of the system. The Third Law states that the entropy of a perfect crystal at 0 kelvins is zero.Rutherford's experiment dealt with determining atomic structure (the location of sub-atomic particles and thenucleus), not the laws of thermodynamics, so choice D is eliminated. The Second Law of Thermodynamics,choice B, is the best choice.
88.
39.
90.
91.
t2.
- rpyright O by The Berkeley Review@ 197 Section VIII Detailed Explanations
94. Choice A is correct. There are a few methods to solve this problem, of which three are listed. It is suggested
that you adopt the method that is quickest for you. It is not a bad idea to learn multiple methods to reach a
soluiion for att of the questions you practice with. This helps to broaden your understanding of the concepts
being tested.
Method one requires taking a weighted mathematical average of the volumes and temperatures:
The final total volume of water is 128 mL. This means that25% t4s!l of the final solution originates from128mL
the solution that starts at 36'C. This also means that 75% t4gll of the final solution originates from the128mL
solution that starts at 64'C. The weighted average is calculated as follows:
0.25 (36"C) + 0.75 (64'C) = 9"C + 48"C = 57'0'C'
This makes choice A the best answer.
A second approach is the intuitive approach. Intuition is often aided by drawing a diagram, as you do in
physics. The sketch below shows the change in heat for the reaction and the math involved:
96 rnL at 64'C (higher Ti.,iri"t)
Difference in T is 28'C Difference in T is set as 4x
3x
4x=28'C :. x=7"C
Tfinal = 64'C - x = 64'C - 7"C = 57'C
or
Tfir,ul = 36'C + 3x = 36"C + 21"C = 57'C
32 mL at 36'C (lower Tir.,iriut)
This leads to a values of.57'C again, so choice A results from this method as well'
The last approach is to equate the heat lost by the warmer solution with the heat gained by the cooler
solution. If no heat is lost to the environment, then E"661int + Ehs6ling = 0'
Ecooling = -Eheating
mCAT"oolir1g = - mCAT6uu6r1t
Canceiing the C from both sides and plugging in for AT yields:
mAT6oolirr, = - mAThearin, + 96 g (64 -Tmail = - 32 8(36 - T6pf = 32 g(Tiinul - 36)
96 g $a- Tmat) = 32 g(Tmat - 36) + 3 (64 -Trinat) = Tfinal - 36
I92- 3 Tfinal = Tfinal - 36 "' 228 = 4 Tfinal "'
57 = Tfinal
WeIl what do you know? The correct answer is choice A yet again. Each method has its unique advantages, so
the best method is the one with which you feel most comfortable'
Copyright @ by The Berkeley Review@ l9a Section VIII Detaited Explanatior
95.
96.
97.
98.
99.
Choice C is correct. Knowing that w - -PAV, the work done on the system is calculated by substituting thevalues for P and AV. Because the answer choices are in units of liter.atmospheres, pressure must be convertedfrom torr into atm. This is done as follows:
P = 950 torr 1J-4tm-) = 1.25 atm'760 torr'The value for work is found by substituting w = -PAV as follows:
w = - (1.25 atm)(2 L) = -2.50liter.atmospheres
The best answer is choice C.
Choice B is correct. The only mathematical equation to know for the Carnot cycle is that the change in internalenergy (AE) equals the sum of the heat energy (q) and the work energy (w): AE - q + w (which can also bewritten as AE = q - PAV). All of the values you need are given in easy-to-use units, so plug-and-chug to yourheart's content. The mathematics should follow:
LE=22.2kI -24.4kI = -2.2kJ.Make sure to pick B to optimize your gratification quotient.
Choice D is correct. The equation you need is: AG = AH - TAS. No numerical values are given for AH, T, or AS,so AG cannot be assigned a definite numerical answer. Since AH and AS are both given as positive, then the freeenergy change (AG) is:
AG=(+#)-T(+#)A large value for T would make the value of AG negative, while a small value for T would make the value ofAG positive. The answer depends on T, so choose D. The solution to this question represents the reasoningbehind Table 8.1
Case ResultAS positive, AH negative Favorable at all temperaturesAS positive, AH positive Favorable at high temperaturesAS negative, AH negative Favorable at low temperaturesAS negative, AH positive Never favorable at any temperature
The term "spontaneous" may be used in lieu of "favorable". Spontaneous implies that the value for the freeenergy change (AG) is a negative value. The term spontaneous also implies that the reaction is favorable inthe forward direction.
Choice B is correct. A gas is the most random phase of the three common phases of matter (solid, Iiquid, andgas), and a solid is the most ordered of the three common phases. Therefore, in changing from a gas into a solid,the atoms are becoming more ordered. AS is a measure of change in randomness for the system, and since therandomness of the system is decreasing, the value of AS must be a negative value. Choice B is the correctanswer.
Choice B is correct. To expand, a gas requires the addition of heat to the system. This makes expansion of a gasan endothermic process and eliminates both choice A and choice C. Because the gas is becoming more random asit expands (it occupies a larger volume once expanded), the expansion process is entropically favorable. Thebest answer is choice B. When the system becomes more random, the change in entropy (AS) is positive.
AV = V6in61 - Vi.,itial = 2.4L - 0.4L = 2.0 L
-opyright @ by The Berkeley Review@ r99 Section VIII Detailed Explanations
100. Choice C is correct. All of the statements entail phase-change processes, so the following table is presented forquick reference:
Process (Phase Change) Enthalpy Change
Melting (Solid to Liquid) Endothermic: Small positive valueFreezing (Liquid to Solid) Exothermic: Small negative value
Evaporation (Liquid to Gas) Endothermic: Semi-large positive value
Condensation (Gas to Liquid) Exothermic: Semi-large negative value
Sublimation (Solid to Gas) Endothermic: Large positive value
Deposition (Gas to Solid) Exothermic: Large negative value
It requires more energy to convert a liquid to a gas (vaporization) than it does to convert a solid to a
(fusion), because all of the intermolecular forces must be broken. Heat is absorbed during vaporization (
endothermic process). This makes statement I a valid statement. Choices B and D are eliminated. Thismeans that statement II is invalid. \A/hile more heat is involved in a phase change between a solid and a
than a phase change between a liquid and a gas, the key word in statement II is "released." Energy isduring sublimation, not released. Statement II is invalid. Heat is released when a liquid freezes into a
because fueezing is an exothermic process and heat is released in exothermic processes. The release offrom the system to the surroundings warms the environment. This makes statement III valid, and makesbest answer choice C.
Copyright @ by The Berkeley Review@ 200 Section VIII Detailed
$ection IXKinetics
Beaction Katesa) Observed Reaction Rateb) Reaction Rate Experi:mentc) Reaction Orderd) Rate Constant and Rate Lawe) Reaction Order Experimentf) Typical Data and Graphsg) Half-life
Keaction Flechanismsa) Ceneral Plechanism Typesb) Energy Diagramsc) Catalysis
by Todd Bennett
(J
(A
500 600Wavelength (nm)
Ist order reaction
tPl
B
time time
Specia};ztn;g in MCAT Preparation
Kinetics Section GoalsBe able to intgrpret kinetic information from graphs.
aB
The sJudy of kinetics involves the use of graphs that show concentration versus time and graphsthat show rate versus time. You must recogniz-e typical graphs for zero-order, first-order, and s"ecdnd-order reactions. Many questions on the MCaf 6iinply-reciuire that you recogni ze a graphand read,information from it.
Know the significance of the rate-determining step.A11 kinetic data in a multi-step reaction are based on the rut*4"t"*itring step. th* ,ut"-d"tur*ilrlr,sstep of a reaction is the slow-est step in the overall reaction process. Th"e rale-determining step haithe greatest activation energy (trdnsition-state energy) of all the steps in the reaction"patliway.
Know how to determine the rate equation and rate law.The rate "quutioffie reactants thut uff".(all of the ieactants in the rate-determining step). The rate law is found by setting the rate of thereaction equal to the rate constant times the reactants in the rate-determinins step. For a first-orderreaction, the rate law is: rate = kIReactant]. The rate Iaw is determined bv"isolitine each reactantand observing how the rate changes when the concentration of that parti'cular reac"tant is altered.These questions require thai-you analyze rate data from a'table of different trials.
Understand the colrelation=Detr,yeen temperature, ac.tivation gnergy, and rate.'I'hereact.ionrateisdeterminedbytheactivationenergyotemperature system is increased, there is more energy av6ilable to overcome the act"ivatibn barrier.With this energy, the reaction can proceed more rap-idly, because more reactants can overcome theactivation barrier per unit time. The followin! gfaphs are ones that you must recognize:
hboli()
Irl
aC)
;JQC)
+itrd)
z
"v Be able to determine quickly the concerrtration at any time for first-order decav.
'v
Thes-e guestions require the use of the half{ife, which is defined as the period of time required forone-haff of a material to decompose (or react). For a first-order reaction, the half-life is iconstantvalue, no matter what the concentration of the reactant may be.
Know the effect of a catalyst on the reaction rate.A catalyst (or enzyme in biological reactions) forms a complex with the transition state of the rate-determinine step in a reaction.- The complex is more stable!. and thus of a lower enersv. ln esqence.determining step i1a reaction.*The comnplex is more stable, and thus of a lower
of a reaction bv lowerine the enerev 1e\,a catalyst lowers the activationrore stable, and thus of a lower energy. Lt essence,ion by lowering the energy level of the transition-a catalyst lowers the activation energy of a reaction by lowering the energy level of the transition-
state complex. In doing so, the rate of the reaction is decreased. A catalystis not consumed duringthe course of the reacti5n.
Transition state
Reaction coordinate
Activation energyrequired for reaction
Kinetic energy
the course of the reaction.
General Chemistry Chemical Kinetics Introduction
Kinetics, from a chemistry perspective, is the study of how fast a reactionproceeds and the conditions that affect the speed of the reaction. In its simplestform, chemical kinetics is the study of the rate at which a product is formed orthe rate at which reactants are consumed. In kinetic sfudies, the disappearance ofreactants is often monitored, in terms of their known spectral data and physicalproperties. Products are harder to monitor, given that there are no productspresent when a reaction commences. A problem that often arises is decidingwhich kind of spectroscopic signal to use to monitor product formation: visiblelight, ultraviolet light, or electromagnetic radiation of some other frequency.
As we saw in the equilibrium section, chemical reactions move simultaneously inthe forward and reverse directions. Chemical kinetics is concerned with theforward reaction rate, the reverse reaction rate, and the overall reaction rate.\Alhen monitoring a reaction, we observe the overall reaction rate. From theoverall reaction rate, we can ascertain information about the forward and reversereaction rates. It is also possible to monitor one component in the mixture byincorporating an isotopic label. By observing the rate of label incorporation,information about either the forward rate or the reverse rate can be obtained.The main reason for studying reaction rates is to be able to determine themechanism by which a reaction proceeds from reactants to products.
In organic chemistry, choosing between the various nucleophilic substitutionmechanisms is based on kinetic data. If the concentration of the nucleophileinfluences the rate of the reaction, then it is assumed that the rate determiningstep involves the nucleophile attacking the electrophile to form the transitionstate that evolves into product. This is referred to as an Sy2-reaction mechanismand is common in organic chemistry. If changing the concentration of thenucleophile shows no effect on the rate of the reaction, the reaction is said tofollow an SytrI-reaction mechanism. Both mechanisms shows a rate dependence onthe electrophile concentration. The number given in the descriptive name of themechanism refers to its reaction order.
In biochemistry, enzymatic behavior is monitored through the study of chemicalkinetics. Observing the reaction rate over time reveals features of the enzymeand whether or not it is saturated. A good foundation in chemical kinetics from ageneral chemistry perspective makes Michaelis-Menten kinetics (studied inbiochemistry) easier to understand. Conversely, if you have a strongunderstanding of Michaelis-Menten kinetics from a conceptual and mathematicalperspective, then reviewing chemical kinetics in general chemistry will be mucheasier. We will intertwine examples of chemical kinetics from a few fields, sothat we can get a more universal understanding of reaction rates and of theaspects of a chemical reaction that influence these rates.
In reviewing chemical kinetics, we will focus on experimental studies that tell usabout the concentration of a component in a reaction as a function of time. Bystudying changes in this concentration over time, we can deduce informationabout the order of the reaction. Knowing the reaction order, in furn, can give usinformation about the reaction mechanism. If the mechanism is valid, then areaction can be manipulated to control the rate of formation and the distributionof products. A catalyst influences the rate of a reaction, but it does not affect theproduct distribution.
Copyright @ by The Berkeley Review 203 Exclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction Rates
ffi-s- ffi1iii;1'1ffiui.i$illr1ii1.*11li"1it4r liii';,,,,;,",,,'',-.,,*,;,i1,r,
Observed Reaction Rate
The rate for any reaction can be measured as either the rate of disappearance of areactant or the rate of apPearance of a product. Given that the ratio of a reactantto a product is not always one to one, we need a way to account for stoichiometn,when studying reaction rates. As a general rule, Equation 9.1 d.escribes the ratlof reactant consumption relative to the rate of product formation for a systemwith no stoichiometric difference between products and reactants.
A[Products] _At
AIReactants](e.1)
At
Given that products form as reactants are consumed, there is a negative sign inthe equality shown in Equation 9.1, which is useful for determiniig the ra"te orformation of a product when the rate of consumption of a reactant is known.
Example 9.1Given that s2o32-(aq) in a 0.50-liter flask is consumed at the rate of 0.00g0 molesper second, what is the formation rate of SaO52-(aq)?
252O32-(aq) + I2(aq)
--> SaO5z-(aq) + 2I(aq)
A. 0.0080 MS
B. 0.0160 MS
c. 0.0320 a4s
D. 0.0160 -lM
SolutionThe product (sao62-) appears at half the rate at which the reactant (szo::-rldisappears, because the stoichiometric relationship between the two species is 1, :2. It is given that s2o32- disappears at a rate or o.oos moles/0.S liters per secoruiwhich equals 0.0160 molar per second. This means that S4O52- appears at a ra@of 0.0080 molar per second, making choice A a terrific and correct choice. ChoieD should have been eliminated immediately, because it has incorrect units.
Example 9.2o2(g) appears as a reaction product at the rate of 18 torr per second. \zvhat G furate of appearance for SO2(g) in the following reaction?
2 SOs(g) --+ 2SO2G) + Oz(g)A. +36 torr.s-1B. +18 torr.s-lC. +9 torr.s-1D. -18 torr.s-1
SolutionThe so2 product appears at twice the rate at which the 02 product appearsto the 2 : 1 stoichiometric relationship between the two products (soz andThis means that the so2 product appears at a rate of z i 18 torr.s-1 = 36 tor:This makes choice A the correct answer.
- !.I L?J \ '*-i'r ,- t bc /\'\
s
Copyright @ by The Berkeley Review 204 The Berkeley
General Chemistry Chemical Kinetics Reaction Rates
U'5lr
a')U
oU
'jJd*r
c)U
oU
The rate of a reaction is determined by the rate-determining step (the slowest stepin the reaction mechanism). This is because the slowest step in a reactiondetermines the observed rate for the overall reaction. The rate is dependent onlyupon the concentration of the reactants involved in the rate-determining step.However, we can observe the rate from any of the reactants or products/ eventhose that are not involved in the rate-determining step. In addition to knowingnumerical relationships between formation rates and consumption rates, it is alsoimportant to be able to interpret graphical representations of reactioncomponents. Drawn in Figure 9-1 below are two graphs, each depicting theconcentration of a species as a function of time for a generic, multi-step catalyzedreaction. The shape of the graph depends on the reaction order and the rate-determining step of the mechanism, so the graphs drawn are not universal.
TimeFigure 9-1
Because the curves in the graph on the left get flatter with time, the rate must bedecreasing as the reaction proceeds. Products are formed during the reaction, sothe product concentration starts at zeto, builds rapidly at first, and then slowlylevels off to a steady concentration (once the reaction reaches equilibrium).Reactants are consumed during the reaction, so the reactant concentration startshigh, drops rapidly at first, and then slowly levels off to a steady concentration(once the reaction reaches equilibrium).
Catalysts present in the beginning of the reaction are part of the transition stateduring the reaction, and then are regenerated at the end of a reaction. Thismeans a catalyst starts with some concentration, drops rapidly at first, stays closeto zero for most of the reaction, and then slowly increases back to its originalconcentration (once the reaction reaches equilibrium). Intermediates are notpresent in the beginning of the reaction. They are in their highest concentrationduring the reaction and then are in diminished quantity at the end of a reaction.This means an intermediate starts with zero concentration, increases rapidly atfirst, stays at a steady concentration for most of the reaction, and then slowlydecreases back to zero concentration (once the reaction reaches equilibrium).When the intermediate concentration remains constant, it is known as a steadystate. The graphs in Figure 9-1 reflect all of these conditions.
These graphs are also seen in biochemistry, where the reactant is referred to as asubstrate, the catalyst is an enzyme, and the intermediate is th.e enzyme-substratecomplex. Free enzyme is regenerated as the reaction reaches its endpoint and theenzyme-substrate complex eventually disappears. Other graphs frombiochemistry kinetics that you may recognize include the rate of a reaction as afunction of concentration and rate of reaction as a function of time in Figure 9-7.
Intermediate
Catalyst
Time--------+
Copyright @ by The Berkeley Review 205 Exclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction Rates
Experimental Study of Reaction RateRates are observed empirically in laboratory studies, so they are subject toenvironmental influences and human error. As such, we indicate the values withan error bar. One technique used in the lab to determine the rate of a reactioninvolves observing product formation or reactant consumption via ultraviolet-visible (UV-visible) spectroscopy. In cases where there is no uv-visible peak toobserve, rates can be obtained using a second technique. Aliquots can becollected at uniform intervals and analyzed using gas chromatography. Thisrequires removing some of the reaction mixture while the reaction is stillproceeding, but the amount is generally so small that it has a negligible effect onthe rate of the reaction. A bigger problem is that the reaction may continue toreact in the syringe or pipette after it has been removed from the original flask.
To avoid this problem, the sample of solution is quenched upon removal (toprevent further reaction). once the samples are quenched, they are analyzed bya quantitative technique, such as gas chromatography (GC) or nuclear magneticresonance (NMR). If the concentration is analyzed at uniform intervals, it ispossible to plot the data and fit a curve to the points. The graphs shown inFigure 9-2 represent data obtained from gas chromatography done on selectedaliquots. The peaks can be integrated to quantify the compounds in the mixture.
Retention time
Retention timeFigure 9-2
The first and last peaks (of the three) in Figure 9-2 represent reactants, becausethey decrease with time. The middle peak represents a product, because it grorr-:with time. The retention time is the time it takes the sample to travel through ihecolumn. It is not important in this particular example.
I'6
CJ
1x':U)
.9
I.:(n
E
{
I
;!=.
Retention
Copyright @ by The Berkeley Review 206 The Berkeley Revier [.'L
General Chemistry Chemical Kinetics Reaction Rates
UV-visible spectroscopy can be done in uitro, so timing and quenching are not aproblem. When carrying out a reaction in a cuvette, the reaction can be placedinto an UV-visible light spectrophotometer. It is best to pulse the sample withlight, rather than subject the sample to continual bombardment, because the lightfrom the detector beam adds energy to the system. An increase in the energyresults in an increase in reaction rate and consequently inaccurate data. Figure 9-3 shows absorbance as a function of wavelength at different time intervals duringa reaction, using a UV-visible spectrometer to collect data.
ProductIntermediate
Catalysta1l_l
Intermediatel^l
Wavelength(nm)-+
Figure 9-3
Wavelength(nm)+
Wavelength(nm)+
Wavelength(nm)+
Copyright @ by The Berkeley Review 207 Exclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction Rates
The change in height of each curve follow a pattem similar to the one in theintensity graphs in Figure 9-2. In UV-visible spectroscop/, peaks are analyzedfor height, rather than area. The peaks from the UV-visible spectrometer aremeasured in absorbance and can be mathematically converted in concentrationusing Beer's law, Equation 9.2 (also listed as Equation 1.3):
Absorbance = e[C]I, (e.2)
where e is the molar absorbtivity constant (also known as the extinctioncoefficient), [C] is the concentration of the species, and I is the length of pathwaythrough which light travels through the cuvette.
It is impossible to identify each peak in Figure 9-3 until t = 2, which is why the
peaks are labeled with question marks in the t = 1 entry. The biggest change inabsorbance is observed between t = 0 and t = 1, and the magnifude of the changein absorbance decreases during each subsequent interval. This means that thereaction rate was fastest during the initial period, but that it slowed down as thereaction proceeded. This is typical for reactions that are not zero-order. Lookingat absorbance as a function of time leads to the graphs seen in Figure 9-1.
The best peak to monitor is the peak that shows the greatest change in height(this reduces the effects of any errors). The apex of the peak should be monitoredrather than the area under the curve. In measuring the height, it is assumed thatthe peaks are all symmetrical. In UV-visible spectroscoPy, the wavelength at
which the apex of the peak occurs is referred to as )"*o* (the wavelength ofmaximum absorbance).
Example 9.3
Which of the following methods is the BEST way to study the rate of a chemicalreaction?
A. Taking samples at regular intervals and immediately placing them into a gas
chromatographerB. Taking samples at regular intervals, immediately quenching each one, and
then placing them into a gas chromatogtapherC. Carrying out the reaction in a cuvette exposed to photons that continually
irradiate the sample, and then analyzing the transmitted light for intensityD. Carrying out the reaction in a cuvette exposed to photons that periodically
irradiate the sample, and then analyzing the transmitted light for intensity
SolutionAs mentioned in the text, it is best to analyze the sample in aitro, therebyavoiding problems associated with transfer. This permits an accurate recording
of the time of reaction. Even with expeditious quenching of a sampling, the
timing is not perfect. This eliminates choices A and B. When analyzing a sample
by irridiation, cure should be taken not to energize the solution more than is
necessary. By continually irradiating the sample, energy is being added to the
system, so it does not behave as it normally would. Reaction rates are going tobe higher, due to the input of energy. From an energetics perspective, the ideal
scenario is periodically to irradiate the sample at the lowest possible frequency(at the l,*u* for the species with the lowest energy absorbance), so that minimalenergy is added to the system. This is not always possible, as is the case whenthe peak intensity is low. The best answer is choice D.
Copyright @ by The Berkeley Review 2o4 The Berkeley Revier
General Chemistry Chemical Kinetics Reaction Rates
Reaction OrderThe order of the reaction is determined by the number of reagents involved inthe rate-determining step. There are three reaction orders to consider: zero, first,and second. A zero-order reaction proceeds at a constant rate, independent ofreactant concentration. The rate law for a zero-order reaction is rate = k. Atypical case where zero-order kinetics is observed is during a catalyzed reaction,where the reaction proceeds at the turnover rate of the catalyst. Even with theaddition of more reactant, the reaction rate cannot change, because the catalysthas a finite number of reactive sites. A first-order reaction depends only on onereactant (like the 5111 reaction). The rate law for a first-order reaction is rate =ktR]. This means that the rate changes linearly as the concentration of thereactant R is changed. A second-order reaction depends on two reactan_ts (likethe 5512 reaction). The rate law for a second-order ieaction is rate = k[R12. thismeans that the rate changes exponentially as the concentration of the reactant Ris changed.
The reaction order with respect to a specific reactant can be masked, however.For instance, in a second-order reaction, if one of the rate-influencing reactants isin significantly higher concentration than the other rate-influencing reactant, thenreaction appears to be first-order, dependent on the reactant in lowestconcentration. By saturating a solution with all of the reactants except for onethat is involved in the rate-determining step, the reaction can become pseudo first-order. Something similar to this is seen in enzyme kinetics, when substrateconcentration is larger than enzyme concentration. The enzyme is safurated, sothe reaction obeys zero-order kinetics relative to the substrate. Additional"cubstrate does not increase the reaction rate. The rate remains constant with theaddition of substrate. The rate laws and half-life equations for three differentreaction orders are listed in Table 9.L below.
Zero-order First-order Second-order
Rate law: rate = k rate = k[A] rate = k[A]2
Half-life: tr = [A]o
22k '" -lrr2 -0.6932kk t12 k [A]o
Table 9.1
Example 9.4What is the reaction order for the following one-step reaction?
2 No2(g) -+ 2 NO(s) + Oz(g)
{.,...Zero-ord,erB,-rFirst-orderC. ,'second-orderd. Pseudo zero-order
SolutionBecause the reaction is a one-step reaction, the rate-determining step (and theonly step) involves two molecules of nitrogen dioxide colliding. The reactiondepends on two reactants, making it a second-order reaction. Choice C is thebest answer.
Copyright @ by The Berkeley Review 2o9 Exclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction Rates
Example 9.5The half-life is independent of the concentration of reactant for what order ofreaction?
A. Zero-orderB. First-orderC. Second-orderD. The half-life is independent of the concentration for all reactions.
SolutionReferring to Table 9.L, we see that only with a first-order reaction is the half-lifeconstant. This means that choice B is correct. It is important to be able to seenew information and process it in a simple manner. Each reaction order has itsunique features. For zero-order reactions, the rate is constant, so it takes lesstime for each subsequent half-life, because the quantity for the half-life decreasesover time. For first-order reactions, the half-life is constant, which is why youoften hear the phrase "half-life for a first-order decay process." Second-orderreactions are known for their rapid reaction rate at first, but a drastic decrease inrate after just a short time.
Example 9.6Given the following table equating reaction time with concentration of reactant,what is the order of reaction?
lAol Time
1.000 M 0 sec
0.500 M L5 sec
0.250 M 45 sec
0.125 M 105 sec
A. Zero-orderB. First-orderC. Second-orderD. Third-order
! .. 1 ltt, f,. "-tfL C.I
/z)ka)-_'ll
.>
SolutionThe first half-life, from 1.000 M to 0.500 M, takes 15 seconds. The second half-Life,from 0.500 M to 0.250 M, takes 30 seconds. The third half-life, from 0.250 M to0.125 M, takes 60 seconds. In this case, as the concentration of reactant decreases.the value of the half-life is increasing. This means that the concentration isinversely proportional to the half-life. This is true for a second-order reaction, so
choice C is the best answer.
fn terms of mechanisms, reactions can be only first-order or second-order. Othsobserved orders are phenomena associated with catalysts and relativeconcentrations. First-order reactions entail one molecule breaking apart, and arereferred to as dissociatiae reactions. A dissociative reaction involvesdecomposition in its rate-determining step. Second-order reactions entail hromolecules colliding, and are referred to as associatiae reactions. An associatirereaction involves the combining of two molecules (via collision) in its rate-determining step.
Copyright @ by The Berkeley Review 2to The Berkeley Revier
General Chemistry Chemical Kinetics Reaction Rates
Rate Constant and Rate Law
The rate constant is the term by which the concentrations of reactants in the rate-determining steps are multiplied to get the rate law. The rate law is derived bydetermining the reactants that affect the reaction rate, and multiplying this valueby the rate constant. Rate constants and rate laws are empirical values, obtainedby observing the initial rate of a reaction under several different conditions. Rafeconstants haae aarying and odd units! The rate of a reaction is affected bytemperature (the reaction rate increases with increasing temperature), activationenergy (the reaction rate decreases with increasing activation energy), catalysts,solvent (solvents affect the transition state stability), collision frequency, collisionorientation, and the concentration of the reactants in the rate-determining step.The rate constant must account for all of these factors, except concentration of thereactants. Equation 9.3 is used to determine the rate constant.
O _ O "-E""I/RT
(e.3)
A is the Arrhenius constant; it takes into account collision orientation andfrequency. Not all collisions between reactants result in reactions. The activationenergy (Eu") is the energy required to get through the transition state. R is theenergy constant given in units of energy per mole.temperature. It is either 8.314joules per mole.kelvins or 1,.987 calories per mole.kelvins.
Example 9.7
If the rate law for the following reirction is found to be: rate = k fiCl2l [HCCI3],what are the units for rate constant (k)?
Clz(g) + CHCI3(g) --+ HCI(g) + CCla(g)
A. N{ry'
1
/M's1,
ru/'.t
SolutionThe rate of the reaction has units of molar per second. The following equationcan be used to solve for units:
Rate = k/lcltlHCClal units: M = ky'MM)=k{ff
Solving for the units of the rate constant yields: k = -$- = +- .
{ffi.' /M's
It may seem counterintuitive for the rate of a reaction to depend on the squareroot of a concentration, but because the rate-determining step may be the secondor third step of a multiple-step reaction, the rate law for the reaction may involvethe concentration of intermediates. If the equation using the intermediates ismanipulated in such a way that reactant concentrations replace intermediateconcentrations, there are sometimes square root factors involved. The point hereis that the square root is merely an artifact of the mathematical calculations, andit shouldn't upset you. The best answer is choice C.
S
B.@S
C.
D.
Copyright O by The Berkeley Review 2tl Exclusive MCAT Preparation
General Chemistry Chemieal Kinetics Reaction Rates
Experimental Study of Reaction OrderReaction order can be determined experimentally in one of two ways. In oneexperiment, the concentration of a reactant is monitored over time, and the decayrate is matched with reaction order data. This works well for simple cases. Zero-order reactions occur at a constant rate, so a uniform decay trend in reactants canbe observed within a few data points. First-order reactions occur at a graduallydecreasing rate, so a gradually diminishing decay trend in reactants can beobserved within a few data points. second-order reactions occur at a rapidlydecreasing rate, so a rapidly diminishing decay trend in reactants can beobserved within a few data points. However, to determine the reaction ordermore precisely, the initial rate is observed as initial concentrations of reactantsare systematically varied. The rate law for a reaction is most often obtained inthis way. If the reaction rate remains constant when a reactant concentration isvaried, then reaction rate does not depend on that reactant. Consider the data inTable9.2 for a hypothetical reaction between one mole of A and one mole of B.
Trial Initial Rate IAI IB1
I 3.07 x 10-3 M.s1 0.10 M 0.10 Mil 6.11 x 10-3 M.s-i 0.20 M 0.10 MilI 1.23 x 10-2 M.s-1 0.20 M 0.20 M
Table 9.2
To evaluate the effect of a reactant on the reaction rate, we start with the genericrate law: rate = k [A]a[B]b. Upon comparing Trial I with rrial II, we
"get the
following results:
Raterl -klAlnalBhP = 6.11 x 10-3M/s _k[0.20]a[0.10]b
Rate 1 k tAhaFlP 3.07 x ro-3 rvr/s k [0.10]a[0.10]bThe ratio of the rates is approximately 2 (6.11 divided by 3.07is roughly equal to2), k cancels out in the numerator and denominator, and [0.t0]o cancels out in thenumerator and denominator. This leaves the following relationship:
2 =lo.zolu = Za ... a = 1
[0.10]a
We see that the rate doubles when [A] is doubled, so the rate depends on [A]1.This means that the rate law can be modified to: rate = k [A][B]b. Uponcomparing Trial II with Trial III, we see that the rate doubles when [B] isdoubled, so the rate depends on [B]1. The rate law for the reaction is thus: rate =ktAltB]. This means that the reaction is first-order with respect to A, first-orderwith respect to B, and second-order overall. This technique is applicable inorganic chemistry, as for example when determining the mechanism for a
nucleophilic substitution reaction. The data in Table 9.3 describes the kinetic."observed for a generic nucleophilic substitution reaction.
Trial Initial Rate lNucleophilel lElectrophilel
I 7.82 x 10-4 M.s-l 0.10 M 0.10 MII 1.17 x 10-3 M.s-1 0.10 M 0.15 Mil 1.54 x 10-3 M.s-1 0.20 M 0.20 M
Copyright @ by The Berkeley Review 212
Table 9.3
The Berkeley Revier
General Chemistry Chemical Kinetics Reaction Ratcs
The numbers may not seem as simple as they were in the previous example, butthe procedure is the same. Upon comparing Trial I with Trial II, we see that therate increases by a factor of 1.5 when [Electrophile] increases by a factor of 1.5, sothe rate depends on [Electrophile]1. This ii expected, because the rate of allnucleophilic substitution reactions depends on the electrophile concentration,whether the reaction proceeds by an Sy1-mechanism or an Sp2-mechanism.Upon comparing Trial I with Trial III, we see that the rate doubles when both[Electrophile] and [Nucleophile] are doubled. We know that there exists arelationship between reaction rate and [Electrophile] that is directly proportional,so the doubling of the reaction rate when comparing Trial I with Trial III can beattributed to the doubling of [Electrophile]. The reaction rate only doubles, so itdoes not depend on [Nucleophile]. The rate law is: rate = klElectrophile], so thereaction proceeds by an Sp1-mechanism.
For Examples 9.8 through9.12, please refer to Table 9.4 below.
Trial Initial Rate IX] IY]
I 3.57 x 10-4 M.s-1 0.10 M 0.10 MII 1.43 x 10-3 M.s-1 0.10 M 0.20 Mn 3.59 x 10-4 M.s-1 0.20 M 0.10 M
ry L.44 x 10-3 M.s1 0.20 M 0.20 M
Table 9.4
Example 9.8\Mrat is the reaction order with respect to Reactant X?
A. Zero-orderB. First-orderC. Second-orderD. Third-order
SolutionThe order with respect to Reactant X is found by observing the reaction rate as
[X] changes. This is done by comparing Trial I with Trial III (or Trial II with Trialry). Because the reaction rate doesn't change when [X] doubled, it shows nodependence on the concentration of X. The order with respect to Reactant X isthe power to which the concentration is raised, which in this case is zero. Withrespect to Reactant X, the reaction is zero-order. The best answer is choice A.
Example 9.9Which of the following changes would MOST increase the reaction rate?
A. Doubling [X] and decreasing the temperature by 10'CB. Doubling [X] and increasing the temperature by 10'CC. Doubling [Y] and decreasing the temperature by 10'CD. Doubling [Y] and inoeasing the temperature by 10"C
SolutionThe rate does not depend on Reactant X. This eliminates choices A and B.Increasing the temperature provides more energy to the system, so that moremolecules can overcome the activation barrier. This means that the reaction rateincreases as temperature increases. The best answer is choice D.
t .,, 'a
7'
4/.2'
'('-7"
Copyright @ by The Berkeley Review 2t3 Exclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction Rates
Example 9.L0What is the overall reaction order?
A. Zero-orderB. First-orderC. Second-orderD. Third-order
SolutionThe reaction rate does not depend on Reactant X, given that changes in [X] do notchange the reaction rate. The order with respect to Reactant Y is found bycomparing Trial I with Trial II (or Trial III with Trlal IV). The rate quadrupleswhen [Y] is doubled, so the rate depends on [Y]2. The rate of ttre reuciiondepends on two moles of Y, so the reaction is second-order with respect to Y.The overall reaction depends on two molecules of Y and no moles of X, sooverall, the reaction is second-order. The best answer is choice C.
The reaction is second-order, which means that the rate-determining stepinvolves two molecules of Y coming together. This same question can be askedin a more convoluted manner, by asking about the mechanism and thecomponents in the rate-determining step. Example 9.11 is seemingly differentfrom Example 9.10, although both questions are asking about the order of thereaction and the rate law.
Example 9.11Which of the following statements accurately describes the reaction mechanism?
A. The reaction mechanism is associative, and the rate-determining stepinvolves one mole of Y.
B. The reaction mechanisminvolves two moles of Y.
C. The reaction mechanisminvolves one mole of Y.
D. The reaction mechanisminvolves two moles of Y.
SolutionThe order with respect to Reactant Y is found by comparing Trial I with Tria-l U(or Trial III with Trial IV). The rate quadruples when [Y] is doubled, so the raredepends on [Y]2. The rate law for the reaction is thus: rate = k[Y]2. This mear$that the rate-determining step involves two moles of Y, and the reaction rs
second-order. Second-order reactions are associative, so the best answer is choiceB. Although mechanisms can never be proven, from the data you camhypothesize that the reaction proceeds by a multi-step mechanism. For examp,ire"if the reaction is X + Y --> P, then the mechanism must have more than one stm-If the reaction is two-step, then the mechanism is probably:
Firststep: Y + Y -+ Y2
rndstep: Y2 + X -+ Y + P
rate-determining step (slowest step)
is associative, and the rate-determining step
is dissociative, and the rate-determining step
is dissociative, and the rate-determining ster
Copyright @ by The Berkeley Review 214 The Berkeley
General Chemistry Chemical Kinetics Reaction Rates
A.
Example 9.12If the reaction of X with Y is monitored by visible spectroscopy, then which of thefollowing graphs is expected for the absorbance of light by Reactant X as afunction of time? The graphs are labeled with t1, t2, and t3, which represent theelapsed time for the reaction, where t3 > t2 > t1.
SolutionWhether we monitor Reactant X or Reactant Y, the reaction obeys second-orderkinetics. Just because reactant X is not in the rate-determining step does notmean it is unreactive. It is consumed at the same rate as Reactant Y, if the twospecies react in a one-to-one stoichiometric fashion. For a second-order reaction,consumption of reactants (and product formation) is most rapid initially, becauseas the reaction proceeds, the reactant concentration diminishes, causing the rateto decrease. The greatest change in concentration should be observed betweentimes t1 and t2. This eliminates choice A (which describes zero-order behavior)and choice B, which is erratic behavior. The change between times t2 and t3should be substantially less than the change between times t1 and t2, so choice Dis more indicative of second-order kinetics. Choice C describes first-orderkinetics.
Graphs contain a great deal of information, and your ability to extractr.rLformation from graphs can be of great benefit on the MCAT. Chemical kineticsentails several graphs, from which trends in reactivity can be extrapolated.Example 9.12 shows this. The information in the answer choices for Example9.72 can be rewritten in a more traditional form of concentration as a function oftime. This results in a graph typically seen in chemical kinetics.
OJU
-oLU)
,a
U
G-oLqs
l
I(.)(J
sHoa
-o
U
€-8
Wavelength (nm)--->Wavelength (nm) ------>
Wavelength (nm)+ Wavelength (nm) -->
Copyright @by The Berkeley Review 215 Exclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction Rates
Typical Rate Data and Graphs
Typical graphs in chemical kinetics involve reactant concentration as a functionof time during the course of a chemical reaction. The rate data for zero-order,first-order, and second-order reactions have distinctive characteristics, becausethe concentration of reactant changes in a unique way over the course of eachkind of reaction, and the reactant concentration affects the rate differently foreach reaction order.
Zero-order decay is recognized when the concentration of the reactants decreasesin uniform increments against uniform time increments. Drawn in Figure 9-4below are typical data in a typical graphic display for a zero-order reaction:
CJ
to.5dl.{
dOJU
oU
Concentration1.00 M0.75 M0.50 M0.25 M0.00 M
Figure 9-4
In Figure 9-4, the data in the table match the graph on the left. The concentrationdecreases by the same amount, 0.25 M, during each time interval. This meansthat the reaction rate is constant over time, as seen in the graph on the right.
First-order decay is recognized when the concentration of the reactants decreases
in a logarithmic fashion against uniform time increments. It may be easier torecognize a first-order reaction by the constant half-life. Drawn in Figure 9-5
below are typical data in a typical graphic display for a first-order reaction:
q,)
&'t6tr
0)
EU
Time01
2J4
Concentration1.00 M0.52 M0.28 M0.16 M0.10 M
Figure 9-5
In Figure 9-5, the data in the table match the graph on the left. The concentrationdecreases by a gradually smaller amount, 0.48 M, then 0.24 M, then 0.12 M,during each time interval. When the time decreases in uniform increments, the
concentration changes by half as much each time. This means that the reactionrate is decreasing with time, as seen in the graph on the right.
Time+ Time---+
Time-+ Time+
Copyright @ by The Berkeley Review 216 The Berkeley Review
General Chemistry Chemical Kinetics Reaction Rates
Second-order decay is recognized by the drastic drop in concentration initiallyfollowed by smaller decreases in concentration againsi uniform time increments.when the time decreases in uniform increments, the concentration changes bysubstantially decreasing amounts each time. It may be easier to recognize asecond-order reaction by an increasing half-life. Drawn in Figure 9-6 below aretypical data in a typical graphic dispray for a second-order r"uJtion'
/,
.5
!q.)
U
oU
Time01
2J4
Concentration1.00 M0.35 M0.25 M0.18 M0.14 M
Figure 9-6
ln Figure 9-6, the data in the table match the graph on the left. The concentrationdecreases significantly in the first intervat, fut tne decrease is substantiallysmaller during each subsequent interval, 0.65 M, then 0.10 M, then 0.07 M. whenthe time decreases in uniform increments, the concentration changes by a muchsmaller amount much each time. This means that the reaction ratJ is dlcreasingexponentially with time, as seen in the graph on the right.A kinetics graph from-biochemistry that should look familiar involves velocitylthe rate of a catalyzed reaction) as a function of substrate concentration for anenzymatic process. Because there is a finite amount oI enzyme,and each enzymehas a finite number of active sites, there is a finite number of reactants that maybe reacting at any given time. At high reactant concentrations, the reaction rateappears to be constant (it is equal to a value that depends on concentration ofenzyme/ and the turnover rate of the enzyme.) At lowlr reactant concentrations,the rate depends on the reactant, becausethe enzyme is not saturated. Figure 9-Zshows a typical graph from Michaelis-Menten enzvme kinetics:
UUc,)
ISubstrate
Figure 9-7
Time_-+ Time_>
Exhibits zero-order kinetics athigh substrate concentration.
'lhe reaction reaches Vrr.,u* whenenzyme is safurated with substrate.
Copyright O by The Berkeley Review 217 Exclusive MCAT Preparation
General ChemistrY Chemical Kinetics Reaction Rates
Half-LifeHalf-life is defined as the period of time it takes for the concentration of a
reactant to decrease to a value that is one-half of its original value' This means
that after one half-life, only fifty percent of the initial value remains' Graphs of
concentration as a function of time demonstrate the concept of half-life' The
graphs in Figures 9-8,9-9, and 9-L0 show half-life dependence on concentration.
Zero-order decay involves a constant rate of decay, so the time required to carry
out the first hali-life (from L00% to 50%) is double that of the time required to
carry out the second half-life (from 50% to 25%). As seen in Table 9'1, the half-
life dlminlshes with concentration. Figure 9-8 presents data that show the
relationship ofhalf-tife to reactant concentration for a zero-order reaction:
25%
boi
(g
qJtr(6
tra)
d
bD
L(6(.,]
Uc.)bo(d
OJUti0)Fi
bo
(!
0)li(!ti0)
bo
ts 50%(c
ooG)bD ttrol-(6 Lr tt
0.)Ulra)
tt=4*z 4.0
+, -) -*LI'L
; 6.0t'- 1*i z.o
F--q+Lt-.v; 7.s
t,- j -*; 2.O
t--) *; 4.0
+,-) 4; 6.0+.-t-:>r l- At 8.0
Concentration1.0000 M
0.s000 M
0.2500 M
0.1250 M
0.0625 M
Time0
Figure 9-8
First-order decay involves a logarithmically decreasing rate of decay, but the-
time required to carry out the firit naff-tlfe (irom 100% to 50%) is equal to that of
the time required to carry out the second half-life (from 50% to 25%). As seen in
Example 9.i, the half-liie is constant. Figure 9-9 presents data that show the
retationship of half-tife to reactant concentration for a first-order reaction:
Half-life decreases as theconcentration decreases.
Time
Concentration1.0000 M
0.5000 M
0.2500 M
0.1250 M
0.0625 M
Time0
Half-life remains constant as
the concentration decreases"
25%
Copyright Oby The BerkeleY Review 2la
Figure 9-9
The BerkeleY
General Chemistry Chemical Kinetics Reaction Rates
second-order decay involves a rapidly decreasing rate of decay, so the timerequired to carry out the first half-life (irom 100% to SoU,; is half tirat of the time1e9u1red_
t9 garry out the second half-life (from 50% to 2s6/.). As seen in Example9.6, the half{ife increases as the reactant concentration decreases. Figure 9-t0presents data that show the relationship of half-life to reactant concentration for asecond-order reaction:
50%
bod
(6
fi
(!Irq)
(d!
bo
k6U)
CJ
AJUkc,)
F.
tr= '52
tr=1,
tt=22
tt=42
Concentration1.0000 M
0.5000 M
0.2500 M
0.1250 M
0.0625 M
Half-life increases as theconcentration decreases.
Time
C1 = (os-kt (e.4)
$'here C1 is the concentration at time t, Co is the initial concentration, t is time,ajld k is the rate constant. Equation 9.4 canbe manipulated to derive the half-tifefor a first-order reaction. The half{ife is calculated using Equation 9.5.
Figure 9-10
It is.important that you be able to interpret data from both a graph and a table. Auseful skill on the MCAT is the ability io recognize trends q.J.dy when scanningnumerical data' It is also important to recognize the effect of concentration onthe half-life of a reaction. This information cin be used to determine the reactionorder, support a proposed reaction mechanism, or eliminate a possible reactionmechanism.
Because half-life is constant Jor first-order processes, most halflife questions onthe MCAT involve decay (a first-order reaction). Equation 9.4 is used to calculatethe concentration of a species that undergoes first-order decay at any time:
C1 = fog-kt
!t = "-kt and at t= trl,,Ct = 1 ... 1 -
"-ktry, * 2 -ekfrt,co ''' C., 2 z
Taking the natural log of both sides yields: ln2 =kty, :. ty, =
Equation 9.5 relates the rate constant, k, and the half{ife for a first-order decayprocess.
(e.s)
ln 2 = 0.693kk
tt1, = 0.693 t
Copyright @ by The Berkeley Review 2t9 Exclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction Rates
Most half-life questions deal with first-order processes. Rather than using theequations to calculate values, it is often easier to cut a concentration in halfseveral times, until the period of time is completed. For instance, after three half-lives, the amount of a material decreases from 100% to 50"h, from 50% to 25"/o,
and finally from 25oh to 725%. This technique is far easier than doing longcalculations. Examples 9.13 and 9.14 demonstrate how to cut the concentration inhalf sequentially.
Example 9.13The concentration of some floral scent is 100 ppm. If the half-life for itsdecomposition is 4 minutes, how much time elapses before there is only L pPm inthe air?
A. 2o.o minutes 1 oc' -) t; '' '. t d ' ; i
B. 26.6 minutesC. 42.2 minutesD. 53.2 minutes
SolutionA quick method to arrive at the solution involves calculating successive half-lives. The decrease taken one-half life at a time is:
100 -+ 50 -r 25 ;-{2
12L -+ { -+2 'r 4 ,:
$- -r 19 -+ less than 1.08 1-6 -.)
After seven half-lives (seven arrows), the concentration reaches a value that is
less than 1 pp*, so the elapsed time is just less than seven half-lives. Because the
concentration after six half-lives is greater than 1 ppm, the total time is more than
six-half lives. Six half-lives is 24 minutes and seven half-lives is 28 minutes, so
the time is more than}4but less than 28 minutes. The best answer is choice B.
Example 9.14If a drug decomposes according to first-order \inetics, what is its concentraticnafter one hour, if the half-life is 21 minutes? ' .,. ''
A. z6.\%ofitsoriginalconcentration i ilt'" i''' I i'"'- -
B. 24.0oh of its original concentrationC. 13.3% ofits original concentration
'' ' 'i : 'r: :
D, 11.7%of its originalconcentration I o..:, ;, ,.. ;,,1,.-;
-1 ! l" ' !
Lr
Solution 7 t2 'q /^Because the decay obeys first-order kinetics, the half-life is constant. Within oru
hour, given a half-life of 2L minutes, just under three half-lives will transpileThe deirease taken one halflife at a time is: 100% to 50"/o, then 50% to 25%, then
25% to a value slightly greater than 12.5%. The best answer is choice C.
Copyright @by The Berkeley Review 220 The Berkeley
General Chemistry Chemical Kinetics Reaction Mechanisms
il. ffin ffiffi#ffiGeneral Mechanism TypesA reaction mechanism is the stepwise pathway that a reaction is believed to take,and it is proposed based on energetic data, kinetic data, and structuraldifferences between reactants and products (bonds and chirality in particular).Mechanisms account for the steps in the reactant-to-product transformation.There are one-step reactions that are said to have concerted mechanisms. In manycases/ however, the reaction mechanism involves more than one step. Todetermine the mechanism for a multi-step reaction, we start by determining thecomponents in the rate-determining step. If we know the components that affectthe rate, we know the components in the slowest step of the reaction.
If the reaction is first-order, then we know that the slowest step involves onemolecule breaking apart. If the reaction is second-order, then we kt ow that theslowest step involves two molecules coming together to form a bond. Whenconsidering kinetic data to support a mechanism, the rate information appliesonly to the rate-determining step of the reaction. Figure 9-11 shows gtuphi totrate as a function of reactant concentration for some simple cases:
A--+ B + COne-step reaction
Dissociative mechanism
rate = k [A]
(First-order kinetics)
A+B ->COne-step reaction
Associative mechanism
rate = k tAltBl(Second-order kinetics)
2A+B+COne-step reaction
Associative mechanism
rate = k [A]2
(Second-order kinetics)
EZ
EJ
tAl
tAl
Figure 9-11
The first reaction in Figure 9-11 involves the dissociation of the reactant in itsrate-determining step (its one and only step). The second reaction in Figure 9-11involves the association of two different reactants in its rate-determining step (itsone and only step). The reaction is first-order with respect to each reactant, butsecond-order overall. The third reaction in Figure 9-11 involves dimerizationthe association of two identical reactants) in its rate-determining step. The
reaction is second-order with respect to the reactant, which results in the rate as a:unction of reactant concentration being exponential.
Copyright @ by The Berkeley Review 221 Exclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction llechanisms
Not all reactions are concerted. Reactants can form intermediates (or activatedcomplexes), which then go on to form products. There may be individual steps
in the mechanism that are independent events in the overall Process.Mechanisms can be disproven or supported, but nevel proven. For this reason,
mechanisms are best thought of as educated guesses at the reaction pathwav.
Figure 9-l-2 shows the graph for rate as a function of reactant concentration for a
multi-step reaction;
A+B -L-ck_1
C+D -9+ E+BEaz
bohqJ
CJ
c,)CJ*rrli
Eactl )Eact2 .'.
Step 1 is the rds
Rate = k1 [A] [B]A + D -* E (overallreaction)
Two-step reaction
Rate varies with rds
OJ
(!!
U(dOJt/
tAl
nl :lE- EI-Jl lltcl tDl
Figure 9-12
The reaction in Figure 9-12 involves the association of two reactants in both
mechanistic steps. Either step could potentially be rate-determining. The step
with the higher activation energy is the slower, and thus rate-determining, step.
It is important that you understand the graphs to the extent that you can
interprei data from them. Free energy diagrams are a common way to describe
the energetics of the reaction mechanism. The term "teaction coordinate" refers
to the stepwise mechanism. The energetics are used as the evidence to propose
that the first step in the reaction is rate-determining.
Kinetic data are the basis for most mechanistic proposals. For instance, consider
Reaction 9.1. We believe Reaction 9.1 is a one-step reaction, because the rate-
determining step (and only step) involves the two nitrogen dioxide reactant-'
colliding to form a bond. Because the reaction shows second-order kinetics, a
concertJd, bimolecular mechanism is proposed as the best explanation of such
behavior. The rate constant for the reaction is k1, implying that it is the forward
rate constant of the first step. The subscript describes the step and the direction'
For instance, if we were given ak-3, we could assume that it was the rate
constant for the back-reaction of the third step.
2 No2(g) Nzo+(e)
Reaction 9.1
According to the rate law, the reaction rate is equal to k1[NO2]2 for Reaction 9'X'
This can Le supported by observing the rate of the reaction as a function of
reactant concentration. This was seen in Example 9.L2. Examples 9.15 and 9'16
are more specific cases where kinetic data can be used to support a proposed
mechanism.
Reaction coordinate
Copyright O by The Berkeley Review 222 The Berkeley Revier
General Chemistry Chemical Kinetics Reaction Mechanisms
Example 9.15The mechanism for the reaction that follows is believed to be two consecutivebimolecular reactions.
Reaction: NOz(g) + CO(g) -----+
Mechanism: NO2(g) + NO2(g) *NOs(g) + CO(g)
->\Alhat is the rate law for this reaction, assuming thatcorrect?
A. k(PNerXP6s)B. k(PNsr)2(Pco)ir.k1n516r;2
5. klnNrer;
NO(g) + Co2(g)-'--'! 'i '1."":
NO(g) + NO3(g) (slow) - \NOz(g) + CO2(g) (fast
this reaction mechanism is , ... ! '
SolutionThe rate depends on the first step, because it is the slowest step in the reactionmechanism (rate-determining step). The first step of the mechanism involvestwo No2 molecules colliding together to form No3 and No. The rate dependson the reactants in the rate-determining step. The rate therefore depends otr t*omolecules of No2 gas. odd as it may seem, Co (carbon monoxide) has no effecton the rate of this reaction. The addition of CO does not increase the rate of thereaction according to the mechanism provided. This makes choice C correct.observations about the rate of a reaction can be used to disprove mechanismsfrom an analytical standpoint. If one were to find that increasing the partialpressure of CO in the reaction mixture resulted in an increase in the rate of thereaction, then the mechanism as proposed would be invalid. Mechanisms shouldbe able to predict the reactivity patterns of reactions before the reaction begins.
Example 9.16For the reaction in Example 9.15, what would the expected rate law be had thereaction had occurred in one step?
A. k(Px1er)(Pco)B. k(PN1sr)2(Pqs)
C. k(Prvoz)2
D. k(P51gr)
SolutionHad the reaction taken place in one step, it would have depended on the tworeactants colliding with one another in the transition state. This means that therate law of the reaction would be k(PIrJ9r)(Pgg), choice A. To confirm this rate1aw (and mechanism), it would be expected that increases in either of the tworeactants would result in a linear increase in the rate of the reaction. Again, thisobservation could either support or disprove the proposed one-step reactionmechanism. on your exam, you may have to make some predictions about therate of a reaction based on its proposed mechanism. The simple rule is that if thereactant is in the rate-determining step, then the rate of the reaction dependsdirectly on the concentration of that reactant.
Copyright @ by The Berkeley Review 223 Exclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction Mechanisms
Energy Diagrams and Activation Energy
Activation energy is the threshold energy required for a reaction to transpire. Itis the energy difference between the starting materials and the transition state
(apex of the energy diagram). The higher the energy of the transition state (also
referred to as the actiaated complex), the greater the activation energy, and thusthe more slowly the reaction rate takes place. The highest activation energy of a
multi-step reaction represents the rate-determining step in the overall reaction.There are two popular energy diagrams that you may recall. Figure 9-13 is the
traditional graph (often found in organic chemistry textbooks) that relates the
energy of the system to the pathway of the reaction. Energy must be absorbed bythe system in order for it to ascend the energy diagram. The lower the positionoccupied by a species on an energy diagram, the more stable it is.
The greater the Eu., (activation
energy), the slower the reaction
rate, becaus" L - 4u-Eact'/RT.
Figure 9-13
Figure 9-1,4is a bar graph of sorts, where the number of molecules in the system
is plotted against their energy. The area under the curve represents the sum oithe molecules in the system. Figure 9-14 shows that at any given time, not aImolecules have the same energy within the system. These two popular graphumethods to depict activation energy are presented below:
-Tr
Threashold energy requiredfor reaction to take place
Tz>TrA greater precentage ofmolecules are beyond thethreshold energy at Ttthan at T1.
Kinetic energy Eact
Figure 9-L4
AtT2, the average energy of molecules is greater than at T1. Consequentlr-" dmore molecules have enough energy to overcome the activation barrier fsr !reaction. This means that more molecules react at T2 than T1, resulting in a
reaction rate at higher temperatures.
T2o0)
Uood
otsrOJ
-oE
z
Reaction coordinate
Copyright @by The Berkeley Review 224 The BerkeleY
General Chemistry Chemical Kinetics Reaction Mechanisms
Example 9.17Using the graph below, find the value of Ex61.
slope =&s't-
-10
-11
3.00; 10-3 3.25 * 10-3 3.50 * 10-3 fcrA. 1.6 x 105 J/moleB. 1.6 x 105 J/moleC. 7.6 x104 J/moleD. 1.6 x 103 J/mole
SolutionBy taking natural log of the equation: k = A
"-Eactlp1, we get the following:
lnk = -Eact a 1'''4R
It looks like the equation of a line (y = rn* + b, where m is the slope, and b is they-intercept), so if we plot ln k as a function of T1, then the slope depends onactivation energy and the y-intercept is the Arrhenius constant. By multiplyingthe slope of the line by - R, we get the activation energy. The slope of the line is:
slone= -11-(-6) - -5 - -5 --z --zx\}a' 3.2s x 10-3 - 3.00 x t0-3 0.25 x L0-3 2.5 x 10a 104
Multiplying the slope by -1 and & yields:
Eact = - (2xrca1x R = 2x 104 x 8.3 = 1.6 x 105, choice B.
Example 9.18\ trhich of the following does NOT always affect the rate of a reaction?
A. Changing the temperatureB. Adding a catalystC. Lrcreasing the volume by adding solventD. Adding a reactant
SolutionThe equation for rate is rate = k lreactants]rds, where rds means rate-determiningstep. If the reactant being added is not involved in the rate-determining step,then it does not influence the rate. This is observed in dissociative reactions (likethe Sp1-reaction), where the reaction rate is determined by the dissociation ofjust one molecule. The best answer is choice D. Changing the temperaturechanges the rate directly. This eliminates choice A. Adding a catalyst lowers theactivation energy and thus increases the reaction rate. This eliminates choice B.
Increasing the volume reduces the concentrations of all reagents, including thereactants in the rate-determining step. This eliminates choice C. Increasingvolume of a gas phase reaction also results in a decreased reaction rate.
)1 -8d*_9
Copyright @ by The Berkeley Review 225 Dxclusive MCAT Preparation
General Chemistry Chemical Kinetics Reaction Mechanisms
Effect of Catalyst
Catalysts are added to a reaction to lower the activation energy (energy of thetransition state) and consequently to increase the rate of reaction. Catalysts reactwith the reactants to form an activated complex of lower energy than the normaltransition state for the uncatalyzed reaction. Biological catalysts are referred toas enzymes. An enzyme lowers the rate of a biological reaction by forming anenzyme-substrate complex. Enzymes are a little more specific than catalysts inthat they also help to align the molecules into the correct orientation for reacting.Enzlzmes cause products to have a specific chirality. Catalysts are not consumedin the reaction. Catalysts do not make a reaction more or less favorable, becausethey increase the rate of the reverse reaction and the rate of the forward reactionequally.
Example 9.19\Atrhich of the following statements is NOT true about catalysts?
A. Catalysts do not react in the reaction pathway.B. Catalysts are not consumed in the reaction pathway.C. Catalysts change the reaction pathway.D. Catalysts allow the reaction to react more quickly.
SolutionThe question asks for a "NOT true" statement. This means that you are searchingfor the false statement in the answer choices. Catalysts do react in the reactisrpathway; otherwise, they would not lower the activation energy of a reaction- ffthis seems strange, think about any acid-catalyzed mechanism from organicchemistry. The acid protonates a reactant to increase its reactivity. A goodexample is the protonation of a carbonyl to make the OH group. Uporprotonation, the carbonyl becomes more electrophilic. Once the reaction fucomplete, the proton comes off of the carbonyl. This makes statement A a falmstatement. To be sure about choosirg A, remember that catalysts are regeneraat the end of a reaction, so they are not consumed in the reaction. Choice B itherefore true. Catalysts do change the reaction pathway (to one ofenergy), so choice C is true. Catalysts lower the activation energy soreactions may proceed more quickly. This makes choice D true. The onlychoice is choice A.
Example 9.20In which of the following values is the concentration of a homogeneousNOT directly involved?
A. The equilibrium constant (K)B. The rate constant (k)C. The reaction rateD. Both the equilibrium constant (K) and the rate constant (k).
SolutionA catalyst (homogeneous or heterogeneous) does not affect the thermodof a reaction. By definition, catalysts change only the reaction rate for areaction. A catalyst is therefore not involved in the equilibrium constdThis makes choice A correct. A catalyst is definitely involved in thebecause it is involved in the rate constant.
Copyright O by The Berkeley Review 226 The Berkeley
KineticsPassages
1O Passages
80 Questions
Suggested Kinetics Passage Schedule:I: After reading this section and attending lecture: Passages I, III, IV,VII, & X
Grade passages immediately after completion and log your mistakes.
II: Following Task I: Passages II, V & VI, (25 questions in JO minutes)Time yourself accurately, grade your answers, and review mistakes.
III: Review: Passages VIII, IX & Questions 75 - 8OFocus on reviewing the concepts. Do not worry about timing.
I. Reaction Kate from Visible Spectroscopy
II. Michaelis-Menten
III. Reaction Rate from Graph
ry. Catalyst Turnover Rate
V. flalf-life Calculations and First-Order Decay
VI. Nucleophilic Substitution Rate
VII. Cobalt-Exchange Mechanism
VIII. Oxygen-Transfer Mechanism
IX. Chelating Effect
X. Proposed Mechanism
Questions not Based on a Descriptive Passage
Kinetics Scoring Scale
Raw Score MCAT Score
70-80 l5 t5
53-69 10 t2
37 52 7 -9
26-56 4-6t-25 I -5
(L -7)
(B - 14)
(15 - 2l)
(22 - 28)
(2e - 35)
(36 - 44)
(45 - 5r)
(52 - 58)
(5e , 65)
(66 - 72)
(73 - 80)
Passage I (Questions 1 - 7)
Reaction 1 represents a generic reaction between tworeactants to form two moles of a single product.
A+B -* 2C
The reaction is monitored over time using a spectrometerwith a wavelength range from 200 nm to 800 nm. The threespectra in Figure I represent the reaction initially, thereaction after one minute, and the reaction after two minutes.
te
-o!q.o
550
Wavelength (nm)
550
Wavelength (nm)
400 550 700
Wavelength (nm)
Figure 1
The absorbance for each peak can be converted intosolute concentration using Equation 1:
Abs = elCll
Equation 1
The term Abs refers to the absorbance of light, e is themolar absorbtivity, [C] is the concentration in solution, and Iis the path length ofthe cell. The peak appearingjust below700 nm is found to belong to the product, C.
*1 fo monitor the reaction at constant wavelength, it
a would be BEST to lock the setting at:
A. 470 nmB. 545 nmC. 625 nm'D.l 695 n-
Copyright O by The Berkeley Review@
ta)
Cd
k
.o
2 , The peak at l"*u* = 684 is BEST described as:
.r'A . a product.-.8 . a reactant.
{. acatalyst.,t'. unintermediate.
3. Which of the following statements is NOT trueregarding absorbance?
,K. As the length of a cuvette increases, the absorbance/- ,///tncreases.
t 6. fne concentration is directly proportional to' absorbance.
.-Clfyte absorbance always increases when the<-L- temperatureincreases.rt . Molar absorbtivity constants are specific for each
individual compound.
4. The rate of reaction547 nm would be:
.1 ,f
O.AAbsAt
.. B) -AAbtt. at
AAbs
' At x lClinit
O. _ AAbs
At x lClinit
found by the peak atmonitoring_"1
^ the rate of reaction?/t1\ '/ A. Products appear at a constant rate throughout the
reaction.,/X Products appear at a gradually decreasing rate
tbroughout the reaction.
(e) ProOucts appear at a gradually increasin g ratethroughout the reaction.
D. Reactants disappear at a constant rate throughoutthe reaction.
Which of the following statements is TRUE regarding
What CANNOT be determined from the experiment?
4 fn"rate ofdisappearance ofA.{. therate of appearance of CC. The concentration of B after two minutes-f .-tne order of the reaction with respect to A
7 . Which assumption is NOT true? a.tlI ^ -' t
r(' .lf .. \ ;'
6.
'A[A]_ IA[C] \'ir-rN'7 C ^P f "-''\
C q |,'i'?r''',g
lL.-1
Tb
/ ,{. _2 AIAI = AlCl -r' 7 F
V/ Ar Ar
/i--iorcr +1 at 2Lt F1 4 r' ,.^\:')/ I..,/,-;i\ AtBl - Atcl
I v./Ll--.' At At
;- 1 *.r-*ot "\ir
r"'-f, r\^11y)e^'\
II
229 GO ON TO THE NEXT PAGE
Passage ll (Questions 8 - 14)
The kinetics of enzymatic processes obeys the rules ofMichaelis-Menten kinetics. Michaelis-Menten kinetics forenzymatic processes is based on enzymatic reactions being
carried out in two steps. The first step is the rapid formation
of the enzyme-substrate complex. The second step is the
slow step, in which the enzyme is regenerated and the product
is formed. A general reaction mechanism is shown in Figure
1.
E+S =k1- ES k2- E+Pk-1
Figure 1
where E is the enzyme, S is the substrate, ES is the enzyme-
substrate complex, and P is the product.
The first step is a rapid equilibrium, so the [ES] remains
constant, unless the substrate concentration diminishes to alevel where it is comparable to the enzyme concentration. Atthis point, the rate becomes dependent on the substrate.
Figure 2 illustrates this behavior:
Figure 2
The reaction obeys first order kinetics until the enzyme
is saturated and the enzyme-substrate concentration remains
constant. The rate of the reaction obeys Equation 1, where
the k2 is the rate constant for the rate-determining step, and
the tESl is found using the steady state approximation.
Rare = k2[ES] [ES] =tEltsl
(\.::-\z) . rsr
Equation L
The relationships between the rate data can be found
through simple algebraic manipulation.
8 . The enzyme-substrate complex can be classified as
which of the following?
ffi Aproduct
X d,reactant
-A.--A transition state
41. A" intermediate
odg
5ocB()
[Substrate] +
Copyright O by The Berkeley Review@ 23o
13.
GO ON TO THE NEXT P
Ti-".*
9 , Which is NOT true about the role of an enzyme?
-* ; k helps to align the reactants correctly.-61 ttlowers the energy of the transition state.tfl nincreases the reaction rate.
G:1, makes the reaction more exothermic.
10. When the catalyst is saturated, what is the reaction order
with respect to substrate?
cJ'[]]Zero-orrder-'.B. First-order
C. Second-order
'D. Third-orrder
1 1. Given the following equation for the reaction rate, wh,m
is true when the reaction rate (v) is equal to one-half cd
the maximum reaction rate (vtnu^)?
V.a"IS]"-kM*tsiA . The reaction is first-order; [S] = kv.B. The reaction is first-order; Vmax = kM'
C . The reaction is zero-order; [S] = ky.D . The reaction is zero-order; Vmax = kM'
Which graph BEST shows
time?[ES] and [E] as a function uff'
,{;
12.
tdoU
tESl----- fEl
-
The rate for an enzymatically catalyzed process
obeys the laws of Michaelis-Menten kinetics is:
-A;-'rate = kl lEnzyme]f . rate = k2 [Enzyme]C. rate = kl [Enzyme'substrate complex]'D.]}ate
= k2 lEnzyme'substrate complex]
Time+
i-"*
1-4f. Which graph accurately shows substrate concentrationas a function of time for an enzymatic process that
, B.-ia)
!
U)
II
(,)jj!v6.o?
Time_> Time _->
Time--+ Ti-"-
,cpyright @ by The Berkeley Review@ 231 GO ON TO THE NEXT PAGE
Passage lll (Questions 15 - 21)
lf the rate constant, k, is known for a standard reaction ata series of temperatures, the Arrhenius constant, A, andactivation energy,8461, con be determined using Equation l:
lnk= Eaa!+lnA
Equation 1
The Arrhenius constant takes into account collisionfrequency and the orientation of the molecules. For instance,in an Sy2-reaction, not all collisions result in a reaction.Only the collisions oriented with the nucleophile aligned toattack the electrophile from the backside of the leaving groupresult in a reaction. Although the rate of a reaction canchange with varying reactant concentrations, the rate constantfor the reaction is constant throughout the reaction.
Reaction 1 is a one-step chemical reaction that wascarried out in CCl4 solvent (an aprotic non-polar solvent) todetermine the relationship between temperature and the rateconstant (k). The gases were bubbled into two separatesolutions of carbon tetrachloride to a saturation point. Thetwo solutions were then mixed, and the reaction wasobserved.
SO:(g) + NO(g) -+ SOz(g) + NOz(e)
Reaction L
Table I shows the results of six trials carried out atvarious temperatures on the Celsius scale.
T ('C) r (x-r)T
k log k
0 3.66 x 10-3 l.8l x 1O-l -6.10
25 3.36 x 10-r 3.46 x 10-5 -4.46
3-5 3.25 x 10-3 1.35 x 10-a -3.87
45 3.14 x l0-3 4.98 x 10-a -3.30
55 3.05 x 10-r 1.50 x 10-3 -2.82
65 2.96 x 10-3 4.87 x l0-3 -2.3t
Table L
The values in the chart contain slight errors, due tomechanical error within the UV-visible spectrophotometerused to measure the reactant and product concentrations. Thevalues in this chart show a correlation between the reactiontemperature and the rate constant for the reaction.
15. As temperature increases, which of the lollowing areobserved?
,. A. k increases: reaction rate increases.' B. k increasesl reaction rate decreases.'C. k d""r"uses; reaction rate increases.
P . kdecreases; reaction rate decreases.
Which of the following represents log k as a function of1rTA.
/
17. Which of the following rate equations appliesReaction 1?
..4. rate = k[SOf]
-ry. rare = klNol(_Q,/ rate = klSOrllNOl
76, rate = k[SO:]2
18. When the solvent for the reaction is changed, the rate
,n varies in a predictable manner. Which of the following
/ q$ables is MOST affected by the change in solvent?-
/ l, / nctivation energy'lf.
Temperature
C . Reactant concentration
D. The R value
19. When plotting ln k as a function of 1, the slope of the,/) T
A. Eu.t
B. - Eact
g. Eact
-.R\_7D, '- Eact\'-J' R
Copyright @ by The Berkeley Review@
1
T
1
T
IT
1
T
Yt\\n
Why is the initial rate observed in reaction studies?
A. As a reaction proceeds, it slows. The same periodin the reaction must be compared to be consistent.
As a reaction proceeds, its rate increases. The sameperiod in the reaction must be compared to be
consistent.
C . Only the initial rate can be accurately measured.
D. Any rate can be measured, but the initial one ismost convenient.
21, To measure the rate of a reaction, which of the
lollowing CANNOT be observed?
4 fnerate of appearance of product
{fhefute of disappearance of reactant
G2fn" change in intermediate concentration over time
_p-: The change in reactant and product concentration ara function of time
232 GO ON TO THE NEXT P
!
Passage lV (Questions 22 - 28)
Heterogeneous catalysis differs from homogeneouscatalysis in the phase of the catalyst. As implied by the term"heterogeneous," the catalyst in heterogeneous catalysis isnot evenly distributed through the solution. A commonexample involves hydrogenation of an alkene using platinummetal as a catalyst. The catalysis takes place only on thesurface of the platinum, and not throughout the solution.This means that the rate of reaction varies with the surfacearea of the catalyst, not with its concentration.
In homogeneous catalysis, the catalyst is evenlydistributed throughout the solution, and the catalysis takesplace evenly throughout the solution. A common exampleof homogeneous catalysis is the addition of hydronium ion tosolution to assist in the acid catalysis of ketal formation froma ketone. The hydronium ion is evenly distributed insolution. In homogeneous catalysis, the rate of the reactionvaries with the concentration of the catalyst.
To determine the strength of a hydrogenation catalyst,the Reaction 1 was employed using four different catalysts.The turnover rate was observed and recorded for each of thecatalysts. The turnover rate is defined as the rate of thereaction divided by the number of active catalytic sites. Theunits for turnover rate is reactions per second at the catalyticsite.
CH2CH3 H3C\ /CH2CH?
H> 6-C<HHrC-\ /
C:C/\Hsc
H2-----'+cat
/\H:C
Reaction 1
Table I below lists the catalyst versus the turnover rate:ssociated with each of the four catalysts:
Catalyst Turnover rate (rxn)
(Me3P)3RhCl 8000
Pd/PdO2 26,500
Pt/PtO2 35,000
(Me5C5)2IrCO 6500
Table 1
The experimental values are obtained from reaction rate;iudies using UV-visible spectroscopy. The reaction rate is:it'ided by the surface area of each catalyst, which is best,rproximated by the shape of the material on which the-rtalysts is plated. In industrial processes, the catalysts are,lten applied to the surface of an inert material, such as
:raphite. This is done to maximize the surface area of the:ateria{ while keeping it a material large enough to filter:.rt of solution easily at the end of the experiment.
-opyright O by The Berkeley Review@
22. Which of the following graphs BEST represents the
Time+
llPl
E.
1tPl
2 3. Which catalyst canconcentration to obtain
,./. lvtqvyrRhcl#.'Yanao2c. Pr/PrO2 tp. (Me5C5)2IrCO
in the LOWESTbe useda set rate?
,t
i^lt{
24. For a catalyst in an industrial system, the number ofactive sites affects the turnover rate. What can beconcluded about the maximum turnover rate of onecatalytic site, if the catalyst contains three catalytic sitesbut the catalyst is NOT saturated with reactant?
A. The maximum turnover rate of one catalytic site isless than one-third of the observed turnover rate,assuming the three catalytic sites react equally.
,.B. The maximum turnover rate of one catalytic site isgreater than one-third of the observed turnover rate,
_*-. assuming the three catalytic sites react equally.
t.' 9,r'Tne maximum turnover rate of one catalytic site isexactly one-third of the observed turnover rate,assuming the three catalytic sites react equally.
The turnover rate of one catalytic site is exactlyone-third of the observed turnover rate, assumingthe three catalytic sites do NOT react equally.
Time
-->Time+
233 GO ON TO THE NEXT PAGE
25. Which of the following is NOT an example ofheterogeneous catalysis?
X. tJt*)zCO + H2 with Pd/PdO2 -+ @3C)2CHOH
fi,rhzCCOCl + C6H6 with AlC13 -+ C6H5COCH3l-,-/,./ ".1. uzCCH2+H2 with Pt/PtO2 -+ H3CCH3
D. CHa + CO with (Me3P)3RhCl -r H3CCHO
2 6. What is always TRUE about a catalyst?
d. A catalyst is always consumed in a reaction.
4. A catalyst is always produced in a reaction.
(9,' A catalyst is always regenerated during a reaction.
nry. A catalyst is always altered in a reaction.
27 . If a catalytic site has a binding affinity for the product,
what will be observed over time?
X Sotnthe turnover rate and the overall reaction rate
lncrease.
Both the turnover rate and the overall reaction rate
decrease.
C. The turnover rate increases, while the overallreaction rate decreases.
D. The turnover rate decreases, while the overallreaction rate increases.
2 8. Why is a heterogeneous catalyst applied to the surface
of an inert solid for industrial reactions?,.<)
/ A/ To maximize the surface areal_-,'- B. To increase the concentration
C . To make the catalyst more buoyant
. D. To make the catalyst less soluble
/4.l-/
Copyright @ by The Berkeley Review@ 234 GO ON TO THE NEXT
Passage V (Questions 29 - 35)
First-order decay processes follow a predictable pattern ofdecreasing concentration over time. The concentration at any
given time can be calculated using Equation 1, if the initialconcentration, the time, and the rate constant for decay are allknown. Equation I applies only to a first-order decay
process.
Ct = cie-kt
Equation 1
where C1 is the concentration at time t, Ci is the initialconcentration, k is the rate constant, and t is time.
The halflife for a first-order decay process is constant
over the course of the decay. If the half-life is known, then
the rate constant for a first-order decay process can be found'
as shown in Figure I below:
Ct - Cie-kt ... 9! = "-k,Ci
att= tr/., Ct = I'"C1 2
- ^-ktrr-
- -ktlz
kttlz= -lnl = ln22
t. - ln2 -0.693tlz ttlz
Figure 1
The half-life can also be determined from the
constant using the same relationship. In summary, the
constant and the half-life for a first-order decay process
inversely proportional.
29. Which of the following graphs does NOT depictorder decay behavior?
A. B.
L2
ln L2
$LNTime+
L-I
Time
-l>
A
I
AtRl
At
C.
tAtPl
At
Time+
Time -+
D.
tFo
EkOr
3 0. If k is doubled, then the half-life of a first-order decay:
A . increases by a factor of e2.
B. increases by a factor of2.C. decreases by a factorof 2.
D. decreases by a factor ofe2.
31. What is TRUE about the value of half-life for a
reaction?
A. The value of the half-life is constant in first-orderreaction, while it decreases with time in a zero-orderreaction.
B. The value of the half-life is constant in first-orderreaction, while it increases with time in a zero-orderreaction.
C. The value of the half-life is constant in both zero-and first-order reactions.
D. The value of the half-life decreases with time inboth zero- and first-order reactions.
3 2. What are the unitsreaction?
A. sec
g. Isec
c. [M]'sec
D. tMlsec
of the rate constant for a first-order
3 3. What period of time is required to reach the point in a
first-order decay where only 20Vo of the original reactantremains?
A. t=ln5xkB. t=lnlx k
5
c. t= kln5
D. t= ln5k
-opyright @ by The Berkeley Review@ 235 GO ON TO THE NEXT PAGE
34. What must be TRUE about a reaction that fits the graphbelow?
[Reactant] ----------t>
The reaction is second-order at high [Reactant] and
first-order at low [Reactant].
The reaction is first-order at high [Reactant] and
second-order at low lReactant].
The reaction is zero-order at high [Reactant] and
first-order at low [Reactant].
The reaction is first-order at high [Reactant] and
zero-order at low [Reactant].
3 5. Which of these reactions has a constant half-life?
A. One step: X + X -+ X2
B. One step: X + Y -+ X-YC. Two steps; X + Y -+ Z + Q-+ W + YD. One step: X -t A + B
tg*
A.
B.
C.
D.
Passage Vl (Questions 36 - 44)
In both organic and inorganic chemistry, one of the mostcommon mechanistic questions is, "Does the leaving groupleave first, or does the nucleophile attack first?" In organicchemistry, the two reaction mechanisms are referred to asSyl and Sy2. In inorganic chemistry, the two reactionmechanisms are referred to as associative (5y12-like) anddissociative (SNl-like). A researcher ran the same
nucleophilic substitution reaction four times at the sametemperature, varying the concentration of reactants with eachtrial. Table 1 below lists the data for the four trials:
lCHsSNal lCHsCHzIl Rate (M/sec)
0.10 M 0.15 M 2.IO x 0-)0.10 M 0.25 M 3.52 x g-s
0.20 M 0.20 M 5.58 x 0-)0.15 M 0.30 M 6.33 x 0-r
Table 1
CH3SNa + CH3CH2I -+ CH3CH2SCH3 + NaI
Reaction L
The data in Table I were collected by monitoring theformation of thio ether in Reaction 1 by UV spectroscopy.The rate of the reaction is assumed to equal the rate offormation of the thio ether. There were no major sideproducts observed during the reaction. The data were repeatedin subsequent trials of the experiment. The reproducibility ofthe data helps to substantiate the conclusions of theexperiment.
3 6. If methanol were used in place of sodium methylsulfide, then the reaction would show which of thefollowing changes under identical conditions?
A . The reaction rate would increase.
B. The reaction rate would stay the same.
C . The reaction rate would decrease.
D . The reaction rate would exactly double.
3 7. What would the rate be, if the reaction were run withthe following concentrations?
[CH3SNa] = 0.25 M [CH3CH2I] = 0.10 M
A. 2.10 x 10-5 t*l/,*
B. 3.s2 * 16-s M1/""
C. s.s8 * 1s-s Ml/ec
D. 6.33 * 1s-s tMl{ec
3 8. What is the reaction order of Reaction 1?
A. Zero-orderB. First-orderC. Second-order
D. Third-order
Copyright O by The Berkeley Review@ 236 GO ON TO THE NEXT P
3 9. What would the apparent reaction order be with the
following concentrations?
[CH3SNa] = 1.25 M [CH3CH2I] = 0.05 M
A. Zero-order
B. First-orderC. Second-order
D. Third-order
4 0. The following reaction is said to proceed b1' a
dissociative mechanism. If the concentration clCo(en)2NHjCl2+ were to double, and the concentratic,m
of NH3 to triple, the new reaction rate would Le
changed in what way?
Co(en)2NH3Cl2+ +NH3 -+ Co(en)2(NHZ)23+ +Ct'
A. It would be twice as large.B. It would be three times as large.C . It would be four times as large.D. It would be six times as large.
41. Which of the following does NOT increase the rate Can associative concerted reaction?
A . Increasing the temperatureB. Adding solventC . Adding one of the reactants
D. Using a less viscous solvent
42. What is the rate constant for the ethyl iodr'iesubstitution reaction listed in Table 1?
A. l.4o * 1s-: Ml/ec
B. 3.ls * 1s-r MJ/r".
1.40 x 1o_3 r4rl.r""
3.15 x to-3 t/trl.r""
4 3. The reaction order of the dissociative mechanism is:
A. zero-order.
B. first-order.C. second-order.
D. third-order.
4 4. Which of the following statements is INVALIDrespect to the rate of an Spl reaction?
A. The reaction rate varies with elecconcentration.
B. The reaction rate varies with nucleoconcentration.
C . The reaction rate varies with temperature.D. The reaction rate varies with solvent.
C.
D.
Passage Vll (Questions 45 - 51)
Mechanisms for chemical reactions are either supported
or disproven by data. For a mechanism to be accepted by the
scientific community, it must be capable of predictingreactivity and of holding form, even under varying reactionconditions. Two common procedures for evaluating the
validity of a mechanism are isolating an intermediate in the
reaction and studying the effect on the reaction rate when
varying the concentration of components. The rate depends
on only the rate-determining step, so any altering of the
concentrations of species involved in the rate-determiningstep results in a change in the reaction rate. Conversely, ifthe species you alter is not in the rate-determining step of the
mechanism, then the rate is not affected. Reaction I belowhas been analyzed by kinetics:
)+ - a+ -
[(NHr)sCoCl]'- + OH--+[(NH.r):CoOH]'' + Cl
Reaction 1
Figure I lists the currently accepted three-stepmechanism for Reaction 1:
Step I (fast):
[(NH3)5CoCl]2* + OHk1---!l tNH, )ac o (N H2 ) c I I
+ + H2 o
Step II (slow):
[(NH3)aCo( NHz)Cl l+ 5 I (NH:)+CoNH212* + Cl-
Step III (fast):)+ kr )+
[(NH:)+CoNHz]- + H2O-+[(NHr)sCoOH]-
Figure 1
The rate data were gathered by observing the formation ofintermediates and products over time. The concentration ofintermediates and products is found by monitoring the
absorbance of various species using UV-visible spectroscopy.
The peak for the intermediate grew initially and then remained
constant until the end of the reaction, when it decreased tozero. The peaks for products also grow over time.
-l 5. What is the rate-determining step in the proposedmechanism?
,{_, Step Ip;' Step II
.{. Step IIID. Both Step I and Step II are rate-determining.
--l'll '.'-. " '-'lt' *" t;:; "^i''' r'i
46. Addition of water to the reactibn would have whateffect?
A . It would increase the reaction rate.
--fi ft would have no effect on the reaction rate.
C . It would decrease the reaction rate.
D . It would make the reaction third-order.
Copyright @ by The Berkeley Review@ 237 GO ON TO THE NEXT PAGE
4 7. Which of the following kinetic relationships is predictedfor Reaction 1 using steady-state approximation?
A. kr tt$H3)5CoC112+11Ou-1- (k-r + kz) [[(NH:)+Co(NH2)Cl]+l [HzO]
B. kr ttNH:)sCoC112+1;OH-1
= (k-r + kz) [[(NH:)+Co(NH2)Cl]+l
C. kr tt(NH:)sCoC112+11oH-l= k-r [[NH:)aCo(NH2)Cl]+l [Hzo]+ kz [[NH:)aCo(NH2)C1]+l
D. kr ttNH:)sCoCq2+1;oH-l= k-r [[NHl)aCo(NH2)Cl]+l
. . ,\ kz [[NH:)+Co(NHz)Cl]+l [Hzo]
iI,1\\{:i' '\.:-"''
-{.t+ fir-r :..,.,\J' -\ or
4 8. Which of the following statements is TRUE about the
relative acidity of the cobalt compound?
1 A.r'pKa NH3 > pKa [(NH3)5cocl]2+1.'. Bli'pKuNH3 = pKa [NH:)scocl]2+1.
C. pKu NH: < pKa [$H3)5CoC1]2+1.D. There is no comparison between the two.
4 9. Which of the following statements is TRUE with regard
to the rate of reaction for the reaction presented in the
._passage?
t' The rate iincreases as both th.e pHth.e fiHiand temperature
lncrease.
& tne rate decreases as both the pH and temperature
increase.
-C. The rate increases as the pH increases, while the
rate decreases as the temperature increases.
D. The rate decreases as the pH increases, while the
rate increases as the temperature increases.
5 0. A catalyst does all of the following EXCEPT:
-k. affect the rate-determining step'
.d. increase the reaction rate.
€. lower the activation energy.
7 flinrr"use the equilihrium constant.t.'
i ..., ,
5 1 . What are the units for koOr in the rate equation, if rate =k665 [[$H3)5CoC1]2+l [OH-] ?
A. [M]2 secl
B. [M] sec-l
C. sec-l
D. [M]-1 sec-1
,+
Passage Vlll (Questions 52 - 58)
The mechanism for a reaction can be inferred from thekinetic data. The kinetic data determine the reactants thataffect the rate of a reaction. The reactants in the rate-determining step can be used to predict a mechanism to matchthe kinetic data. Ifone ofthe reactants does not affect the rateof the reaction, then the reaction must be a multi-stepreaction. A one-step reaction shows a rate dependence on allof the reactants. Reaction 1 was observed at varying initialconcentrations for the reactants. The temperature was heldconstant for each trial.
2 SO2(g) + Oz(g) -+ 2 SO3(g)
Reaction 1
Table 1 lists the initial reaction rate for four trials ofReaction 1, carried out at varied concentrations.
initiar *" (y) Psoz ittit Poz itrit
2.17 x 11-2 0.60 atm 0.40 atm
2.I3 x g-2 0.60 atm 0.80 atm
8.66 x g-2 1120 atm 0.40 atm
8.',72 x g-2 1.20 atm 0.80 atm
Table 1
The reaction order in terms of each reactant can bedetermined from the data Table l. Because oxygen shows noeffect on the reaction rate, the reaction is said to be zero-orderwith respect to oxygen gas. The fact that oxygen is zero-order simply means that oxygen is not involved in the rate-determining step of the reaction. The same results wereobserved when the reaction was repeated at differenttemperatures. The reactions rates were greater at highertemperatures, but the reaction order remained the same forboth sulfur dioxide and oxygen gas.
52. Which of the following accurately reflects thecorrelation between the initial rate and the concentrationof sulfur dioxide?
A.
oL
o!
H
[Soz] lsozl
[Soz] lSozl
Copyright @ by The Berkeley Review@ 258, GO ON TO THE NEXT PAGE
5 3 . Addition of a catalyst to a reaction results in which ofthe following?
A . An increase in the forward reaction rate onlyB . An increase in the amount of SO3 formed
C . An increase in both the forward and reverse reactionrates
D . An increase in the amount of both SO2 and 02
5 4. The mechanism for the reaction is MOST likely to be
which of the following?
A. Fast: SOz(g) + Oz(g) -+ SO3(g) + O(e)Slow; SOz(g) + O(g) -+ SO3(g)
B. Slow: SOz(g) +OzG) -+ SO3(g) +O(g)Fast: SO2(g) + O(s) -+ SO3(g)
C. Fast: SOz(e) + SOz(e) -+ SO3(g) + SO(e)Slow: SO(g) + OZ(g) -+ SO:(g)
D. Slow: SOz(e) + SOz(g) -+ SO3(g) + SO(g)Fast: SO(g) + OzG) -+ SO3(g)
5 5. What is the relative change in concentration over tir:r:between oxygen and sulfur trioxide?
^ AlOzl _ , .., A[SO:J
At At
D A[Ozl _ "A[SO:1At At
". A[Oz]
=..,. 1 AISO:IAt 2Lt
D. Alot __ r A[So:]At2Lt
5 6. As the reaction proceeds, what is observed for the tota-pressure of the system?
A . The total pressure increases at a constant rate.
B. The total pressure decreases at a constant rate.
C. The total pressure increases at a graduall1decreasing rate.
D. The total pressure decreases at a gradual11decreasing rate.
5 7. Which of the following changes does NOT increase the
rate of a reaction?
A . Increasing the temperature of the reaction
B . Decreasing the volume of the container
C. Adding sulfur dioxide to the mixture
D . Adding oxygen to the mixture
5 8. When the same experiment was carried out at elevated
temperatures, it was observed that the rate began to vary
with oxygen concentration. How is this best explained?
A. The mechanism of the reaction changed so that
oxygen was now involved in the rate-determining
step.
B. The mechanism of the reaction changed, so that
oxygen was no longer involved in the rate-
determining steP.
C. The bond energy of oxygen is lower at higher
temperatures.D. The bond energy of sulfur dioxide is greater at
higher temperatures.
-opyright @ by The Berkeley Review@ 239 GO ON TO THE NEXT PAGE
Passage lX (Questions 59 - 65)
The rate at which a metal takes on ligands depends on
several factors. The strength of the coordinate covalent bond
between the metal and the original ligand (leaving group)
plays a major role. The frequency of collision between the
new ligand and the transition metal complex also plays a
role. Unique to transition metal substitution chemistry is the
chelating effect. The chelating effect is observed when a
ligand is polydentate (possesses more than one lone pair to be
shared with the metal), in which case the second coordinate
covalent bond forms more rapidly than the first. This is to
say that once one site has been coordinated, the chance forcollision of a second lone pair with the transition metal has
increased over that of ligands free in solution. The net result
is that it is easier for a polydentate chelating agent to form
subsequent bonds to a metal than its first bond' Polydentate
chelating agents have two or more lone pairs to share.
A researcher set up an experiment to study Reaction 1, in
an effort to support the concept of the chelating effect. Aseries of hexa-coordinate metals with three sites bound by a
polymer and the other three sites bound by phosphine ligands
was exposed to a compound containing three aminefunctional groups (H2NCH2CH2NHCH2CH2NH2). The
reactant has a visible absorption at l,*u^ = 519 nm. The
product transition metal complex has a visible absorption at
fnru* = 465 nm. The rate of the reaction is monitored by
visible spectroscopy. Reaction 1 is drawn below:
*r*"G{Ps,,PRr
"r*atratr*atratrY
3PR3 +
HzN
Reaction 1
The rate of the reaction can be monitored by either the
disappearance of reactant or the appearance of product' Anormal substitution reaction on a transition metal complex
shows either first-order or second-order kinetics. In either
case, the observed rate gradually decreases over time' The
chelating effect predicts that the rate of formation of
phosphine actually increases with time for a short period'
5 9 . Which method is NOT effective for monitoring the
reaction rate?
A . Monitoring the appearance of polymer-MC4H13N3
B . Monitoring the disappearance of PR3
C . Monitoring the disappearance of C4H13N3
D . Monitoring the disappearance of polymer-M(PRs):
o"','*(|}r*yrl
60. Which of the following is TRUE for the observedreaction rate of a generic reaction?
A. The observed reaction rate increases over timeduring an endothermic reaction.
B. The observed reaction rate is greater, if a higherenergy product is formed.
C. The observed reaction rate is greater, if a higherenergy transition state is formed.
D. The observed reaction rate is greater, if a lowerenergy transition state is formed.
6 1.. How can it be explained that the exchange rate for the
first phosphine is slower than that of the second and
third phosphines?
A. The second and third phosphines bind the metalmore tightly, due to increased steric hindrance afterthe first phosphine has left.
B. The second and third phosphines bind the metalmore tightly, due to reduced steric hindrance afterthe first phosphine has left.The nucleophile can more easily displace the second
and third phosphines, due to the chelating effect.
The nucleophile cannot displace the second and
third phosphines as easily, due to the chelatingeffect.
62. It in one reaction the R group of the phosphine ismethyl (CH:), what is observed when the methyl isreplaced by ethyl (CH2CH3)?
A. The reaction rate increases with the ethyl, because
steric hindrance ofthe leaving group has increased.
B. The reaction rate decreases with the ethyl, because
steric hindrance of the leaving group has increased.
C . The reaction rate increases with the ethyl, because
steric hindrance of the leaving group has decreased.
D. The reaction rate decreases with the ethyl, because
steric hindrance ofthe leaving group has decreased.
63. The FASTEST reaction is observed with which of the
following conditions?
A. A one-step reaction where a strong bond is brokenand a weak bond is formed.
B. A one-step reaction where a weak bond is brokenand a strong bond is formed.
C . A two-step reaction where step I requires breaking a
strong bond and step II forms a strong bond.
D . A two-step reaction where step I requires breaking a
strong bond and step II forms a weak bond.
c.
D.
Copyright @ by The Berkeley Review@ 240 GO ON TO THE NEXT PAG'S
64. Which of the following statements CANNOT be true
aboutReaction 1?
L Adding methyl groups to the nitrogens of thetriamine species reduces the reaction rate.
II. Steric hindrance forming the transition statecomplex increases the reaction rate.
m. Increasing the temperature lowers the energy levelof the transition state.
A. I onlyB. II onlyC. IandtronlyD. IIandlllonly
6 5. Which of the following does NOT increase the reaction
rate?
A. Using one bidentate ligand in lieu of twc'
monodentate ligands
B. Adding a catalyst that weakens the leaving group
bond to the transition metal
C . Adding solvent to a saturated solution
D. Increasing the temperature of the solution
Passage X (Questions 66 - 72)
The following reaction can be carried out in the gas
:hase. Gas phase mechanisms are very different fromsolution phase mechanisms, in that the solvent can be
rnvolved in the transition state in a solution phase reaction.
C:O(e) + HzO(e) -+ CO2(g) + Hz(e)
Reaction ITwo chemists propose mechanisms to explain the
reactivity of the overall reaction. Chemist I proposes the
tollowing two-step mechanism.
kt4 ^+ ^-- ^U -\/ -U.
Step I:
Step I:
Step II:
H-o-H :c(J:H
\
H
H^/\)
Steprl: VUt- _,+l-
k2H-O.r+J + O:C:O
,HI--*-
\/ '
ar'@'r: &^
Mechanism IChemist II takes issue with the mechanism proposed by
Chemist I, stating that charged molecules are highly unstable
in the gas phase. As an alternative, Chemist II proposes the
followin g th,ree-step free radical mechanism :
H-Oc-o
H
H
+ | + O:C:oH
,-gQe;-9i* H-6:o + H--o
H-O)v,Step III: zz.-$c:o
H
Mechanism IIBoth mechanisms have their respective strengths and
weaknesses. Key features to observe include the bond
dissocig(on energy and the stability of intermediates.
, -\,/K. ttin"rearrangement step in Mechanism I is slower than*
the addition step, what is true about the reaction?z1l\
(iL.'tt't" reaction is one-step, and rate = kt [HzO][CO].
..El The reaction is one-step, and rate =kZlHZCOZl.d. ttre reaction is two-steps, and rate = kt[HzO][CO].
.}. The reaction is two-steps, and rate =kZIHZCOZ).
67 . A catalyst plays what role in the reaction mechanism?
/s It increase the value of K"q.
*f It decreases the value of AG.
,.V_It increases the value of AH.
""D ,. It decreases the value of the activation energy.
Copyright O by The Berkeley Review@
anction of time forj*. Wtri.h graph accurately shows the product concentration
first-order reaction?
;t',El \
Timenulf'v Time
li
Ek
k
If Step II is the rate-determining step in Mechanism I,
then all of the following predictions are valid EXCEPT:
A. Labeled oxygen in CO would not be found in CO2.
B. Adding H2O would increase the rate, because
lHzCOzl would increase.
C . Removing CO would decrease the rate, because
IHzCO zl would decrease.
D. If the temperature were held constant, the ratewould decrease as the reaction proceeds.
With which observation would BOTH mechanisms
agree?
.A. Doubling the [H2O] doubles the rate'.-8. Carrying out the reaction in ether solvent would
- vield the same results.
CE)n t the temperature increases, the reaction ratelncreases.
"D. The K.o is greater with a more stable intermediate.
6 9. Mechanism II would be preferred over Mechanism Ibecause:
4. carbocations are stabler than free radical carbons.
# anions are highly stable in the gas phase.
C . it requires less energy to break a covalent bond in a
- heterolytic fashion than a homolytic fashion.
/D free raiicals are stabler in the gas phase than ions'
7 0. Intermediates are defined as:
I 'd unstable with no lifetime.
t€?unstable with a short lifetime.'fu nightly stable with no lilerime.D. highly stable with a short lifetime.
rlIIi
71.
1)
241 GO ON TO THE NEXT PAGE
Questions 73 through 80 are NOT based on a descriptivepassage.
7 3. As AG6s1 increases, which of the following statements
is TRUE?
I. A larger temperature is needed to obtain the samereaction rate.
II. Fewer molecules have the amount of energynecessary to overcome the activation bamier.
m. A catalyst further increases AGact.
A. I onlyB. II onlyC. III onlyD. I and II only
7 4. For a first-order reaction with k = .0693 sec-l, what isthe half-life, given ln 2 is 0.693?
A. 10 seconds
B. 1 second
C. 0.10 secondsD. 0.01 seconds
7 5. For the hypothetical reaction listed below, the rate ofdisappearance of A is 0.450 IA.
sec
2A -+ B + C
The rate of appearance for B is which of the following?
A. 0.225 Msec
B. 0.4s0 ]4sec
c. 0.e00 aasec
D. -0.4s0 !Lsec
76. Which of the following is NOT true about an
intermediate?
A . The intermediate is in its highest concentration justafter the start of the reaction.Intermediates have finite lifetimes.An intermediate is in its highestconcentration when it is formed afterdetermining step of the reaction.
D. An intermediate is in its highestconcentration when it is formed beforedetermining step of the reaction.
B.C. possible
the rate-
possiblethe rate-
Copyright O by The Berkeley Review@ 242 GO ON TO THE NEXT PAGT
Time+
7 7. Given the following data, what is the concentration ofXattimet=3?
Time txt (M)
0 1.00
I 0.56
2 0.35
4 0.25
A. 0.33 MB. 0.31 Mc. 0.30 MD. 0.28 M
7 8 . Which of the following graphs BEST represents [B] as a
function of time for the following reaction?
2A -+ B + C
B.
1E
A.
1E
C.
Fts1
79. The
A.B.C.D.
rate-determining step in a two-step reaction is:
always the first step.
always the second step.
always the fastest step.always the slowest step.
Time+
Time+ Time+
8 0. The HIGHEST intermediate concentration is found withwhich of the following graphs?
A.
tsoH]{)0)
t!Reaction Coordinate -=>
Reaction Coordinate ---+
Reaction Coordinate *
Reaction Coordinate -t>
- :r-right @ by The Berkeley Review@ 243 ENOUGH KINETICS FOR NOW!
1.8 2.A 3.C 4.B 5.86.D 7.C 8.D 9.D 10.A
11. A 12. D 13. D 14. A 15. A16. A 11. C 18. A 19. D 20. A21. c 22. A 23. C 24. B 25. B26. C 21. B 28. A 29. C 30. C31. A 32. B 33. D 34. C 35. D36. C 31. B 38. C 39. B 40. A41. B 42. C 43. B 44. B 45. B46. C 47. C 48. A 49. A 50. D51. D 52. D 53. C s4. D 55. D56. D 57. D 58. A 59. B 60. D61. C 62. A, 63. B 64. D 65. C66, D 67. D 68. D 69. D 70. B]t. A 72. C 73. D 74. A 75. A76. C 17. D 78. C 79. D 80. C
Kinetics Passages Answers
J
3.
4.
5.
6.
n
Choice B is correct. At 625 nm, there is no peak throughout the experiment, so choice C should be eliminate;immediately. The largest initial peak of the three remaining choices occurs at 545 nm. The peak at 695 nm at i= 2 is also large, but it is difficult to monitor product formation, because it is not possible to calibrate themachine in the beginning, given that initially there is no peak for products. It is best to monitor the large=:peak associated with a reactant, because at that wavelength, there is the greatest absorbance. The greater theinitial absorbance, the greater the change in absorbance. This leads to the best results when monitoring thechange in absorbance. The best answer is choice B.
Choice A is correct. Because the peak at684 nm is growing as the reaction proceeds, the peak is best describecas belonging to a product. The best answer is therefore choice A. A reactant peak disappears over time, as see:lwith the peaks at 470 nm and 547 nm. A catalyst peak remains fairly constant throughout the course of lh.e
reaction. An intermediate peak grows and then holds steady for a while before gradually dropping off to zer;at the end.
Choice C is correct. The absorbance is directly proportional to the cell length, the molar absorbtivity constar.:and the concentration according to the equation. This makes choices A and B true. The molar absorbtir-if'constant is specific for each compound, so choice D is true. Only choice C can be false. The absorbance can \rar-;with tempetature as the compound and concentration vary, but not necessarily in a predictable manner. Th.a
change is not known, so the word "always" makes choice C invalid.
Choice B is correct. The peak at 547 nm belongs to a reactant, as is shown by its gradual decrease ---.
concentration over the course of the reaction. The rate at which the peak at 547 nm diminishes is equal to th.rate of disappearance of reactant, which is the negative of the rate of appearance of product. We defin.reaction rates according to the formation of product, so the reaction rate is the negative of the change r:absorbance. The value for reaction rate should always be positive or zero. The concentration and absorbanctare directly proportional, so their changes should be proportional. The best answer is choice B.
Choice B is correct. Products appear at a gradually decreasing rate, because the reaction rate slows over tir::euntil it reaches zero. The products show their biggest change in concentration initially, and the magnitude c:the change decreases with the reaction time. The best answer is choice B. The rate slows because the reacfic:slowly runs out of reactant as it approaches equilibrium.
Choice D is correct. The order of the reaction cannot be determined without running the reaction at least tu'ic=The order refers to the dependency of the rate on the reactant concentrations. The concentration of one of th;reactants must be varied between two separate trials to determine its effect on the rate. All other variabtre.must be held constant. The best choice is D.
Choice C is correct. The product appears at twice the rate at which the reactants disappear, because of the i-to- 2 ratio of reactants to the product in the balanced equation. This factor is important in most instances, but 1:
this particular question, the answer can be found based solely on the negative sign. Because the product (C
appears while the reactants (A and B) disappear, the rates of change for compounds A and B relative ::compound C must differ by a negative sign. A negative sign is present in every answer choice except for choice Cmaking choice C the best answer.
8. Choice D is correct. The enzyme.substrate complex is detectable by spectroscopy, meaning it has a measurab-elifetime, so it cannot be a transition state. Transition states are short-lived species that cannot be detecteaChoice C is eliminated. The enzyme.substrate complex is present neither in the beginning nor at the end of tLrreaction, meaning that choices A and B are eliminated. It is best described as an intermediate, so the besranswer is choice D.
Copyright @ by The Berkeley Review@ 244 Section IX Detailed Explanations
9. Choice D is correct. The enzyme is a catalyst, and like all catalysts, it lowers the activation energy (theenergy of the transition state complex in the rate-determining step) and thus increases the reaction rate. Thiseliminates choices B and C. An enzyme is special in that it is chirally active, and thus it is responsible foraligning the reactant molecules correctly. This eliminates choice A, because it is a valid statement. Catalystsdo not affect the thermodynamics of the reaction, as they are not present in the reactant or product. This meansthat the reaction cannot be made more exothermic by an enzyme. Thus, choice D is a false statement, andtherefore the best answer.
Choice A is correct. Because the rate does not change upon the addition of substrate once the enzyme is
saturated, the reaction obeys zero-order kinetics (which is confirmed by the fact that there is no change in rateover time). It is alluded to in the last paragraph of the passage that the reaction has lost its first-orderbehavior by becoming zero-order. It is best if you were to select choice A. Zero-order reactions have constantreaction rates, which according to Michaelis-Menten terminology is known as V*u*.
Choice A is correct. This is a classic Michaelis-Menten kinetics question, which is best solved throughmathematical substitution and manipulation. The calculation is as follows:
V=Vmax= V*u*[S]2 ky+ [S]
1= [S] +ku+[S] =2[S] =k1s=[S]2 krra + [S]
The reaction is first-order, because the enzyme is not saturated when the reaction rate is one-half of V-u*, so
the reaction rate depends on the substrate concentration (tS]). The math shows that ky = [S], making choice Athe best answer choice.
Choice D is correct. The enzyme.substrate complex is an intermediate. Therefore, it should not be seen in thebeginning and at the end of the reaction. It should be present only during the course of the reaction. Thiseliminates choices A and C. The enzyme starts and finishes with the same quantity; but because it is convertedto the enzyme.substrate complex during the reaction, its concentration dwindles to zero in the middle of thereaction. This is best described by choice D. From the graphs, it can be seen that the intermediate is in a
steady-state concentration only during the middle of a reaction. This is why the steady-state approximationonly applies during the middle of a reaction, once equilibrium between the intermediate and reactants has beenestablished.
Choice D is correct. The relevant information is given directly in the passage, so your answer should be theobvious choice of answer D. If you chose not to read the passage, then you should at least know that thereaction when the enzyme is saturated has its second step as its slowest, (and thus rate-determining) step.
Choice A is correct. Because the catalyst is initially saturated, the reaction proceeds according to zero-orderkinetics, which can be viewed as uniform decay at a constant rate. This is seen with all four answer choices.
The reaction evolves slowly from zero-order to first-order, meaning that no sharp changes should be observed.This eliminates choice D. Choice C should be eliminated, because the erratic rate behavior shown at the end
of the reaction would not be observed. Choice B should be eliminated, because the reaction rate should be
constant, and then gradually slowly down, not speed up. The graph that best shows zero-order kinetics slowlyevolving into first-order kinetics is choice A.
X5. Choice A is correct. From the equation, as T increases, the value for 1€ du.r"u."r. This means that the valueRT
o1 - Eact increases. Because ry is directly proportional to ln k, ln k must also increase, as does k' TheRT RT
ultimate conclusion is that k increases with T. The reaction rate also increases with T, so answer A is correct.
16. Choice A is correct. Equation 1, the rate equation, is the equation of a line where y is ln k and x is T-1. Thiseliminates choices C and D. As log k increases (as you read down the chart), the value of T-r increases' Thismakes graph A correct.
10.
1.1.
L2.
13.
14.
- rpyright O by The Berkeley Review@ 245 Section IX Detailed Explanations
17.
18.
79.
20.
2'1..
.rn
Choice C is correct. It is stated in the passage that the reaction is a one-step reaction, so the rate depends onboth reactants. The reaction is second-order when there are two reactants and the mechanism is
"ott."itud. Th"
correct answer is choice C, rate = k[SOs][NO].
Choice A is correct. The temperature and R constant are independent of any other factors. This eliminateschoices B and D. Tlre concentration of all solutes (including the reactants) is affected by changing the solvent,but only by the quantity of solvent added, not the iype of solvent. In one liter of solution, the concentration isthe same, independent of the solvent. This means that changing solvent may not necessarily change theconcentration of reactants. The question emphasizes a change in the type of solvent, not a change in the amoulllof solvent (concentration). This makes choice A the best answer. The activation energy is affected, because thesolvent can solvate the transition states and thus stabilize or destabilize the transition state. Changing thestability of the transition state changes the activation energy, so changing the solvent affects the reaclionrate. If you need a tangible example to support this notion, consider nucleophilic substitution reactions. -A-
polar/protic solvent favors an S511-reaction while a polar/aprotic solvent favors an Sy2-reaction.
Choice D is correct. Viewing Equation 1 as the equation for a line, Lr k is the y-term and T-1 can be considered tc
be the x-term. The slope is - Ea"t and the y-intercept is ln A. The question asks for the slope, so choose D is tlte'Rbest answer.
Choice A is correct. The reaction rate depends on the concentration of the reactant, so the reaction ra:E
decreases as the reaction proceeds, because the reactant concentration decreases over the course of the reactionThe fact that the rate decreases eliminates choice B, In reality, the rate at any given time can be measurejaccurately, but precisely the same time interval in the course of the reaction must be measured in each trial f:tthe values to be comparable. In practicality, the rate is determined after the concentration has been measur*jfor the duration of a reaction. The rate at any point in the reaction is thus easy to measure, This eliminateschoice C. The initial rate is chosen primarily because that time can most accurately be repeated in subseque,r:trials, which assures that the time of the reaction is the same in each comparative trial. This makes choice -rt
the best choice.
Choice C is correct. The rate of a reaction is defined as the rate of appearance of product and/or the rate :ndisappearance of reactant. This means that either the rate of appearance of product or the rate :'tdisappearance of reactant can be monitored to determine the reaction rate. This eliminates choices A and 5The intermediate concentration should remain fairly constant in a multi-step reaction; but in a one-sie:reaction, there is no intermediate formed. This means that choice C is an invalid statement and thus --=
correct answer selection. Choice D is eliminated, because by definition, the rate of a reaction is the change mreactant and product concentration as a function of time.
Choice A is correct. If the catalyst is the limiting reagent in the rate-determining step, then the cata-,dregulates the rate of the reaction. The reaction can go only as fast as free sites on the catalyst open up. Ii:mmeans that the reaction proceeds at a constant rate (the rate at which the catalyst can reopen catalytic s;lenm
after binding a reactant molecule) until the reactant begins to be depleted. The product forms at a constant :Muntil the reaction reaches a point where the catalyst is no longer the limiting reagent. At this point, it follroaffis;
standard kinetics by gradually slowing to a stop. The graph that shows this best is choice A. Choice D sl -nom;
a change in rate that is too abrupt. The reaction gradually slows, rather than abruptly stopping. Cho::Eshows a reaction that gradually speeds up, then abruptly stops. There is no common scenario that r,,;produces such results. Choice B is just tossed in there to complete the four choice per question requiremen:should be eliminated immediately.
23. Choice C is correct. The catalyst that can be used in the lowest concentration is the one with the gre
turnover rate. The passage does not discuss the lifetime of a catalyst, so only the rate of reaction nee:considered. Table 1 lists the turnover rate of the four choices. The fastest turnover rate is 35,000 reactiors I
second, so the catalyst that can be used in lowest concentration and still have a fast reaction rate is Pt :
The best answer is choice C. The rate of a catalyzed reaction depends on catalyst concentration and tu::rate.
Copyright @ by The Berkeley Review@ 246 Section IX Detailed
,/l Choice B is correct. If all three catalytic sites react equally, then it can be assumed that each site has aturnover rate of one-third that of the observed turnover rate for the catalyst. Because the catalyst is notsaturated, the observed turnover rate is less than the maximum turnover rate for the catalyst. Combining theserelationships yields the following:
Maximum site rate = I Maximum catalyst tumover rate > I Observed catalyst turnover rate = Observed site rate
This assumes that the maximum turnover rate per catalytic site is equal to or less than one-third of themaximum tutnover rate for the catalyst. Choice A can be eliminated, because the maximum tumover rate of onecatalytic site is greater than, not less than, one-third of the observed turnover rate. Because the maximumturnover rate per catalytic site is therefore greater than one-third of the observed turnover rate for thecatalyst when it is not saturated, the best answer is choice B. The maximum tumover rate of one site equals one-third of the observed turnover rate only when the catalyst is saturated. Because the catalyst is nof saturated,choice C can be eliminated. If the sites do not react equall/, then no solid conclusion can be drawn. Thiseliminates choice D. Pick choice B.
Choice B is correct. A heterogeneous catalyst does not evenly distribute itself in solution. This is mostcommonly observed when the catalyst is in the solid phase and the reaction is either a gas phase or solutionphase reaction. Aluminum trichloride is a Lewis acid that dissolves into solution evenly to react throughoutthe solution with the acid chloride. The three metal catalysts listed as other choices (PdlPdO2, Pt/PtO2, and(Me3P)3RhCl) are from Table 1 of heterogeneous catalysts, so they are all examples of heterogeneous catalysts.The best choice is answer B.
Choice C is correct. A catalyst by definition is neither consumed or produced in a reaction, so choices A and Bare eliminated. A catalyst is regenerated in the reaction, which means that the correct choice is answer C. Acatalyst is not altered in a reaction, unless it undergoes an undesirable reaction that renders it useless. Theword "always" appearing the answer choice should get your attention, but because it is present in all of theanswer choices, we can disregard any concems.
Choice B is correct. If the product has a binding affinity for a catalytic site, then it does not release from thecatalytic site as easily. This means that the product competes with the reactant for the catalytic site. Lessfree catalyst is available for the reactant as more of the product is formed. Because increasingly more productis formed as the reaction proceeds, the catalytic turnover rate slows as the reaction proceeds. This means thatboth the reaction rate and the turnover rate slow as the reaction proceeds. That makes choice B correct. This isknown as feedback inhibition.
Choice A is correct. The tumover rate for a heterogeneous catalyst in the solid phase depends on its surfacearea. The more surface area, the more active sites available to reactant. To maximizethe surface area, a solidcatalyst can be powdered. An alternative to powdering the catalyst is to apply a thin coating of the catalystto the surface of an inert solid. As mentioned in the passage, carbon is often chosen as the inert solid, becausemetals can be reduced onto the surface of carbon through electroplating. Choose A as the best answer to thisquestion. The other advantage to plating a metal catalyst onto the surface of an inert solid is the ease withwhich the catalyst can be removed (filtered) from solution once the reaction is complete. A large chunk of solidcan be removed from solution rather easily.
Choice C is correct. For a first-order reaction, the rate does in fact decrease with time, so choice A accuratelydepicts a first-order reaction. For a first-order reaction, the half-life remains constant, so the decay of reactantand formation of products is greatest initially, and slowly the rate tapers off. The concentration is based on anexponential relationship, so the graphs of both the disappearance of reactant and formation of productappears as they do in choices B and D. In choice C, the graph shows a steady rate throughout the process/which is true for zero-order'reactions, but not for first-order reactions. The best answer is choice C.
Choice C is correct. Because the product of the half-life and rate constant (t, / r* k) is ln 2 (a constant), the half-
life and rate constant are inversely proportional to one another. This implies that if the rate constant isdoubled, the half{ife is cut in half. The best answer is choice C.
25.
26.
27.
28.
29.
30.
:i
'il
'il
,iLl
ii
I
I
I
I
!
l
Copyright @ by The Berkeley Review@ 247 Section IX Detailed Explanations
31.
32.
34.
35.
36.
38.
JJ.
37.
Choice A is correct. One notable feature of a first-order reaction is that the half-life remains consta-n:
throughout the reaction. This eliminates choice D. The rate for a zero-order reaction is constant, so it take=
more time to go from 100% reactant to 50% reactant than it does to go from 507o reactant to 25"h reactant. Thi'means that for a zero-order reaction, the half-life is decreasing with time. The best answer is choice A.
Choice B is correct. Because the product of the rate constant and the half-life is a constant, the units of the rate
constant must cancel the unit of half-life. The unit of half-life is simply time, so the units of the rate constar:must be inverse time (sec-1). The best answer is choice B.
Choice D is correct. Twenty percent is one-fifth, so in the derivation in the passage, simply substitute one-fijrfor one-half. The math is shown below, which leads you to select choice D'
Q = "-kt ... e-kt = 1
Ci5h:r 1= -kt ... -hl = kt ... ln5 = kt
55
Choice C is correct. At low reactant concentration, the half-life of the reaction remains constant, whicimplies that the reaction is first-order at low concentrations. At higher concentrations, the half-life increase.
as the reactant concentration increases. Because a zero-order reaction has a constant rate, the first fifty percen:
of reactants takes longer to decay than the next twenty-five percent (the second half-life). This means that as
the concentration increases, the half-life increases for a zero-order reaction. The graph shows zero-orde:behavior at higher reactant concentrations. The best answer is choice C. You may recall from enzyme kineti.-that at high substrate concentration, the reaction is zero-order (saturated), but at lower substrat"concentrations, the reaction becomes first-order.
Choice D is correct. A constant half-life is associated with a first-order reaction. A first-order reactio:depends on just one reactant in its rate-determining step. In choices A and B, there is one step, but they ea.lhave two reactants, so both of them are second-order. In choice C, there are two steps, but both steps invol"-=
two reactants, so no matter which step is rate determining, the reaction is second-order. In choice D, thers ''one step and just one reactant, so the reaction is first-order.
Choice C is correct. Because the overall reaction is second-order, there is a first-order rate dependence on Li.nucleophile. Methanol is not as good a nucleophile as the methyl sulfide, so the reaction rate should decrea-with methanol. To support this idea, methyl sulfide is a stronger base than methanol, so methyl sulfide is .stronger lone pair donor than methanol. This makes choice C the best answer.
Choice B is correct. The reaction is second-order, so rate = k [CHSSNa]ICHgCHZI]. The concentration values
given in this problem correspond to the values given in the second row of the chart within the passage/ excep:
in the .".r"r"e order (which does not matter in this case, since the reaction is first-order with respect to bo-:substrates). Rate = k [0'25][0.10] = 3.52x 10-5, so answer B is the right choice'
Choice C is correct. It is stated in the passage that the reaction is a substitution reaction, which depends c:either one or two reactants. This makei it either an 5511 reaction, and thus a first-order reaction, or an Saireaction, and thus a second-order reaction. To determine the dependence on both reactants, hold ti-=concentration of one of the reactants constant, and observe how the rate changes with an increase in th.concentration of the other reactant. Comparing the data for the first two trials, [CHgCHzI] is increased br- 'factor of five-thirds, while ICHSSNa] t"miit t constant. The rate between the first two trials also increases i -
a factor of five-thirds. This implies that the reaction is first-order with respect to CH3CH2I. There are r':two trials to compare where [CFI3CH2I] stays constant. This is where intuition comes into play. From Trial 1 :;Trial 4, tCHgCH2Il increases by a factor of two, while [CH3SNa] increases by a factor of 1".5. If the reaclic:depends only on [CHgCHZI], then the rate will increase by a factor of two, while if the reaction depends c:boih, tne rate will triple. The rate does in fact triple, so the reaction rate depends on both reactants' T-::
reaction is second-order, meaning it is an Sp2 reaction. Choose C'
,-lnst -- k
Copyright @ by The Berkeley Review@ 24A Section IX Detailed ExPlanations
39.
40.
4't.
42.
43.
44.
45.
46.
Choice B is correct. According to Table 1, the reaction is second-order; but since [CH3SNa] is much larger than[CH3CH2I], we can disregard the reactant in high concentration ([CH3SNa]) in the rate equation. The normalrate law for a second-order reaction is rate = k [CHeSNa][CH3CH2I]; but because [CH3CH2I] << [CH3SNa],the rate law is g"iven as rate = k66s [CH3CH2I], because throughout the duration of the reaction, the [CH3SNa]stays relatively constant so it does not act as a variable. It becomes engulfed by kous. Thus, even though thereaction is truly second-order, it appears to be first-order due to the vast difference in concentrations. This isreferred to as pseudo first-order. The best answer is B.
Choice A is correct. Since the reaction mechanism is dissociative, the rate of the reaction can be only as fast asthe reactant that dissociates. This also means that the rate of the reaction depends on the concentration of thespecies that dissociates. In this case, Co(en)2NH3C12+ is the only reactanf in the rate-d.etermining step, sodoubling its concentration will in tum double the reaction rate. The new reaction rate is twice as large as thereaction rate was before doubling the reactant concentration. The nucleophile NH3 has no direct effect on therate. Choose A for success and correchress.
Choice B is correct. Increasing the temperature of a system increases the reaction rate, by providing thereactant molecules more energy to overcome the activation barrier. Choice A is thus eliminated. Addingsolvent lowers the concentrations of all species in solution, including the two reactants in the rate determiningstep. Adding solvent decreases the reaction rate, so choice B does not increase the reaction rate. The bestanswer is choice B. Adding a reactant in the rate determining step increases the reaction rate. The reaction isconcerted (one-step), so all reactants are involved in the rate determining step. This means that addition of onereactant increases the reaction rate, so choice C is eliminated. Using a less viscous solvent in a one-step reactionallows the molecules to move through solution faster, and thus collide more frequently. An increase in collisionfrequency causes an increase in reaction rate, so choice D is eliminated.
Choice C is correct. According to the data in Table 1, the reaction order for the ethyl iodide substitutionreaction is second-order, so the rate is proportional to the product of reactant concentrations. This makes therate law for the reaction equal to k [CH3SNa][CH3CH2I]. The rate constant can be solved for by substitutingany row of data points from the table. Choosing the first row of data points yields: 2.70 x 10-5 = k [0.15][0.10],and solving for k yields:
. 2.10 x ro-s ll4-k _ -.^-..-- /s = 1..40 x 10_3[M]_1s_1,choiceC.
0.015 M2
Choice B is correct. A dissociative reaction mechanism is synonymous with the 51111 reaction mechanism inorganic chemistry, which is known to be a first-order reaction. In inorganic chemistry, a dissociativemechanism means a ligand dissociates (leaves) first before another one attaches to the central metal. Choice Bis the best answer.
Choice B is correct. The rate of an 5511 reaction depends exclusively on the electrophile concentration, since theelectrophile must dissociate (the leaving group must leave) before a nucleophile can attack. The dissociationreaction of the electrophile is the rate-determining step, so answer choice B is not a valid statement. Thenucleophile has no influence on the rate of an 51.11 reaction. The rest of the choices are valid with respect to therate of an S1rJ1 reaction.
Choice B is correct. As stated in the passage, the slowest step of the three steps is Step II. A reaction canproceed only as fast as the slowest step in the mechanism, so the rate-determining step must be Step II. ChoiceB is the best answer.
Choice C is correct. Adding water to the reaction mixture would push the Step I reaction backwards, generatingmore reactants ([(NH3)5CoCl]z+ and OH-) and consuming the intermediate product ([(NH3)aCo(NHZ)C1]+).The rate-determining step, Step II, depends on the concentration of the product from Step I[(NH3)aCo(NH2)C1]+. Thus, a decrease in products from Step I slows down the rate-determining step, therebydecreasing the overall reaction rate. The best answer is choice C. Choose C for sensations of correctness andsatisfaction.
Copyright @ by The Berkeley Review@ 249 Section D( Detailed Explanations
47. Choice C is correct. The steady state approximation says that after an initial period of time, balance withinthe reaction takes place, and the concentration of any intermediates stay relatively constant (in a steadr-state). This means that the rate at which the intermediate is produced is equal to the rate at which it isconsumed. The intermediate in this case is [(NH3)aCo(NH2)C1]+, which is held constant by being in the rate-determining step. It is formed by the forward reaction in Step I and consumed by both the reverse reaction ofStep I and the forward reaction of Step II. Water is involved only in the reverse reaction of Step I, so watermust appear only with k,1. This eliminates all choices except C, which is the correct choice. In choice C, therate law for the reaction that forms the intermediate is set equal to the sum of the rate law for the reversereaction of Step I and the forward reaction of Step II.
Choice A is correct. Because of the bulky metal group attached to NH3, the lone pair is withdrawn from thenitrogen, which increases the acidity of the protons bonded to nitrogen. This is verified by Step I, where thehydroxide anion (OH-) deprotonates a proton from the NH3 that is attached to the cobalt. Hydroxide anion isnot strong enough to deprotonate ammonia normally, so the cobalt complex must have increased the acidity oiammonia. Increased acidity results in the lowering of the pKu, so choose A for best result. The metal acts like a
Lewis acid, and the ammonia acts like a Lewis base.
Choice A is correct. As pH is increased, the [OH-] is increased, so the equilibrium of Step I is pushed to theproduct side, resulting an increase in [[(NH3)aCo(NH2)Cl]*1. As [[(NHg)+Co(NH2)Cl]+l increases, the rate o:the reaction increases. So, as the pH increases, it can be concluded that the rate of the reaction increases. -{temperature increase always affects the reaction rate, so pick A.
Choice D is correct. A catalyst does not increase the equilibrium constant (product yield). A catalyst increase-.only the ease and rate at which the reaction proceeds to the products. That is, it lowers the activation energl'which speeds up the rate-determining step, which increases the reaction rate. Choice D is not characteristic o:a catalyst.
Choice D is correct. The units for rate are molar per second, and the units for concentration are molar. Solvinlfor the units of ko6, gives us M-1s-1, which makes choice D the best answer.
kobs = rate M/ 1l- /sec - /sec _ I
product of concentrations M2 M M.sec
Choice D is correct. Table 1 shows that when the partial pressure (and thus the concentration) of sulfur dioxideis doubled, and the partial pressure of oxygen remains constant, then the rate of the reaction is quadruplecThis means that there is an exponential relationship between the initial rate and sulfur dioxide. This is bes:shown in choice D,
Choice C is correct. A catalyst lowers the activation energy for a reaction. According to the principle c:microscopic reversibility, the same pathway that is taken for the forward reaction is also taken for the reversereaction. This means that a catalyst lowers the activation energy for both the forward and the rever:€reactions by the same amount. Consequently, the forward and reverse reaction rates are both lowered, makir.;choice C the best answer.
Choice D is correct. Because the reaction rate does not depend on oxygen gas, 02 cannot be involved in the ratt-determining (slowest) step of the reaction mechanism. This eliminates choices B and C. Because the ra.-depends on the square of the partial pressure of sulfur dioxide, SO2 must be in the rate-determining (slon'es:step twice. This is demonstrated by the slowest step involving two molecules of SO2. This eliminates choi;tsA, making the best choice D.
Choice D is correct. For every one oxygen molecule lost, two sulfur trioxide molecules are gained. This mea::.that the equality between the two rates must contain both a negative sign and a factor of 2. Choices A and lare both eliminated because they both lack a negative sign in the equality. The magnitude of the rate --r
consumption of oxygen is one-half that of the magnitude of the rate of formation of sulfur trioxide, so choice Dis the best answer.
48.
49.
50.
51.
52-
53.
54.
55.
Copyright @ by The Berkeley Review@ 2so Section IX Detailed Dxplanationr
56. Choice D is correct. Because three gas molecules lie on the reactant side of the equation and only two gasmolecules lie on the product side of the equation, the total pressure of the system decreases as the reactionproceeds forward. The reaction is fastest initially, so the prlrrrrr" shows the greatest reduction in the earlystages of the reaction. As the reaction slows, so does the rate at which the total pressure of the systemdecreases. The total Pressure decteases, but at a gradually decreasing rate, until it comes to rest. Choice D isthe best answer.
Choice D is correct. The rate of the reaction increases when the temperature increases, so choice A is valid andthus eliminated' Decreasing the volume of the container increases the concentration of the reactants, so therate increases' Choice B is valid and thus eliminated. Because sulfur dioxide shows a dependency on the rateof the reaction, choice C is valid and thus eliminated. As shown in Table 1, there is no eflect on the rate whenoxygen is added. Choice D is the best answer.
Choice A is correct. The only way that the rate could depend on oxygen would be if the rate-determining stepinvolved oxygen. At the temperature of the original experiment, oxygen was not involved in the rate-determining step. Adding a catalyst sometimes lowers one step of a reaction enough so that it is no longer therate- determining step. In this case, the increased temperature gives rise to an altogether different mechinism,where oxygen is now involved in the rate determining step. This is sometimes seen with thetmodynamic versuskinetic control reaction mechanisms, where a change in the system temperature changes the mechanism thereaction proceeds by. The best answer is choice A.
Choice B is correct. To monitor the rate of a reaction, one must monitor either the appearance of product ordisappearance of reactant. The products are polymer-MCaH13N3 and PR3, so the reaction can be monitored bythe appearance of either one. Choice B suggests monitoring by the disappearance of PR3, which wouId, notwork. The reactants are CaH13N3 and polymer-M(PR3)3, so it is correit to monitor the reaction by theirdisappearance. Choice B is the best answer.
Choice D is correct. An endothermic reaction absorbs heat, so that over time, the reaction cools the solution. Asthe solution cools, the reaction rate decreases, resulting in a lower observed reaction rate. This eliminateschoice A. A higher energy product is associated with an endothermic reaction (or less exothermic reaction), sothe rate decreases, because the heat energy in the system decreases. Choice B is eliminated. A higher energytransition state requires more activation energy, so the rate decreases. This eliminates choice C. The correctanswer must be choice D because of the lower activation energy associated with the lower energy transitions ta te.
Choice C is correct. The ligand replacing the phosphines is a tridentate ligand. Therefore, after the newligand attaches, the second and third lone pairs of the tridentate ligand can more easily bind the metal anddisplace the phosphine. This is known as the chelating effect and makes choice C the best answer choice. Thechelating effect is greatest in associative (SN2-like) mechanisms where there is a rate depend.ence on thenucleophile. The chelating effect is weakest in dissociative (Syl-like) mechanisms where there is no ratedependence on the nucleophile.
52. Choice A is correct. The ethyl substituent is larger than the methyl substituent, so the triethyl phosphine isbulkier than the trimethyl phosphine. This makes the triethyl phosphine a better leaving gto,rp. The rate ofthe reaction increases with the ethyl substituents, making choice A the best answer. Whether the reactionmechanism is associative or dissociitive, the leaving gtonp affects the reaction rate.
63. Choice B is correct. Any time that a strong bond must be broken it slows the reaction rate. A one-step reaction isfaster than a two-step reaction as a general (but not absolute) rule. The lowest activation energy ii associatedwith the breaking of a weak bond. The best answer is choice B.
54. Choice D is correct. By adding methyl groups onto the nitrogens of the amine, the steric hindrance in thetransition state increases and thus the reaction rate decreases. This makes statement I a true statement. Thefact that steric hindrance in the transition state reduces the reaction rate makes statement II false (nof true).Changing the temperature affects only the free energy in the system and does not affect the energy levels of anytransition states. This makes statement III false (not true). The correct answer is therefore choice D.
58.
59.
50.
5't_.
Copyright @ by The Berkeley Review@ 2|i1 Section IX Detailed Explanations
55. Choice C is correct. Using one bidentate ligand instead of two monodentate ligands increases the reaction rate,
because of the chelating effect. It is easier for the second lone pair of a bidentate ligand to bind the central
metal than it is for a second ligand to attach to the metal. Choice A is thus eliminated. Adding a catalyst
that weakens the leaving group's bond to the transition metal lowers the transition state energy and thus
increases the reaction tut".- Cnoi.e B is therefore eliminated. Adding solvent lowers the concentration of all
species in solution, so the concentration of any species in the rate-determining step is reduced' The rate of the
reaction decreases (or remains the same, if it remains saturated), so choice C is the choice that does not inctease
the reaction rate. Increasing the temperature of a reaction always increases the reaction rate. Choose C for
best results.
Choice D is correct. Regardless of which step is the rate-determining step in the reaction, according to
Mechanism I, the reaction is a two-step reaction, so choices A and B are eliminated. On top of that, the
question discusses two steps (rearrangement and addition), so it can't be one-step. Because the second step is
*t"-d"tur*ining, the rate law for the reaction is k2 times the reactant in step two (H2CO2). Choice D is the
best answer.
Choice D is correct. A catalyst does not affect the thermodynamic values of a reaction; it affects only the
reaction rate and the transition state energy. Choices A, B, and C are all thermodynamic values, while choice
D involves a kinetic value. This eliminates choices A, B, and C, while making choice D the only possible
choice. The lower the activation energy (transition state energy), the greater the rate of the reaction. This
makes choice D the best answer.
Choice D is correct. Choice A would represent the product formation, if the reaction proceeded at a constarLt
rate; but a constant reaction rate describ", u ,uro-order reaction, not a first-order reaction' Choice B describes
the consumption of reactant in a first-order reaction. Choice C does not describe any obvious typical grap:
associated with kinetics in chemistry. It shows exponential growth, but such things are more associated witpopulation genetics than chemical kinetics. I-n choice D, the product concentration is building with time, bu:
ine rate at whicn it builds (slope of the tangent) is gradually decreasing, indicating that the reaction -slowing as it proceeds. A gradually diminishing reaction rate is observed with a first-order reaction, so the
best answer is choice D.
Choice D is correct. In the gas phase, free radical carbons are more stable than carbocations, so choice A is 'false statement. If choice A were a true statement, it would support Mechanism I over Mechanism II (nl:
Mechanism II over Mechanism I), so choice A should be eliminated. Ions in general are not stable in the gl'phase, so anions (being a sub-group of ions) are not stable, making choice B a false statement. Choice C requirt=
previous knowledge io be elimlnated. More energy is required to break a bond in a heterolytic fashic:
iresulting in ions) t[an to break a bond in a homolytic fashion (resulting in free radicais). Choice C should b"
eliminatJd. If you didn't know this, then move on to choice D. Chemist II states that charged molecules ar:
highly unstable in the gas phase, so choice D must be valid, because free radicals are more stable in the g;s
phase than ions are.
Choice B is correct. Intermediates are considered to be unstable relative to products and reactants, so thev a-:
not collected in high concentration. This eliminates choices C and D. A transition state is defined as unstab.r*
with no measurable lifetime, so choice A can be eliminated. An intermediate is said to have a short lifetrn"'
so choice B is the best answer.
Choice A is correct. Labeled oxygen in the reactant (CO) can still be found in the CO2, because carbon dioxict r;
the only reactant that contains oxygen, so all oxygen atoms in the reactants (labeled or not) will be founc ::
carbon dioxide. This means that lhoice A is invaiid. Adding H2O causes more intermediate to form, so --:c
reaction rate increases with the increasing intermediate concentration. This makes choice B a valid statem.:r;;
Removing CO causes less intermediate to"form, so the reaction rate decreases with the decreasing intermed:":n*
concentralion. This makes choice C a valid statement. As the reactant concentration decreases, the rate m';n
decrease, because there are fewer collisions between reactant molecules as the concentration decreases. lbsmakes choice D a valid statement. Choice A is the best answer'
66,
6/.
68.
69.
70.
7'1..
Copyright @ by The Berkeley Review@ 252 Section IX Detailed ExPlanations
72. Choice C is correct. Doubling the [H2O] may or may not change the reaction rate, depending on whether wateris involved in the rate-determining step. Choice A is probably eliminated. Carrying the reaction out in adifferent solvent (or in this case, using a solvent), definitely changes the reaction, meaning that one mechanismmay be preferred over the other. This eliminates choice B. Because the rate constant varies directly withtemperature, it is always true that as the temperature is increased, the reaction rate increases. Choice C isthus the best answer. The stability of the intermediate affects the reaction rate, but should have no bearing onthe Ksq for the overall reaction. This eliminates choice D. The Ksq for a reaction depends on the stability ofthe products and reactants.
/5.
77-
74.
75.
/o.
Choice D is correct. Be careful not to use thermodynamics when looking at kinetics in this case. AG" = -RT ln K,and a reaction is favorable if AG is negative, according to thermodynamics. This is AGact, however, and that isthe activation energy. As the activation barrier becomes greater, fewer molecules can overcome the activationbarrier, and more energy (heat) is needed to obtain the same reaction rate. Statements I and II are both correct.A catalyst lowers the energy needed to form the transition state, which reduces the value of AGq61. It does notincrease the value of AG661, so statement III is eliminated. Answer choice D is consequently the best answerchoice.
Choice A is correct. For a first-order reaction, the half-life is found as follows:
tl" = h2'k
As given in the problem, the value of ln 2 is 0.693 and k is 0.0693 sec-1, so this is just a plug-in mathematicalproblem.
u^= 0'693 =loseconds'' 0.0693 s-l
The best answer is choice A.
Choice A is correct. Using the equation for the reaction, it is easiest to follow changes in concentration for eachcomponent in the reaction.
2A-)B+C- 0.450 +0.225 +0.225
The rate of disappearance of A is twice the rate of appearance of B and C, The rate of appearance of B must behalf of the rate of disappearance of A, which is 0.225 molar per second. Select answer choice A for correctnessrewards.
Choice C is correct. The intermediate is in its highest concentration just after the start of the reaction, becauseit is formed from the reactant but is not yet used up to form the product. As the reaction proceeds towardcompletion, the intermediate is depleted. Choice A is valid. Intermediates are defined as species that havefinite lifetimes and exist only during the course of the reaction. If the lifetime is too small to be measured, thenthe species is referred to as a transition state. Choice B is therefore valid. An intermediate builds up when itcollects before the rate-determining step. This makes choice D valid and choice C invalid, which means thelatter is the best choice.
Choice D is correct. To solve this problem, you must see the trend in the data. The rate of decomposition isgreatest initially, and it slows drastically with time. The data describe a second-order process, where 0.25 isthe concentration after 4 minutes. Because the half-life constantly doubles, the first half-life is 1 minute 20seconds and the second half-life is 2 minutes 40 seconds. The trend shows that significantly more X decomposesbetween 2 minutes and 3 minutes than decomposes during the period between 3 minutes and 4 minutes. Thismeans that the [X] at the 3-minute mark is less than the midway point between 2 minutes and 4 minutes. Thecorrect answer must be less than 0.30 M, making choice D the best answer.
Choice C is correct. B is a product, so the [B] is initially zero. This eliminates the graphs in choices A and B.Equilibrium is represented by a flat slope, but it is reached gradually and is not the sharp point indicated inthe graph in choice D. The graph in choice C is the best representation of the reaction.
78.
Copyright @ by The Berkeley Review@ 2s3 Section IX Detailed Dxplanations
79.
80.
Choice D is correct. The rate-determining step of a reaction is always the slowest step in the reaction. Theslowest step in a two-step reaction can be either the first or second step. The best answer is choice D.
Choice C is correct. Choices A and B can be eliminated, because there is no intermediate (both reactions areone-step). The deeper nadir (low point) between the two transition states is associated with the greaterconcentration of the intermediate. This makes choice C the best answer.
Copyright O by The Berkeley Review@ Section D( Detailed
Redox Reactions
$ection Xa)
b)c)
d)
Determining Oxidation StatesOxidation and ReductinnBalancing Kedox Keactionsi. by the Bridge Methodii. ' by the Half-Cells l{ethodVariations on BalancingElectroChemistry
by Todd Bennett Voltage and Energya) Energeticsb) flalf-Reaction Potentials (nMf')c) Cell Potentialsc) Free trnergy Change
Hlectrochamical Cellsa) Definitions and Terminologyb) Calvanic Cellc) trlectrolytic Celld) Concentration Effects on Voltagee) Nernst Eguation
Kedox Applicationsa) BatLeriesb) Electrical Devicesc) Chemical Applicationsd) Electrolysise) trlectroplatingfl Calvanizing
M(s): -:-> M.,*2*
Oxidation occursat the anode
l{r.d2* -+ M{s)
Reduction occursat the cathode
I2EKITAMYL)R.E.v.r.E.-ff
Speii alizing : in MCAT Prep aration
Electrochemistry Section GoalsKnow the definitions of terms used in oxidation-reduction chemistrv.I" "-rd"til"^-*d".t1""
.h"",trtry;lrery are tw-o half-reactions whi.ch together form the overall
"l*.t.onll.insfer reaction. The sfecies losing electrons is b_eing oxidized ind is referred to as the
iiiiiing ogn f . The species gainin'g electrons dUeing reduced and is referred to as the oxidizing agent.
xidationstatesofatom$withinthemolecu1es,determinewhich;i.* ii;r gui""d electrons'and,which atom has lost electron?, uld then.set tne,stoigliglgtlfg.""iii.l""tirr init t[" **" number of electrons lost by one reactant equals lhe number of electrons;;i";J bt the other reactant. This balances the elections jn the reaction. The atoms and chargesil""iifr,i balance, and this is often accomplished by adding either hydroxide anions or hydroniumcations along with water to the reaction.
Be able to balance a redox equation rn neutral aqueous s
Be able to cell and an electrol cell.
in the wire.
Understand the behind the Nernst esuation.
Be able to determine the quantity o@
Understand how a frrnctions.Batteries are essentially reversible cells that discha.rge voltage when
for ion eichange.
6s a51e to calculate the voltage for redox reactions, given the half-cell potentials.The voltaee for a redox reaction is the sum of the half-reaction voltages. You do not multiply thevoltage biany factor to balance elecirons, because voltage is measuret jn terms of a.setrtumber ofelectr?ns.' Foi the free enerqv the number of electrons becomes important, because the free energy
; of m"o'les, not electrons. The equation for the free energy is: AG" = -nFE".is expressed in terms
Be able to draw an electrochemical cell and label its components.lectrons)oCCurSattheanodeandreduction(gainof
eleitrons) occurs at the cathode. This means that electron flow-(the opposite. of current) is from
"".a" io i"thode. Ttre counter-flow of anions is from cathode solutionio anode solution, througha membrane or salt bridge.
A r"|"""i.. *tl ir.haracterized by a favorable redox reaction that releases energy. An electrolytic."ff ii .nuru.terized by an unfavorible redox rea-ction that requires the addition of
-enprgy,. { Salvanic
cell has a solid wire oi a voltmeter in the wire, while an electrolytic cell has a battery (applied voltage)
@tratesthatvoltagedependsnotonlyonthestandardreaction,butalso;; th";;;.;ritiations of the products and reictanis. As is true irf,all reactions, an electrochemical;;;;tilG;o*"r itt..uuringly'less favorable urrtil.il reaches an equilibrium state, ryherep dryI_",Tlltl;h;;; ";;"i.hu.ts"
in cfidcentration. At equilibrium, no en6rgy is exchanged, so the reaction ig."r"ol.i" ;; ih; .-"il li .to lo.rger functionall The Nernst equat'iirn_calculates the voltage of a cellund6r any conditions, taking in"to account this concentration dependence'
o? M"tut cations plate out on the surface of the cathode (electron-rich electrode) yhfn.a current is;;;li"j i;; i"lt. fnit process is known as clectroplating, a special case of electrolysis. The voltage;iFii;J *ften Jectiopl'ating must be great enough to fdrce ah unfavorable reactioh to proceed.
can be
O gf Understand the similarities between circuits and electrochemlcal cells.:l'current,whileinchemistrythesearereferredtoaselectyochemicaI( iitii.iiii
^"aA;;;;;-i";,ies?ectiuelv. You should recognize these sy'nonyms and know how chemical
'p;;liili;;a1h; i"i.tuntt a'ffect its'physical values (elg., electron"affinity and reduction potential).
General Chemistry Electrochemistry Introduction
Electrochemistry addresses the features of chemical reactions in which electronsare transferred. Noting changes in oxidation state and accounting for electronsare both important observations to be made in an oxidation-reduction reaction.The number of electrons lost by the molecule being oxidized equals the numberof electrons gained by the molecule being reduced. The reactant being oxidizedexperiences an increase in oxidation state that is reflected in the number ofelectrons lost. Likewise, the reactant being reduced experiences a decrease inoxidation state that is reflected in the number of electrons gained. These factscreate (or constitute) the foundation that makes it possible to balance anyoxidation-reduction reaction.
when electrons are exchanged in an oxidation-reduction reaction, largequantities of energy are involved. Of the different types of chemical reactionsthat may occur (including oxidation-reduction, acid-base, substitution,precipitation, composition, and decomposition), oxidation-reduction reactionsgenerate the greatest change in free energy and enthalpy. As a consequence ofthe iarge amount of energy involved in oxidation-reduction reactions, they arethe chemical source of energy for many household devices. Most fuels andpower sources rely upon oxidation-reduction reactions. Common examplesinclude hydrocarbon fuels for combustion engines, and oxidation-reductionreactions involving the transition metals used in batteries. Keeping track of theenergy associated with oxidation-reduction reactions is imperative to chemists,but actually harnessing the energy unleashed by these powerful reactionsinvolves the application of physics and engineering principles as well.
To hamess the energy of electron transfer in chemical reactions, electrochemicalcells are designed to separate the oxidation half-reaction from the reduction half-reaction. Electrons are transferred through a wire from the molecule releasingthe electrons to the molecule gaining the electrons. The energy associated withelectron flow can be converted into mechanical work, heat, light, andtranslational energy for particles. The role of a chemist in all of this is to design afuel source that efficiently generates electron flow. This is done with galvaniccells and batteries. We will address the difference between a galvanic cell and abattery within the core of this section. To understand electrochemistry, one mustunderstand how to generate electron flow and determine the energy and voltageassociated with this process. Electromotive force determines the cell voltage,which can be converted into an amount of energy that is proportional to thenumber of electrons involved in the overall oxidation-reduction reaction.
Electrochemistry also entails reactions that naturally occur in an oxygen-richenvitonment. Oxygen in the air oxidizes (rusts away) most metals with which itcomes in contact. Natural oxidation is facilitated in the presence of moisture or aconducting solvent. We shall consider chemical processes that work against thenatural deterioration of materials due to oxidation in air. Galvanizing involvesplacing a more reactive species in the environment, so that this more reactivespecies (the sacrificial metal) is the one oxidized by aft, rather than the materialbeing protected. Repairs to an oxidized metal can be made by electroplating,which lays a coat of a conducting material onto its surface. This works only if thematerial can conduct electricity. Oxidation-reduction chemistry can also be usedto form pure liquids and gases. This procedure is referred to as electrolysis, andwe will address it briefly. Electrochemistry, in summary, is the part of chemistrythat looks primarily at reactions and energetics of electron transfer.
Copyright @ by The Berkeley Review 257 Exclusive MCAT Preparation
General Chemistry Electrochemistry Redox Reactions
Redox:Keactions ":,Determining Oxidation States
Each atom within a molecule shares electrons with the neighboring atoms towhich it is bonded. The oxidation state of the atom results from electronicbookkeeping. Oxidation states are based on an all-or-nothing approach toelectron-sharing. Within a bond, the more electronegative atom is assumed totake all of the electrons, while the less electronegative atom gets none of thebonding electrons. Determining the oxidation state of an atom within a moleculerequires comparing the electronegativity of the atom of interest with theelectronegativity of all the atoms to which it is bonded. Bonding electrons areviewed as being completely associated with the more electronegative atom. Theoxidation state is a sum of these bonding values. Drawn in Figure 10-1 below arethe oxidation states for a series of molecules.
o-tt -
'+l l+'+q+
H.I U u:t- tlHydrogen: +1Chlorine: -1 +1 +1 = +1Oxygen: -1 -1 = -2
Hydrogen: +1Sulfur: +1 +1 +1 +1 = +4Oxygen: -1 -1 = -2
Figure 10-1
Hydrogen: +1Oxygen: -1 +0 = -1
Oxygen usually has an oxidation state of -2 (except in molecular oxygen, whenit's 0, and in peroxides, when it's -1) and hydrogen usually has an oxidation stateof +1 (except in hydrides, when it's -1, and hydrogen gas/ when it's 0). Forexampie, in the compound H3PO3, H is +1 and O is -2. In order for H3PO3 to beneutral, the sum of the oxidation states must be zero. The oxidation state of Pmust be +3. In addition, be aware that alkali metals (group I metals) usuallyhave an oxidation state of +1, alkaline earth metals (group II metals) usually havean oxidation state of +2, and halogens usually have an oxidation state of -1(except in oxyacids and molecular halogens). The oxidation state of sulfur inSOCI2 is +4, because oxygen is -2 and chlorines are -1 each. To be neutral, theoxidation state of sulfur must be +4. The sum of the oxidation states of the atomswithin a molecule must equal the charge of the molecule.
Example 10.1In which of the following compounds is the oxidation state of phosphorus theGREATEST?
A. P4O6 ' 1'
B. P4 t'
C, .PH3, D. POC13 '
SolutionUsing the idea that hydrogen carries an oxidation state of +1, chlorine -1, andoxygen -2,'the oxidation state of phosphorus can be found in each compound. LrP4O6, the oxidation state of P is +3. In P4 (a pure element), the oxidation state ofP is 0. In PH3, the oxidation state of P is -3. Finally, in POCI3, the oxidation stateof P is +5. The oxidation state of phosphorus is highest in POC13, choice D.
Copyright @by The Berkeley Review 254 The Berkeley Review
General Chemistry Electrochemistry Redox Reactions
Oxidation and Reduction
Oxidation-reduction reactions involve the transfer of electrons from the reducingagent to the oxidizing agent. The consequence of electron transfer in a chemicalreaction is a change in the oxidation states of at least two atoms. If the oxidationstate increases due to the loss of electrons, then that process is oxidation. If theoxidation state decreases due to the gain of electrons, then that process isreduction. In every redox reaction, there must be both oxidation and reduction.Remember the mnemonic acronym LEOGER, which stands for lose electronsoxidation, gain electrons reduction. Listed below are some common terms toreacquaint yourself with in the electrochemistry section.
Oxidation: A loss of electrons by an atom, resulting in an increased oxidation state.
Reduction: A gain of electrons by an atom, resulting in a decreased oxidation state.
Oxidizing agent: The resctant doing the oxidizing, getting reduced itself in the process.
Reducing agent: The reactant doing the reducing, getting oxidized itself in the process.
Example 10.2Which of the reactions below represent oxidation-reduction?
I: HBr(g) + Al(s) -+ Hz(g) + AlBr3(s)II: HCIOa(aq) + Ti(OH)2(aq) + HCIO(aq) + TiQ(s) + 2H2O(l)III: Na2CO3(aq) + HCl(aq) -+ H2O(l) + CO2(g) + NaCl(aq)
A. I onlyB. II onlyC. I and II onlyD. I and III only
SolutionIn Reaction I, the oxidation state of hydrogen changes from +1 (in HBr) to 0 (inH2), so hydrogen is reduced. Aluminum changes from 0 (in Al) to +3 (in AlBr3),so aluminum is oxidized. This makes Reaction I an oxidation-reduction reaction.In Reaction II, the oxidation state of chlorine changes from +7 (in HCIOa) to +1(in HCIO), so chlorine is reduced. Titanium changes fuom +2 (in Ti(OH)2) to +4(in TiO2), so titanium is oxidized. This makes Reaction II an oxidation-reductionreaction. This makes choice C correct. Reaction III is an acid-base reaction inwhich no oxidation states change. Protons are transferred, but not electrons.
Example L0.3In the following reaction, which compound is the oxidant (oxidizing agent)?
HCl(aq) + KMnO4(aq) + FeCl2(aq) -+ MnCl2(aq) + FeCl3(aq) + H2O(l) + KCI(aq)
A. HCI(aq)B. KMnOa(aq)C. FeCl2(aq)D. MnCl2(aq)
SolutionOxidizing agents (oxidants) get reduced, thus causing oxidation. Mn in KMnO4has an oxidation state of +7. Once reduced, Mn (in MnCl2) has an oxidation stateof. +2. Manganese is reduced f.rom +7 to +2 in the reaction. Oxidizing andreducing agents are reactants. The reactant containing Mn is the oxidizing agent.This makes choice B, KMnO4, correct. In the reaction, Fe is oxidized from +2 to+3, making FeCl2 (the reactant that contains iron) the reducing agent.
Copyright @ by The Berkeley Review 259 Exclusive MCAT Preparation
General Chemistry Electrochemistry Redox Reactions
Balancing Redox Reactions (Balancing Electrons, Charges, and Atoms)Balancing redox reactions may be accomplished using either the bridge methodor the half-cell method. The bridge method involves connecting atoms that havechanged oxidation states, and determining the number of electrons that havebeen exchanged. The half-cell method involves breaking the reaction into twosub-reactions: oxidation and reduction. With either method, you start bydetermining the oxidation states of atoms within the reactant and productmolecules, you balance the reaction electronically by having equal numbers ofelectrons involved in the oxidation and reduction half-reactions, and you balancethe charges and atoms last. Balancing oxidation-reduction reactions entailsfollowing a basic recipe, shown below, which holds true for both the half-cell andthe bridge methods.
First: Determine the oxidation states of the atoms.
Second: Identify how many electrons are involved in the oxidation half-reactionand how many electrons are involved in the reduction half-reaction.
Third: Balance the number of electrons.
Fourth: Balance the charges on each side of the equation by adding H+ to theside with either excessive negative charge or insufficient positive charge(or OH- to the side with either excessive positive charge or insufficientnegative charge).
Fifth: Balance the atoms by adding water to the appropriate side of thereaction.
Bridge Method (Connect-the-Reactants Method)
Balancing by the bridge method entails connecting the oxidizing agent to itsreduced product and the reducing agent to its oxidized product, thereby creatinga so-called 'bridge'. The electron count in each bridge must be the same. Toaccomplish this, the bridges are cross-multiplied, as shown below:
Consider the following reaction: Cr(s) + Cu2+ -+ Cr3+ + Culs;
First: The oxidation state of Cr is 0, while the oxidation state of chromium inCr3+ is +3. The oxidation state of copper in Cu2+ is +2, while theoxidation state of Cu is 0.
Second: Identify how many electrons are involved in the oxidation half-reactionand how many electrons are involved in the reduction half-reaction.Connect like reactants and products by a bridge.
Oxidized by 3e-
,*^Cr(s) +C.ur+ + Crr++Cg1s)-rl n l'
Reduced by 2e-
Third: To balance the electrons, cross-multiply the two bridge half-reactions.The oxidation half-reaction is multiplied by 2 and the reduction half-reaction is multiplied by 3.
Following these three steps yields the following balanced reaction:
2Cr(s) + 3Cu2+ -+ 2Cr3+ + 3Cu(s)
This particular example is straightforward, and likely could have been done byinspection. More challenging cases involve the balancing of charges and atoms,which is accomplished by adding either hydroxide or hydronium, (depending onthe pH), and then water.
Copyright @ by The Berkeley Review 26() The Berkeley Review
General Chemistry Electrochemistry Redox Reactions
Consider the balancing of the following reaction in a basic solution:
VO3-(aq) + Ti(OH)2(aq) -+ VO(s) + Ti(OH)a(aq)
First: The oxidation state of vanadium in VO3- is +5. The oxidation state ofvanadium in VO is +2. The oxidation state of titanium in Ti(OH)2 is +2.The oxidation state of titanium in Ti(OH)a is +4.
Second: Identify how many electrons are involved in the oxidation half-reactionand how many electrons are involved in the reduction half-reaction.Connect like reactants and products by a bridge.
Reduced bv 3e-
+5 +2 +2 +4
VO3-(aq) + Ti(OH)2(aq) -->
VO(s) + Ti(OH)a(aq)
O"rClr"d by r"-
Third: To balance the electrons, cross-multiply the two bridge half-reactions.The oxidation half-reaction is multiplied by 3 and the reduction half-reaction is multiplied by 2.
2VO3-(aq) + 3Ti(OH)2(aq)--'> 2VO(s) + 3Ti(OH)+(aq)
w,Fourth: The net charge on the left side of the equation is 2 (-7) = -2, while the net
charge on the right side of the equation is 0. Because the reaction iscarried out in base, hydroxide is added to balance charges. To balancethe charges, 2 OH- are added to the right side, making the net charge oneach side of the equation -2. This yields the following:
2VO3-(aq) + 3Ti(OH)2(uq) -+ 2VO(s) + 3Ti(OH)a(aq) + 2OH-(aq)
Fifth: To balance atoms, count the H atoms on both sides. There is an excess
of 8 H atoms (74 - 6 = 8) on the right side of the equation, so 4 H2O areadded to the left side of the equation to balance the hydrogen atoms.This leaves the following overall balanced equation:
aH2O(l) + 2VO3-(aq) + 3Ti(OH)z(uq) -r 2VO(s) + 3Ti(OH)a(aq) + 2OH-(aq
Example 10.4In acidic solution, what is the balanced form of the following reaction?
BiO3-(aq) + Mn2+1aq; -+ Bi3+1aq; + MnO4-(aq)
A. 8H+(aq) +3 BiO3-(aq; +2Mn2+(uq) -+ 5Bi3+1aq; +2MnO4-(aq) +4H2O(l)B. 14H+(aq) +5BiO3-(aq; +2Mn2+1a9) -.> S Bi3+(aq) +2MnO4-(aq) +7H2O(l)C. 12H+(aq) +5BiO3-(aq; +2Mn2+1a9) -+ SBi3+1aq; +2MnO4-(aq) +6H2O(l)D. 1,4H*(uq) + 3BiO3-(aq; +2Mn2+1aq) -+ gBi3+1aq; +2MnO4-(aq) +7H2O(l)
SolutionFirst, the electrons must be balanced. The oxidation state of Bi changes from +5to +3 (Bi is reduced by 2 electrons), while the oxidation state of Mn changes from+2 to +7 (Mn is oxidized by 5 electrons). To balance the electrons, the bismuthcompounds must be multiplied by 5, and the manganese compounds must bemultiplied by 2. This eliminates choices A and D. Charge must also balance. Inchoice B, the reactant side charge is +14 -5 +4 = +13 and the product side chargeis +15 -2 = +13. The charges balance, so choice B is the best answer.
Copyright @ by The Berkeley Review 26t Exclusive MCAT Preparation
General Chemistry Electrochemistry Redox Reactions
Half-Cell Method (Separate the Half-Reactions Method)Balancing by the half-cell method entails separating and independentlybalancing the half-reaction of the oxidizing agent to its reduced product, and thehalf-reaction of the reducing agent to its oxidized product. The two half-reactions are balanced independently. The electron count in each half-reactionmust be equal. To accomplish this, the half-reactions are cross-multiplied. Thismethod is shown below:
Consider the following reaction: MnO4- + Zn -+ MnO2 + Zn(OH)az
First: The oxidation state of manganese in MnO4- is +7, while in MnO2, theoxidation state of manganese is +4. The oxidation state of zinc in Zn is0, while in Zn(OH)a2- th" oxidation state of zinc is +2. Using thisinformation, the half-cell reactions can be drawn:
Second: Identify how many electrons are involved in the oxidation half-reactionand how many electrons are involved in the reduction half-reaction.
MnOa- + 3 e- -+ MnO2Zn -+ Zn(Of1n2- + 2e'
Third: To balance the electrons at six for each half-cell reaction, the reductionhalf-reaction must be multipliedby 2 and the oxidation half-reactionmust be multiplied by 3.
Reduction half-reaction:Oxidation half-reaction:
2 (MnOa- + 3 e- -+ MnO2)3 (Zn -+ Zn(OlH1nz- + 2 e)
This leads to the electronically balanced equation:
2MnOa- + 6e- -r 2MnO23Zn -+ 3Zn(OH)az- + 6e-
Fourth: For the reduction half-reaction, the net charge on the left side of theequation is -8, while the net charge on the right side of the equation is 0.
For the oxidation half-reaction, the net charge on the left side of theequation is 0, while the net charge on the right side of the equation is-12. Because the reaction is carried out in base, hydroxide is added tobalance charges. To balance the charges, 8 OH- are added to the rightside in reduction half-reaction, and 72 OH- are added to the left side inoxidation half-reaction. Net charge does not have to equal zero on bothsides of a reaction, but it must be the same on both sides. This yields thef ollowin g half -reactions :
2lvlnOa- + 6e- -+ 2MnO2 + 8OH-3Zn + 12OlH- -+3 Zn(OH)+2- + 6e-
Fifth: To balance atoms in the reaction, count the H atoms. There is an excessof 8 H atoms on the right side of the reduction half-reaction, so 4H2Oare added to the left side to make the hydrogens balance. The half-reactions are added to complete our balancing process, yielding the finalbalanced equation.
Reduction: 4H2O +2MnO4- + 6e- -+ 2MnO2 + 8OH-Oxidation: 3Zn + 12OH- -+3 Zn(OH)+2- + 6e-
Adding the two balanced half-reactions leads to the overall balanced oxidation-reduction reaction:
4H2O + 2MnO4- + 3Zn + 4OH- -+ 2MnO2 + 3 Zn(OH)a2-
Reduction:
Oxidation:
Reduction:
Oxidation:
Reduction:Oxidation:
Copyright @ by The Berkeley Review 262 The Berkeley Review
General Chemistry Electrochemistry Redox Reactions
Variations on Balancing (Compounds with Multiple Redox Sites)
Balancing redox reactions is easy when you follow the procedure just outlined inthe previous section. Flowever, there are a few pitfalls of which you should beaware. They involve compounds with multiple atoms undergoing oxidation orreduction. Reaction 10.1 and Reaction 10.2 are examples of reduction half-reactions with multiple atoms within a molecule being reduced.
Cr2O72- + 6 e- -+ Cr2O3
Reaction 10.1
.,-('-4...3:1B. 1:3C.3:2D. 2:3
H2O2 + 2e- -+ 2 OH-
Reaction 10.2
Example 10.5What are the coefficients after the following reaction is balanced in base?
lH2o2 + Vo2+ -.> vO2+ * ott-A. H2O2 + 2vo2+ -+ 2YO2+ + 2 OH- + 2H2OB. H2O2 + 2vO2+ + 2 OH- -+ 2YO2+ + 21120C. 2H2o2 + 2vo2+ -s 2Yo2+ + 2 oH- + 2H2oD. 2H2O2 + 2vo2+ + 2 OH- -+ 2Yo2+ + 2H2O
SolutionThe oxidation state of oxygen inH2O2 is -1, while in_OH-, the oxidation state ofoxygen is -2. The oxidation state of vanadium in VO2+ is +4, while in VO2+ theoxidation state of vanadium is +5. Two oxygen atoms per peroxide molecule are
reduced, so the peroxide reduction half-reaction requires two electrons (as shownbelow). The vanadium oxidation half-reaction releases one electron.
H2O2 + 2e- -+ 2OH-
Upon cross-multiplying the half-reactions, the ratio of peroxide to vanadiumoxide dication 1VO2+) is 1. :2, eliminating choices C and D. Although it seems
from the reduction half-reaction that hydroxide should be on the product side ofthe overall reaction, the charges, oxygen atoms, and hydrogen atoms do notbalance in choice A. This eliminates choice A. Choice B is correct, because
hydroxide is added to the reactant side to balance charges. This cancels out thehydroxide formed upon the reduction of peroxide.
Example 10.6VVhat is the ratio of silver species to aluminum species in the following reaction?
AgOH(aq) + Al(s) + Ag(s) + AlO2-(aq)
,'.-\
SolutionIn this example, the silver atom is getting reduced by one electron, and thealuminum atom is getting oxidized by three electrons. This means that balancingyields a ratio of three silver species (AgOH) to one aluminum atom, Al, choice A.
You may have experienced Example 10.6 firsthand (or more appropriately, "firsttooth"), if you have ever bitten into a piece of aluminum foil with a toothcontaining an old silver filling. This experience is rather jolting.
Copyright @ by The Berkeley Review 265 Exclusive MCAT PreParation
General Chemistry Electrochemistry Voltage and Dnergy
ff$lffiiilffiffiiiffiilEnergetics
Electrochemistry studies the transfer of energy in oxidation-reduction reactions.Energy is released in the form of electrical flow, and is harnessed in the form ofvoltage (joules per coulomb). Current is the result of a potential difference(voltage) between two points that have an electrically conducting mediumbetween them. This is what causes the sensation when you bite on aluminumfoil with a silver filling! The task of interest now becomes converting whatcauses this less than pleasurable jolt into a productive form of energy (such as
heat, light, or mechanical work). The first step involves determining the voltagethat is generated by an oxidation-reduction reaction.
Electrons flow from the species with the lower electron affinity to the specieswith the greater electron affinity. However, we do not calculate the energetics ofthe reaction based on electron affinity difference. We use electromotive force(emf). Electromotive force is a voltage, and exhaustive tables have beenproduced that list the emf values associated with hundreds of half-reactions.
Half-Reaction Potentials (Standard emf Values Relative to Hydrogen)
The energetics of half-cells are measured relative to the reduction of hydroniuminto hydrogen. The reduction of two protons (H+) to form hydrogen gas (H2) isdefined as the reference standard, and assigned an emf of zero volts. Anycompound that can be reduced more favorably than a proton has a positivereduction potential. Likewise, any compound for which reduction is lessfavorable than a proton has a negative reduction potential. Oxidation values are
relative to the oxidation of hydrogen gas, H2(g). Any compound that can be
oxidized more favorably than a hydrogen gas has a positive oxidation potential.Likewise, any compound for which oxidation is less favorable than hydrogen gas
has a negative oxidation potential. Table 10.1 is an abbreviated list of some ofcommon reduction half-reactions and their corresponding emf values.
Half Reaction E'(V) Half Reaction E" (v)
Au3+ + 3e- --> Au 1.50 Fe2+ + 2e- -+ Fe -0.44
Cl2+2e- -> 2CI- 1.36 Cr3+ + 3e- -+ Cr -0.73
Pd2++2e- + Pd 0.99 Zn2+ +2e- --+ Zn -0.76
As+ + 1e -+ Ae 0.80 Al3+ + 3e- -+ Al t.66
Cu2++2e- + Cu 0.34 Me2+ + 2e- -+ Ms nan
2H+ + 2e- -+ Hl 0.00 Na+ + 1e -+ Na -2.77
Ni2+ + 2e- -+ Ni -0.23 K++1e- -+ K -2.92
Table L0.1
Most tables list reduction half-reactions, rather than oxidation half-reactions. The
greater (more positive or less negative) the emf value, the more favorable the
reduction half-reaction. This table can be used to determine the relative strengthof an oxidizing agent. Of the reactants listed in Table 10.1, Au3+ is the strongestoxidizing agent, because it is the species that undergoes the most favorablereduction half-reaction. Conversely, of the products listed in Table 10.1, K is the
strongest reducing agent, because it is the species that undergoes the mostfavorable oxidation half-reaction. Reactions in Table 10.1 can be reversed toshow oxidation half-reactions. \rVhen the half-reaction is reversed, the sign of the
emf changes, but the magnitude does not.
Copyright @ by The Berkeley Review 26'4 The Berkeley Review
General Chemistry Electrochemistry Voltage and Energgr
Cell Potential (Voltage Associated with Redox Reactions Pairs)
The cell potential for an oxidation-reduction reaction is the sum of the oxidationhalf-reaction potential and the reduction half-reaction potential. This is shown asEquation 10.1below:
treacfion = treduction + €oxidation (10.1)
The values for t are listed in terms of voltage, which is independent of thenumber of electrons in the reaction. Voltage is defined as joules per coulomb, sothe number of electrons in a reaction is not pertinent, given that a coulomb ofcharge is the same amount, no matter how many electrons are exchanged in areaction. This is to say that a coulomb represents a fixed number of electrons"\zVhether a coulomb of charge is gathered one electron at a time or three electronsat a time, there is a fixed amount of energy per coulomb. The emf value for theoxidation-reduction reaction from Example 10.6 is determined by a standardprocedure, as shown below:
AgOH(aq) + Al(s) -+ Ag(s) + AlO2-(aq)
The half-cell reactions are:
Al(s) -+ Al3+(aq) + 3e- Ag+(aq) + 1e- -+ Ag(s)
From Table 10.1, the t" for each reduction half-reaction is:
Al3+1aq; + 3e- -+ Al(s) -1..66V Ag*(uq) + 1e- + Ag(s) + 0.80 V
The A1 half-reaction must be reversed to fit the overall reaction, because Al isoxidized in the reaction. When reversing the Al half-reaction, the emf sign isreversed too. However, it should be noted, when you multiply a reaction by aninteger, you do not multiply the emf by an integer. Reduction potential is anintensive property that does not change with conditions. This means that voltagedoes not depend on the number of electrons in the reaction. The following valueis determined for the silver-aluminum oxidation-reduction reaction:
Oxidation: Al(s) + A13+1aq) + 3e-
Reduction: 3 Ag+(aq) + 3e- -+ 3 Ag(s)
Overall: 3 Ag+(aq) + Al(s) -+ 3 Ag(s) + AI3+1aq; 2.46V
Emf values need not be memorized; but in general, the greater the electronaffinity, the greater the reduction potential. Likewise, the lower the ionizationenergy/ the greater the oxidation potential.
1.66V0.80 v
Example 10.7\A/hat is the emf for the following oxidation-reduction reaction?- .
, (r''l/ re ")lCu(s) +Cl2(g) -+CuCl2(aq) '\" t* ']t
A.2.38V C, ii +?-( Ct.8.7]0V
, Lt-Le
,L: ''j '\-l
t-.3
' t2l.-\l+ t'z
-. t ' O-"-$t.ozvD. 0.68 V ^ .-J
SolutionThe two half-reactions are: CIZ(g) + 2e- -+ 2 Cl-(aq) and Cu(s) -+ Cu2+(aq) +2e-. According to Table 10.1, the emf for the reduction of chlorine gas is + 136V,and the emf for the oxidation of copper metal is - 0.34V. This means that theoverall reaction voltage is 1.36 - 0.34 = 1.02 V. Choice C is the best answer.
Copyright @ by The Berkeley Review 265 Exclusive MCAT Preparation
General Chemistry Electrochemistry Voltage and Energy
Free Energy Change (Conversion from Voltage to Energy)
in the same manner that energy is exchanged when bonds are broken andformed, energy is exchanged when an electron is transferred from one atom toanother. F{owever, unlike electromotive force (cell potentials), the energvassociated with an electrochemical cell depends on the number of electrons. Free
energy is considered in units of joules per mole of electrons. The energy permole of electrons can be determined from the cell voltage. Equation 10.2 is theequation for free energy in an electrochemical cell:
AG=-nF€
where F =96,500 C per mole and n = electrons per reaction
(10.2)
rJ'.t'Y
)(-
A positive electromotive force (cell voltage) is associated with a favorableoxidation-reduction reaction, while a negative free energy change (AG) isassociated with a favorable oxidation-reduction reaction.
Example 10.8How much work can be done by a7.60 V cell in which there is one electron in the
oxidation half-reaction and from which 1.00 mole of electrons flow?
A.0.772x10:l -(,.)'JL, .,'V. 1.286x105J \t ' *'.\ )t'-
/'Y^t(c)4554 x 10:JD. 1.882x 1057
SolutionThe units for work are joules (newton'meters). By keeping units in mind,calculations can be made far more easily. This is even truer in physics wherethere are more units. Work is a form of energy, so the equation to use in solvingthis question is:
w=nFtThe sign is ignored, because all of the answer choices are positive numbers.Substituting into the equation yields the following:
w =lmoleelectronsx 96,500 C x 1.60I = (96,500x 1.60)Jmole electrons C
This multiplies out to be just less than 1.6 x 1d joules, which makes choice C the
best choice. The important conceptual message here is that energy is
proportional to both cell voltage and the number of electrons involved in the
oxidation-reduction reaction.
Copyright @ by The Berkeley Review 266 The Berkeley Revieu
General Chemistry Electrochemistry Electrochemical Cells
mffi$ff#$ffiffi$ffifffi[xii $Definitions and Terminology
Electrochemical cells convert energy produced in a chemical reaction into electric
flow (current). This is accomplished by separating the oxidation half-reactionfrom the reduction half-reaction, and connecting the two half-reactions usingsome conducting material. Rather than generate heat energy, the reactiongenerates electrical flow from the reducing agent to the oxidizing agent.
Electrical flow is a form of energy, just as heat, light, and mechanical work are
forms of energy.
There are some standard terms that hold true for all electrochemical cells. These
terms are often a source of confusion, because physicists, chemists, and
biochemists focus on different aspects of electricity and electrochemical cells, so
each scientist defines the terms slightly differently. A primary goal of this
section is to develop a generic set of definitions that can be used in several
related areas of discussion about electricity. The following terms are
fundamentally defined to hold true for all electrical circuits, electric fields, andelectrochemical cells.
Oxidation occurs at the anode; therefore, electrons flow away from the anode.
Reduction occurs at the cathode; therefore, electrons flow towards the cathode.
CeIIs are cyclic; therefore, ions must flow to balance the charge dffirence caused by
electron flow.
Anions migrate towards the anode, and cstions migrate towards the cathode'
Physicists often concern themselves with the flow of charge through a wire(current or electron flow), so they assign charges to the anode and cathode of abattery according to the type of flow that the poles of the battery induce'Electrons flow from the anode to the cathode, so a physicist reaches the
conclusion that the anode of a battery carries a negative charge (repelling the
electrons) and the cathode of a battery carries a positive charge (attracting the
electrons). Conventional usage among physicists denotes the cathode of a
battery by a positive sign, and the anode is denoted by a negative sign.
Biochemists focus on the flow of ions through electrical fields (gel electrophoresis
occurs between the charged plates of a capacitor). They assign charges to the
anod.e and cathode of a capacitor according to the type of ions they attract.
Anions migrate to the anode because of the anode's positive charge buildup (the
anode haslost electrons). A biochemist concludes that the anode of a capacitor
carries a positive charge (attracting anionic molecules). Cations migrate towards
the cathode because of the cathode's negative charge buildup (the cathode has
gained electrons). A biochemist concludes that the cathode of a capacitor carries
a negative charge (attracting cationic molecules).
Physicists and biochemists define the anode and cathode differently, because
they are looking at different circuit elements (battery versus a chargingcapacitor). Both perspectives are valid. They each are specific for their topic ofinierest. In electrochemical cells, we have both electron flow and ion migration,so we need a more general set of rules. To achieve that, we start by presenting
the cathode and anode according to a multi-disciplinary perspective.
Cathodes have a positive core and accumulate negative charge on their surfaces
as current flows. Anodes have a negative core and accumulate positive charge
on their surfaces as current flows.
Copyright @ by The Berkeley Review 267 Exclusive MCAT PreParation
General Chemistry Electrochemistry Electrochemical Cells
Figure 10-2 shows the circuit notation for a battery and capacitor from thisuniversal perspective.
Cathode
Anion migration
Cation migration
Anode
Figure 10-2
Figure 10-2 shows the flow of electrons through a wire, the migration of ionsthrough a field, the poles of a battery, and the plates of a charging capacitor. Thesignificant electricity terms are addressed, so any observations in Figure 10-2 areuniversal for all electrical devices. The universal rules for electrical circuits,electric fields, and electrochemical cells are thus:
1. Electrons flow from the anode to the cathode.
2. Cations migrate to the cathode.
3. Anions migrate to the anode.
The physical makeup of an electrochemical cells allows for the transfer ofelectrons from the reductant (material being oxidized) to the oxidant (materialbeing reduced). This is accomplished by placing the oxidation half-reaction inthe anode and the reduction half-reaction in the cathode. Figure 10-3 shows ageneric electrochemical cell, where the reactions and components are defined.
p.5
-o0)bo!6-U0)
@oo.
late
@@@;@@'@
@.@@'@'@@
3PlElectric field of a capacitor
plate
M(s) + Mo^'*
Oxidation occursat the anode
M."d2* + M(s)
Keduction occursat the cathode
Copyright @ by The Berkeley Review 268
Figure 10-3
The Berkeley Revie*
General Chemistry Electrochemistry Blectrochemical Cells
The salt bridge (or porous membrane in some cells) permits the flow of ions(specifically anions). Anions flow in the direction opposite to the direction of theelectron flow, in order to balance out the distribution of charge associated withflowing electrons. The salt bridge completes the circuit of the cell. Without a saltbridge, or some form of anion transfer, the circuit is incomplete and the cell couldnot produce energy. By definition, a cell has specific components that carry outspecific tasks. Reduction occurs at the cathode, and oxidation occurs at theanode. In electrochemical cells where both electrodes involve a reaction with a
cation and metal, cathodes plate out (cations in solution are converted into a
metal coating on the surface of the electrode), and anodes dissolve (the metalcoating on the surface of the electrode is converted into cations that dissociateinto solution). Salt bridges are not made of salt. They contain an aqueoussolution, held in place by a membrane, through which ions can diffuse from onehalf-cell into the other half-cell.
In addition to cell layout, there is also standard cell short hand to consider whenwriting the equation of a cell. When reading cell short hand from left to right, itgoes from anode to cathode, and reactant to product within each half-reactionelectrode. It follows the format shown in Figure 10-4:
Moxidation(s) lMoxidatior.,2* (uq yM) I lMreduction2* (uq yM) lM.u4rr.1i61(s)
Figure 10-4
The oxidation reaction in Figure 10-4 is Moxidation(s) to M6ai6n1ior,2*(uq), and thereduction reaction is Mys6rslie.r2*(uq) to M1g6r.1i.n(s). The molarity of the cationin each cell is given with the individual cell.
Example 10.9A cell is composed of Zn metal in 0.10 M Zn2+1aq) solution in one half-cell andCu metal in 0.10 M Cu2+(aq; solution in the other half-cell. The metal plates areconnected by a wire, and the solutions are connected by a salt bridge in thestandard cell manner. After a certain amount of time, it is expected that:
A. zinc metal dissolves away.B. copper metal dissolves away.C. both zinc metal and copper metal dissolve away.D. zinc cation precipitates out as zinc metal.
SolutionThe first task is to determine which metal is being oxidized and which cation isbeing reduced. This requires using the Table 10.1. The reduction half-cellpotential for Cu2+1aq) + 2 e- -+ Cu(s) is 0.34V, and the reduction half-cellpotential for Zn2+1aq) + 2 e- -+ Zn(s) is -0.76 V. For a positive cell potential, zincmust be oxidized and copper cation must be reduced. Because zinc is beingoxidized, zinc metal dissolves away. This makes choice A the best answer.
There are two types of electrochemical cells we shall considered. They aregaloanic (a spontaneous cell with E' > 0) and electrolytic (a non-spontaneous cellwith E" < 0 and an applied voltage present to power the cell). Galvanic cellsrelease energy in the form of electrical flow. Electrolytic cells are used for thestorage of electrical potential, as is seen when charging a battery. Electrolyticcells are also used in electrolysis (used to purify gases) and electroplating (usedto purify metals and coat conducting surfaces). Because the voltage can becontrolled, the rate of plating (dependent on the current) can be controlled.
Copyright @ by The Berkeley Review 269 Exclusive MCAT Preparation
General Chemistry Electrochemistry Electrochemical Cells
Galvanic CellGalvanic cells discharge voltage, which means that they harness the energy of afavorable oxidation-reduction reaction. The oxidation-reduction reaction of agalvanic cell has a negative AG and a positive emf. By convention in chemistry,galvanic cells are drawn with the anode on the left and the cathode on the right.Electrons flow naturally from left to right (anode to cathode) through a wireconnecting the two electrodes, because the reaction is favorable. A salt bridgemust be present if spectator anions are to migrate from the cathode to the anode.The spectator anion is often chloride or nitrate, because of their high solubilityand lack of base properties. Drawn in Figure 10-5 below is a galvanic cellcomposed of a zinc anode and copper cathode with a salt bridge.
Zn(s)-> 711 ct'*- t"tt'
Copper dication (Cu2*) isreduced in the cathode.
Figure 10-5
over the life of the cell, the anode dissolves away and the cathode plates out.Salt bridges are devices that allow anions to flow. Higher quality electrochemicalcells employ a membrane that selects for spectator ion flow exclusively.
The voltage of a cell is calculated by summing the reduction half-cell potentialand the oxidation half-cell potential, as shown in Equation 10.1. To increase thevoltage of a galvanic cell, the concentrations of the ions in the cathode solutioncan be increased or the concentration of the ions in the anode solution can bedecreased. The more a reaction can proceed in the forward direction, the greaterits voltage. A change in cation concentration changes voltage only slightty. Toincrease voltage significantly, multiple cells are aligned in series. This is whvseveral batteries are hooked up in a row in many battery-operated devices.
Zinc metal (Zn(s)) isoxidized in the anode,
A.6(E-s--D. e
Example 10.10How many of the galvanic cells shown in Figureattain a potential of approximately 5.5 volts?
SolutionThe reduction half-reaction potentials (emf values) for zinc dication and coppe:dication are - 0.76Y and + 0.34V respectively, according to Table 10.1. The ce]-l
voltage is 0.34 - (- 0.76) = 1.10 volts. To have a voltage near 5.5, five cells areneeded, and they should be aligned in series. This makes choice B correct.
10-5 must be aligned in series to
!.. :
<:L tJ :\'\
1'tul
Salt bridge
Copyright @ by The Berkeley Review 270 The Berkeley Review
General Chemistry Electrochemistry Electrochemical Cells
Electrolytic CellLike all electrochemical cells, an electrolytic cell has two half-cells that areseparated to allow electrons to flow from the site of oxidation to the site ofreduction. However, an electrolytic cell is based on a chemical reaction going inthe thermodynamically unfavorable direction. Energy must be added to theelectrolytic cell to accomplish this. This means that electrolytic cells must beconnected to a voltage source. The sequence of circuits in Figure L0-6 shows twobatteries opposing one another, which is analogous to an electrolytic cell. If twobatteries are set opposing one another, then current (and electron flow) aredetermined by the relative strength of the voltage sources. The greater voltagesource dictates the direction of current and electron flow. This is the reason whya voltage source is added to an electrolytic cell to oppose the natural flow ofelectricity from the chemical reaction.
oar---+ I c| '-lI'rl
all"
Without the voltages, the directionof the electron flow is uncertain.
o 4V
r=
Zinc dication (Znz*) isreduced in the cathode.
4V
-r F
| :]LJF2V
The greater voltage dictates thedirection of the electron flow.
Copper metal (Cu(s)) isoxidized in the anode.
@4V
iEP*c _ZVa c _2Va
The poles of the lower voltage battery Electrons flow in a counterclockwiseare reversed by the forced electron flow. direction, charging the lower battery.
Figure 10-6
In an electrolytic cell, the applied voltage must exceed the natural voltage inorder to force the reaction to proceed in the reverse direction. An electrolytic cellis the reverse of a galvanic cell, with electrodes interchanged because of theapplied voltage. If the applied voltage is not high enough, then the reaction doesnot proceed in the reverse direction, and no charge is stored. Figure 10-7 showsthe same half-reactions (except now they are reversed) as seen in Figure 10-5.
Cu(s) --------+6.12+
o
{
Figure 10-7
Eapplied )
Copyright @ by The Berkeley Review 27t Exclusive MCAT Preparation
General Chemistry Electrochemistry Electrochemical Cells
Concentration Effects on Cell Voltage
Galvanic cells eventually die out, so reactant and product concentrations must
affect cell voltage. There are three ways to address this: first by observation of a
battery-operated device (a flashlight is a good example) second by using Le
Chateiieis principle, and finally by more rigorous mathematical calculation
(using the Nernst equation). If one observes the light emitted by a flashlight over
iime,"the intensity is constant for several hours before the light dims and then
rapidly diminishes to zero after a few minutes. Figure 10-8 shows light intensity
(a direct measure of cell voltage) as a function of time'
*ta0-stsb
boy
5uCJa
-oU
Figure 10-8
A good explanation for this observation is provided_by Le Chatelier's principle.
As more reactant is added to a reaction, the reaction becomes more favorable, so
it produces more energy. Likewise, as more product is added to a reaction, the
reiction becomes less f-avorable, so it produces less energy. Over the course of a
reaction, reactants are converted into products, so the reaction becomes
gradually less favorable until it reaches equilibrium, whete it stops. once at
Jquilibrium, the reaction can release no more energy. As a galvanic cell runs,
reactants (cathode cations) are consumed and products (anode cations) are
formed. This lowers the favorability, the energy, and the voltage of the cell' To
maximize cell voltage, the reactants must be maximized and the products must
be minimized. In fact, no products are needed to get the reaction to start.
Example L0.11"
Which of the following cells has the GREATEST voltage?
A. Zn(s) I 0.10 M Zn2*(uq) I I 1.00 M Cu2+1aq; I Cu(s)
B. z"i'j r 1.00 M zS.ail I I 0.10 M Cu2+1aq) I Cu(s)
C. Ni(s) l0.10MNi2+1aq) ll 1.00MAg+(aq) lAg(s)D. Ni(s) I 1.00MNi2+1aq; ll0.10MAg+(aq) lAg(s)
SolutionThis question can be answered by first using Table 1"0.1 to-get emf values and
then considering Le ChAtelier's principle. The zinc oxidation half-reaction
produces +0.76V, and the copp"t reduction half-reaction produces +0.34 V'
bverall, the zinclcopper cell produces +1.10 V. The nickel oxidation half-
reaction produces +O.lZg V, and the silver reduction half-reaction produces +0'80
V. Overall, the nickel/silver cell produces +1.03 V. This eliminates choices C
and D. To decide between choice A and choice B, the concentrations must be
considered. The greater voltage is found in the cell with the most reactants and
the fewest products.
Time
--+
Copyright @by The BerkeleY Review 272 The BerkeleY Review
General Chemistry Electrochemistry Electrochemical Cells
In the zinc-copper cell, copper cation is a reactant (it's being reduced), and zinccation is a product (it's being formed from the oxidation of zinc metal). The cellwith the greatest voltage is the cell with more copper cation and less zinc cation.This is choice A. As a point of general interest, because concentration affects thevoltage, emf tables are measured starting at standard conditions of 25"C,1 atm,and 1.00 M concentration of ions. And because temperature affects theequilibrium, it affects the voltage. That's why when flashlight batteries appear tohave died, they work again for a short time after the flashlight is switched off fora brief interval to cool down. As the batteries cool, the reaction is no longer atequilibrium. Equilibrium changes with temperature.
Electrochemical potential can be generated when the anode and cathode are at
different concentrations. This creates what is known as a concentration cell.
Concentration cells have no common application, but the effect of concentrationis observed as a cell loses reactant and gains product. Using the Nernst equation,it is possible to calculate the effect of concentration on the voltage of anelectrochemical cell. This is observed in the generation of action potentials inphysiology, but we will not address that here. The key fact is that as a cell runsdown, voltage drops, because reactants are depleted.
Nernst EquationThe effect of half-cell concentrations on voltage can be quantified according tothe relationship between the reaction quotient (Qr") aod the equilibrium constant(K"O). For an oxidation-reduction reaction with K"O greater than 1.0, if there are
mor'e reactants than products, then the cell voltage'is greater than the standardcell voltage. If there are fewer reactants than products, the cell voltage is less
than the standard cell voltage. In a general sense, the following estimate is valid:
Zn(s) I 0.10M Zn2+(uq) I I 1.00MCu2+1aq; I Cu(s)
cell voltage is: 1.10 V + a little bit
Zn(s) I 1.00M Zn2+1aq) I I 0.10MCu2+1aq; I Cu(s
cell voltage is: 1.10 V - a little bit
The Nernst equation can be used to determine the exact value of the quantityrepresented by the phrase "a little bit" in the relationships above. The Nernstequation is derived from standard thermodynamic principles, The energy of anyreaction, including electrochemical reactions, is the energy released as thereaction proceeds from its starting point to equilibrium. You should recall thatAG" is RT lnKeq (the energy of reaction starting at standard conditions), and Q11
is the reaction quotient at initial conditions. This means that the followingrelationship holds true:
Energyinitial-to-equilibriu*=Energylnit.-to-standa.4+Energystandard-to-equil.
This equation translates in a free energy relationship as:
AGoverall = AG'+ RT lnQrx
Substituting - nFt for AG in the equation yields:
-nFt6vs12ll = -nFt" + RT lnQr* (because AG = -nFt)
Upon dividing both sides of the equation by - nF, we get Equation 10.3:
toverall = t' (10.3)RT lr-,er*nF
Copyright O by The Berkeley Review 273 Dxclusive MCAT PreParation
General Chemistry Electrochemistry Electrochemical Cells
Converting from natural log to base-10log is done by multiptying by 2.3, whichconverts Equation 10.3, the Nernst equation, into Equation 10.4:
toverall = t" 2.3RI log e.*nF
By substituting 8.314 for R,298 for standard T, and 96,500 for F, we get a workingequation, Equation 10.5:
toverall = t' - log Qrx (10.s)
Equations 10.4 and 10.5 are derivations of the Nernst equation, which puts aquantitative value on the effects of concentration on cell voltage. The equationlooks intimidating, but the likelihood of using it in a calculation on the MCAT isminimal. Conceptually, however, it is useful for determining relative voltages ofdifferent electrochemical cells. For instance, in the two zinc-copper examples inanswer choices A and B in Example 10.11, Q is 0.1 in choice A and e is 10 inchoice B. There are two electrons in the reaction, log 10 is equal to 1.0, and log0.1 is equal to -1.0. This means that the voltage of the cell in choice A is 1.10 +0.03 = 1.13 v and the voltage of the cell in choice B is 1.10 - 0.03 = 1..07 v. Because0.03 is small compared to the standard emf for the cell, the conceptual methodusing Le ChAtelier's principle is more than adequate for arriving at an answer.The small magnitude of the difference in voltage as concentration goes from 1.00M to 0.10 M also explains why cell voltage remains relatively constant during thelifetime of a galvanic cell. \Arhile ion concentration affects the voltage, it does notsignificantiy affect the voitage until the ratio of products to reactants is eitherextremely high or extremely low. The Nernst term for concentration is on themillivolt scale, making it useful in cell physiology, but not in general chemistry.
Example 10.1,2
What is the voltage an electrochemical cell with an anode of zinc metal in 0.01 MZn(NO3)2 (aq) and a cathode of silver metal in 1.00 M AgNQ (aq)?
A. 2.42VB. 2.30 Vc. 7.62VD. 1.50 V
SolutionFirst, we must balance the oxidation-reduction reaction:
Zn(s) + 2Ag+(aq) + 2Ag(s) + Zn2+1aq1
The standard cell potential for the reaction is found using values from Table 1.0.1
and substituting into Equation 10.1:
E" = 0.80 - (-0.76) = 1.56 VThe actual cell voltage is slightly higher than 1.56, because there is a higher cationconcentration in the cathode than in the anode. The exact value is found usingEquation 10.5:
toverali = t" 1zn2+unn6"1
[Ag+.utnc,d"]2= 7.56 y - 0.059 log0.0l
2 "t2toveral = 1.56 - 0.03 (1og0.01)V = 1'.56 - 0.03(-2)V =1.56 + 0.06V = 1.62\-
Choice C is the best answer.
(10.4)
0.059n
Yt"t
Copyright @ by The Berkeley Review 274 The Berkeley Review
General Chemistry Electrochemistry Redox Applications
rcffix::::iffiHHlilfiflti$ffsBatteries
Batteries are essentially a galvanic cell or a group of galvanic cells in series.Batteries convert electrical potentiai energy into direct current. By using areversible oxidation-reduction reaction, a battery can act as either a galvanic cellor an electrolytic cell. As the battery discharges current (releases voltage), it isacting as a galvanic cell. By applying voltage to a battery, it can be recharged.As the battery absorbs current (recharges voltage), it is acting as an electrolyticcell. Batteries have a membrane that is highly selective, so that only spectatorions can pass. By keeping the two electrodes (half-reactions) separate, it ispossible to discharge and recharge a battery for many years. Evenfually, entropyand chemical side-reactions will deteriorate the battery; but until that time, it canbe used as a reliable power source. The lead battery found in most cars can bedischarged and recharged roughly 2000 times.
Example 10.13\A/hat is the cell potential of the following reaction used in a car battery?
Overall reaction:
Pb(s) + PbO2(aq) + H2SO4(aq) -------+ 2PbSOa(aq) + 2H2O(1)
Half-reactions:
PbO2(aq) +3H++HSO4(aq) +2e- -+
PbSOa(aq) +2H2O(1) e" = 1'.69Y
PbSOa(aq) + H+ + 2 e- --------+ Pb(s) + HSOa-(aq)
A. 1.02 VB. 1.34 Vc. 2.04vD. 4.08 V
t" = -0.35 V
SolutionThe second half-cell reaction must be reversed, so that when it is added to thefirst reaction, the sum is the overall reaction. This means that the emf must also
be reversed. The cell potential for the battery is 1..69 - (-0.35) = 2.04 volts, choiceC. To achieve 12 volts total (as most standard car batteries provide), six cellsmust be aligned in series. This is why there are six cells in water-based batteries,to which you must add distilled water if you have a maintenance battery.
The dry cell batteries with which we are most familiar (used to power flashlights,radios, and other devices) are alkaline-based. They employ the oxidation of zincmetal coupled with the reduction of manganese dioxide (MnO2). A graphite rodis the conducting material through which electrons are transferred. A pastemade of MnO2 in NH4CI surrounds the graphite rod in the interior of thebattery. This paste in tum is surounded by insulated zinc connected to a cap at
the end of the battery. If the battery is placed into a circuit, then the cap isconnected to the graphite rod, resulting in the transfer of electrons from the zincto the manganese dioxide, through the circuit. Graphite is a good conductorbecause of its conjugation, but it has an internal resistance of roughly 15 O' As aresult, batteries heat up when they are in use. The emf is not significantlyaffected, but the equilibrium of the reaction is.
Copyright @ by The Berkeley Review 275 Exclusive MCAT Preparation
General Chemistry Electrochemistry Redox Applications
Electrical Devices (Conversion of Electrical Flow into Work or Energy)Some common devices are designed to convert electrical flow into other forms ofenergy, such as light, heat, and mechanical work. Knowing the basic schematicsof how such devices operate will help you on the MCAT. Listed below are sometypical examples of the conversion of electrical flow into other energy forms.
Fluorescent Tubes
Fluorescent tubes work by creating a potential difference between two plates (i.e.,one plate builds up a positive charge by losing electrons as the other plate buildsup a negative charge by gaining electrons). A gaseous ion between the plates isaccelerated towards the oppositely-charged plate (and away from the like-charged plate). Acceleration increases the kinetic energy of the particle, until itcollides with another gas particle. Some of the ion's kinetic energy is transferredto the other particle, which absorbs this energy by exciting an electron. Whenthis excited electron relaxes to its ground state, a photon is emitted from the gasparticle in the tube. The energy of each photon is random, although the averageenergy of an emitted photon depends on the density of the gas, the length of thetube, and the plate charges. The inner surface of the tube is coated with either a
fluorescing or phosphorescing agent to convert ultraviolet radiation into visiblelight.
Fluorescent tubes use altemating current. If direct curent were used, then theplates would always carry the same charge, and the ion in the tube wouldaccelerate in one direction until striking the oppositely-charged plate. Afterreaching the plate, the ion would no longer move. The tube would be finished,and thus only a flash of light would be produced. In order for the tube tofunction, the ion must move back and forth, which occurs as the plates reversecharges because of the alternating current. This means that fluorescent tubesactually produce pulsing light, much like a strobe lamp. The frequency is too fastfor the human eye to detect, so the bulbs appear to produce continuous light.
Example 10.14It dangerous to use fluorescent lighting in saw mills, because the tubes emit lightat a frequency that could:
A. deterioratelubricants.B. decompose wood.C. be in phase with the saw blade.D. initiate combustion.
SolutionThe fluorescent bulb emits light that is periodic. Under certain circumstances,the spinning saw blade in a saw mill could be in phase with the light. The resultwould be an optical illusion that would make the spinning saw blade appear tobe stationary. Needless-to-say, spinning saw blades that appear stationary to theeye can be dangerous to the touch. This makes choice C the best answer.
Heating Coils
Conventional heaters function by having a coiled wire through which currentpasses. A fan blows air across the surface of the hot coils, removing heat energyfrom the coils via convection. Electrical flow is converted into heat by theresistance in the wire. Each coil is thin, so resistance is high. Coils are employedto maximize surface area. Heat is transferred only at an interface betweenmediums, so more surface area allows for more heat transfer.
Copyright @ by The Berkeley Review 276 The Berkeley Review
General Chemistry Electrochemistry Redox Applications
Incand es c en t Light Bulb s
Incandescent bulbs convert electrical flow into light by passing current through aresistor in a vacuum. The thermal energy builds up in the resistor, but it cannotbe dissipated through convection or conduction in a vacuum (due to the absenceof a medium). The only way to release energy to the environment is throughradiation of light causing the hot bulb filament to glow. The bulb emitselectromagnetic radiation of many wavelengths.
Incandescent bulbs are spherical to maintain their structural stability. The bulbscontain gas at low pressure, so the design prevents the atmospheric pressurefrom crushing the bulb. Gas-filled bulbs are typically more tubular thanevacuated incandescent bulbs. An added benefit is the inert nature of a vacuum,which protects the filament from oxidation. If preventing oxidation were theonly consideration, some incandescent bulbs might be filled with an inert gas,such as nitrogen or a noble gas. This is not the case. If an incandescent bulbwere filled with an inert gas, then the increase in temperature would cause thegas to expand, which could rupture the bulb.
Example 10.15why is the filament of the typical incandescent bulb thin and made of tungsten?A. It is thin to maximize resistance, and tungsten is used because of its low
thermal coefficient of expansion.B. It is thin to maximize resistance, and tungsten is used because of its high heat
of fusion.C. It is thin to maximize resistance, and tungsten is used because of its low
ionization energy.It is thin to minimize resistance, and fungsten is used because of its highsublimation point.
SolutionA thin filament maximizes resistance, and thus removes more energy from thecurrent. This eliminates choice D. Tungsten is chosen because it has the highestmelting point of any conducting metal. The temperature of the filament isextremely high, so a filament with a high melting point must be used to avoidhaving the filament melt and break the circuit. A high melting point is associatedwith a large enthalpy of fusion, so choice B is the best answer. A material with alow melting point would be used in a fuse. A fuse is designed to break whencurrent exceeds a certain value, which is why metals used to make fusesgenerally have a low melting point.
Electric Motors
Electric motors function by means of a magnetic field induced by electrical flow.When current flows through a loop, a magnetic field with specific orientation isgenerated. By allowing the induced field to interact with an existing stationarymagnetic field, a torque can be exerted upon the loop (depending on theorientation of the fields). The loop rotates to align with the existing stationarymagnetic field (rotating up to a maximum of 180'). When the current is reversed,the induced magnetic field reverses, and thus opposes the existing stationarymagnetic field. Torque is again generated, so the loop rotates 180" to realign. Byrepeating this process (using alternating current through the loop), the loopspins, generating mechanical work that can turn the axis of a motor. The loopmay actually be a series of loops that make up a solenoid.
D.
B.- P;'ra
Copyright @ by The Berkeley Review 277 Dxclusive MCAT Preparation
General Chemistry Electrochemistry Kedox Applications
Chemical Applications (Common Usage of Redox Reactions)
In addition to being responsible for most of the energy used to power householddevices and heavy equipment, redox chemistry is also useful in the purificationof materials, the production of catalytic surfaces, and cosmetic applications inautomobile detailing and jewelry plating. The surface of any material that canconduct electricity can be the site of an electrochemical reaction. The differentprocesses involve different reactants and materials for the electrodes.
Electrolysis
Electrolysis involves the application of a voltage (addition of electrical energy) tocarry out an overall unfavorable process. This is what an electrolytic cell does aswell. The goal of electrolysis is not to store charge, however, but to generate lessfavorable compounds. It is commonly used to obtain pure samples of gases(such as oxygen and chlorine) and reduced metals that are normally oxidizedunder standard conditions (such as sodium and calcium). Shown below areReaction 10.3, the electrolysis reaction of water to form oxygen and hydrogengas; and Reaction 10.4, the electrolysis reaction of hydrochloric acid to formhydrogen gas and chlorine gas:
2 H2O(l) -+ 2HzG) + 1O2(g)
Reaction 10.3
2 HCl(aq) ------+ 1 Hz(g) + 1Cl2(g)
Reaction 10.4
Because hydrogen ion is reduced in both Reaction 10.3 and Reaction 70.4, it formson the cathode (reduction occurs at the cathode), and the bubbles of hydrogengas can be collected as they rise from the cathode. Chlorine gas and oxygen gascan also be collected from the anode. Generally, the anode and cathode plates inan electrolysis apparatus are made of an inert material that can conductelectricity. Materials commonly used include carbon (in its graphite allotrope)and platinum. Figure 10-9 below is the basic schematic of an electrolysisapparatus as applied to Reaction 10.3:
4H+(aq) +4e-+ 2HzG) 2 H2O(l) ----> O2(g) + 4 H+(aq) + 4 e
Figure 10-9
To carry out an electrolysis reaction, the applied voltage must be great enough toovercome the negative voltage of the unfavorable redox reaction. If anexcessively large voltage is applied to the cell, then other unfavorable reactionscan take place, resulting in multiple products being formed. The ideal scenario isto apply a voltage that is slightly in excess of the absolute value of the voltage forthe unfavorable oxidation-reduction reaction. Because the half-reactions bothhave water as the reactant, they need not be separated into half-cells.
H2(g) forms at O2(g) forms atthe cathode the anode
Copyright @ by The Berkeley Review 274 The Berkeley Review
General Chemistry Electrochemistry Redox Applications
Example 10.16When a voltage is applied to the following reaction in the electrochemical cellbelow, what is true of the gases that form on each electrode and the solution?
2 H2O(l) + 2 Cl-(aq) -+ 1Hz(g) + Ci2(g) + 2 OH-(aq)
A. HZ(g) bubbles form on the left electrode, Cl2(g) bubbies form on the rightelectrode, and the pH of the solution gradually increases.
B. H2(g) bubbles form on the right electrode, C12(g) bubbles form on the leftelectrode, and the pH of the solution gradually increases.
C. H2(g) bubbles form on the left electrode, Cl2@) bubbles form on the rightelectrode, and the pH of the solution gradually decreases.
D. HZ(g) bubbles form on the right electrode, Cl2(g) bubbles form on the leftelectrode, and the pH of the solution gradually decreases.
SolutionBecause the applied voltage is drawn the way it is, then according to convention,electrons are considered to be flowing from right to left. As a result, reductiontakes place on the left electrode, and oxidation takes place on the right electrode.Hydrogen gas forms bubbles on the left electrode, and chlorine gas formsbubbles on the right electrode. This eliminates choices B and D. Becausehydroxide anion is formed as a by-product, the solution becomes basic over time,causing the solution pH to increase. This makes the best answer choice A.
ElectroplatingElectroplating is the process of reducing ions from solution onto the surface of aconducting material. Reduction occurs at the cathode, so the cathode gains a thinfilm of the reduced metal on its surface. Practical applications of this includejewelry, chrome-plated car parts, and platinum-plated carbon matrices used incatalytic processes. Electroplating can be used to convert your copper necklacewith medallion into a beautiful gold-plated necklace with medallion (increasingyour desirability at the local discotheque). The copper necklace is placed at theelectrode pole in the cathode cell, which is filled with a solution of gold cations.Gold is reduced and consequently plates out onto the surface of the copper.
Electroplating requires a voltage source so that the current (and thus rate ofplating) can be controlled. Electroplating is used to maximize the surface area ofa catalytic metal, such as platinum and palladium. Catalytic metals are oftenplated onto the surface of graphite, so that all of the molecules of the preciousmetal are on the surface and involved in catalysis. This maximizes the utility(surface area) and makes recovery easier (a large solid can be filtered more easilythan a small solid). Because the carbon matrix of graphite contains extensiveconjugation, graphite conducts electricity, making it a material for use as anelectrode. You may recall from organic chemistry that hydrogenation reactionsemployed platinum metal plated on a carbon support, Pt(C).
Copyright @ by The Berkeley Review 279, Exclusive MCAT Preparation
General Chemistry Electrochemistry Redox Applications
GalvanizingMetals exposed to the environment oxidize over time. This is especially truewhen they are exposed to salt water, because salt water conducts electricitybetter than fresh water and air. To prevent against this, a metal can begalvanized. Galvanizing involves the addition of a more reactive metal (knownas the sacrificial metnl) to be preferentially oxidized over the metal beingpreserved. Consider a steel-hulled boat, for instance. The boat rests in salt waterfor most of its life, so it oxidizes at a rapid pace. To avoid this, the steel isgalvanized by adding a metal that is easier to oxidize than iron (the majorcomponent of steel). A thin, reactive metal plate is added to the surface of thesteel hull to prevent the iron in steel from rusting away. Because iron conductselectricity, the galvanizing plate can be placed anywhere, as long as it's in directcontact with the iron and is not submerged in the water.
Example 10.17\z\4rich of the following metals can be used to galvanize steel?
A. IronB. NickelC. PotassiumD. Ztnc
Solution\Alhen galvanizing steel, the goal is to protect the iron. The metal added must bemore reactive than iron, which has an oxidation potential of +0.44 V, according toTable 10.1. Iron is the component in steel that needs to be protected, so ironcannot be added to protect iron from oxidation. Choice A is eliminated. Nickelis not reactive enough, with an oxidation potential of +0.23 V, so choice B iseliminated. Potassium is too reactive, with an oxidation potential of +2.92Y.Potassium would explode in water, and while this would minirnize the oxidationof iron in steel, as a general rule, exploding boats are not as effective as non-exploding boats! This eliminates choice C. Ztnc is slightly more susceptible tooxidation than iron, with an oxidation potential of +0.76 V, so zinc is a goodgalvanizing materiai. Choice D is the best answer.
Copyright @ by The Berkeley Review 2ao The Berkeley Review
ffic ffi ##ffi.,'ffiffiiilffi##HEes'
I.
II.
III.
IV.
V.
VI.
VII.
VIII.
IX.
X.
XI.
XII.
XIII.
XIV.
Water-Based Redox Reaction
pFI Meter and Calomel Cell
Metallurgy
Electrochemical Titration
Redox Titration
Purification of Metals
Calvanic Cell
Photochemical Cell
Batteries
Electrolysis
Electrolysis of FICI
Standard Cells
Corrosion Prevention
Electroplating Experiment
Questions not Based on a Descriptive Passage
Electrochemistry Scoring Scale
Raw Score MCAT Score
84 - 100 15-1566-85 ro-t247 -65 7 -934-46 4-6t-55 t-3
(r -7)
(B - 14)
(15 - 20)
(2r - 26)
(27 - 33)
(54 - 40)
(4r - 47)
(48 - 54)
(55 - 60)
(6t - 67)
(68 - 74)
(7s - 8r)
(82 - 87)
(88 - e5)
(e4 - roo)
Passage I (Questions 1 - 7)
Many oxidation-reduction reactions take place in water.Water provides the medium through which electrons maytransfer. The presence ofions, cations in particular, enhancesthe conductivity of water. Reactions 1 through 4 are someoxidation-reduction reactions that take place in water, whichacts as a reactant:
Br2(g) + HzO(l) -+ HOBr(aq) + HBr(aq)
Reaction 1
cl2(g) + Hzo(l) -+ Hocl(aq) + HCl(aq)Reaction 2
$(s) + H2Orl) + KOH(aqt + lHztglc neactii' I 'iNa(s) + HzO(l) -+ NaOH(aq) + lHz(g)
Reaction 4 2
In Reactions I and 2, the water is not being oxidized orreduced. The water counteracts the charges ofthe halides. InReactions 3 and 4, the hydrogen of water is being reduced toform hydrogen gas. A correlation can be drawn betweenionization energy of the species and the electromotivepotential associated with oxidation of the species. The lowerthe ionization energy, the more favorable the oxidation. Thefirst ionization energy for each ofthe non-aqueous reactants islisted in Table 1.
Compound Ionization Energy (,[f')B12(g): 987.2
cl2(g): 1251
K(s) 418.1
Na(s) 495.9
Table 1
The same relative reactivity of the species observed inwater can be observed in other solvents. but not with thesame absolute energy values as those found in water solvent.
1 . Which of the following processes does NOT involve theloss of an electron?
-Af Oxidation.B-': Ionization€'i Conversion of oxidation state from -1 to 0Dl , Electron affinity
I
L>-.-'
2. Which of the following reactions is the MOSTfavo,rable (has the MOST positive value for E')?
$. 2 K(s) + Br2(g) -+ 2 KBr(aq) ',.' ''l "] ' i -
B- '. 2 Na(s) + Br2(g) -+ 2 NaBr(aq)c. 2 K(s) + Cl2(g) -+ 2 KCl(aq)'D. 2 Na(s) + C12(g) -+ 2 NaCl(aq)
Copyright O by The Berkeley Review@ 243 GO ON TO THE NEXT PAGE
3. What is the oxidation state of chlorine in both HOCIand HCI?
,,4.. +l in HCI and +1 in HOCIp-+l in HCI and -l in HOCIC. -l in HCI and +l in HOCI
-V.'" -t in HCI and -1 in HoCl
What is the ionization energy of Rb metal?
A. The value is greater than 1000 kJmole
B'. The value is less than 1000, but greate.r than 495.9kJ.
moleC'. The value
kJ.mole
4.
is less than 495.9,but greater than 418.7
value
- lr'ri!ina
'-i ""^-"
.-...' / i
5 . What forms whefiithium iretal is added to water?
A. No reaction is observed.B. Lithium hydroxide (LiOH) and lithium hydride
.-(LiH) form.
, .
Q. 'Only lithium hydroxide (LiOH) forms.D. Only lithium hydride (LiH) forms. j
: t \,' '''r\', ': '
6 . In the reaction of potassium metal in water, what are
the oxidizing and reducing agents?
4L the oxidizing agent is K (potassium), and the' , reducing agent is H2 (hydrogen gas).y'. fne oxidizing agent is H2 (hydrogen gas), and the
reducing agent is K (potassium).Gt',,The oxidizing agent is K (potassium), and the" )reducing agent is H2O (water).
, Dr" The oxidizing agent is H2O (water), and the
reducing agent is K (potassium).
Which of the following statements must be TRUEabout redox chemistry in water?
I. Metals are more easily oxidized than non-metals.
II. Metals in water form metal oxides and metalhydroxides.
m. Non-metals in water form metal oxides and metalhydroxides.
:,., ,'''1 A ." I only
,{ IrontyC j I and II onlyD. II and III only
is less than 418.7 kJmole
. .'l
t Li'L.
Passage ll (Questions 8 - 14)
A pH meter is an instrument that uses two electrodes to
measure the potential difference between a solution ofunknown pH and a standard solution of known pH. Because
the potential difference between the solutions is dependent
upon the [H+], it is possible to determine the pH from the
voltage of the cell. A typical standard cell employed in the
electrode of a pH meter is the calomel half-cell, which isshown as Reaction 1 below:
Hg2C12(s) + 2e- --> 2HpOt + 2C1-(aq)
\ st ReactionPl -1.1.
The standard E'""1 for the calomel half-cell is 0.285 V'
Combining this with the Nernst equation provides the
equation used to calculate pH. The Nernst equation is:
Eobserved = E"cell - 0'0592log [H+]
Equation L
Using Equation 1, it is possible to calculate the pH of a
solution from the observed voltage of the cell, because the
electrode of the pH meter contains the calomel cell' When
in contact with the calomel half-cell through a porous
membrane, the unknown solution experiences a potential
difference. The electrode probe of a pH meter is shown in
Figure 1.
++To circuit with voltmeter
-'t'- ^\ L\.!,''-
Reference half-cell
Porous bulb
Figure IGetting an accurate reading of the pH requires that the
bulb be neutralized and dried before it is placed into an
unknown solution. To preserve the lifetime of the electrodes,
they are stored in saturated chloride solutions' Cells must be
refilled at least three times a year to prevent crystallization ofthe salt caused by the evaporation of water.
8. A pH meter would yield what reading in a 0.010 M HCI
solution? .i..
A. .01
B. I \/'rj"" 2p.7
Copyright @ by The Berkeley Review@ 2A4 GO ON TO THE NEXT PAGE
g . To reduce the reading on a pH meter, which of the
following can be added to a solution?
r: A. HCI(aq)B. NH3(aq)
C. NaOH(aq)
D. KCI(aq)
*\-'; ', \! '_' = 1
10. Which of the following would be observed after base is
added to a solution monitored by a pH meter?--a----\
*-- A . Both Eobserved and pH would increase.
-#Eo6r".us6 would increase and pH would decrease'
--t:-E66ssrvg6 would decrease and pH would increase'
*Sf Both Eobserved and pH would decrease'
t.v
I 1 . If the E'ss11 were actually 0.300 V rather than 0.285 V'
what would be true of the pH measured by the pH
meter, assuming that no correction is made?
A. The pH determined by the pH meter would be too
high, because the E"sell in calculation is too high'
-3f The pH determined by the pH meter would be too.. high, because the E"ss11 in calculation is too low'
, C i The pH determined by the pH meter would be too''-./ low, because the E's.11 in calculation is too high'
"F. The pH determined by the pH meter would be too
low, because the E"..11 in calculation is too low'
12.In the calomel half-cell, which of the following is
TRUE?
-"A-. Chlorin e inHg2Cl2(s) is reduced.
.B{, Chlorine inHg2Cl2(s) is oxidized.
{-34' Mercury in Hg2Cl2(s) is reduced.
" D. Mercury inHg2Cl2(s) is oxidized.
13. What is the pH of a solution that shows Eobserved =
0.699 V? : :
+ '::. i)
,i:CoL.' .. a 2
: 7.0-' t l; -t o
.
I
ABC
/"DL--
.0.7 0'
.1.0
. 4.1
i.14. Which of the following does NOT affect the pH of the
SOlutiOn? .-, /*- '-t 'l\ - . '
. i t.. 'A . A chang'e in temPeiature
-ry: A change in the volume of water solvent
A change in the tyPe of solvent
A change in the position of the electrode in
solution.
Passage lll (Questions 15 - 20)
Metallurgy is the process by which metals are purified.The metals are often isolated as a pure salt (usually throughprecipitation), and the salt is purified by either electrolysis oran oxidation-reduction reaction (single displacement). Listedbelow are industrial methods for purifying gold, zinc,mercury, silicon, and phosphorus.
Gold: Impure samples of gold ores can be treated withaqueous cyanide anion in the presence of oxygen gas. Othercontaminants in the ore can be filtered out. The gold cyanidecomplex can then be reduced by zinc metal, leaving pure goldbehind. ,eRx I: 4 Au(s) + 8 CN-(aq) + OZG) + H2O(l) -+ -2
4 Au(CN)2-(aq) + 4 OH-(aq)
Rx II: Zn(s) + 2 Au(CN)2-(aq) -+ Zn(CN)a2- + 2 Au(s)
Zinc: Zinc sulfide can be mined and then purified byexposure to a hot stream of oxygen gas. The zinc oxide thatforms can be treated with finely powdered carbon at 700"C togenerate carbon monoxide and pure zinc metal.
Rx I: 2 ZnS(s) + 3 6zG) -+ 2ZnO(g+ 2 SOite)Rx II: ZnO(s) + C(s) -+ Zn(s) + CO(g)
Mercury: Like zinc, mercury sulfide can be mined and thenpurified by exposure to a hot stream of oxygen gas. Themercury oxidethat forms can be heated.
99 drive off 02 gas.
RxI: 2 HgS(s)+3Oz(g)--) 2HgO(s)+2 SO2(g)
Rx II: 2 HgO(s) + heat -+ 2 Hg(s) + O2(g) t'i
'v:.
Silicon: Silicon oxide (known as sand) can be treated withfinely powdered carbon at 1350'C to yield carbon monoxideand pure silicon (a liquid at that temperature).
Rx I: SiO2(l) + 2 C(s) - flQ + 2 CO(e) (1350'C)
Phosphorus: Calcium phosphate (the major component oftooth enamel) can be treated with silicon dioxide (sand) andfinely powdered carbon (charcoal) to yield calcium silicate,carbon monoxide, and pure phosphorus (occurring naturallyin the tetra-atomic state).
Rx I: 2 Ca3(POa)2(s) + l0 C(s) + 6 SiO2(s) -+6 CaSiOr(s) + 10 CO(e) + P4(s)
The multi-reaction processes require some form ofpurification between the two reactions. Often, filtering theproduct mixture through a molecular sieve works well. Thisremoves any impurities that are too large to fit through thepore, or too small to be caught, depending on the design onthe sieve. Solid oxides are assumed to be pure.
\S. Wtrat is the role of CN- in the first reaction in thepurification of gold?
) L. The cyanide serves as a reducing agent.
9.. The cyanide serves as an oxidizing agent.'C. The cyanide serves as a Lewis acid.D . The cyanide serves as a Lewis base.
Copyright @ by The Berkeley Review@
), ,; '-t'l
I 6. What is the oxidation state of gold in Au(CN)2-?
{\
c^*4
iI
i' ',' '
co(e)I orrnr -+ co(e)2 -"-+ Si(l) + Oz2-(e)
-r Si(l) + 2O2-(e)
.x1"+3Gz*r-c. -1
D. -3
17 . In the purification of zinc metal, which of the followingstatements must be true?
I. Sulfur is reduced in the overall process.
III. Zinc is reduced in the overall process.
m. Oxygen gas is the oxidizing agent.
.-.k"," llonly- B. III only
.E . I and II onlyD..l II and III only
18. What is the reduction half-reaction in the purification ofsilicon?
A. C(s) +2e- -+---B:" C(s) + 2 e- +
.€; SiO2(s) +2e-alllsio2(s) + 4 e-
19. Which reactions involves the oxidation of oxygen?
, .-*- The purification of gold, zinc, and silicon l' ( B. Thepurificationofmercury - 2 * Q
\
-t7 The purification of gold, zinc, and mercuryD. The purification of phosphorus
2 0 . In the purification of mercury metal:
A. mercury is reduced by one electron when HgO isheated.
..-{:' mercury is oxidized by two electrons when HgO isheated.
,C . mercury is oxidized by two electrons when HgS is
"-. _ treated with oxygen gas.
,j n )sutfur is oxidized by six electrons when HgS istreated with oxygen gas.
245 GO ON TO THE NEXT PAGE
Passage lV (Questions 21 - 26)
To test the Nernst equation, a student titrates the cathode
cell of a zinc-lead galvanic cell with potassium dichromate(K2Cr2O) under acidic conditions. The lead dication forms
an insoluble precipitate (PbCr2O7(s)) with the dichromate
dianion. As the lead concentration lowers, the cell potential
decreases. Reaction I is the redox reaction of lead dication
with zinc metal.
Pb2+(aq) + Zn(s) -) Zn2+1aq7 + Pb(s)
Reaction 1
Figure 1 shows the titration apparatus and the two half-cells
of the electrochemical cell. A voltmeter measures the
potential difference throughout the duration of the reaction.
Figure 1
The voltage of the cell drops almost negligibly until just
before the equivalence point, where it drops rapidly. Once the
lead has been completely precipitated, excess dichromate
anion can be reduced, as shown in Reaction 2.
14 H+(aq) + Cr2O72-1aq) + 3 Zn(s) -r3 Zn2+1aq1 + 2 Cr3+1aq; + 7 H2O(l)
Reaction 2
The following voltage data can be used to determine the
exact voltage at any point during the titration. Beforeequivalence, the voltage depends on the reaction of lead
dication and zinc metal. After equivalence, the voltage
depends on the reaction of dichromate anion and zinc metal.
Pb2+(aq) + 2 e- -r Pb(s) E'= -0.13 V
Zn2+1aq1 + 2 e- -+ Zn(s) E' = -0.76 V
Cr2OfGq)+ 14 H+(aq) + 3 e- -+ 2 Cr3+1aq; + 7 H2O(i)
E'= +1.33 V
Copyright @ by The Berkeley Review@ 246 GO ON TO THE NEXT P{['M
Equation 1 can be used to determine the exact voltage at
any point during the titration prior to the equivalence point.
Equation 2 canbe used to determine the exact voltage at any
point during the titration after the equivalence point.
Eobs = Ecell - o'059 1o, 4n [pb2*]
Equation I
C)oo
€
Eobs = Ecerr - Q.p 6rtzn2if Ic:?:12
lCr2C,72-l
Equation 2
Figure 2 graphs the voltage as a function of time for the
experiment.
mL 0.10 M C9O72-(aq) added +
Figure 2
21. The reduced voltage prior to reaching the equivalenc-point can be attributed to which of the following?
A. The precipitation ofPb2+(aq) from solutionB. Competing reduction half-reactions betu ee:
Pb2+(aq) and Cr 20 72- (aq)
C. A redox reaction with Pb2+(aq) andCr2Ol-@qt
D . Competing oxidation half-reactions betu e::Pb2+(aq) and. Cr 20 72- (aq)
22. The higher maximum voltage following the equivalen:'r
point can be attributed to which of the following?
A. Pb2+(aq) has a lower oxidation potential tl-'ar
Cr2of-@q).B. Cr2O12-(aq) has a lower oxidation potential ili,n
Pb2+(aq).
C . Pb2+(aq) has a higher reduction potential l:-ur
Crzo*-@q).D. Cr2O12-(aq) has a higher reduction potential t"ilrn
Pb2+(aq).
23. Given that zinc sulfate and lead sulfate are used in thereaction, what is the role of sulfate anion?
A . To be oxidizedB. To be reduced
C . To precipitate excess dichromateD. To migrate through the salt bridge
2 4 . Why should the circuit be left open until the start of thetitration?
A . To prevent the cell from charging voltageB. To prevent the cell from exchanging electrodesC. To prevent the lead and zinc from crossing through
the salt bridgeD. To prevent the cell from depleting voltage
2 5. What must be true about the zinc half-cell?
A. It is the oxidation half-cell found at the anode.B. It is the oxidation half-cell found at the cathode.C. It is the reduction half-cell found at the anode.D. It is the reduction half-cell found at the cathode.
26. The observed voltage reaches a minimum when:
A . [Pb2+]in i1> lCr2O12-ladded.B. [Pb2+]in i, = ]Cr2Of-ladded.C. mL 0.1 M Pb2+init > mL 0.1 M Cr2O72-added.
D. mL 0.1 M Pb2+;nir = mL 0.1 M Cr2O72-added.
Passage V (Questions 27 - 33)
Oxidation-reduction (redox) titration can be used toascertain the concentration of transition metal cations insolution. Other techniques, like ultraviolet spectroscopy,visible spectroscopy, and precipitation may also work. Theoptimal method is solution-dependent. A redox titration issimilar to an acid-base titration, except that the titrant is anoxidizing or reducing agent. For a cation such as titanium+2, an oxidizing agent can be added quantitatively to oxidizethe titanium dication to a +4 charge. It is critical to balancethe redox equation to determine the stoichiometric ratio of thetitrant to the reactant in the solution.
A student fills three flasks with iron sulfate solution, so
that each flask may be titrated by an oxidizing agent. Theoxidation state of the iron is initially +2. After titration, theiron is oxidized into a +3 cation. This means that theoxidizing agent removes one electron from each iron dication.Iron sulfate is fairly soluble in water, so it completelydissociates when mixed into water: The contents of each ofthe three flasks is listed in Figure 1.
Flask 1: 1.52 grams FeSO4 is added to enough water tomake 100 mL solution.
Flask 2: 1.52 grams FeSO4 is added to enough water tomake 200 mL solution.
Flask 3; 3.04 grams FeSO4 is added to enough water tomake 100 mL solution.
Figure 1
The molecular mass of FeSO4 is 152 grams per mole.A 20.00-mL aliquot from each of the three flasks is titratedwith 0.020 M KMnOa(aq) solution in three separatetitrations. The unbalanced equation for the oxidation-reduction reaction that takes place between permanganateanion and iron dication is shown in Reaction 1.
Fe2+1aq; + MnO4-(aq; --) F"3*1uq) + MnO(s)
Reaction 1
MnO4-(aq) is a purple solution, which causes some
difficulty in reading the volume of titrant in the burette. Theproduct MnO(s) is a brown solid that precipitates from thesolution. The Fe2+1aq; solution is clear. The titrations can
be monitored by observing the color change of the solution.When the equivalence point is reached, the titration is stopped
and the volume of titrant added is recorded. The concentrationof the initial iron dication solution can be determined, if theconcentration of the titrant solution is known. Conversely,the concentration of the titrant solution can be determined, ifthe concentration of the initial iron dication solution isknown. Equation I is used to determine the concentration ofeither species, when only one concentration is known.Deviations from the formula are found with reactions thathave different coefficients.
eq x Ms61u1;6n x Vsolution = eq x M1i116nt x Vtitrant
Equation 1
where eq is equivalents, M is molarity, and V is volume.
Copyright @ by The Berkeley Review@ 247 GO ON TO THE NEXT PAGE
27 . The endpoint of the titration of ferrous ion (Fe2+) by
permanganate (MnOa-) could be detected by which of
the following?
A . The clear solution turning and remaining brown
B. The brown solution turning and remaining clear
C . The violet solution turning and remaining clear
D. The clear solution turning and remaining violet
28. lf a 20.00-mL sample from flask 2 requires a 17'50-mL
aliquot of Na2Cr2O7(aq) solution of unknown
concentration to reach the endpoint, then a 20.00-mL
sample from Flask 3 would require which of the
following?
A . A 70.00-mL aliquot of Na2Cr2O7(aq) solution
B . A 25.00-mL aliquot of Na2Cr2O7(aq) solution
C . A 17.50-mL aliquot of Na2Cr2O7(aq) solution
D . A 8.75-mL aliquot of Na2Cr2O7(aq) solution
2 9. Which of the following is the reducing agent in the
titration of Flask I with KMnO4(aq) solution?
A. KMnO4(aq)
B. MnO(s)C. Fe2+(aq)
D. Fe3+(aq)
30. Which of the following is the oxidizing agent in the
titration of Flask 2 with Na2Cr2O7(aq) solution?
A. Na2Cr2O7(aq)B. CrzOl(s)C. Fe2+(aq)
D. Fe3+(aq)
31. Which of the following is the ratio of Na2Cr2O7(aq) to
Fe3+1aq; in the titration of Flask 2, where CtzOf-@q)is reduced into CrZO:(s)?
I3
6
A. 1
B. 1
C,2D. I
Copyright @ by The Berkeley Review@ 2AA GO ON TO THE NEXT P.A.ffi
3 2 . Given the values for the reduction half-cells of Fe2+ and
Fe3+, what is the E' for the oxidation half-reaction ofthe titration ofFlask l?
Fe3+ + 3e- -+ Fe
Fe2+ + 2e- -+ Fe
A. +0.76 VB. +0.40 Vc. -0.40 vD. -0.76 V
3 3 . The BEST container for the titration would be made ofwhich of the following materials?
A. Aluminum metalB. GlassC. Copper metalD. Polyethylene
-0.04 v-0.44v
Passage Vl (Questions 34 - 40)
Most precious metals can be found in nature in the formof an ore with an oxide coating. An ore is defined as an alloyof two or more metals homogeneously mixed. The oxidelayer on the outside of the ore protects the inner core fromoxidation. This occurs because metal in the core cannottransfer electrons through the oxide coating. To isolateselected precious metals, the ore is extracted in its entiretyand then selectively treated to separate the metals, often inion form.
Gold, for instance, is purified by first converting it toAuCl4- and then selectively precipitating the AuCl4- anionwith some cation. The gold in the anion is next reduced togold metal and thus is purified. The reduction of goldrequires the application of current to the solution to plate outthe gold onto the cathode (which must be an inert electrodefrom which gold can be removed easily).
Recently, scientists have aimed their research attechniques useful for mining precious metals from sea water.The plan involves running a microscopic net lined withsequestering agents that selectively bind target metals. Theselectivity can be carried out by cation size or bindingstrength of the ligands in the sequestering agent.
Once the sequestering agent has bound the metal, it canbe isolated and the metal can be released by lowering the pHdrastically. The free precious metal cations can then bereduced by applying a current to the solution and plating themetal onto the cathode. Table I shows the standard reductionpotentials for some precious metals in water:
Ag+(aq) + I e- + Ag(s) 0.80 VPt2+(aq) + 2e' -+ Pr(s) 1.20VAu3+1aq1 + 3 e- -+ Au(s) 1.50 V
Table 1
3 4. Gold trication could be recovered in the form of goldmetal by adding which of the following to the solution?
A. Cl2(g)B. Ag+(aq)C. Zn(s)D. Pt2+(aq)
3 5. Because the Cl-(aq) anion is NOT oxidized by theAg+(uq) cation, it can be safely assumed that thereduction potential of Cl2(g) is which of the following?
A. Less than -0.80 voltsB. Between -0.80 and 0.0 voltsC . Between 0.0 and 0.80 voltsD. Greater than 0.80 volts
Copyright O by The Berkeley Review@ 2A9 GO ON TO THE NEXT PAGE
3 6. What is the oxidation srate of gold in AuCl4-?
A. -3
B. 0C. +1
D. +3
3 7. Since silver is in a column of the periodic table thatexpresses filled d-shell stability when electrons fill theorbitals, what electronic configuration is observed withsilver cation (Ag+)?
A. [Kr]Ss2+48B. [Kr]Ss1+49C. 1rr1ss14610D. Jrrl+610
38. A sequestering agent should contain which of thefollowing structural features?
A . Atoms with no lone pairs of electronsB. Atoms able to donate electron pairsC. Atoms without completed octet valencesD. Atoms that are good Lewis acids
3 9. Which of the following is the mass percent of gold inAuCl4-?
A. Less than25VoB. Between 25Vo and5j%oC. Between 50Vo andl5VoD. Greater than'|SVo
4 0. Which of the following reactions would produce theLOWEST (or most negative) voltage?
A.
B.
C.
Ag+(aq) + Cl- -+ Ag(s) + l.rrrrt,Pt2+(aq) + 2 Cl- -r Pr(s) + Cl2G)
Au3+1aq; + 3Cl--+ Au(s) + 3Ctz(g)2
D. All of the above reactions would require the samevoltage.
Passage Vll (Questions 41 - 47)
The galvanic cell, by definition, is a cell utilizing a
favorable redox reaction that releases energy in the form ofelectron flow. The cell is composed of two half-cells: an
oxidation half-cell and a reduction half-cell. The oxidationhalf-reaction takes place in the anode half-ce1l, and overallelectrons are released from the anode half-cell. The reductionhalf-reaction takes place in the cathode half-cell, and overallelectrons are absorbed at the cathode half-cell. Figure 1
shows a standard galvanic cel1.
Anode Cathode
Figure L
Cu2+1aq; + Ni(s) + Cu(s) + Ni2+(aq)
Reaction L
The overall voltage of the cell is the sum of the half-cellvoltages for the two component reactions. Equation I shows
this calculation.
E'cell=E'1s6+E'.xEquation I
Also by definition, a galvanic cell has a positive cellvoltage overall. Reduction is defined as the gain ofelectrons,so reduction occurs at the cathode of a battery.
4 1. In the galvanic cell shown in Figure 1, oxidation occurs
ati
A. the anode, because electrons flow from the anode to
the cathode.
B. the cathode, because electrons flow from the anode
to the cathode.C. the anode, because electrons flow from the cathode
to the anode.
D. the cathode, because electrons flow from the
cathode to the anode.
4 2. Which of the following half-reactions could occur at the
anode in a galvanic cell?
A. Cl2(g) + 2e- -+ 2 Cl-(aq)
B. Na+(g) + 1 e- -+ Na(s)C. VO3-(aq)+3e-+VO(s)D. Cr2O3(s) -+ Cro+2-(aq) + 3 e-
Salt bridge(allows for anion
Copyright @ by The Berkeley Review@ GO ON TO THE NEXT PAGT
4 3 . Which of the following combinations does NOT resultin a galvanic cell when the reduction half-cell carries an
E" value greater than zero?
A. E'oxidation < 0; lE"1g6u61isnl > lE'6ai6stionlB. E'oxidation > 0; lE"1s6us1;enl > lE'61i6stionlC. E"oxidation < 0i lE"166us1;6nl < lE'616s11sn1
D . E'oxidation > 0; lE'166us1i611 < lE'611631i6n1
4 4 . What is NOT true about a galvanic cell?
A. The electron flow is from the anode to the cathode'
B. Reduction occurs at the cathode.
C . The voltage of the cell decreases as the cell runs in
the discharging direction.D. Ecell starts greater than zero and finishes when
E""1 is less than zero.
45. What BEST describes the initial composition of a
galvanic cell?
A. The cathode half-cell contains a cation in a high
oxidation state, and the anode half-cell contains a
metal with a low ionization energy.
B. The cathode half-cell contains a cation in a high
oxidation state, and the anode half-cell contains a
metal with a high ionization energy.
C. The cathode half-cell contains a cation in a louoxidation state, and the anode half-cell contains a
metal with a low ionization energy.
D. The cathode half-cell contains a cation in a louoxidation state, and the anode half-cell contains a
metal with a high ionization energy.
4 6. What is the ratio of reducing agent to oxidizing agent in
the following reaction?
MnO+-(aq) + Al(s) + MnO(s) + AlO2-(aq)
A. 3:5B. 5:3C. 2:3D.3:2
47 . What must occur after the is closed the cell
shown in Figure 1?
I. Copper metal builds up on the cathode.
II. The overall cell voltage is greater than 0.00 V.
Itr. Anions flow through the salt bridge from th;cathode to the anode.
A. I onlyB. I and II only
C . II and III onlyD. I, II' and III
Passage Vlll (Questions 48 - 54)
Photoelectric cells function by absorbing photons and
converting photon energy into electrical flow (electricity).The electricity generated is then collected and stored as apotential difference in the form of a cell (battery). The
electricity is generated when the photon strikes an ionizingplate (an electrode) and discharges an electron from the
ionizing plate. The electron flows through the wire from one
electrode to the other electrode where it is then stored. The
potential builds up across the two plates. This potential
difference is known as capacitance. A sample photoelectriccell is shown in Figure l.
(Ionizing plate) (Storage plate)
Figure 1
The incident photon must have an energy high enough toovercome the electrical potential for the cell; otherwise, the
cell cannot build up charge difference. Cells are chosen witha negative (unfavorable) E' value so that upon the addition ofthe energy, they build up a charge that can discharge in a
favorable manner. Equation 1 shows the conversion between
free energy and cell voltage:
AG = -nFE'where F = 96,500 C and E' = Ered * Eox.
Equation 1
The efficiency of the cell is determined by comparing the
energy of the incident photon relative to the stored energy ofthe cell. An efficient cell dissipates little or no energy.
4 8. An electrolytic cell:
A. produces energy by way of a favorable oxidation-reduction reaction.
B. produces energy by wayoxidation-reduction reaction.
C . stores energy by way of a
of an unfavorable
favorable oxidation-reduction reaction.
D. stores energy by way of an unfavorable oxidation-reduction reaction.
4 9. For a photochemical cell to work, the energy of the
photon absorbed must exceed:
A. the AG for oxidation at the anode.
B. the AG for the overall oxidation/reduction reaction.
C. the ionization potential for the anode.
D. the electron affinity for the cathode.
e+
Copyright @ by The Berkeley Review@ 291 GO ON TO THE NEXT PAGE
50. According to the diagram of the cell, photons of lightdirectly cause which of the following?
A . The anode to be oxidizedB. The anode to be reduced
C . The cathode to be oxidizedD . The cathode to be reduced
5 1 . The BEST material for the anode (ionizing plate) wouldbe which of the following?
A. A metalB. A metal oxideC. A non-metalD. A non-metal oxide
5 2. The overall conversion of energy in a photoelecric cellis from:
A. potential energy to electrical energy.
B. electrical energy to potential energy.
C. radiation energy to electrical energy.
D. radiation energy to potential energy.
5 3. When the cell discharges, it will:
A. dissipate the most energy initially, because it has
the highest voltage at the start.
B. dissipate the least energy initially, because it has
the highest voltage at the start.
C . dissipate a constant amount of energy, because the
cell has a constant voltage throughout the
discharging period.
D. dissipate an increasing amount of energy, because
the anode is dissolving throughout the dischargingperiod.
5 4. What is true about the current, I, at points a, b, and c in
the following circuit?
A. Ia>Ib>IcB. Ic>Ib>IuC. Ic>Iu>IbD. IU>Ic>Iu
t2v
Passage lX (Questions 55 - 60)
Dry cell batteries are common in many householdproducts such as flashlights and radios. They contain noaqueous solution through which ions migrate, but insteadcontain a gelatinous paste of concentrated aqueous NHaCl. Itis important that the electrons can easily flow through themembrane. Dry cell batteries are convenient because of theirsmall size and long lifetime. A similar battery is the alkalinemanganese cell, which exploits the same reaction, but in a
basic medium. In a basic medium, the battery employs thelollowing two hal f-cel I reactions:
Anode: Zn + 2 OH- -+ ZnO(s) + H2O + 2 e'
Cathode: MnO2 + lUzO + 1 e- -+ I MnZO: + OH-
In lieu of manganese oxide, the cells can also use silveroxide and mercury oxide. In acidic medium, the dry cellbattery employs the following two half-cell reactions.
Anode: Zn -+ Zn2+ + 2 e-
Cathode: MnO2 + NH4+ + I e- -+ l}r/rn2Q + | HzO + NH3
The overall cell potential (E") for the dry cell battery isslightly greater than 1.50 V. The nickel-cadmium battery isa rechargeable battery as well. The nickel-cadmium battery(also known as the nicad battery) is found in calculators andelectric shavers. The following two reactions are employedfor the nickel-cadmium battery:
Anode: Cd + 2 OH- -+ Cd(OH)2 + 2 e-
Cathode: NiOOH +H2O + 1 e- -+ Ni(OH)z + 1 OH-
A typical nickel-cadmium battery has a fairly lowvoltage. The outer wall of the anode solution is initially thecadmium metal, but as the cell runs, the cadmium dissolvesaway. Figure 1 shows a cross section of a typical nicadbattery. As drawn, the outer casing contains zinc metal (inlieu of cadmium) in a basic potassium hydroxide solution.The cathode is made of HgO (NiOZH in typical nicadbatteries), which reduces on the surface of a stainless steelelectrode. The electrons flow to the top of the cathode. Theinsulation holds the cell together.
Figure 1
Copyright @ by The Berkeley Review@
lnsulation
Solution of HgO in a basrcmedium of Zn(OH)2/KOH
Solution of Zn powder in KOH
.#?,'-'@rltiiiLiitllllllllllliiltititiiiiii jiNf {linnni
All of the cells function in a similar manner, allowingsubstitutions to be made to achieve different voltages. Thelifetime of a battery is determined by the quantity of species.
5 5. What is the overall balanced reaction for a nickel-cadmium battery?
A. Cd + 2 NiO2H + 2 OH- -+ Cd(OH)2 + 2 Ni(OH)2B. Cd + 2 NiO2H + 2H2O -+ Cd(OH)2 + 2 Ni(OH)2
C . 2 Cd+ NiO2H + 2 OH- + 2 Cd(OH)2 + Ni(OH)2D . 2Cd + NiO2H +2H2O -+ 2 Cd(OH)2 + Ni(OH)2
5 6. What is the oxidation state change for manganese in theZnlMnO2battery?
A. Mn goes from +4 to +6B. Mn goes from +4 to +3C. Mn goes from +2 to +6D. Mn goes from +2 to +3
5 7. Which of the following reactions represents the cathodereaction in a zinc-mercury oxide battery at pH = 10?
A. Zn-)Zn2+ +2e-B. Zn +2OH-+ZnO(s) +H2O+2e-C. HgO + 2 NH4+ + 2 e- -+Hg + H2O + 2 NH3D. HgO+ H2O +2 e- -) Hg+2 OH-
5 8. Which of the following reactions CANNOT take pla;:at the anode?
A. 2 FeO + 2 OH- -r Fe2O3 + H2O + 2 e-
B. Ag2O + H2O + 2e- -+ Ag(s) + 2 OH-C. Ti(OH)2 + 2 OH- -+TrO2+H2O +2e-D. V2O3 + 4 OH- -+ V2O5 + 2H2O + 4 e-
5 9. The electrons in a dry cell battery flow from:
A . the outer casing to the surface of the stainless :::f-'cap.
B . one wall of the core to the other wall of the core
C . bottom of the insulator to the top of the stair:l:l'.steel cap.
D . from the stainless steel cap to the wall of the cc:;
60. After time, what is observed with the nickei-cadn--inbattery?
A. Electrons build up in the anode.
B. Cadmium metal builds up in the anode.
C . Nickel hydroxide builds up in the anode.D. Cadmium hydroxide builds up in the anode.
292 GO ON TO THE NEXT P.\GM
Passage X (Questions 61 - 67)
One type of electrolysis referred to as electropLating, is atechnique used either to purify a gas or to obtain a pure solidof an easily oxidized metal. For instance, sodium metal and
chlorine gas can be obtained by treating molten sodiumchloride salt with a current of substantial voltage. Theapplied voltage must be greater than the voltage released upon
the oxidation of sodium metal to sodium cation, coupled withthe reduction of chlorine gas to chloride anions. If aqueous
sodium chloride is used, hydrogen gas is obtained rather than
sodium metal, because less voltage is required to reduce protic
hydrogen cations than sodium cations. Table 1 shows the
voltages associated with selected reduction half-reactions:
Reaction emfClz(e) + 2e- -+ ZCl' 1.36 volts
BrZ(1) + 2e- -+ 2Br 1.09 volts
Cu2+ + 2 e- -+ Cu(s) 0.34 volts
2H+ + 2e- -+ H2G) 0.00 volts
Ni2+ + 2 e- -+ Ni(s) -0.23 volts
Zn2+ + 2 e- -+ Zn(s) -0.76 volts
Mn2+ + 2 e- --r Mn(s) -l 18 volts
Table 1
A standard electrolysis apparatus is shown in Figure 1.
The anode is the right electrode, and the cathode is the leftelectrode in conventional electrolytic cells. There is no salt
bridge required when both electrodes exist in the same cell.
Figure 1
Reduction occurs at the cathode, while oxidation occurs
at the anode. The voltage necessary to run the cell gradually
increases over time, because the concentrations change as the
cell runs. Equation 1, the Nernst equation, is used todetermine the cell voltage as the concentrations vary, where Qis the ratio of the anode cation concentration to the cathode
cation concentration.
Eobserved - E'cell - 0'059 1ot qn
Equation 1
61. When aqueous sodium chloride undergoes electrolysis,at25"C, what are the gases formed?
A. Na(g) and Cl2(g)B. Hz(e) and Clz(e)C. Na(e) and OzG)D. Hz(e) and O2(g)
Copyright @ by The Berkeley Review@ 293 GO ON TO THE NEXT PAGE
62. In an electrolytic cell with a molten salt:
A . the boiling point of the electrode must be less than
the melting point of the salt.B. the melting point of the electrode must be less than
the melting point of the salt.C. the boiling point of the electrode must be greater
than the melting point of the salt.D. the melting point of the electrode must be greater
than the melting point of the salt.
6 3 . When must a molten salt be used rather than an aqueous
solution?
A . When the oxidation potential of the cation is largerthan the oxidation potential of hydrogen in water
B. When the oxidation potential of the cation is less
than the oxidation potential of hydrogen in waterC . When the reduction potential of the cation is larger
than the reduction potential of hydrogen in water
D. When the reduction potential of the cation is less
than the reduction potential of hydrogen in water
64. What is TRUE about the left electrode of the standard
electrolysis cell that releases hydrogen and chlorinegases from aqueous hydrochloric acid?
A . The left electrode is the anode, where H2(g) forms.
B. The left electrode is the anode, where Cl2(g) forms.
C. The left electrode is the cathode, where H2(g)
forms.D. The left electrode is the cathode, where Cl2(g)
forms.
6 5 . Which of the following reactions would NOT be
observed when the indicated voltage is applied?
A. NiCl2(l) + Ni(s) + Clz(e) from 1.75 VB. ZnClZ{0) -+ Zn(s) + Cl2(g) from 2.25 YC. NiBr2(l) + Ni(s) + Brz(l) from 1.60VD. ZIBTZQ) + Zn(s) + BrZ(l) from 1.50 V
66. Why does copper metal plate from CuCl2(aq), whilenickel metal does NOT plate from NiCl2(aq)?
A. E'red(Cu2+; > E'rcd(H+; > E'rcd(Ni2+)
B. E"1s611q;2+; > E"1sd(H+; > E"1sd(Cu2+)
C. E"red(Cu2+; > E'1sd(Ni2+; > E'1gd(H+)
D. E'1g61H+) > E"red(Cu2+; > E'rcd(Ni2+)
6 7. Which of the following electrolysis reactions can be
carried out in an aqueous solution, as well as in a
molten salt?
A. CuBr2(l) -+ Cu(s) + Brz(l)B. MnCl2(l) -+ Mn(s) + Cl2G)C. ZnCl2Q) -+ Zn(s) + Clz(g)D. NiBrz(l) -+ Ni(s) + Br2(l)
Copyright @ by The Berkeley Review@ 294 GO ON TO THE NEXT P.{C,'M,
Passage Xl (Questions 68 - 74)
Pure gases can be obtained through a process known as
electrolysis. Electrolysis is the cleaving of a compound
using an applied voltage. A typical example is the
conversion of water liquid into hydrogen gas and oxygen gas.
Electrolysis can also be used to form metals from cationic
salts. Figure 1 is a schematic layout of an electrolysis cell
used industrially to make chlorine gas from aqueous sodium
chloride:
Figure 1
The oxidation of chloride anion must use less voltage
than the oxidation of the oxygen of water; otherwise, the
reaction must be carried out in the absence of water. Sodiun
transfers electrons from the cathode of the applied voltage irthe lower electrolytic region to the hydrogens of water in the
upper galvanic region. Because the voltage is reducin-E
sodium, the applied voltage must be high enough to reduce
sodium cation and oxidize chloride anion' Heat energf is
released when the sodium metal reduces the hydronium ion r.'
water. The appropriate half-cell reactions for the electro1l s.'
apparatus shown in Figure I are listed below:
Clz(g) + 2e- -+ 2 Cl-(aq) +1.36 V
Na+(aq) + 1e- +Na(s) -2.71V
H+(aq) + 1e- -+ trrtt, 0.00 V
6 8. Overall, what occurs in the cell as written?
A . Na is oxidized, and Cl is reduced.
B. Na is reduced, and Cl- is oxidized.
C . H+ is oxidized, and Cl- is reduced.
D . H+ is reduced, and Cl- is oxidized.
I .'rtrlt
1n,,,,
69. What must be true regarding the applied voltage in
Figure 1?
A. It must lie between 0 and 1.36 volts.
B . It must lie between 1 .36 and 2.71 volts'C. It must lie between 2.71 and 4.07 volts.
D. It must be greater than 4.07 volts.
70. The overall reaction is:
A. NaCl(aq) + HCI(aq) -+ Hz(g) + Na(s) + Clz(g)
B. Na+(aq) + H2O(1) -+ HzG) + Na(s) + Oz(g)
C. NaCl(aq) + H2O(l) + Hz(g) + NaOH(aq) + Clz(e)
D. NaCl(aq) + Hz(g) + HzO(l) + Na(s) + Clz(g)
71. Which of the following statements describes sodium
cation in the electrolysis cell?
I. Sodium cation is more favorably reduced than
hydronium cation.
II. Sodium cation is reduced in the salt solution
segment of the cell, because it can cross the
membrane into mercury metal'
Itr. The [Na+] in the NaCl(aq) solution that enters the
electrolytic cell is greater than the [Na+] in the
NaCl(aq) solution that leaves the cell.
A. I onlyB. I and II onlyC. II and III only
D . I, II, and III
7 2. Electron flow is defined as moving:
A. from anode to cathode in both the battery and the
electrolytic cell.B. from cathode to anode in both the battery and the
electrolytic cell.C. from anode to cathode in the battery, and from the
cathode to the anode in the electrolytic cell'
D. from cathode to anode in the battery, and from the
anode to the cathode in tbe electrolytic cell.
7 3. In an electrolytic cell, what must be TRUE about the
cell and battery voltages?
A. Both values are positive; Ebutt"ry > E.*.
B. Both values are positive; Ebutt"ry < E.*.
C. Ebattery > 0 and E.* < 0; Ebattery > lErxl'
D. EUatt".y > 0 and E.* < 0; Ebattery < lErxl'
Copyright @ by The Berkeley Review@ 29s GO ON TO THE NEXT PAGE
74. Which of the following is the STRONGEST reducing I
agent?
A. Na+(aq)
B. Na(s)C. Cl-(aq)D' Hz(e)
Passage Xll (Questions 75 - Bl)
A researcher sets up an electrochemical cell, Cell 1 inFigure 1 below, by placing a strip of zinc metal into a
solution of 0.10 M ZnSO4 in the left half-cell and a strip ofcopper metal into a solution of 0.10 M CuSO4 in the righthalf-cell. The two metal strips are connected by a copperwire connected to a voltmeter. The half-cells then areconnected by a string soaked in NaZSO+(aq).
The researcher next creates a second cell, Cell 2 in Figure2 below, by placing a strip of copper metal into a solution of0.001 M CuSO4 in the left half-cell and a strip of coppermetal into a solution of 1.00 M CuSO4 in the right half-cell.
The two metal strips are connected by a copper wireconnected to a voltmeter. The half-cells are then connectedby a string soaked in Na2SO4(aq).
{: Figure
in a concentration cell is the equilibrating of concentration inthe two half-cells. The first cell is driven by the standard cellpotential. Listed in Table 1 below are the reductionpotentials for selected standard half-cells:
Half-reaction E'.oll
As+ + le- -+ As 0.80 vCu2+ + 2e- -+ Cl 0.34 VFe3+ + 3e- -+ Fe -0.04 vNiz+ + 2e- -+ Ni -0.23 V
FeZ+ + 2e- -+ Fe -0.44 V
CrJ+ + 3e- -+ Cr -0.73 VznZ+ + 2e- --> Zn -0.76 VAIJ+ + 3e- -+ Al 1.66 V
Table 1
Copyright O by The Berkeley Review@
75. In Cell I, the anode and electron flow are correctlydescribed by which of the following statements?
r'l--d, The anode is the zinc electrode, and the electron
___-$o* is from right to left.(-n. the anode is the zinc electrode, and the electron
flow is from left to right.C. The anode is the copper electrode, and the electron
flow is from right to left.D. The anode is the copper electrode, and the electron
flow is from left to right._ ei1 1..,, ,." r\ r
.j*--t -! ? ,. '' i '' .. 1''-1
7 6 . Cell I is BEST described as which of the following?
A. An electrolytic cellBi A concentration cell
CelAgalvaniccell-'-- ''"' 'i' " -'--ri'
-D; AchargingcellL.t -i t, \ t.' l-/
iq
\ -.7..*: lf the string in Cell I were replaced by a copper wire.{ whichi-f the following would be the cell potential foi
i-r Cell I ?\ '/ A. 1.46v
5fi'rovc. 0.76 vD. 0.00 v
J :1 --'-'t,1 ,-.."--'1. ^T "'"'
4,' 4: ! a{ "1 {'
')''t r*--r''i r' i '
-i:
f,
-. --.- Cell 2 is closed, EXCEPT:,\ )\/ -ffilectrons flow from left to right, gradualii
, slowing, until to no electron flow is observed.
/ qB-. The mass of the left copper electrode decreases
while the mass of the right copper electrodeincreases.
C . The copper cation concentration in the ngh:electrode decreases to a value of roughly 0.50 M.
.-D . The copper cation concentration in the left electro'tre
increases to a value of approximately 1.00 M.
A. ouof the following are observed, once the switch ir
7 9. Given that the cell potential can be calculated b1 :'tfollowing formula, what is the cell potential in Cell I -
E = E'- o.?tn tor q
A. 0.177 VB. 0,089 Vc. 0.517 VD. 0.429 V
Salt bridge
i'
296 GO ON TO THE NEXT P.\G'U
8 0 . What is the standard cell potential for Cell 1?/-''-(A. 1.10 vB. 0.42 Vc'. 2.20v"6. -0.+zv
81. What are the coefficients when the following equation is
balanced?
H2O(l) + ClO3-(aq) + Mg(s) -+Cl-(aq) + Mg2+1aq; + OH-(aq)
x.2'-N e6-:zp.3
II2
I
2->l:2:43->l:3:33->2:3:63->1:3:6
ior"t;o
2\^ro
-{C
,' .1''. ,.!Q'1,,
ltrt-* n -..&1 "j
-t) ( t c':. -"",4"uiJ --i
"', J 1..
tl{+..- -r"r '2 c)
\;, ci -i '-?
ct-+
?ci r
,q 7tr.-.) |
)l .2M<'''
.- .;+'!.1"
Copyright @ by The Berkeley Review@ 297
7ol{'
Passage Xlll (Questions 82 - 87)
With items made from metal, corrosion is a majorproblem. Corrosion is the air-induced oxidation of metal inthe presence of moisture. Water serves as a solvent throughwhich anions and cations may flow, to assist in carrying out
the oxidation-reduction process. Several techniques have been
developed to minimize, and even eliminate, the effects ofcorrosion. The techniques of concern are galvanizing,coating, and cathodic protection (the insertion of sacrificialmetal). Each has its benefits and drawbacks.
Galvanifing: Galvanizing is the process of mixing twometals of different oxidation potentials. One of the metals
is oxidized preferentially over the other, leaving behind the
other metal in its elemental form. For instance, iron can
be galvanized with zinc so that zinc oxidizes away, leaving
the iron. If zinc were not present, iron would be oxidizedand eventually weaken.
Coating'. Coating is the process of plating the surface of ametal with a material that are not oxidize. For instance,
'chromium, tin, and aluminum form oxide coatings through
which electrons cannot transfer. The coating serves to
insulate the metal from the environment.
Cathodic protection: Cathodic protection involves running
a wire through the material and connecting the end of the
- wire to an easily oxidized metal. The sacrificial metal isoxidized, leaving the guarded metal intact. This technique
is often used to protect long iron pipelines.
To apply these techniques, the electromotive potentials
for each component in the mixture must be known. Table Ilists the emf for favorable half-reactions of selected metals.
Ca(s) -) Ca2+1aq) + 2e- 2.81V
Me(s) -+ Mg2+(aq) + 2e' 2.36V
Al(s) --) Al3+(aq) + 3e- 1.66 V
Mn(s) -+ Mn2+(aq) + 2e- 1.18 V
Pd2+1aq) + 2e- -+ Pd(s) 0.99 V
Zn(s) -) Zn2+(aq) + 2e- a.75Y
Fe(s) -+ Fe2+(aq1 + 2e' o.44 V
Cu2+(aq) + 2e- -+ Cu(s) 0.34 V
Ni(s) --) Ni2+(aq) + 2e- 0.28 V
Pb(s) -+ Pb2+(aq) + 2e 0.13 V
Table I
The emf for an oxidation or reduction half-reaction not
listed in Table I can be determined, if two or more reactions
that sum to that half-reaction are known. Hess's law is used
to determine AG" values, but not the emf values' The overall
AG" value is found by summing the AG' values for the
component reactions. The AG" for each half-reaction is found
from the emf vaTtes given. The emf for the overall reaction
is found using Equation 1.
AG'= -nf€'
Equation IGO ON TO THE NEXT PAGE
,su.
,A11
To calculate the emf for the oxidation of Cr to Cr3+, the
AG" for oxidation of Cr to Cr2+ and Cr2+ to Cr3+ may be
used. The two oxidation half-reactions, and theircorresponding AG" and emf values are shown below:
Cr(s) -+ Cr2+1aq; + 2e- e' = +0.89V AG" = -2(0.89)f
Cr2+(aq) -+ Cr3+1aq; + le- e" = +0.41 V AG" = -(0.41)f
The emf for the oxidation of Cr to Cr3+ is calculatedfrom the AG" for oxidation of Cr to Cr3* as follows:
cr(s) -) cr3+1aq; + 3e- AG" = -2(0.89)r + -Q.4Dr
€. = -AG _-Gl.78f + (-0.41D)
=z.t9,F = 0.73 Vnf 3ry 3r
and its corresponding e"
for the oxidation of iron
Fe2+1aq; -+ Fe3+1aq; + 1e- e" = -0.77 Y
K"?tn!a>:!g!l)3
*. 2(+0.441- (0.77)/.
-l
.' ? .)(+0.44) + 2(-0.71)\_-,/
3
,. (+0.44) -2(-0.17)'3
8 3. Given that Cu2+(aq) + e- -+ Cu+(aq) has e" = 0.15 V,what is the free energy change for the followingreaction?
Cu(s) -+ Cu+(aq) + e-
A. AG" = (2(-0.34) +0.15))fB. AG' = (2(-0.34) -(0.tsDrC. AG'= (2(0.34) +(0.15))fD. AG" = (2(0.34) -(0.15)r
8 4. Which of the following is NOT possible as a sacrificialcoating to protect the indicated metal?
A. Aluminum as the sacrificial metal to protect ironB , Zinc as the sacrificial metal to protect copper
$,,:'rZinc as the sacrificial metal to protect iron
1D./Nickel as the sacrificial metal to protect zinc
Given the following reactionvalue, what is the emf valuemetal to iron trication?
n
Copyright @ by The Berkeley Review@ 294 GO ON TO THE NEXT PAGE
1 ) , ,,.1- .,'' r' "' '. - i,. ,.
t
$S. Wtrictr of the following reactions is NOT favorable as
.,' written?
(e. Rtfsl + Cr3+1aq; -+ Cr(s) + Al3+(aq)Xun(r) + Pd2+1aq; + Pd(s) + Mn2+1aq;
<7 ZnG) + Ni2+1aq; -+ Ni(s) + Zn2+1aq1
D. Pd(s) + Cu2+1aq; -+ Cu(s) + Pd2+1aq; '
Y; - :. ''., /
(',.,'' ' ' r:-
I-!
8 6. Which of the following graphs represents the change in
the mass of zinc metal and iron metal as a function oftime in a galvanized steel sample?
-----Zn
-PeA. B.
8 7. Which of the following would NOT form a protectrve
oxide coating on copper?
A. AluminumB. ChromiumC. TinlI. Palladium
Time--+
Time
-.>
Time+
Passage XIV (Questions 88 - 93)
A student sets up an experiment to determine the effectsofthe cell potential, concentration, and cunent on the degree
of electroplating. In the first experiment, the student sets upsix solutions with the following contents:
Solution I: 0.10 M AgNO3 (aq)
Solution II: 1.00 M AgNO3 (aq)
Solution III: 0.10 M ZnSO4 (aq)
Solution IV: 1.00 M ZnSO4@q)
Solution V: 0.10 M CuSOa (aq)
Solution VI: 1.00 M CUSOa (aq)
Into each solution, an inert electrode weighing exactly5.00 grams is inserted. A steady current of electricity isapplied to each electrode for the duration of one minute for allsix solutions. The electrode is then removed and the mass ismeasured. The degree of plating can be determined bysubtracting the original mass of the electrode from the finalmass of the plated electrode. The mass for each electrode afterit is removed is listed below:
Solution I: 6.187 grams
Solution II: 6.182 grams
Solution III: 5.360 grams
Solution IV: 5.358 grams
Solution V: 5.347 grams
Solution VI: 5.350 grams
The difference between the solutions in the mass afterplating of the electrode is attributed to both the difference inmolecular mass for each element, and the number of electronsrequired to reduce each cation. The degree of plating can be
calculated, if the current and the duration of time are bothknown. The conversion from coulombs to moles involvesFaraday's constant (96,500 coulombs per mole).
The procedure in this experiment can be used to formpure samples of metal. The metal coating adheres to theelectrode until it is removed using the application of someforce. This technique is used in the purification of gold.Electroplating is also used in the application of protectivemetal coatings to materials. Chromium can be applied tometal surfaces in this manner.
88. If an equal current is applied for an equal duration oftime to separate flasks containing the aqueous ionsgr2+,7n2+, Ag*, and Au3+, which ion yields theMOST precipitated metal by mass?
A. Cu2+B. zn2+c. Ag*D. Au3+
Copyright @ by The Berkeley Review@ 299 GO ON TO THE NEXT PAGE
Solution IV
Time
-l>
8 9. Why is it NOT possible to electroplate onto a plasticmaterial?
A. The plastic melts at the high temperatures requiredfor electroplating.
B. The plastic is oxidized by the current.C. Plastic is an insulator and does not conduct
electricity.D. Plastic is too malleable to allow for electroplating.
9 0. Which of the following graphs shows the relationshipof mass as a function of time for Solutions I and IV?
91. Electroplating results from:
A . the oxidation of a metal into a cation.B. the reduction of a metal into a cation.C . the oxidation of a cation into a metal.D . the reduction of a cation into a metal.
9 2. Which of the following is a desirable property of a
metal to be used to coat a structural beam that will be
exposed to moisture?
A. Its cation has a large positive reduction potential.B. The metal has a large positive reduction potential.C. Its cation has a large positive oxidation potential.D. The metal has a large positive oxidation potential.
93. Approximately what mass would be expected for a
solution of Cd2+, if it were exposed for one minute tothe same current and electrode used in the experiment?
A. A mass less than 5.347 grams
B. A mass greater than 5.350 grams, but less than
5.358 grams
C. A mass greater than 5.360 grams, but less than6.182 grams
D . A mass greater than 6. 182 grams
Time+
Time+
uro4j
Time ------>
To calculate the emf for the oxidation of Cr to Cr3+, theAG' for oxidation of Cr to Cr2+ and Cr2+ to Cr3+ may beused. The two oxidation half-reactions, and theircorresponding AG' and emf values are shown below:
Cr(s) -+ Cr2+1aq; + 2e- e" = +0.89V AG' = -2(0.89)f
Cr2+1aq; -+ Cr3+1aq; + le- e' = +0.41 V lC" = -(0.41)f
The emf for the oxidation of Cr to Cr3+ is calculatedfrom the AG' for oxidation of Cr to Cr3+ as follows:
cr(s) *) cr3+1aq; + 3e- AG' = -2(0.89)f + -(0.4Dr
e. = -AG _-(-1.78f +(0.41D) _2_Jg-f-=0.13ynf 3F 3f
lC. Giu"n the following reaction and its corresponding e'
A. value, what is the emf value for the oxidation of iron
/f' metal to iron trication?- Fe2+(aq) -+ Fe3+1aq; + le- e' = -0.7'7 Y
o,. 2(+0.44) + (-0.77)
J
O. 2(+0.44)-- (-0.77)
r )G0.44) + 2(-0.77),_>2 3
o. G0.44)-2(-0.17)3
83. Given that Cu2+(aq) +e- + Cu+(aq) has €'= 0.15 V,what is the free energy change for the followingreaction?
Cu(s) -+ Cu+(aq) + e-
A. AG' -- (2(-0.34) +0.15))fB. AG' = (2(-0.34) -(0.ls))fC. AG" = (z(0.34) +(0.15))fD. AG" = (2(0.34) _(0.15)r
., 8 4 . Which of the following is NOT possible as a sacrificial/ | coating to protect the indicated metal?
A. Aluminum as the sacrificial metal to protect ironB . Zinc as the sacrificial metal to protect copper
llZinc as the sacrificial metal to protect iron
6E.TNickel as the sacrificial metal to protect zinc\_/
Copyright @ by The Berkeley Review@ GO ON TO THE NEXT PAGE
I
*s . Which of the following reactions is NOT favorable as
written?""4-
^ '
, A. Alisl + Cr3+1aq; -+ Cr(s) + Al3+1aq)-#tvtn(r) + Pd2+(aq) + Pd(s) + Mn2+(aq){ Zn1s1+ Ni2+(aq) -+ Ni(s) + Zn2+1aq1
D. Pd(s) + Cu2+(aq) + Cu(s) + Pd2+1aq)
ii / -!
8 6. Which of the following graphs represents the change inthe mass of zinc metal and iron metal as a function oftime in a galvanized steel sample?
-----zn
-FeA. B.
87. Which of the following would NOT form a protectiveoxide coating on copper?
A. AluminumB. Chromium.C. TinII. Palladium
q
a
c.
1
z
Time --> Time+
Time-'+ Time+
Passage XIV (Questions 88 - 93)
A student sets up an experiment to determine the effectsofthe cell potential, concentration, and current on the degreeof electroplating. In the first experiment, the student sets upsix solutions with the following contents:
Solution I: 0.10 M AgNO3 (aq)
Solution II: 1.00 M AgNO3 (aq)
Solution III: 0.10 M ZnSO4 (aq)
Solution IV: 1.00 M ZnSO+ (aq)
Solution V: 0.10 M CuSO4 (aq)
Solution VI: 1.00 M CuSOa (aq)
Into each solution, an inert electrode weighing exactly5.00 grams is inserted. A steady current of electricity isapplied to each electrode for the duration of one minute for allsix solutions. The electrode is then removed and the mass ismeasured. The degree of plating can be determined bysubtracting the original mass of the electrode from the finalmass of the plated electrode. The mass for each electrode afterit is removed is listed below:
Solution I: 6.187 grams
Solution II: 6.182 grams
Solution III: 5.360 grams
Solution IV: 5.358 grams
Solution V: 5.347 grams
Solution VI: 5.350 grams
The difference between the solutions in the mass afterplating of the electrode is attributed to both the difference inmolecular mass for each element, and the number of electronsrequired to reduce each cation. The degree ofplating can becalculated, if the current and the duration of time are bothknown. The conversion from coulombs to moles involvesFaraday's constant (96,500 coulombs per mole).
The procedure in this experiment can be used to formpure samples of metal. The metal coating adheres to theelectrode until it is removed using the application of someforce. This technique is used in the purification of gold.Electroplating is also used in the application of protectivemetal coatings to materials. Chromium can be applied tometal surfaces in this manner.
8 8. If an equal current is applied for an equal duration oftime to separate flasks containing the aqueous ionsgr2+,7n2+, Ag*, and Au3+, which ion yields theMOST precipitated metal by mass?
A. Cu2+B. Zn2+c. Ag*D. Au3+
Copyright @ by The Berkeley Review@ 299 GO ON TO THE NEXT PAGE
Time
->
8 9. Why is it NOT possible to electroplate onto a plasticmaterial?
A. The plastic melts at the high temperatures requiredfor electroplating.
B. The plastic is oxidized by the current.C. Plastic is an insulator and does not conduct
electricity.D. Plastic is too malleable to allow for electroplating.
9 0. Which of the following graphs shows the relationshipof mass as a function of time for Solutions I and IV?
91. Electroplating results from:
A. the oxidation of a metal into a cation.B. the reduction of a metal into a cation.C . the oxidation of a cation into a metal.D . the reduction of a cation into a metal.
9 2. Which of the following is a desirable property of a
metal to be used to coat a structural beam that will beexposed to moisture?
A. Its cation has a large positive reduction potential.B. The metal has a large positive reduction potential.C. Its cation has a large positive oxidation potential.D . The metal has a large positive oxidation potential.
93. Approximately what mass would be expected for a
solution of Cd2+, if it were exposed for one minute tothe same current and elechode used in the experiment?
A. A mass less than 5.347 gramsB. A mass greater than 5.350 grams, but less than
5.358 gramsC . A mass greater than 5.360 grams, but less than
6.182 gramsD . A mass greater than 6. 182 grams
Time -----J>
Time+
Solution I
Time+
Questions 94 through 100 are NOT based on a
descriptive passage.
94. Tarnish on silver is attributed to Ag2S forming on the
surface. By placing silver in an aluminum pan andadding aqueous baking soda (NaHCO3), the tarnish can
be removed. Which of the following equations showsthat reaction taking place?
A. Ag2S(s) + 2NaHCO:(aq)+ Hz(e) + 2 Ag(s) + NazS(s)
B. Ag2S(s) + 2 NaHCO3(aq)-+CO2(g)+2Ag(s)+S(s)
c. Ag2S(s) + 2 Al(s) -+ 2 Ag(s) + AlzS(s)
D. 3 Ag2S(s) + 2 Al(s) -+ 6 Ag(s) + Al2S3(s)
9 5. Given that AG' = -RT ln K and AG' = -nFE", which ofthe following equations holds true?
A. -RT ln K = nFE'
B. lnK=nFE'RT
C. lnK__nFE'RT
D. lnK=-BT-nFE"
9 6. What are the coefficients for the following oxidation-reduction reaction?
Cr(s) + HI(aq) -> HZ(g) + CrI:(s)
4.28.2c.2D.4
.,)
3
3
3
2
424
If a solution of 1.0 M ZnCl2 were exposed to a 9.65
ampere current for 1,000 seconds, how many grams ofzinc metal would plate onto the cathode?
(F = 96,500 coulombs ;mole electrons
A. 3.269 grams ZnB. 6.538 grams ZnC. 9.807 grams ZnD . 13.016 grams Zn
What is the cell potential for the following cell? The
reduction potentials of nickel and magnesium are -0.23
Y and -2.37 V, respectively.
Mg(s) + Ni2+(aq) + Ni(s) + Mg2+1aq;
A. 5.20 VB- 4.28 Vc. 2.60 vD. 2.t4 Y
6
6
J
6
97.
98.
Copyright @ by The Berkeley Review@ NOW THAT WASN'T THAT BAD]
99. Which of the following graphs represents voltage as afunction of time for a galvanic cell obeying the Nernstequation?
E = E" - 0.059 1ot qn
B.
100. Aluminum metal is formed according to the Hallprocess. The reaction for the Hall process is drawn
below:
2 Al2O3(solution) + 3 C(s) -+ 4 Al(s) + 3 COz(g)
What must be true for this reaction?
A. Aluminum oxide is reduced.B. Carbon is reduced.
C. Aluminum metal is oxidized.D. Carbon dioxide is oxidized.
l.D 2.C 3.C6.D 7.C 8.C
11. C 12. C 13. D16. B 17. D 18. D2t. A 22. D 23. D26. D 21. D 28. A31. D 32. C 33. B36. D 37. D 38. B41. A 42. D 43. C46. B 47. D 48. D51. A 52. D 53. A56. B 57. D 58. B61. B 62. D 63. D66. A 67. A 68. D71. C 12. A 73. C't6. c 11. D 78. D81. D 82. A 83. D86. A 87. D 88. C91. D 92. A 93. C96. A 97. A 98. D
4.D 5. C9.A 10.A
14. D 15. D19. B 20. D24. D 25. A29. C 30. A34. C 35. D39. C 40. A44. D 45. A49. B 50. A54. C 55. B59. A 60. D64. C 65. D69. D 10. C74. B ',75. B19. B 80. A84. D 85. D89. C 90. A94. D 95. B99. D 100. A
A.
to
ttl"
Time+ Time+
Time+ Time-+
Electrochemistry Passage Answers
)
Choice D is correct. The loss of an electron from a compound is referred to as oxidation, so choice A is a validanswer. When an element loses one electron, the process is referred to as ionization (the ionization energy isthe energy required to lose an electron if you recall), so choice B is valid. To convert an oxidation state from -1
to 0 requires losing an electron, because the new oxidation state is more positive (less negative). This makeschoice C valid. The correct choice must be answer D. The electron affinity is the energy released when anelement gains an electron, meaning that choice D does not involve the loss of an electron.
Choice C is correct. Potassium is less electronegative than sodium, so potassium gives off an electron moreeasily. This means that the oxidation potential for potassium is more positive than the oxidation potentialfor sodium. Chlorine is more electronegative than bromine, so chlorine gains an electron more easily. Thismeans that the reduction potential for chlorine is more positive than the reduction potential for bromine. Thebest reaction (most favorable and with the largest cell voltage) is the reaction of potassium with chlorine.This makes choice C the best answer.
Choice C is correct. The oxidation states of hydrogen and oxygen are +1 and -2, respectively. The oxidationstate of chlorine can be found by looking at the difference in the two compounds in question. In HCl, theoxidation state of chlorine must be -1, to keep the overall compound neutral. In HOCI, the oxidation state ofchlorine must be +1, to keep the overall compound neutral. This makes choice C the correct answer choice.
Choice D is correct. From the periodic table, it can be determined that Rb has a lower ionization energy thaneither potassium (K) or sodium (Na). This results in a value for the ionization of Rb that is less than that ofpotassium or sodium, so it is less than either 495.9 or 418.7 kJ per mole. The best answer is choice D, less than478.7 kJ per mole.
Choice C is correct. Because lithium is in the same column of the periodic table as sodium and potassium, thereaction of lithium with water should produce similar products, but with different energy. Both sodium andpotassium when added to water produce metal hydroxide and hydrogen gas. No metal hydride forms, sochoices B and D are eliminated. There is a reaction, so choice A is eliminated. Although both hydrogen gas
and lithium hydroxide are produced, "only lithium hydroxide" is the best answer. Choose answer C.
Choice D is correct. Potassium metal is oxidized by the water, and because reducing agents get oxidized,potassium must be the reducing agent. This eliminates choices A and C. Water is a reactant, not hydrogen gas/
so choice B is eliminated. The correct choice is answer choice D.
Choice C is correct. Metals have a lower ionization energy than non-metals, so metals are more easilyoxidized. This can be confirmed from the highly positive reduction potentials of non-metals and the highlypositive oxidation potentials of metals. Statement I is therefore true. From the reactions listed, metalsdefinitely form metal hydroxides. Several metals form the oxides when the water is removed. Although thepassage does not provide enough information, calcium and magnesium are examples of metals that form bothoxides and hydroxides in water. This makes statement II true. From the reactions of bromine and chlorine, itcan be inferred that non-metals do not form metal oxides and metal hydroxides. The best answer is choice C.
Choice C is correct. The pH of the solution is found using the equation: pH = -log [H*]. In this question, theconcentration is 0.010 M, so pH = -1og (.01) = -log 1,0-2 = 2. Pick choice C for optimum satisfaction.
Choice A is correct, Reducing the pH requires making the solution more acidic. Increased acidity results froman increase in the [H+]. Only the addition of HCI(aq) (choice A) would result in an increase in the [H+]. ChooseA. Answer choices B and C are bases, so when they are added to solution, the result is a decrease in the [H+].Choice D is a neutral salt, meaning that the only affect it has is to dilute the solution, which also results in adecrease in the [H+].
J.
4.
5.
6.
n
8.
9.
Copyright @ by The Berkeley Review@ 301 Section X Detailed Explanations
13.
10.
11.
12.
15.
't6.
17.
1.8.
Choice A is correct. As base is added, the concentration of hydronium decreases. The passage provides thefollowing equation: Eobserved = E'cell - A.0592log [H+]. as [Ui] decreases, the pH increases, which eliminateschoices B and D' As [H+] decreases, the log [H+] (which equals -pH) gets more^negative. This means a largernegative number is subtracted, thus making the number more positive. This makes the value of Eo6ru.rr"6increase. Pick choice A today and smile tomorrow.
Choice C is correct. The pH read by the pH meter is adjusted linearly using the Nernst equation. plugging thevoltage values into the Nernst equation yields: Eobserved = E"cell - 0.0SgZ 1og [H+]. If the ?eal value foi-E"""11 isactually higher than the,value being plugged in, then a value that is too large is being subtracted to getEobserved. This means that the [H+] is too high, so the calculated pH is too low. This ii best explained inanswer choice C.
Choice C is correct. From the equation in the passage, it can be seen that Hg2Ct2(s) is gaining electrons, whichis defined as reduction. This eliminates both choice B and choice D. In Hg2Cl2(s;, the oxidation state o{mercury is +1, so mercury in the reactant has an oxidation state of +1 and 0 in the product. Mercury is the atombeing reduced in the reduction half reaction. Chlorine has an oxidation state of -1 befor" and after the reaction.This means that chloride is simply a spectator ion, and it does riot get involved in the oxidation-reductionreaction. ln light of this revelation, the correct answer choice for this question must be choice C.
Choice D is correct. Again, Equation 1 from the passage must be used. Eo6r"rved = E"cell - 0.01gzlog [H+] withthe Eo6rurrr"4 is 0.699V yields:
0.69e = 0.285 - .0s92(log tH*l) + 0.414 = - .ossz(log tu+l)
- 0.414 - -log[H+] = pH ... pH = 0.414 = 47.4 = Q _ 70.0592 0.0592 5.92 6The closest answer choice is 7 , so you really should pick D.
Choice D is correct. A change in temperature always shifts the equilibrium of a reaction. The equilibrium oithe reaction affects the [H+], so pH of the solution is affected by a change in the temperature. ihoi"" A is avalid statement. Because the pH of the solution is dependent on the hydronium concentration ([H3O+]), achange in the volume of solvent changes the [H3O+], and thus, changes the pH. Choice B is a valid statement.The degree to which a proton can dissociate into solution depends on the solvent, so the [HaO*] and thereforepH vary with the type of solvent as well. Choice C is a valid statement. The solution is homogeneous, so ncmatter where the electrode probe is positioned in solution, it reads the same [HgO*], so pH d,oei not vary wit].position of the electrode. Choice D is an invalid statement, so it is the best choice.
Choice D is correct. The cyanide anion is a ligand (lone pair donor) to the gold cation. The gold cation formswhen oxygen gas oxidizes the gold metal. The cyanide ligand is simply acting as a Lewis base. The best ansn,e:is choice D.
Choice B is correct. The sum of the oxidation states of the atoms in the compound must add up to the overa-,charge on the compound. The cyanide ligand carries a -1 charge, so the gold cation must carry a +1, in order fc:the sum of the two cyanide ligands and one gold cation to have a -1 charge. The correct answer is choice B.
Choice D is correct. Sulfur goes from a -2 oxidation state in zinc sulfide to a +4 oxidation state in sulfur dioxicieso sulfur has been oxidized. This means that statement I is not true. Zinc goes from a +2 oxidation state in zin:sulfide to an oxidation state of 0 in zinc metal, so zinc has been reduced. This means that statement II is tnieOxygen gas gets reduced by oxidizing the sulfur in the first reaction, so oxygen gas is the oxidizing agent. Th:-.means that statement III is true. The correct answer is choice D.
Choice D is correct. Silicon is reduced from an oxidation state of +4 to an oxidation state of 0. The silicon --.therefore reduced by a total of four electrons, making choice D the only reasonable choice. The reaction does r.c:need to be balanced; the four electrons are enough to distinguish the best answer.
't4.
Copyright O by The Berkeley Review@ Section X Detaited Explanations
19.
20.
Choice B is correct. Oxygen is only oxidized when heat is applied to mercury oxide. All of the reactions listedshow oxygen in a reduced form in the final product, except when it exists as oxygen gas. The best answer ischoice B.
Choice D is correct. Mercury is not oxidized in the first reaction; it remains with an oxidation state of +2. Thiseliminates choice C. Sulfur goes from -2 in HgS to +4 in SO2, so sulfur is in fact oxidized by six electrons. Thismakes choice D the best answer. Mercury is reduced by two electrons in the second reaction, so choices A and Bare both wrong.
21-.. Choice A is correct. As we read in the passage, the_ dichromate anion precipitates out of solution with leaddication as an ionic solid. As the concentration of Pb2+ decreases, the obierved voltage decreases, because thedenominator of the Q term in the Nernst equation becomes smaller, making the overall value larger. The resultis that the cell voltage decreases. This is best explained as choice A. The last three choices should beeliminated, because Pb2+1aq; and Cr2O72-(aq) do not react with one another by any redox reaction, onlyprecipitation.
Choice D is correct. The lead-zinc galvanic cell has an E"g"11 of -0.13 - (-0.76) = 0.63 V. The dichromate-zincgalvanic cell has an E"qs[ of 1.33 - (-0.76) = 2.09 V. Regardless of the concentration difference between the zinccation and the other species in solution, the E"C"lt for the dichromate-zinc cell is so much greater than thelead-zinc cell that the E66sstr"4 is greater for the dichromate-zinc cell. The reason is that the reductionpotential of dichromate (1,.33 V) is much higher than the reduction potential of lead (-0.13 V). The best answeris choice D. Choices A and B can be eliminated, because dichromate (Cr2O72-) and lead dication (fbz+, do notoxidize, according to the data presented. Even if they do, their presence in the cathode half-cell requires thatthey undergo reduction rather than oxidation. From the data presented, choice D is true, while choice C isfalse.
Choice D is correct. The sulfate anion is a spectator ion that balances out the positive charge in each cell. Thelead cation is reduced at the cathode so the [Pbz+, is decreased; while zinc metal is oxidized at the anode, sothe lZn2+l is increased. This means that the sulfate anion must flow from the tathode to the anode through thesalt bridge to keep the net charge in each half-cell balanced. The best answer is therefore choice D.
Choice D is correct. When the switch is closed, the circuit is complete, and the cell begins to produceelectricity. As the cell produces electricity, the [fUz+, slowly deireases. This prevents the iccuratemeasurement of the amount of lead present in the cathode half-celI, so determination of the exact equivalenceis not possible. Given the answer choices, the best answer is choice D, because the voltage is depleted. ChoiceB is a throw-away choice, and regardless of whether the circuit is open or closed, the lead and zinc can possiblymigrate through the salt bridge, eliminating choice C.
Choice A is correct. Zinc metal is being oxidized into zinc cation (zinc is losing electrons), so choices C and D areeliminated. Oxidation occurs at the anode, so choice A is the best answer. You should recall that oxidationoccurs at the anode and reduction occurs at the cathode.
Choice D is correct. The observed voltage reaches a minimum when the concentrations of lead and dichromateare closest to zero. This occurs when the moles of Pb2+ present initially equals the moles of. Cr2O72- added.When volume is multiplied by molarity, the result is moles of solute. Only choice D lists when the moles of thetwo species are the same. Pick D and be a star-studded chem trooper.
'J
27. Choice D is correct. The endpoint of titration occurs when the clear Fe2+ solution tums and remains purple fromthe excess MnO4-(aq) in solulion following the complete oxidation of Fe2+ by MnOa- (KMnOa is an oxidizingagent). MnO is a solid, so the solution does not turn brown, but a brown precipitate forms on the bottom of theflask. Over-titration with KMnO4 results in the purple color remaining in solution. It tums clear only if theKMnOa reacts. The lingering of the purple color indicates that the reaction is complete. Pick D.
23.
24.
25.
26.
Copyright @ by The Berkeley Review@ 303 Section X Detailed Explanations
28. Choice A is correct. Flask 3 has twice as much solute by weight and half the volume of solution compared to
Flask 2. Relatively, as measured in grams per liter, the following is true:
MFIask, = fi:#*bLffi xMFIask, = ? "MFhsk2 = 4 xMFlask2
2
Flask 3 has four times the concentration of Flask 2, as measured in grams per liter. It is given that Flask 2
requires a 17.50,mL aliquot of Na2Cr2O7(aq) to reach equivalence, so Flask 3 (containing 20.00 mL solution)
re{uires four times the volume of titrant (4 x 17.50 mL) = 70.00 mL Na2Cr2O7(aq) solution. Flask 3 requires a
70.00-mL aliquot, so choice A is the correct answer.
Choice C is correct. Since we know that KMnO4 is the oxidizing agent in-the reaction, Fe2+ must be the reducing
agent. Based on this, you must choose C. The Fe2+ ion is oxidized to Fe3+, so it causes reduction, thus defining
itself as the reducing agent of the reaction.
Choice A is correct. In Flask 2, FeSO4 is titrated with a Na2Cr2O7(aq) solution. Na2Cr2O7(aq) acts similariy
to KMnO4 and is also a great oxidizer, so it must be the oxidizing agent. Choose A'
choice D is correct. First, you should write out the unbalanced equation:
Fe2+ + Na2Cr2o7 -+ Fe3+ + Cr2o3
From here, the first step is to balance the electrons in the two half-reactions. The ferrous (+2) cation loses one
electron, while the dichromate gains three electrons per chromium (and thus six electrons per compound). To
balance the electrons, the ferroui and ferric (+3) cations must both be multiplied by a factor of six.
6Fe2+ + 1, Na2Cr2o7 --> 6Fe3+ + 1 Cr2o3
This solves the question and allows you to choose answer choice D'
On the MCAT, you should stop at this point and not waste time going further. Because this is a learning
environment, however, we wid continue to balance the reaction completely. From here, the next step is to
balance the charges on each side of the reaction. The reactant side has a +12 net charge, while the product side
has a +20 net chirge. To balance the charges requires that an additional +8 be added to the reactant side of the
equation. This is done by adding 8 H+ to the left side of the reaction.
29.
30.
31.
The final step is to balance the atoms in the equation. There are seven oxygen atoms on the reactant side and
only three ot, ttn" product side. The product side requires four more oxygen atoms, so to balance the atoms, 4H2O
must be added to the product side:
8H++ 6Fe2+ +1Na2Cr2O7overall charge = +20
-+ 6Fe3++1Cr2O3+2Na+overall charge = +20
-) 6Fe3+ + 1 Cr2O3 + 2Na+ + 4H2Ooverall charge = +20
BH++ 6Fe2+ +1Na2Cr2O7overall charge = +20
The ratio of Na2Cr2O7(aq) to Fe3+1aq) is confirmed to be L : 6. This correlates to answer choice D' The ratio can
be found quickly Uy .o-pur*g the'oxidation and reduction counts. Fe is oxidized by one electron, while the
chromium compounds aie red"uced by three electrons per chromium (from +6 to +3). Because there are two
chromium atoms in Na2Cr2O7, the total amount of electrons necessary to reduce the compound is six (2 x 3
electrons). This makes the ratio 1 : 6'
Choice C is correct. The oxidation half-reaction of Flask 1 is Fe2+ -+ Fe3+ + 1e-
Component reactions: Fe2+ + 2e- -+ Fe -0.44 VFe -+ Fe3+ + 3e- 0.04 V
When the two reactions are added together, the overall oxidation half-reaction is found. This means that the
E" for the half-reaction can be found by summing the two voltages. E" = -0'44 + 0.04 = -0.40 V. Pick C'
Choice B is correct. The best container is made of the least reactive material. Glass (SiO2) is already in a
fully oxidized form, so it is unreactive with respect to oxidation-reduction chemistry. Glass is thus the best
maierial in which to carry out the titration. Aluminum readily oxidizes, coPper slowly oxidizes, and
polyethylene oxidizes over time in air. The best answer is B'
32.
33.
Copyright O by The Berkeley Review@ 3o4 Section X Detailed ExPlanations
34. Choice C is correct. Gold trication (Au3+, has a reduction potential of 1.50 V, so almost any compound that canbe oxidized can reduce gold trication. The Cl2(g) has chlorine atoms with oxidation states of 0, so they won't beoxidized (lose electrons to be cations). Ag+ is oxidized in its cation form, so Ag+ cannot be oxidized any further.Pt2+ is already oxidized as it is, so it cannot be oxidized any further. The best choice is zinc metal (Zn(s))which readily loses two electrons. Pick C.
Choice D is correct. The reduction potential for silver cation is 0.80 volts. This means that because chlorideanion is nof reduced to Cl2(g), the oxidation potential of Cl- to CIZ(S) is less than -0.80 volts. Reversing thismeans that the reduction potential of C12(g) to Cl- is greater than +0.80 volts, which is selection D.
Choice D is correct. The chlorine atoms in the compound have an oxidation state of -1 each. This means thatfor the overall charge of the compound to be -1, the oxidation state of the gold must be +3 (4 (-1) + (+3) = -t),choice D.
Choice D is correct. Silver (Ag) has an electronic configuration of [fr]Ss1+d10 when neutral because of thefilled d-shell stability associaied with the d10 electronic configuration. The electron is lost from theoutermost shell when the silver becomes a cation. This would be the 5s1 electron that silver loses, whichleaves an electronic configuration for Ag+ of: [Kr]4a10. Pick D.
Choice B is correct. A sequestering agent must bind a metal cation, so it must be capable of donating a lone pairof electrons to a metal cation in a ligand fashion. Sequestering agents donate electron pairs, so they are Lewisbases. Atoms without filled octet valences can accept electrons, which makes them good Lewis acids. Thiseliminates both choice C and choice D. Choice B is the best answer.
Choice C is correct. The mass percent of Au in AuCl4- can be solved as follows:
Mass Au 197 = 1'97 > l. efiminatins choicesMass AuCl4- 197 + a (35.5) 197 +1'42 2
ting choices A and B.
L97 = W- < ?7-9 = 3- ehminating choice D.07 + a Q5.5) 339 360 4
This leaves only choice C, which you should pick for best results. The actual value can be approximated as
follows:197 -797 -ahttle = 591, -alittle=59%-alittle.339 333 999
Choice A is correct. Because chlorine is oxidized in each answer choice, the question is really which cation has
the lowest reduction potential, since the metal cations are reduced in each case. The lowest reductionpotential, according to the chart, is found with Ag+. This makes choice A the correct choice.
Choice A is correct. In all galvanic cells, oxidation occurs at the anode, so electrons are lost from the anode and
hence flow from the anode to the cathode. The best answer is choice A.
42. Choice D is correct. Oxidation occurs at the anode, and the loss of electrons occurs with oxidation, so choice Dmust be the best answer. It is only in choice D that electrons are lost.
49. Choice C is correct. In a galvanic cell, the E1s4r61ion * Eoxidali6l must be greater than 0 volts. If Eo*i4u1ion is
greater than 0 volts, the overall voltage is positive, so choices B and D are galvanic cells, meaning they are
eliminated. The only way that E."11 can be less than zero when Er"4rrction is positive, is when E6ai4uliqn is anegative value with a magnitude greater absolute value than Ereduction. The best answer is choice C.
44. Choice D is correct. In all cells, electron flow is from the anode to cathode, and reduction occurs at the cathode.
This makes both choices A and B true in a galvanic cell, so they can be eliminated. Choice C is true according tothe Nernst equation, because the cell eventually dies out once the voltage has dropped. The voltage neverdrops below iero, however, so choice D is not true. The best answer is choice D, because the cell stops when the
voltage equals zero (not when it is less than zero).
35.
36.
3/.
38.
39.
40.
41.
Copyright O by The Berkeley Review@ 305 Section X Detailed Explanations
47.
45.
46.
48.
49.
50.
51.
Choice A is correct. The reduction of the cation in the cathode is most favorable when it starts at a highoxidation state. This eliminates choices C and D. The anode should contain a metal that is easily (favorably)oxidized, if the cell is to be galvanic. This describes a metal with a low ionization energy. The best answer isthus choice A.
Choice B is correct. To answer this question, the reaction must first be balanced. The oxidation and reductionhalf-reactions are shown below.
Oxidation: Al(s) + AlO2-(aq)+3e Reduction: MnOa-(aq) + 5 e- -+ MnO(s)To balance the reaction from an electronic standpoint (at fifteen total electrons transferred in each half-reaction), 5 Al(s) are needed to react with 3 Mnoa-(aq). Because Al(s) is oxidized, it is the reducing agent,making MnO4-(aq) the oxidizing agent' The ratio of reducing agent to oxidizing agent is therefore 5 : 3. The bestanswer is choice B.
Choice D is correct. At the cathode, reduction occurs, so copper cations in solution plate out onto the surface ofthe electrode' This makes statement I valid. A galvanic *tt Uy definition involvls a favorable reaction thatreleases energy as it runs. The voltage of a favorable reaction is greater than 0.00 volts, so statement II isvalid. To balance the flow of electron charge from anode to catho-de, anions flow from cathode (where thepositive charge diminishes due to reduction) to the anode (where the positive charge increases due tooxidation) through the salt bridge. This makes statement III true. Because all three statements are true, thebest answer is choice D.
Choice D is correct. By definition (and from the passage), electrolytic cells are unfavorablethermodynamically' Energy must be supplied for an electrolyiic cei to run. The electrolytic cell can build uppotential energy that can be harnessed later (the energy can be discharged in the form of a galvanic cell). Thebest answer is therefore answer D.
Choice B is correct. The first sentence of the second paragraph states that the energy of the photon must behigh enough to overcome the electrical potential for the ce[. firis makes choice B correct. Choices A and C aret!9 samg answer just stated differently, so they cannot both be correct on a multiple-choice exam. The electronaffinity for the cathode is usually favorable, so no energy should be required for the gain of an electron by thecathode. The total energy to move the electron
"o*er from the photon and the-affinity for the catfiode(attraction to the storage plate).
Choice A is correct. Photons strike the ionizing plate and emit an electron. The ionizing plate loses an electron,so it has been oxidized. By definition, the anode loses electrons (gets oxidized), *it ltrg choice A the bestANSWEI.
Choice A is correct. The best material is one that can be oxidized very easily. A non-metal and non-metaloxide do not readily lose electrons because of their high electron affinity (and electronegativity). A metaloxide is in a higher oxidation state than the metal, so metal oxides are leis able to share tfieir electrons thanmetals. Metal oxides have already been oxidized. The best answer for this question is a metal, choice A.
Choice D is correct. As is stated in the first paragraph of the passage, a photoelectric cell absorbs a photon,which then emits an electron that flows through a wire in the circuit, ultimately to be stored in a capacitor orbattery' The overall process involves converting a photon into potential energy. A photon is electromagneticradiation, so the best answer is choice D, the converiion of radiation into poten"tial energy. pick choice O If youwant to be outstanding in electrochemistry.
Choice A is correct. If you believe the Nemst equation, then the voltage is constantly dissipating as a cell run-..
]f9 mo_st energy is released initially, because that is when the volta[e (joules per co,"rlomb) ; the greatestThis is best described by answer choice A. If you have any doubt lbout this, think about how a flashlighrslowly becomes dimmer rather than brighter. The change in voltage is minimal, but nonetheless evid.ent.
52.
5J.
Copyright O by The Berkeley Review@ Section X Detailed Exptanations
54.
55.
59.
60.
Choice C is correct. The current at point c should be the greatest, because it has not been through a resistor. The
resistors are in parallel, so the current is split between the two resistors (Ic = Ia + 16). The greatest current flows
through the paihway of least resistance. Yorl muy recall that current and resistance are inversely proportional
to onJ anothlr. The best answer is choice C, because the resistance is less through the resistor through which
current a flows. I is equal to current.
Choice B is correct. The two half reactions for the nickel-cadmium battery are given in the passage as:
Anode: Cd + ZOH- + Cd(OH)z + 2 e-
Cathode: NiOOH + H2O + 1e- -+ Ni(OH)2 + 1OH-
To find the overall balanced equation, the number of electrons in each half-cell must be the same. In the
oxidation half-cell, two electrons are produced, while in the reduction half-cell, only one electron is consumed'
This means that the reduction half-cell must be multiplied by two. This yields the following two half-cells
that upon addition yield the overall reaction:
Anode:
Cathode:
Overall:
The correct answer is choice B.
Cd +?OI{- -+ Cd(OH)2 + }e-
2 NiOOH +2HtO + *e- -+ 2 Ni(OH)z + ?€H-
Cd +2NiOOH +2H2O -+ Cd(OH)z + 2 Ni(OH)2
55. Choice B is correct. ln the reduction half-cell, MnO2 is converted to Mn2O3. In MnO2, the oxidation state of Mn
is +4. In Mn2O3, the oxidation state of Mn is +3. These can be determined by assuming that the oxidation state
of oxygen is -z itr all of the compounds. The oxidation state of Mn goes from +4 to _+3.
Because zinc is oxidized,
*u1gi^"r" must be reduced, making choices A, C, and D incorrect. The correct choice is answer B. Only one
electron is absorbed, so it should be reduced by one electron'
Choice D is correct. Reduction takes place at the cathode, so choices A and B are immediately eliminated'
Because at a pH equal to 10 the solution il basic, the correct answer must be choice D. Ammonium cation (NHa*)
cannot exist at a pH of L0, because it has a pKu value less than 10. It would exist as ammonia (NH3) in a pH =10 solution. The cell is basic, so choice D is best'
Choice B is correct. Oxidation takes place at the anod.e, meaning that the reaction that cannot occur at the
anode is reduction. In choice B, the reictant side of the equation shows electrons, meaning that electrons are
gained in the reaction. A gain of electrons is the definition o? reduction, so choice B cannot occur at the anode.
Choice A is correct. The flow of electrons in any electrochemical cell is defined as being from the anode to the
cathode (from the oxidation half-cell to the ieduction half-cell). The trick to this question,is therefore
deciding where the anode and cathode actually are. The core of the dry cell battery carries 9ut the- oxidation
half-reaction, while the stainless steel cap is lhe electrode on which manganese,is reduced' In the battery
drawing in Figure L, the stainless steel "up
ir defined as- the cathode. This means that the correct answer must
have the electron flow going towards the stainless steel cap. This eliminates choices B and D. The insulator
does not get involvea in tne reaction or the flow of electrons (hence the term "insulator"). This means that
choice C is eliminated and makes the best answer (and only choice remaining) choice A. The electrons flow
from the metal being reduced, which is found in the outer casing (as stated in the passage)'
Choice D is correct. Electrons do not build up in a cell; potential difference builds,tp. T" electrons flow, but
they do not exactly collect. Choice A is therefore eliminated. Cadmium is oxidized in the reaction, meaning
thai cadmium meial disappears and does not build up, whether it's at the anode or the cathode' Choice B is
therefore eliminated. Nlctet hydroxide is a product of the reaction, so nickel hydroxide builds up somewhere'
Nickel hydroxide does not build up at the anode, however, because nickel hydroxide is formed from the
reduction half-reaction, so it builds up at the cathode, according to the two half-cell equations' -
Choice C is
therefore eliminated. Cadmium hydroxide is formed from the oxidation half-reaction, so it builds up at the
anode. The best answer is choice D.
57.
58.
Copyright @ by The Berkeley Review@ 307 Section X Detailed ExPlanations
64.
65.
61.
62.
63.
55.
57.
Choice B is correct. It is easier to form hydrogen gas than sodium gas (consider that hydrogen reduces morereadily than sodium, and that sodium does not readily vaporize),io choices A and C are"eliminated. Thereduction potential of sorijum is negative, while the reduction potential of protons from water is zero. It iseasier to oxidize chlorine than oxygen gas (chlorine is less electronegative), so chlorine gas will form. The bestanswer is choice B.
Choice D is correct. The electrode must remain solid to allow for a reaction to take place on its surface, Theelectrode cannot melt into the molten salt or there would be no plate on which the reduction half-cell couldtake place. Thus, the electrolysis cell would not work. This means that the melting point of the salt must belower than the melting point of the electrode. The potential material for an electrode is therefore limited.The best answer is choice D.
Choice D is correct. A molten salt is used rather than an aqueous solution when the salt is less reactive thanwater. When dealing with water, the reduction half-reaction H2O + 2 e- --> Hz(g) + O2-(uq) and the oxidationhalf-reaction H2O -r 2 H+(aq) + 1 / 2 OzG) + ?_ e- must always be considered. Wi-en the reiuction potential ofthe cation is negative, then it is more favorable to reduce the protons of water. Thus, the reaction must becarried out on the molten salt. This means that a molten salt should be used when the reduction potential ofthe cation is less than the reduction potential of hydrogen in water. Choice D is the best answer.
Choice C is correct. The left electrode, according to the diagram, is the cathode. Reduction occurs at thecathode-, so the product from reduction forms at the cathode. Hydrogen gas is formed when protic hydrogen isreduced, so-hydrogen gas is the correct choice. The best answer is choice C. Chlorine gas is formed upon theoxidation of chloride anions.
Choice D is correct. For an electrolysis reaction to take place, the applied voltage must be greater than thevoltage released when the cell runs in the favorable direction (the siandard potential). Tlie standard cellpotential can be found from the reactions in the chart. The voltage released when chlorine gas oxidizes nickelmetal is 1.36 volts plus 0.23 volts (negative -0.23 volts for nickel). The result is a value oif .Sq volts. Uponadding 1.75 volts, the reverse reaction (unfavorable reaction) transpires. This makes choice A valid. Thevoltage released when chlorine gas oxidizes zinc metal is 1.36 volts plus 0.76 volts. The result is a value of 2.12volts. Upon adding 2.25 volts, the reverse reaction (unfavorable ieaction) transpires. This makes choice Bvalid' The voltage released when bromine liquid oxidizes nickel metal is 1.09 volts plus 0.23 volts. The resultis a value of 1-.32 volts' Upon adding 1.60 volts, the reverse reaction (unfavorableleaction) transpires. Thismakes choice C valid. The voltage released when bromine liquid oxidizes zinc metal ls t.b9 volL plus 0.76volts. The result is a value of 1.85 volts. Upon adding only 1.50 volts, the reverse reaction (uniavorablereaction) cannot transpire. This makes choice D invalid and thus the not true choice. pick D.
Choice A is correct. Nickel cannot plate from an aqueous solution when a voltage is applied, because a protonhas a higher reduction potential than nickel dication. The applied voltage
"urrri", the protons of water rather
than nickel dication to be reduced. The higher the reduction potential,lhe more likeiy a species is reduced,implying that the reduction potential of copper must be greaterlhan the reduction potential of protons. This isbest shown in answer choice A. The answer could also have been found by simply reading the chart.
Choice A is conect. The reaction that can be carried out in an aqueous solution is the reaction where the cationof the salt is more easily reduced than the proton of water. (Does this question sound familiar?) This questionis better reduced to asking :"Which metal has the most positive reduction potential?" The correct choice (andonly positive choice) is answer choice A.
68. Choice D is correct. In the electrolytic portion (lower half) of the overall cell, chlorine gas is released. Thismeans that chloride anion loses an electron to form chlorine gas. Because Cl- (chloride) iJ oxidized, choices Aand C are eliminated' The sodium metal produced in the lower cell ultimately serves to reduce hydronium ionin water in the upper portion (galvanic section). Choice D is correct, because H+ is reduced to H21g; in theoverall cell reaction.
Copyright @ by The Berkeley Review@ Section X Detailed Explanations
69.
70.
71.
n,
74.
75.
76.
Choice D is correct. The applied voltage causes the unfavorable reaction in the lower electrolytic portion ofthe electrolysis cell. The lower reduction and oxidation half-reactions are:
Na+(aq) + L e- -+ Na(t) with Ecell (reducti on) = -2.71'V Cl-(aq) -+1/ ZCl2(g) + 1 e- with Ecell (oxidation) = -1'36 V'
The overall voltage is -4.07 V, so a voltage greater than 4.07 V must be applied to force the unfavorablereaction. The best answer is choice D.
Choice C is correct. Both hydrogen gas (H2G)) and chlorine gas (C12(g)) are released from the cell, so they must
both be products, which eliminites ihoices B and D. In the upper portion of the cell, NaOH(aq) is produced,
which makes choice C the best answer.
Choice C is correct. Because the reduction potential of sodium cation is negative and the reduction potential of
hydronium ion is zero (see the reactions), statement I is a false statement. In the electrolytic portion of the cell,
,odirr* metal is formed, because the electrons are delivered from the cathode into the mercury liquid, and
sodium metal flows through mercury, while hydrogen gas does not. This makes statement II a true statement.
Because there are ro1n" rodirrm cations that are transferred to the upper cell to form NaOH, the [NaCl] entering
the cell must be greater than the [NaCl] leaving the cell. This makes statement III a true statement, and thus
makes choice C the best answer.
Choice A is correct. By definition, reduction occurs at the cathode and oxidation occurs at the anode, in a
battery, a galvanic cell, and an electrolytic cell. Oxidation involves loss of electrons, so electrons must flow
from the ariode. Reduction involves gain of electrons, so electrons must flow fo the cathode. Overall, the flowof electrons is always from anode to cathode, so choice A is the best answer.
Choice C is correct. An electrolytic cell is defined as a cell that carries out an unfavorable reaction by applying
an external voltage. This means that the reaction voltage is negative, so to have the reaction occur, the
battery voltage niust be positive, with the absolute value of the battery voltage greater than the absolute
value tf the rioltage of the reaction. When the two values are summed, the cell voltage will be positive. The
best answer is choice C.
Choice B is correct. The strongest reducing agent is the compound (or element) that is most easily oxidized.
Sodium cation cannot give up ai electron (because it would lose its noble gas electronic configuration), so choice
A can be eliminated. The reactions in Table l" are reductions, so the product formed from the least favorable
reduction is the most favorably oxidized, and thus the strongest reducing agent. This makes choice B the best
answer,
Choice B is correct. The net reaction for Cell 1 is favorable, if the zinc electrode is oxidized and the copper
electrode is reduced (an electrochemical cell is favorable when E'.u11 is positive). Thus, the electron flow must
be from the Zn electrode (anode) to the Cu electrode (cathode). Because the copper electrode is not the anode,
choice C and choice D are both eliminated. Since, by definition, electron flow is always from the anode to the
cathode, and the Zn electrode (anode) is on the left, the electron flow must be from the left. Referring to the
illustration of Cell 1 in the passage, the electron flow is from left to right, so, choice B is the correct answer.
Choice C is correct. Cell 1 does not use energy to produce chemical change, so it is not an electrolytic cell. The
two half-cells do not differ in cation .or,""t i*tion, so it is not a concentration cell. The cell is set up so that
energy from a spontaneous redox reaction is converted into electrical energy (electrical energy is discharged),
which is a galvinic cell by definition. Pick C. "Nernst cell" is a fictitious term'
Choice D is correct. The cell would no longer be complete without the saturated aqueous salt string, which
serves as a salt bridge to complete the circuil Copper wire does not allow for the flow of sulfate anions from
one half-cell to the other half-cell, so the cell potential is 0 volts. Pick D.
73.
77.
Copyright @ by The Berkeley Review@ 509 Section X Detailed ExPlanations
78. Choice D is correct. Cell 2 is a concentration cell, given the different concentrations in each half-cell. Theelectrons flow in such a way that the concentration of Cu2+ in each half-cell becomes equal over time. Whenelectrons flow from the anode, the copper electrode dissolves into solution, resulting in an increased Cu2+concentration in solution and a reduced mass of the electrode. This means that the half-cell with the lowerconcentration of Cu2+ is the anode, and the half-cell with the greater concentration of Cu2+ is the cathode. Thehaif-cell on the left is less concentrated (making it the anode), so electrons flow from the left to the right asdrawn. This makes choice A a valid statement, so it is eliminated. The left copper electrode dissolves awayand thus decreases in mass, while the right copper electrode plates out and thus increases in mass. This makeschoice B a valid statement, so it is eliminated. In a concentration cell, the electrons flow from the lessconcentrated half-cell to the more concentrated half-cell, until the two cells have equal Cu2+ concentration.The electrons flow from the half-cell with 0.001 M Cu2+ to the half-cell with 1.00 N[Cu2+. The Cu2+ ions inthe 1.00 M solution get reduced. The Cu2+ concentration decrease from 1.00 M until both cells are equallyconcentrated. The final concentration is the average of the two initial concentrations, which is 0.5005 M Cuz+.This makes choice C a valid statement and choice D an incorrect statement, so choice D is the best answer.
Choice B is correct. You are given the equation t = t" g$9 log Q, where q = -4t9d"J-.n - tCathodel
Q = 0'001 M = 0.001, n = 2molesof electrons,andg" = g1.000 M
Plugging the values into the equation yields:
r = r" _ 0.059 toge = g _ 0.Q59 log(.0gi) = (_0.0295X_3) = 0.0885vn*2The cell potential in Cell 2 (the concentration cell) is calculated to be 0.089 V, so choice B is the best answer.
80. Choice A is correct. The half-reactions are listed as reduction reactions in the table, so be sure to change thesign for zinc oxidation from positive to negative. The largest positive cell potential for Cell 1 is obtained whenthe zinc is oxidized and the copper cation is reduced. In a cell, there must always be both a reduction half-reaction and an oxidation half-reaction. The numerical value for E'.u11 is calculated from the followingequations and half-cell potentials:
79.
The cell potential of CeII 1 is therefore 1.10 V, answer choice A.
81. Choice D is correct. Starting with H2O + ClO3- + Mg -+ CI- + Mg2+ + OH-, first balance the electrons eitherthrough the bridge method or the half-cell method. Magnesium goes from an oxidation state of 0 to +2, so 2electrons are lost. Chlorine goes from an oxidation state of +5 to -1, so 6 electrons are gained. This needs to bebalanced so that there are 3 magnesiums for every L chlorine.
? H2O+ 1 ClO3- + 3 Mg -+ 1 Cl- + 3 Mg2+ + ? OH-
Choices A and C are eliminated, because they do not have the correct ratio of magnesiums to chlorines. Afterthe electrons, the charges must be balanced. The overall charge on the left side is -L due to the chlorate anionand the overall charge on the right side is +5 due to the three magnesium cations and one chloride anion.
Cu2+ + 2e- -r CuZn --+Zn2+ +2e-
?H2O+1C1O3-+3Mg -)overall charge = -1
?H2O+1ClO3-+3Mg -+overall charge = -1
+0.34 V+0.76Y+1.1"0 V
1Cl-+3lrlg2++?OH-
1Cl+3Mt2++6OH-
overall charge = +5
To balance charge, six hydroxide anions are added to the right side of the equation. This eliminates ansrve:choice B, narrowing the answer down to choice D. The last step of the balancing, if you wish to go on (alihoug:you could stop here), is to balance the atoms. To balance atoms, water is added to the left side of the reaction.
overall charge = -1
There are six oxygen atoms on the right, and only three on the left, so three waters must be added to the ie::side of the equation. Adding three waters results in the following balanced equation:
3 H2O + 1 ClO3- + 3 Mg -+ 1 Cl- + 3 Mg2+ + 6 OH-
Thecoefficientsofthebalancedequationare3:1:3-+1:3:6,sochoiceDisthecorrectanswer.
Copyright @ by The Berkeley Review@ 3lo Section X Detailed Explanations
82. Choice A is correct. By using the chromium sample as a reference, the following mathematical process can beemployed to determine the electromotive force (emfl for the iron system:
Fe(s)
Fe2+1aq)
-) Fe2+1aq; + 2e-
-+ Fe3+1aq; + 1e-
t'= +0.44V
t' = -0.77 Y
AG" = -2(0.44)f
AG'= _(_0.72)f
Fe(s) -+
^ -AG
nfThe best answer is choice A.
Choice D is correct. By using the chromium sample as a reference, the following mathematical process can beemployed to determine the free energy change (AG")for the copper system:
Cu(s) -) Cu2+1aq7 + 2e- t' = -0.34 V AG' = -2(-0.94)f = +2(0.g\fCu2+(aq) + 1e- -+ Cu+(aq) t'= +0.15 V AG" = -(0.15)f
Cu(s) -+ Cu+(aq) + 1e- AG" = +2(0.34)F - (+0.15)f
AG"iculs) -+ cu+(aq)) = +2(0.34)F- (+0.15)7
The best answer is choice D.
Choice D is correct. To be a sacrificial metal, the metal must have a higher oxidation potential than themetal being protected. Aluminum oxidizes more favorably than iron, zinc oxidizes more favorably than copper,and zinc oxidizes more favorably than iron. This eliminates choices A, B, and C. Because zinc oxidizes morefavorably than nickel, protecting zinc with nickel would not be effective. The best answer is choice D.
Choice D is correct. The reaction that is not favorable is the one with a negative reduction potential (e" < 0).
In choice D, the cell voltage is 0.34 - 0.99 = -0.65, which is an unfavorable reaction. The reaction is drawnbelow:
Pd(s) +Cq2*(uq) --+C..t(t) + Pd2+1aq;
Reduction: t' = +0.34 VThe cell potential for the aluminum-chromium (III) cell is 1..66 - 0.73 = 0.93 V. The cell potential for themanganese-palladium (II) cell is 1.18 - 0.99 = 0.19 V. The cell potential for the zinc-nickel (II) cell is 0J6 - 0.23
= 0.53 V. The cell potential for the chromium half-reaction is taken from the passage, while the cellpotentials for the other half reactions (aluminum, manganese, palladium, zinc, and nickel) are all from Table1. The only unfavorable reaction is choice D.
Choice A is correct. Because zinc is the sacrificial metal in galvanized steel, it will decay (oxidize) away firstbefore any of the iron will. The mass of the zinc will steadily decrease to zero before any iron begins to oxidizeaway, choice A.
Choice D is correct. The passage states that aluminum, tin, and chromium can be used to form protective oxidecoatings. If you did not read the passage, then the answer can be found by looking at the reaction betweenpalladium oxide and copper. From the table, the reduction potential for palladium dication (present in PdO) isE' = 0.99 V. The oxidation potential for copper metal is E' = -0.34 V. The cell potential (E') for the followingreaction is thus 0.65 V:
PdO(s) + Cu(s) -+ Pd(s) + CUO(s)
A palladium oxide coating on copper would actually oxidize the copper metal into copper cation rather thanprotect the copper from oxidation. An oxide coating of palladium would be effective, so choice D is correct.
Fe3+(aq) + 3e- AG' = -zQ.aDf + -(-0.77)y
- (-2 x 0.44f - (-0.77r)) _ 2(+0.a9 + (0.77)3q
83.
84.
85.
86.
87.
Copyright @ by The Berkeley Review@ 3ll Section X Detailed Explanations
88. Choice C is correct. Because the charge on the silver cation is +1, twice as many moles of silver metal (Ag) formas both copper metal (Cu) and zinc metal (Zn). Three times as many moles of silver metal (Ag) form as goldmetal (Au). To find the mass formed, the moles formed must be multiplied by the atomic mass of the element.The answer choice with the greatest value from moles times molecular mass is the correct answet choice. Silverhas an atomic mass of 107.9, and it requires one electron per ion, so one mole of electrons produces 107.9 grams ofsilver. Copper has an atomic mass of 63.5, but it requires two electrons per ion, so only 31.75 grams form from one
mole of electrons. Zinc has an atomic mass of 65.4,but it requires two electrons per ion, so only 32.7 grams formfrom one mole of electrons. Gold has an atomic mass of 797.0,but it requires three electrons per ion, so only 65.67
grams form from one mole of electrons. The greatest mass results from the silver cation, so choice C is the best
answer.
Choice C is correct. To carry out electroplating, a current must run through some material suspended in a
solution of ions. The charge builds up on the surface of the material where ions are reduced and thus can
precipitate (onto the surface of the metal). If the material does not conduct electricity, charge cannot build upon iti surface. Plastic is an insulator, not a conductor of electricity, making choice C the best answer. Choice Ashould have been eliminated, because high temperatures are not required for electroplating. Choice B shouldbe eliminated, because a current of electrons causes reduction, not oxidation. Choice D should have been
eliminated, because malleability does not effect electroplating'
Choice A is correct. Because Solution I has a lower concentration of metal than Solution IV, Solution I finishes
before Solution IV. This means that either choice A or choice D is the correct choice. Silver has a greater mass
than copper, and it requires only one electron for complete reduction, so the electrode of Solution I has a greater
initial increase in mass per unit time than solution IV. This can be seen in choice A where Solution I shows a
steeper slope than Solution IV.
Choice D is correct. Electroplating results from the conversion of cations in solution into precipitated metalatoms. To convert a cation to a neutral element, electrons must be added. The gain of electrons is defined as
reduction, so choice D best describes the process.
Choice A is correct. A good protective metal coating is one that does not oxidize away. This means that it mustbe unfavorable for the metal to oxidize awayt which would imply that the metal has negative oxidationpotential. A negative oxidation potential for the metal would result in a positive reduction potential for the
cation (the reverie reaction). This is answer choice A. This also explains why choice D is eliminated. Choice
B is incorrect, because the reduction of a metal to an anion is not known to occur. This not only would not be a
desirable property, but it would not be practical. A large positive oxidation potential for its cation just means
that the cation can further be oxidized, making choice C a bad choice.
Choice C is correct. In the experiment, the cation solutions are exposed to the current for one minute. Like zinc
and copper, cadmium cation (Cd2*) takes two electrons to be reduced. Because cadmium has an atomic mass
greateithan that of both copper and zinc, the mass of the electrode is greater than the electrodes for Solutions
iII, IV, V, and VI. The elec[rode has a mass greater than 5.360 grams (the mass of the heavier zinc plated
electrode), which eliminates choices A and B. To form the same mass as the silver cation, cadmium wouldhave to have twice the atomic mass of silver. The atomic mass of cadmium is less than twice the atomic mass
of silver, so the electrode has a mass less than 6.182 grams (the mass of the lighter silver plated electrode '
This eliminates choice D. The best answer is choice C.
Choice D is correct. Choice A can be eliminated, because both the H+ and the Ag+ are reduced in the reaction a.
given, and nothing is oxidized. An oxidation-reduction reaction must have both an oxidation half-reaction arl:a reduction half-reaction. If choice B were true, then tarnish could never occur, because the reduction of silr'e:
would be carried out by the sulfide anion it binds. Choice C is eliminated, because Al cannot exist in a --oxidation state. The coriect answer is choice D, because the oxidation state of aluminum (which was oxidizejis +3. Pick choice D and feel happy.
89.
90.
91.
92.
93.
94.
Copyright @ by The Berkeley Review@ 312 Section X Detailed ExPlanatiorc
95.
96.
97.
Choice B is correct. Setiing the two equation equal to each other yields: -RT h K = - nF€". This becomes RT ln K= nFt' once the negative signs have been eliminated. Choices A and C are therefore wrong. Dividing both sides
by RT yields choice B: l:rK = dto.RT
Choice A is correct. By simply balancing the atoms in the equation, we find that the coefficients should be 2 : 6: 3 : 2, so choose A. Sometimes it is easier to balance by inspection than to balance by the redox method. Asimple guideline to follow is: If the molecules are all uncharged, balance by inspection.
Choice A is correct. First, convert the current units into coulombs: C = amps x sec. = (9.65 amps)(1000 sec) = 9650C. One mole of electrons carries a charge of one faraday (F), which equals 96,500 coulombs (C), and each mole ofZn dication requires two moles of electrons to be reduced to zinc metal. The molecular weight of Zn is 65.38 g.By substitution:
2.37V-0.23 V
e6s0c fllqlc-e:l/r more zn)/65.98 sraqls zn) = u.run r^
\eo,soo c /\ z mote e- /\ 1 mole Zn I " *
Choice A is the correct answer.
98. Choice D is correct. Magnesium is oxidi zed., and.nickel dication is reduced. The half-reactions for the givencell are as follows:
Mg -+ M** * z"-Ni2+ + 2e- -+ Ni
Because Mg is oxidized, not reduced, the value for the reduction potential must be reversed to +2.37 V. The cellpotential is 2.37 + (-0.23) = 2.L4 V. Choice D is the best answer.
99. Choice D is correct. The Nernst equation shows logarithmic decay of the voltage. The log value is multipliedby a small term, making the decay insignificant until the last few moments of the decay. This is best shown inchoice D. Choice A is close, but the linear decay at the end is inaccurate.
L00. Choice A is correct. Because aluminum and carbon dioxide are products in the reaction, choices C and D areeliminated. Carbon is oxidized to carbon dioxide, and aluminum oxide (A12O3) is reduced to aluminum metal.The best answer is choice A.
Copyright O by The Berkeley Review@ 313 Section X Detailed Explanations
l-General Chemistry Sections VI - X Section Answers
General Chemistry Bubble Sheet(Make five copies, one for each section of the book.)
@@@@@@@@@@@@
@@@@@@@@@@@@@@@
@@@@@@@@@@@@@@@@@@@@
@@@@@@@@
@@@@@@@@@@@@@@@@@@@@
@@@@@@@@@@
73.
74.
t5.
37. @ @ @ @38.@@@@3e.@@@@40. @ @ @ q41.@@@@42. @ @ o @43.@@@@44. @ @ @ @4s.l@@@@l46. @ @ @ @47. @ @ @ @48.@@@@4e.i@ @ @ @s0.@@@@sl.@@@@s2. @ @ o @s3.1@@@@]s4.l@@@@lss.@@@@rs6.@@@@s7. @ @ @ @s8.@@o@
@se.@@@60.@@@@61.@@@@i62.@@@@l63.@@@@64.@@@@6s.i@ @ o @66.@@@@67.@@@@68.@@@@
@6e. @ @ @70.@@@@7t.@@@@72.@@@q
@l@l(Er@)l@l
1.
)3.
4.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.'r)
23.
24.)<
26.
27.
28.to
30.
31.
32.
33.
34.
35.
36.
76.:,
@@@@@@@@@@@@@@@
@@@@@G)@@@@@@@@@
81.
82.
83.
84.
85.
86.
87.
88.
89.
90.
91.
92.
93.
94.
95.
96.
77.
78.
79.
80.
97.
98.
99.
100.
@@@
@@@@@@@@@@
@@@@@@@@@@@@
@@
@@@@@@
ChemistryI Section
t lRawL I Score
EstimatedScaled Score
@@@@@@@@@@@@@@@@@@@@@@@@@@
@@@@@i@@@@@@@@@@@@@
(o@@@I@@@@@
@o@l@@@l@@@l@@@r@@@l@-@ @l
@@@@@@
Copyright O The Berkeley Review The Berkeley ReviewSpecializing in MCAT Preparation
320
ERKELEYR.E.V.I.D.W'"
PEBIODIC TABLE OF TTilE ELEMENTS
IH1.0
2
He4.0
3
Li6.9
4Be9.0
i-B
10.8
6C
t2.0
7
Nr4.0
8
o16.0
9F
19.0
10
Ne20.2
l1Na
23.0
12
Mg24.3
13
AI27.0
t4Si
28.1
l5P
31.0
16
s32.1
l7CI
35.5
18
Ar39.9
19
K39. l
20Ca40.1
21
Sc45.0
22Ti
47.9
23
v50.9
24Cr52.0
25Mn54.9
26Fe
55.8
2'lCo
58.9
28Ni
58.7
29Cu
63.5
30Zn65.4
31
Ga69.'l
32Ge
72.6
JJ
As74.9
34Se
79.0
35
Br79.9
36Kr
83.8
37Rb85.5
38Sr
8'7.6
39Y
88.9
40Zr
91.2
4lNb
92.9
42Mo95.9
43Tc
(98)
44Ru
101.1
45
Rh102.9
46Pd
106.4
4',7
Agt07.9
48
Cd112.4
49
In1 14.8
50Sn
118.7
51
Sbl2l.8
52Te
t27.6
53I
126.9
54Xe
131.3
55Cs
t32.9
56Ba
137.3
<7"'L
Lal138.9
72Hf
178.5
73Ta
180.9
74w
183.9
75Re
186.2
76Os
t90.2
7'lIr
192.2
78Pt
195. I
79Au
197.0
80Hg
200.6
81
TI204.4
82Pb
207.2
83
Bi209.0
84Po
(2Oe)
85
At(210)
86Rn
(222)
87Fr
(223)
88Ra
226.0
89aAc"
227.0
IMRf
(261)
105
Db(262)
106
Sg(263)
107
Bh(262)
108
Hs(26s)
109
Mt(266)
110
Uun(269)
111
Uuu(272)
112
Uub(277)
58Ce
140.1
59
Pr140.9
60
Nd144.2
61
Pm(145):--%
Np(237)
90Th
232.0
9tPa
(231)
92U
238.0
65Tb
158.9
62
Sm50.4
94Pu
(244)
63
Eut52.0
95
Am(243)
68Er
167.3
69Tm
168.9
96Cm
(247)
98cf
(2s 1)
Specializing in MCAT Preparation
![The RIVERA (TM) Tube Rack Series Stereo Amplifiers · 2010-02-15 · The RIVERA (TM) Tube Rack Series Stereo Amplifiers Models: TBR-1, TBR-2, TBR-3 (HAMMER [TM] 120), TBR-5 (HAMMER](https://img.dokumen.tips/doc/110x75/5f69e58a30b75209be206ce2/the-rivera-tm-tube-rack-series-stereo-amplifiers-2010-02-15-the-rivera-tm.jpg)
![[tbr] menu](https://img.dokumen.tips/doc/110x75/568cadc91a28ab186dad27b9/tbr-menu.jpg)
