Chemistry 18 Notes

Table of ContentsCalculus ReviewLinear Regression (Least Squares Method)Chemical Kinetics

Rate 3Theories on Reaction Rate 4Factors Affecting Reaction Rate 4

Methods for Determining the Rate Law 5Dependence of the Concentration of Reactants with Time (Integrated Rate Laws) 7Gas-Phase Reactions 10

Reaction Mechanisms 13Nuclear Chemistry

Type of Nuclear Reactions 17Effect of Radiation on Biological Matter 20

Chemical ThermodynamicsFirst Law of Thermodynamics (Law of Conservation of Energy) 22Thermochemistry 25Second Law of Thermodynamics 29Third Law of Thermodynamics 33

Chemical EquilibriumApproaches to Equilibrium 33Molecular Equilibrium 35Factors Affecting Equilibrium 38

Ionic EquilibriaTheories on Acids and Bases 40Ionization of Water 41Strong Electrolytes 41Weak Electrolytes 42Polyprotic Acids 42Salts 42Common Ion Effect 43

Buffer Solutions 44Acid-Base Indicators 44Acid-Base Titration 45Complex Ion Equilibrium 45Slightly Soluble Salts 45Factors Affecting Solubility 45Precipitation 46Dissolution 46

ElectrochemistryElectrochemical Cell 47

Thermodynamics of Electrochemical Cells 48Electrolytic Cells 51

Faraday's Law of Electrolysis 52Coordination Chemistry

Complex / Coordination Compound 53Ligands 54Nomenclature of Complexes 55Isomerism 56

CHEMISTRY 18The Fundamentals of General Chemistry IIGeoffrey C. LiDepartment of Physical Sciences and Mathematics

Summer || AY 2010-2011 || Section AMonday to Friday: 7:30-9:30 AM RH 119 Notes by Alvin J. Logronio

ii | Chemistry 18: Fundamentals of General Chemistry II

Structural Isomers 56Optical Isomers (Enantiomers) 56

Theories on Bonding 57Valence Bonding Theory (Localized Electron Model) 57Crystal Field Theory 58

Reactions 60Applications of Coordination Chemistry 61

Calculus ReviewDifferentiation: finding the slope of the tangent line - represented by f'(x), dy/dx or y'

Properties 1. y = C y = 0

2. f(x) = xn f (x) = nxn1

Example: f(x) = 4x3 f (x) = 3(4)x31 = 12x2

3. d(u+ v) = du+ dv

Example: f(x) = 3x2 2x+ 4 f (x) = 6x 2

Example: y = 5x 3 4x+

10

x2y = 5 + 4x2 20x3

= 5x 3 4x1 + 10x2

4. d(uv) = udv + vdu or 1d2, 2d1

Example: f(x) = (3x 5)(4x2 3) f (x) = (3x 5)(8x) + (4x2 3)(3)

5. d(uv

) vdu udvv2

orldh hdl

l2

Example: f(x) =4x2 + 3x 5

2x+ 1f (x) =

(2x+ 1)(8x+ 3) (4x2 + 3x 5)(2)(2x+ 1)2

6. Chain Rule

Example: f(x) = (4x2 7x+ 13)2 f (x) = 2(4x2 7x+ 13)(8x 7)

Integration: finding the area under the curve

Indefinite Integration

xndx =

{xn+1

n+1 + C if n = -1ln|x| + C if n = -1

Examples:

1.

xdx =

x2

2+ C

2.

3 dx = 3

dx = 3

x0 dx = 3x+ C

3.

5x2 dx = 5

x2 dx =

5

3x3 + C

4.

3x2 x

4+ 14 dx =

3x2 dx

x

4dx+

14 dx

= 3x3

3 x

2

8+ 14x+ C

= x3 18x2 + 14x+ C

5.

(5 10

v+12

v2

)dv =

5 dx

10v1 dx+

12v2 dx

= 5v 10ln|v| + 121v1 + C

= 5v 10ln|v| 12v1 + C

| 1Geoffrey C. Li Notes by A-Log

2 | Chemistry 18: Fundamentals of General Chemistry II

Definite Integration

ba

f(x) dx

Examples:

1.

42

(x+ 2) dx

=

42

x dx+

42

2 dx

=x2

2+ 2x

42

=

(42

2+ 2(4)

)(22

2+ 2(2)

)= 16 6 = 10

2.

11

(4x3 + 2x2 3x+ 1

2

)dx

=

11

4x3 dx+

11

2x2 dx1

13x dx+

11

1

2dx

= 4x4

4+ 2

x3

3 3x

2

2+

1

2x

11

= x4 +2

3x3 3

2x2 +

1

2x

11

=

((1)4 +

2

3(1)3 3

2(1)2 +

1

2(1)

)((1)4 + 2

3(1)3 3

2(1)2 + 1

2(1)

)=

7

3

3.

2tt

(7t 3 4t1 + 10t2) dt

= 7t2

2 3t 4ln|t| + 10 t

1

12tt

=7

2t2 3t 4ln|t| 10t1

2tt

=7

2

((2t)2 t2) 3 (2t t) 4 (ln|2t| ln|t|) 10 ((2t)1 t1)

=7

2

(3t2

) 3t 4(ln 2tt) 10( t2t2

)=

21

2t2 3t 4ln2 5t1


Linear Regression (Least Squares Method)- to know if there is a linear relation between two variables- gets the best-fit line between a set of points

r = correlation coefficient

ranges from -1 to 1, -1/1 means a straight line

in a calculator: y = A+Bx

A = y int B = slope

y = the value of y when x is...

Example: 20y

x = the value of x when y is...

Example: 20x

Example: x y15 1812.5 1512 14.38 10

13.8 16.56.7 814.6 17.5

r = 0.999228

A = 0.3086

B = 1.176

Equation: y = 1.176x+ 0.3086

Chemical Kinetics- branch of chemistry that deals with rates or speeds of reaction- determines the factors that affect rates and its mechanisms

Rate- in chemistry rate is

rate of appearance of the product = rate of disappearance of the reactant

For the reaction A + 2B C

Rate of reaction =d[C]

dt= d[A]

dt= 1

2

d[B]

dt

ifd[C]

dt(rate C) = 0.1 M/s, then

d[A]

dt(rate A) = 0.1 M/s, d[B]

dt(rate B) = 0.2 M/s

Rate of reaction = 0.1 M/s = (0.1) M/s = 12(0.2) M/s

[B] is changing at the greatest rate

For the reaction: 3X + 5Y 2Z

Rate of reaction =1

2

d[Z]

dt= 1

3

d[X]

dt= 1

5

d[Y ]

dt

[Y] is changing at the greatest rate

For the reaction: N2O5 2NO2 + 12O2

Rate of reaction =1

2

d[NO2]

dt= 2

d[O2]

dt= d[N2O5]

dt

[NO2] is changing at the greatest rate


Quiz:

Given: 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) d[NO]dt

= 0.60M/s

Rate =1

4

d[NO]

dt=

1

6

d[H2O]

dt= 1

4

d[NH3]

dt= 1

5

d[O2]

dt

1:d[H2O]

dt=

6

4

d[NO]

dt=

6

4(0.60M/s) = 0.090M/s

2:d[O2]

dt= 5

4

d[NO]

dt= 5

4(0.60M/s) = 0.075M/s

3:d[NH3]

dt= 4

4

d[NO]

dt= 4

4(0.60M/s) = 0.060M/s

4: Rate of reaction =1

4

d[NO]

dt=

1

4(0.60M/s) = 0.015M/s

Theories on Reaction Rate1. Collision Theory- for a reaction to occur, the reactant molecules must collide - not all collisions are effective

Requirements for an effective collision a. reactants must have the proper orientation Example: CO +NO2 CO2 +NO

Ineffective: CO ON=OEffective: OC ON=O

b. reactants must have sufficient energy to break bonds - activation energy: minimum energy that is required for the reaction to occur, energy barrier that must be overcome - enthalpy (H) has nothing to do with the rate of the reaction

Ea, rate

rate of reaction frequency of collision

krate constant

= psteric factor (orientation)

+ Zfrequency of collision

+ eEaRT

fraction of molecules with sufficient energy

2. Transition State Theory (Henry Eyring)- all reactants pass through a transition state - species in transformation - shown in the peak of an energy profile - highly energetic and highly unstable - therefore, you cannot isolate the transition state - collision is not necessary for the reaction to occur - reaction can occur through bond weakening - contains partial forming and partial breaking of bonds Example: CO + O-N=O [OC--O--N=O] O=C=O + N=O

rate of reaction ease of formation of the transition state

- ease is related with the activation energy Ea, ease of formation

Factors Affecting Reaction Rate1. Nature of Reactants - some reactions are naturally fast while others are naturally slow Example: between Mg and Fe in water, Mg reacts faster


- the activation energy is dependent on the nature of the reactants

2. Surface Area SA, Rate - when there is a larger surface area exposed to collision, the rate of reaction increases

3. Concentration of Reactants Concentration, Rate; Rate [reactants]n

- more molecules are available for collision

Rate Law:

rate = k[reactants]n

k = rate constant

n = order of reaction

- the order of reaction is experimentally determined - it is not equal to the coefficient in the balanced equation

Methods for Determining the Rate Law a. Method of Initial Rates - initial rate: rate at the first instant the reactants react

Example:Given: A+2B C

[A] [B] Initial Rate (M/s)0.010 M 0.010 M 1.2 1040.010 M 0.020 M 2.4 1040.020 M 0.010 M 4.9 104

a. Rate Law

rate = k[A]x[B]y

A :rate 3

rate 1=

k(0.020)x(0.010)y

k(0.010)x(0.010)y

4.9 1041.2 104 =

(0.020

0.010

)x4.08 = 2x

x = 2

B :rate 2

rate 1=

k(0.010)x(0.020)y

k(0.010)x(0.010)y

2.4 1041.2 104 =

(0.020

0.010

)y2 = 2y

y = 1

rate = k[A]2[B]

Overall Order: 2+1=3

Substitute Experiment 1: k =rate

[A]2[B]

=1.2 104

(0.010M)2(0.01M)

= 120 M2s1

Rate = (120 M2s1)[A]2[B]

Double A? 4 Rate Double B? 2 Rate[2A]2[2B]? 8 Rate [2A]2[0.5B]? 2 Rate[0.5A]2[2B]? 0.5 Rate

b. Differential Method

Rate = k[A]n

logRate y

= log kb

+n log[A] mx

Plot log Rate vs log [A]

slope = n

y-int = logk


c. Isolation Method rate = k[A]

x[B]y[C]z

A: at constant [B] and [C]

log rate y

= log k[B]y[C]z b

+x log[A] mx

Plot log rate vs log [A]

slope = x

y-int = log k[B]y[C]z

B: at constant [A] and [C]

log rate y

= log k[A]x[C]z b

+ y log[B] mx

Plot log rate vs log [B]

slope = y

y-int = log k[A]x[C]z

C: at constant [A] and [B]

log rate y

= log k[A]x[B]y b

+ z log[C] mx

Plot log rate vs log [C]

slope = z

y-int = log k[A]x[B]y

Example:[A] [B] Initial Rate (M/s)

0.020 M 0.020 M 8.40 1050.020 M 0.025 M 1.31 1040.020 M 0.030 M 1.89 1040.020 M 0.035 M 2.57 1040.025 M 0.030 M 2.36 1040.030 M 0.030 M 2.83 1040.035 M 0.030 M 3.30 104

rate = k[A]x[B]y

A: at constant [B]

log rate = log k[B]y + x log[A]

Plot log rate vs log [A]

r = 0.99999

B = 0.996 1 = xA = 2.032 = log k[B]y

B: at constant [A]

log rate = log k[A]x + y log[B]

Plot log rate vs log [B]

r = 0.999998

B = 1.99 2 = yA = 0.679 = log k[A]x

rate = k[A][B]2

log k[B]y = 2.0315log k(0.030)2 = 2.0315

k(0.030)2 = 102.0315

k = 10.33 M2s1

log k[A]x = 0.67912log k(0.020) = 0.67912

k(0.020) = 100.67912

k = 10.47 M2s1


Quiz: [HgCl2] [C2O

24 ] Initial Rate (M/s)

0.164 M 0.15 M 3.20 1050.164 M 0.45 M 2.90 1040.164 M 0.90 M 1.15 1030.164 M 1.80 M 4.61 1030.082 M 0.45 M 1.40 1040.041 M 0.45 M 7.25 1050.021 M 0.45 M 3.71 105

rate = k[HgCl2]x[C2O

24 ]

y

[C2O24 ]: at constant [HgCl2]

log rate = log k[HgCl2]x + y log[C2O

24 ]

Plot log rate vs log [C2O24 ]

r = 0.9999

slope = y = 1.99 2

[HgCl2]: at constant [C2O24 ]

log rate = log k[C2O24 ]

y + x log[HgCl2]

Plot log rate vs log [HgCl2]

r = 0.9999

slope = y = 0.995 1

rate = k[HgCl2][C2O24 ]

2

From constant [C2O24 ]:

y-int = 2.76213 2.76213 = log k[C2O24 ]2 2.76213 = log k(0.45)2k = 8.54 103 M2s1

When [HgCl2] = 0.12 M and [C2O24 ] = 0.10 M:

rate = (8.54 103 M2s1)(0.12 M)(0.10 M)2rate = 1.02 105 M/s

Dependence of the Concentration of Reactants with Time (Integrated Rate Laws)- One-Reactant Type of Reactions [A]0 = concentration at time 0 [A]t = concentration at time t

rate = k[A]n

rate = d[A]

dt

d[A]dt

= k[A]n

n = 0 : d[A]dt

= k[A]0

[A]t[A]0

d[A] = kt

0

dt

[A]t [A]0 = k(t 0)[A]ty

= ktmx

+ [A]0b

n = 1 : d[A]dt

= k[A]1

[A]1 d[A] = k dt[A]t

[A]0

[A]1 d[A] = kt

0

dt

ln[A]t[A]0

= k(t 0)

ln[A]t ln[A]0 = ktln[A]t

y

= ktmx

+ ln[A]0 b

Plot ln[A] vs t

slope = -k

y-int = ln[A]0

Plot [A] vs t

slope = -k

y-int = [A]0


n = 2 : d[A]dt

= k[A]2


[A]0

[A]2 d[A] = kt

0

dt

1[A]t

+1

[A]0= kt

1

[A]ty

= ktmx

+1

[A]0b

Plot1

[A]vs t

slope = k

y-int =1

[A]0

n = 3 : d[A]dt

= k[A]3


[A]0

[A]3 d[A] = kt

0

dt

12[A]2t

+1

2[A]20= kt

1

2[A]2t y

= ktmx

+1

2[A]20 b

Plot1

2[A]2vs t

slope = k

y-int =1

2[A]20

Summary of Integrated Rate Laws:

n=0: [A]t = kt+ [A]0n=1: ln[A]t = kt+ ln[A]0n=2:

1

[A]t= kt+

1

[A]0

n=3:1

2[A]2t= kt+

1

2[A]20

Example:Given: 2C5H6 C10H12 [A]0 = 0.0400 M

t (sec) 0 50 100 150 200[C5H6] (M) 0.0400 0.0300 0.0240 0.0200 0.0174

1: Order of Reaction

n=0: Plot [A] vs t

n=1: Plot ln [A] vs t

n=2: Plot1

[A]vs t

n=3: Plot1

[A]2vs t

rn=0 = -0.967

rn=1 = -0.9905

rn=2 = 0.999776

rn=3 = 0.995

The reaction is a second-order reaction.

2: k slope = k = 0.163 M1s1

3: [C5H6]300 x = t; y =1

[C5H6]t

300y =1

[C5H6]300

[C5H6]300 =1

300y= 0.0135 M

4: time when [C5H6]t = (0.20)[C5H6]0

t =1

(0.20)(0.0400)x = 612 s

Half-Life t1/2 - time required to reduce the concentration of the reactant to half of its initial value


t = t1/2 when [A]t =

1

2[A]0

n = 0:

[A]t = kt+ [A]01

2[A]0 = kt 1

2+ [A]0

kt 12= [A]0 1

2[A]0

kt 12=

1

2[A]0

t 12=

[A]02k

t 12 [A]

n = 1:

ln[A]t = kt+ ln[A]0ln

(1

2[A]0

)= kt 1

2+ ln[A]0

kt 12= ln[A]0 ln

(1

2[A]0

)kt 1

2= ln

[A]012 [A]0

t 12=

ln 2

k

t 12independent of [A]

n = 2:

1

[A]t= kt+

1

[A]01

12 [A]0

= kt 12+

1

[A]0

2

[A]0 1

[A]0= kt 1

2

1

[A]0= kt 1

2

t 12=

1

k[A]0

t 12 1

[A]

n = 0 : t 12=

[A]02k

n = 1 : t 12=

ln 2

k

n = 2 : t 12=

1

k[A]0

n = 0 : 1M20 s 1

2M

10 s 14M

5 s 18M

n = 1 : 1M20 s 1

2M

20 s 14M

20 s 18M

n = 2 : 1M20 s 1

2M

40 s 14M

80 s 18M

Examples:

1. Given: order = 1 t 12= 12.0min

100%12 min 50% 12 min 25% 12 min 12.5% 12 min 6.25%

a. 25% remaining: t = 12+12 = 24 min

b. 20% remaining: [A]t = 0.20[A]0

ln[A]t = kt+ ln[A]0 t 12=

ln 2

k

k =ln 2

12.0 min= 0.0578 min1

ln(0.20)[A]0 = (0.0578 min1)t+ [A]0ln

(0.20)[A]0[A]0

= (0.0578 min1)t

t =ln (0.20)[A]0[A]0

(0.0578 min1) = 27.84 min

c. 10% remaining: [A]t = 0.10[A]0

ln[A]t[A]0

= kt

ln(0.10)[A]0

[A]0= (0.0578 min1)t

t =ln (0.10)[A]0[A]0

(0.0578 min1) = 39.84 min


2. Given: order = 2 HI(g) 12H2(g) +

1

2I2(g)

t = 5 [A]0 = 0.90 M [A]5 = 0.68 M

a. time when [HI] = 0.50 M

1

[HI]t= kt+

1

[HI]01

0.68 M= 5k +

1

0.90 M

k =1

0.68 M 10.90 M5 min

= 0.0719 M1min1

1

0.50 M= (0.0719 M1min1)t+

1

0.90 M

t =1

0.50 M 10.90 M(0.0719 M1min1) = 12.36 min

b. half-life when [HI] is 0.90 M

t 12=

1

k[A]0=

1

(0.0719 M1min1)(0.90 M)t 12= 15.45 min

Gas-Phase ReactionsA(g) B(g) + 2C(g)

A B Ct = 0 PA,0 0 0C -x +x +2xt PA,0-x x 2x

PTOTAL = PA + PB + PC

= PA,0 x+ x+ 2x= PA,0 + 2x 2

n = 1 ln[A]t[A]0

= kt

from PV = nRT, P = MRT, C =P

RTTherefore,

lnPA,tRTPA,0RT

= kt

lnPA,tPA,0

= kt

Example:

Given: n=1 C8H18O2(g) 2C3H6O(g) + C2H6(g)t 12at 147C = 80 min PA,0 = 800 mmHg PA,600 =?

lnPA,tPA,0

= kt t 12=

ln 2

k

k =ln 2

t 12

=ln 2

80 min= 0.0087 min1

lnPA,t

800 mmHg= (0.0087 min1)(600 min)

PA,t = (e(0.0087 min1)(600 min))(800 mmHg) = 4.43 mmHg


C8H18O2(g) C3H6O(g) C2H6(g)t = 0 800 mmHg 0 0C -x +2x +xt 4.43 mmHg 2x x

800 x = 4.43 mmHgx = 795.57 mmHg

PTOTAL = 4.43 + 2x+ x

PTOTAL = 4.43 + 2(795.57) + 795.57 = 2391.14 mmHg

4. Temperature Temperature, Rate - increasing the temperature increases the initial energy of the reactants, making it easier to overcome the activation energy - the temperature alters the rate constant T, k

Arrhenius Equation: k = AeEaRT

Ea: activation energy T: absolute tempterature (in K)

R: ideal gas constant (8.314 J/molK) A: pre-exponential factor

k = AeEaRT

ln k = lnA+ ln eEaRT

ln ky

=EaR

1

T mx

+ lnAb

Plot ln k vs1

T

slope =EaR

y-int = ln A

ln k2 =EaR

1T2

+ lnA

ln k1 = EaR 1T1 + lnAln k2k1 = EaR

(1T2 1T1

)y2 y1 = m(x2 x1)

Example:T (C) 0 25 35 45

k(s1) 103 0.0106 0.319 0.986 2.92

Find Ea and A:

ln k =EaR

1

T+ lnA Plot ln k vs

1

Tr = 0.999

slope =EaR

= 10863Ea = 90319 J

y int = lnA = 28.336A = 2.02 1012

Find half-life at 40C: x =1

Ty = ln k

ln k =1

313.15y

ln k = 6.355k = 1.738 103s1

t 12=

ln 2

k

=ln 2

1.738 103 s1= 399 s


Quiz

Given: n=2

Ea = 250 kJ/mol [N2O]t = 0.20[N2O]0 = 0.002 M [N2O]0 = 0.01 M

T1 = 838.15 K t 12= 25.25 hours k1 =?

T2 = 973.15 K t =? k2 =?

1

[N2O]t= kt+

1

[N2O]0t 12=

1

k1[N2O]0

k1 =1

(t 12)([N2O]0)

=1

(25.25 hours)(0.01 M)= 3.96 M1hr1

lnk2k1

= EaR

(1

T2 1T1

)ln

k23.96 M1hr1

= 250000 J/mol8.314 JmolK

(1

973.15 1

838.15

)k2

3.96 M1hr1= 145.03

k2 = 574.32 M1hr1

1

[N2O]t= kt+

1

[N2O]01

0.002 M= (574.32 M1hr1)t+

1

0.01 Mt = 0.696 hrs or 41.79 min

5. Catalyst - substance that is not consumed in the reaction - alters the activation energy 1. positive catalyst: speeds up the reaction by lowering the activation energy 2. negative catalyst: inhibitor: slows down the reaction by increasing the activation energy

a. homogenous catalyst: catalyst in the same phase as the reactants - it provides an alternate pathway with a series of steps with a lower activation energy each - changes the reaction mechanism

- intermediate: species formed at one step but consumed at subsequent steps

Overall Reaction: 2H2O2 2H2O + O2H2O2 + I

H2O + OIH2O2 + OI

H2O + O2 + I2H2O2 2H2O + O2

I: Catalyst

OI: Intermediate

Uncatalyzed Reaction Catalyzed Reaction

Higher activation energy than catalyzed Each step has lower activation energy than the uncatalyzed

Has only one transition state Has one transition state for each step

Has no intermediate Has one or more intermediates

same change in enthalpy


Transition State Intermediate

Peaks in the energy profile Valleys in the energy profile

high energy, cannot be analyzed lower energy, can be analyzed

b. heterogenous catalyst: catalyst in the different phase as the reactants - usually solid - acts through adsorption - through binding sites in the solid's surface Example: CH2 = CH2 +H2

Pt C2H6

Reaction Mechanisms - detailed description of how a reaction occurs - shows the reason why the observed rate orders are different from the coefficient in the balanced equation

Overall Reaction: CO + NO2 CO2 + NO Rate = k [NO2]2NO2 + NO2

k1 NO3 + NO (slow)CO + NO3

k2 CO2 + NO2 (fast)CO + NO2 CO2 + NO

elementary steps - the most basic steps - the orders of reaction are the coefficients in the equation

slowest step - rate determining step - rate law of reaction = rate law of slowest step

Rate = k1[NO2]2

molecularity 1: unimolecular concerted reactions: single step 2: bimolecular 3: termolecular

Concerted Reaction:

CH3Cl +OH CH3OH + Cl

Rate = k[CH3Cl][OH]

- you can never tell with absolute certainty the exact mechanism of a reaction, you can only propose plausible mechanisms

Requirements for a Plausible Proposal: 1. Sum of elementary steps = overall reaction 2. Rate law for the mechanism = observed rate law

Example 1:Overall Reaction: 2NO + Cl2 2NOCl Rate = k[NO]2[Cl2]

Mechanism 1

NO + Cl2k1 NOCl2 (slow)

NO + NOCl2k2 2NOCl (fast)

2NO + Cl2 2NOCl

Step 1 is the rate determining step: Rate = k1[NO][Cl2]

The proposed mechanism is not plausible


Mechanism 2NO + Cl2 NOCl2 (fast)

NO + NOCl2 2NOCl (slow)2NO + Cl2 2NOCl

Rate = [NO][NOCl2]

Methods for Eliminating Intermediates1. Equilibrium Method - since fast steps are faster than the slowest step, they will become in equilibrium before the slowest step finishes

NO + Cl2 NOCl2

K =[NOCl2]

[NO][Cl2][NOCl2] = K[NO][Cl2]

Rate = k2K[NO]2[Cl2] = k[NO]

2[Cl2]

The proposed mechanism is plausible

2. Steady State Approximation - the intermediate will reach a steady state before the reaction finishes

d[int]

dt= 0

d[int]

dt appearance d[int]

dt disappearance= 0

Rate of appearance of NOCl2 = Rate of disappearance of NOCl2

k1[NO][Cl2] = k1[NOCl2] + k2[NO][NOCl2]k1[NO][Cl2] = [NOCl2](k1 + k2[NO])

[NOCl2] =k1[NO][Cl2]

k1 + k2[NO]

Rate =k2k1[NO]

2[Cl2]

k1 + k2[NO]

If k1 >>> k2k1 + k2[NO] k1Rate =

k2k1[NO]2[Cl2]

k1=

k2k1k1

[NO]2[Cl2]

Holds with experimental rate law

If k2 >>> k1k1 + k2[NO] k2[NO]

Rate =k2k1[NO]

2[Cl2]

k2[NO]= k1[NO][Cl2]

Does not hold with experimental rate law

This means that rate2 < rate1

Example 2:Overall Reaction: A + 2B C Rate = k[A]2[B]

A + A A2 (fast)A2 + B AB + A (slow)AB + B C (fast)A + 2B C

Rate = k2[A2][B]


Equilibrium Method:

K =[A2]

[A]2[A2] = K[A]

2

Rate = k2[B]K[A]2

Rate = k2K[A]2[B]

The proposed mechanism is plausible

k1[A]2 = k1[A2] + k2[B][A2]

k1[A]2 = [A2](k1 + k2[B])

[A2] =k1[A]

2

k1 + k2[B]

Rate =k2k1[A]

2[B]

k1 + k2[B]

If k1 >>> k2k1 + k2[B] k1Rate =

k2k1[A]2[B]

k1=k2k1k1

[A]2[B]

Holds with experimental rate law

If k2 >>> k1k1 + k2[B] k2[B]

Rate =k2k1[A]

2[B]

k2[B]= k1[A]

2

Does not hold with experimental rate law

This means that rate2 < rate1

Chain Reaction / Mechanism - occurs at high temperatures or at presence of radiation

Example: H2 +Br2 2HBr

1. Chain initiation:

2. Chain propagation:

3. Chain termination:

Br2 + light 2BrBr +H2 HBr +HH +Br2 HBr +Br ...Br +Br Br2H +H H2Br +H HBr

radical: with unpaired electron

they are reactive

Overall Reaction: propagation steps:

Br +H2 HBr +HH +Br2 HBr +Br

H2 +Br2 2HBr

CF2Cl2 + uv CF2Cl + Cl (chain initiation)Cl + O3 O2 + ClO (chain propagation)ClO + O O2 + Cl (chain propagation)O3 + O 2O2


Quiz

Overall Reaction: I + OCl OI + Cl

OCl + H2O HOCl + OH (fast)I + HOCl HOI + Cl (slow)

HOI + OH H2O + OI (fast)I + OCl OI + Cl

Equilibrium Method:

K =[HOCl][OH]

[OCl][HOCl] =

K[OCl][OH]

Rate = k2K[I][OCl][OH]1

Rate = k[I][OCl][OH]1

Steady State Approximation

k1[OCl] = k1[HOCl][OH] + k2[I][HOCl] [HOCl] =

k1[OCl]

k1[OH] + k2[I]

Rate =k2k1[I

][OCl]k1[OH] + k2[I]

If k1 >>> k2, Rate =k2k1k1

[I][OCl][OH]1

If k2 >>> k1, Rate = k1[OCl]

Nuclear Chemistry

Chemical Reactions Nuclear Reactions

Change that occurs formation/breaking of bondstransformation of the nu-

cleus itself

Subatomic Particles Involved valence electrons all subatomic particles

Energy change lower higher

Radioactivity: Henri Bequerel - spontaneous emission/radiation of particles in order to become stable - radioactive nuclei: radionuclide - spontaneous: occurs naturally

Stability - atoms with low atomic number: n/p ratio is 1:1 - atomic numbers have higher p-p repulsion so they need more neutrons to overcome that repulsion - atoms without 1:1 ratio are unstable and outside the belt of stability - largest atom with a 1:1 ratio is 83Bi

Balancing Nuclear Reactions

AZX

A: mass #: p + n (nucleons)

Z: atomic #: p

- balance both A and Z in nuclear reactions

Example: 147 N +

42 He178 O +11 H


proton: 11p or11H electron:

01e neutron:

10n

Type of Nuclear Reactions1. Radioactive Decay: spontaneous emission of radiation/particles a. Alpha Decay: occurs usually on heavy atoms (Z>82)

Alpha Particle: 42He or

42

Example: 23892 U 23490 Th+42 He

b. Beta Decay: done by nuclei with high n/p ratio p, n, n\p ratio Beta Particle:

01e or

01

Example: 23490 U 23491 Pa+01 e 10n11 p+01 e

c. Gamma Decay: electromagnetic radiation (light), usually accompanies other decays

Gamma Particle: 00

- the above three decays are ionizing radiation and are harmful

Material Alpha passes? Beta passes? Gamma passes?

Paper NO YES YES

Wood NO NO YES

Concrete/Lead NO NO SOME- differences are due to different masses and charges of ionizing radiation

d. Positron emission: done by nuclei with low n/p ratio p, n, n\p ratio

Example: 3015P 3014 Si+01 e 11p10 n+01 e

e. Electron capture: done by nuclei with low n/p ratio p, n, n\p ratio

Example: 4019K +

01 e4018 Ar 11p+01 e10 n

Radioactive Disintegration Series - after one instance of decay, a nucleus can still undergo decay

Example: 23892 U 23490 Th 23491 U 23492 U 23090 Th 20682 Pb

23692 U 20782 Pb 23292 U 20882 Pb

Kinetics of Radioactive Decay - all obey first-order kinetics Activity = kN SI unit: disintegration per second (dps) / Bequerel (Bq) Other unit: Curie (Ci): activity of 1 gram of radium 1 Ci = 3.7x1010 Bq

Integrated Rate Law: ln

NtN0

= kt Half Life: t 12=

ln 2

k


Example:

t 12= 12.8 hrs N = 1.50 mg t = 48 hrs

activity = kN =ln 2

t 12

(mg)

(1 g

1000 mg

)(1 mol

64 g

)(6.02 1023 disintegrations

1 mol

)(1 hr

3600 s

)=

ln 2

12.8 hrs(1.50 mg)

(1 g

1000 mg

)(1 mol

64 g

)(6.02 1023 disintegrations

1 mol

)(1 hr

3600 s

)= 2.12 1014 Bq

(1 Ci

3.7 1010 Bq)

= 5741 Ci

lnNtN0

= ktNt = e

kt(N0)

Nt = e(0.0542 hr1)(48 hrs)(1.50 mg) = 0.111 mg = 1.11 104 g

Mass Defect - calculated mass of nucleons in a nucleus is always greater than the actual mass - the mass is transformed into nuclear binding energy - energy that is released when nucleons combine - energy that is required to separate the nucleons - also a measure of stability of a nucleus - NBE Stability; the nucleus is more intact

Example

2814Si atomic mass = 27.96924

atomic mass = 14 protons

(1.00728 amu

proton

)+ 14 neutrons

(1.00867 amu

neutron

)= 28.22330 amu

m = calculated mass - actual mass = 28.22330 amu 27.96294 amum = 0.25406 amu

E = mc2

E = (0.25406 amu)(3 108)2(

1 g

6.02 1023 amu)(

1 kg

1000 g

)= 3.798 1011 J

nuclear binding energy per nucleon =3.798 1011 J28 nucleons

= 1.357 1012 J/nucleon


Quiz

activity0 = 0.0500 Ci activityt = 0.0310 Ci t = 30.0 hrs

lnNtN0

= kt

k = ln

activitytk

activity0k

t= ln 0.0310 Ci0=0.0500 Ci

30.0 hrs

k = 0.159 hr1

t 12=

ln 2

k

t 12=

ln 2

k=

ln 2

0.159 hr1= 43.5 hours

N96 : lnNtN0

= (0.159 hr1)(96 hours) = 1.5624NtN0

= 0.2166 remaining 78.34% disintegrated

5626Fe atomic mass = 55.92066

atomic mass = 26 protons

(1.00728 amu

proton

)+ 30 neutrons

(1.00867 amu

neutron

)= 56.44938 amu

m = calculated mass - actual mass = 56.44938 amu 55.92066 amum = 0.52872 amu

E = mc2

E = (0.52872 amu)(3 108)2(

1 g

6.02 1023 amu)(

1 kg

1000 g

)= 7.904 1011 J

nuclear binding energy per nucleon =7.904 1011 J56 nucleons

= 1.41 1012 J/nucleon

2. Nuclear Transmutation: bombardment of a nucleus by a particle to form another nucleus

Examples: 147 N +

42 He178 O +11 H 94Be+42 He126 C +10 n

Short-Hand: bombarded (bombarder, new particle) new nucleus

Example: 147 N (, p)

178 O

94Be (, n)

126 C

Transuranium Elements - man-made elements, prepared from Uranium - located after Uranium in the periodic table

Example: 1st Transuranium element to be discovered: 23892 U (n,)

23993 Np

23892 U +

10 n 23993 Np+01 e

3. Nuclear Fission: splitting of a heavy nucleus to form two lighter nuclei

23592 U +

10 n 14256 Ba+9136 Kr + 310n

Nuclear Chain Reaction - when a particle product from a previous fission is used to do another one


- critical mass: minimum mass needed to sustain a nuclear chain reaction

4. Nuclear Fusion: combining of two light nuclei to form a heavier nucleus - produces a higher energy, no radioactive waste - hard form of energy to manipulate

411H + 2

01e 42He

Effect of Radiation on Biological Matter - ionizing radiation can rip off electrons

H2O

radiation H2O+ + eH2O

+ +H2O H3O+ +OH OH starts a chain reaction that affects cells

SI Unit Other Unit Conversion Factor

Activity Bequerel (Bq) Curie (Ci) 1 Ci = 3.7x1010 Bq

Absorbed Dose Gray (Gy) in J/kg rad in 0.01 J/kg 1 Gy = 100 rad

Effective Dose Sievert(Sv) Gy x RBE rem rad x RBE 1 Sv = 100 rem

Example

mass = 75 kg Activity = 90 mCi time = 2 hours

energy = 9.121013 J/disintegration 85% absorbed RBE = 1

Activity in Bq = 90 mCi

(1 Ci

1000 mCi

)(3.7 1010 Bq

1 Ci

)= 3.3 109 Bq

Absorbed Dose =

(3.3109 disintegrations

1 s

)(7200 s)

(9.121013J

1disintegration

)(0.85)

75 kg= 0.246 Gy

0.246 Gy

(100 rad

1 Gy

)= 24.6 rad

Absorbed Dose = Gy x RBE = 0.246Gy 1 = 0.246 Sv

0.246 Sv

(100 rem

1 Sv

)(1000 mrem

1 rem

)= 24600 mrem

Chemical Thermodynamics- branch of chemistry that deals with heat/energy changes- predicts the spontaneity of a reaction

System: part of the universe under studySurroundings: everything not in the system

System + Surroundings = Universe

Types of System1. Open - allows the exchange of matter and energy - nonconservative system - the wall is permeable and diathermal; imaginary wall

2. Closed - only allows the exchange of energy - conservative to matter, nonconservative to energy - the wall is impermeable and diathermal; real wall

3. Isolated - does not allow the exchange of both matter and energy - conservative system - the wall is impermeable and adiabatic; real wall


Properties of the System1. Extensive - properties that are dependent on the amount of matter present Examples: mass, weight, volume, moles2. Intensive - properties that are independent on the amount of matter present Examples: temperature, density, molar mass, boiling point

Intensive Property =Extensive Property 1

Extensive Property 2

Extsys = Ext1 + Ext2 + . . .

Intsys = Int1 = Int2 = . . .

State Function- properties that define the state of the system- they are path independent and only focus on the initial and final states of the system Examples: P, V, T, U, H, G, S

Nonstate Function- properties that are dependent on the path taken by the system Examples: q and w

Example: travelling from QC to UPM paths: jeep, FX, taxi distance is constant distance is the state function fare is the nonstate function

Sign Conventions(+) q = heat is absorbed by the system (-) q = heat is released by the system(+) w = work is done on the system (-) w = work is done by the system

Note, in Chemistry, work is defined as the Pressure-Volume work (compression/expansion)

Heat, q (+) q = heat is absorbed by the system (-) q = heat is released by the system

q = mcT m: mass c: specific heat - the amount of heat needed to raise the temperature of 1 gram of substance by 1C - H2O: 1cal/gC or 4.184 J/gC

q = nCT n: moles C: molar heat capacity - the amount of heat needed to raise the temperature of 1 mole of substance by 1C - c and C are interconvertible

For Ideal Gases: CV = C at constant Volume CP = C at constant Pressure - this is because gases are greatly affected by pressure changes and are compressible - liquids and solids are incompressible and therefore their CV and CP are very close

monoatomic diatomic

CV3

2R

5

2R

CP5

2R

7

2R

CP = CV +R


Work, w (+) w = work is done on the system (-) w = work is done by the system

w = F dx P-V work: P =

F

AF = PA A: cross-sectional area

A dx = dV P: external pressure

w = P dV Following the sign convention: w = PdV = Fdx =

PdV

- the negative sign is added to follow direction, it has no bearing on the magnitude

Two Types of Work

Irreversible Work - actual work that we do wirrev =

PextdV

Irreversible Work Single-Step: from V1 to V2 in one pressure change Two-Step: from V1 to V2 in two pressure changes Three-Step: from V1 to V2 in three pressure changes and so on.... Reversible Work Infinite Steps: from V1 to V2 in infinite pressure changes

Work in One-Step


3. Isothermal (Constant T)

U = 0

U = q + w

0 = q + w

q = w

wirrev = PextdV

wrev = PgasdV

= V2

V1

nRT

VdV

= nRTV2

V1

dV

V

= nRT ln V2V1

4. Adiabatic

q = 0

U = q + w

U = w

U = nCVT

Example

Given:

Helium Gas n = 3 moles Tf = 100 K Ti = 25 Kwrev=? q=? U = ? Assume He is an ideal gas

A: at constant P

qP = nCPT

qP = (3 mol)

(5

2 8.314 J

molK

)(100K 25K)

qP = 4676.63 J

wrev = V2

V1

PgasdV

wrev = PgasV2

V1

dV = PV = nRT

wrev = (3 moles)(8.314

J

molK

)(75K)

wrev = 1870 J

U = q + w

U = 4676.63 J + 1870.65 J

U = 2805.98 J

U = nCVT

U = (3 moles)

(3

2 8.314 J

molK

)(75K)

U = 2805.98 J

B: at constant V

wrev = 0

qV = nCVT

qV = (3 mol)

(3

2 8.314 J

molK

)(100K 25K)

qV = 2805.98 J

U = q + w

U = q + 0 = q

U = 2805.98 J


For Isobaric Systems, qP is called H

U = qp + w

U = H + w

H = U wH = U (PV )H = U + PV

Isocoric: H = U +(PV )

= nCVT + nRT

= n(CV +R)T

= nCPT

Isothermal: H = 0

Adiabatic: H = nCPT

H: enthalpy = U + PV

For Ideal Gases: H = nCPT

Example:

Given:

N2(g) n = 2 moles Vf = 4L Vi = 2LAssume N2(g) is ideal T = 303.15 K

A: reversible

H = 0 U = 0 (isothermal system)

wrev = V2

V1

PgasdV = V2

V1

nRT

VdV

wrev = nRTV2

V1

dV

V= nRT ln V2

V1

wrev = (2moles)(8.314 JmolK

)(303.15K) ln4L

2Lwrev = 3494 J

q = wq = 3494 J

B: against a constant pressure of 2 atm (irreversible)

H = 0 U = 0 (isothermal system)

wirrev = V2

V1

PextdV = PextV2

V1

dV

wirrev = Pext(V2 V1)wirrev = (2atm)(4L 2L)

wirrev = 4 Latm8.314 JmolK

0.08206LatmmolK= 405.26 J

q = wq = 405.26 J

Isobaric Isocoric Isothermal Adiabatic

q qP = nCPT qV = nCVT q = w 0

wirrev

PextdV 0

PextdV

PextdV



wrev PV = nRT 0 nRT lnV2V1

PgasdV

U nCVT nCVT 0 w = nCVT

H nCPT nCPT 0 nCPT

Quiz

Given:

O2(g) n = 2 moles T1 = 313.15 Kadiabatic (q=0) Pgas = 4.50 atm Pext = 800 torr

wirrev = V2

V1

PextdV = PextV2

V1

dV

wirrev = Pext(V2 V1)

wirrev = (800 torr 1 atm

760 torr

)(34.26L 11.42L)

wirrev = 24.04 Latm8.314 JmolK0.08206LatmmolK

= 2435.85 J

U = q + w = 0 + w = w

U = 2435.85 J

V1 =nRT

P

V1 =2 mol 0.08206 LatmmolK 313.15K

4.50 atmV1 = 11.42L

V2 = 3V1 = 3(11.42L)

V2 = 34.26L

U = nCVT

2435.85 J = (2 mol)(5

2 8.314 J

molK

)(T2 313.15 K)

T2 = 254.55 K

Thermochemistry- measurement and calculation of heats of reaction at constant pressure- H is measured

H = qP H = U + PV H = U +(PV ) endothermic: H > 0

exothermic: H < 0

Hrxn = Urxn +ngRT

ng = moles of gaseous products - moles of gaseous reactants

Example

Given:

U = -2648 kJ/mol T = 298.15 K

C4H10(l) +13

2O2(g) 4CO2(g) + 5H2O(l)

ng = 4 132

= 52

Hrxn = Urxn +ngRT

Hrxn =

(2648 kJ

mol

)+

(52

)(8.314

J

molK

)(1 kJ

1000 J

)(298.15 K)

Hrxn = 2654.20 kJmol


Evaluating H1. Calorimetry - apparatus used to measure heat: isolated vessel a. Bomb (Constant V) Calorimeter: qV (U) b. Constant Pressure Calorimeter: qP (H)

qtotal = 0

qrxn + qcal = 0 (dependent on system)

qcal = mcT

= CcalT (m and c are constant - Ccal: Calorimeter Constant)

Finding which exchanged heat to create the working equation

Finding Ccal:

qhot H2O + qtap H2O + qcal = 0

Finding qfus of ice:

qfus + qice H2O + qtap H2O + qcal = 0

Finding qneutralization:

qneut + qacid + qbase + qcal = 0

Example:

Given:

Phenol (C6H5OH(s)) m = 1.800 g Ccal = 11.66 kJ/C

Bomb Calorimeter T1 = 21.36C T2 = 26.37C

qtotal = 0

qcombustion + qcal = 0

qcombustion = qcal = CcalTqcombustion = qV = 58.42 kJ per 1.8 g phenol

U =

(58.42 kJ1.8 g

)(94.12 g

1 mol

)U = 3054.72 kJ

mol

H = U +ngRT C6H5OH(s) + 7O2(g) 6CO2(g) + 3H2O(l)average temp: 298.15 K ng = 6 7 = 1

H = 3054.72 kJmol

+ (1)(8.314

J

molK

)(1 kJ

1000 J

)(298.15 K)

H = 3057.20 kJmol

2. Standard Heat of Formation (Hf) - heat involved in the formation of 1 mole of compound from its elements at standard state (P = 1 atm; activity = 1)


2C(s, graphite) + 3H2(g) +1

2O2(g) C2H5OH(l) Hf = 277.7

kJ

mol

Ca(s) + Cl2(g) CaCl2(s) Hf = 795.8kJ

mol

N2(g) + 2H2(g) +3

2O2(g) NH4NO3(g) Hf = 365.6

kJ

mol

Hf, elements = 0

Hf, N2(g) = 0

Hf, O2(g) = 0

Hf, Cl2(g) = 0

Hf, Br2(g) = 0 (is not a gas in standard state; liquid)

Hf, Cgraphite = 0 (stable allotrope in standard state)

Hf, Cdiamond = 0

Hrxn = nHf , products nHf , reactants

Example

C4H10(g) +13

2O2(g) 4CO2(g) + 5H2O(l)


=

[(4 mol)

(393.5 kJ

mol

)+ (5 mol)

(285.83 kJ

mol

)][(1 mol)

(124.73 kJ

mol

)+ (

13

2mol)

(0kJ

mol

)]= 2878.42kJ

3. Hess' Law of Heat Summation - possible because H is a state function

A B H1B C H2C D H3A D H = H1 +H2 +H3

solid liquid Hfus

liquid gas Hvapsolid gas Hsub

Hsub = Hfus +Hvap

Example

4NH3(g) + 3O2(g) 2N2(g) + 6H2O(l) H = 1531 kJmol

N2O(g) +H2(g) N2(g) +H2O(g) H = 367.4 kJmol

H2(g) +1

2O2(g) H2O(l) H = 285.9 kJ

mol


1

2[4NH3(g) + 3O2(g) 2N2(g) + 6H2O(l)] H = 1

21531 kJ

mol

3[N2O(g) +H2(g) N2(g) +H2O(g)] H = 3367.4 kJmol

3[H2(g) + 12O2(g) H2O(l)] H = 3285.9 kJ

mol

Overall : 2NH3(g) + 3N2O(g) 4N2(g) + 3H2O(l)HT = 1531 kJ

mol 367.4 kJ

mol 285.9 kJ

mol= 1010 kJ

4. Bond Enthalpy - energy required to break 1 mol of a bond - applicable only to gaseous covalent compounds - correlated to atomic size and bond length - bond enthalpy double bond is not equal to twice for that of a single bond - different nature of bonds (single: sigma, double: sigma + pi)

Hrxn = nH, bonds brokennH, bonds formed

Example:

C4H10(g) +13

2O2(g) 4CO2(g) + 5H2O(g)

Using Standard Heats of Formation:


=

[(4 mol)

(393.5 kJ

mol

)+ (5 mol)

(285.83 kJ

mol

)][(1 mol)

(124.73 kJ

mol

)+ (

13

2mol)

(0kJ

mol

)]= 2878.42kJ

Using Bond Enthalpies:

Hrxn = nHf , bonds broken nHf , bonds formed

=

[(10 mol)

(413

kJ

mol

)+ (3 mol)

(348

kJ

mol

)+ (

13

2mol)

(495

kJ

mol

)][(8 mol)

(799

kJ

mol

)+ (10 mol)

(463

kJ

mol

)]= 2630.5kJ

C4H10(g) +132 O2(g) 4CO2(g) + 5H2O(g) H = 2630.5 kJ

5H2O(g) 5H2O(l) H = 5(44.01 kJ)C4H10(g) +

132 O2(g) 4CO2(g) + 5H2O(l) H = 2850.55 kJ

Measuring Hrxn using standard heats of formation is more reliable than using bond enthalp-ies since bond enthalpies change according to the environment of the bond.

Quiz

Given:

C8H18 V = 10.00 mL = 0.688 g/mL Ccal = 7.62 kJ/C

T2 = 66.5C T1 = 23.2 C Average = 25 C Bomb Calorimeter


qtotal = 0

qcomb + qcal = 0

qcomb = qcal= CcalT

(7.62

kJC

)(66.5C 23.2C)

= 329.946 kJ = qv

U =329.946 kJ

(10.00 mL)(0.688 g1 mL

) (1 mol

114 gC8H18

) = 5467.21 kJmol

C8H18(g) +25

2O2(g) 8CO2(g) + 9H2O(l) ng = 8 25

2= 9

2H = U +ngRT

= 5467.21 kJmol

(9

2

)(8.314 103 kJ

molK

)(298.15 K)

= 5478.36 kJmol


5478.36 kJmol

=

[(8 mol)

(393.5 kJ

mol

)+ (9 mol)

(285.83 kJ

mol

)][(1 mol)

(Hf, C8H18

)+ (

13

2mol)

(0kJ

mol

)]Hf, C8H18 = 242.11

kJ

mol

Second Law of Thermodynamics- enables us to predict the driving force of reactions- spontaneous: naturally occuring Examples: rusting, transfer from hot to cold- the second law predicts the direction of a spontaneous processs

S: entropy- measure of the degree of disorder or randomness

dS =

dqrevT

- the natural tendency of things is to obtain disorder- the entropy od an isolated system increases in a spontaneous process - as long as the system absorbs heat, dS - a system with more entropy is more probable

solidS>0 liquid S>0 gas S 0


Ideal Gas System

1. Isobaric 2. Isocoric

dS =dqPT

=nCP dT

TS2S1

dS =

T2T1

nCP dT

T

S = nCP lnT2T1

dS =dqVT

=nCV dT

TS2S1

dS =

T2T1

nCV dT

T

S = nCV lnT2T1

3. Isothermal 4. Adiabatic

dS =dqrevT

= dwrevT

=(PgasdV )

T=PgasdV

T

=nRTV dV

T= nR

dV

VS2

S1

S =

S2S1

nRdV

V

S = nR lnV2V1

dS = 0

Phase Changes (constant T and constant P)

S =

qrevT

=HtransitionTtransition

Chemical Reactions

Srxn = nS

prod nSreac

Selements = 0

Example

Given:

H2O(s) H2O(g) n = 2 moles Tf = 383.15 K Ti = 268.15 KP = 1 atm

Hfus, ice = 6.02 kJ/mol Hvap, water = 40.67 kJ/mol

ice CP = 37.7 J/molK

liquid water CP = 75.3 J/molK

steam CP = 33.9 J/molK

S1 = nCP lnT2T1

= (2 mol)

(37.7

J

molK

)ln

(273.15 K

268.15 K

)= 1.39

J

K


S2 =nHfusTm

=(2 mol)

(6.02 kJmol

)273.15 K

= 44.08J

K

S3 = nCP lnT2T1

= (2 mol)

(75.3

J

molK

)ln

(373.15 K

273.15 K

)= 46.98

J

K

S4 =nHvap

Tb=

(2 mol)(40.67 kJmol

)373.15 K

= 217.98J

K

S5 = nCP lnT2T1

= (2 mol)

(33.9

J

molK

)ln

(383.15 K

373.15 K

)= 1.79

J

K

ST = S1 +S2 +S3 +S4 +S5

= 312.22J

K

Quiz

Given:

Hfus = 9.87 kJ/mol Hvap = 30.7 kJ/mol Svap = 86.9 J/molK

liquid CP = 136.1 J/molK

gas CP = 84.67 J/molK

Ti = 278.75K

Svap =HvapTb

Tb =HvapSvap

=30700 Jmol86.9 JmolK

= 353.28 K

ST = S1 +S2 +S3 +S4

=9870 Jmol278.75 K

+

(136.1

J

molK

)ln

353.28 K

278.75 K+ 86.9

J

molK+

(86.67

J

molK

)ln

373.15 K

353.28 K

= 159.19J

molK

Alternative Basis for Spontaneity

Suniverse > 0

Ssystem +Ssystem > 0

Ssys qsysT

> 0

TSsys qsys > 0at constant P, qP = H TSsys Hsys > 0

Hsys TSsys < 0Gsys < 0

G: Gibb's Free Energy- available energy needed to do useful work

G = Hheat available

TSheat used

G = H TS


G < 0: spontaneous G = 0: equilibrium G > 0: nonspontaneous

Example:

Hfus = 6.02 kJ/mol Sfus =HfusTm

=6.02 kJ/mol

273.15 K

= 0.02204kJ

molK

G = H TS

at 0C : G = 6.02kJ

mol (273.15K)

(0.02204

kJ

molK

)= 0

kJ

mol

EQUILIBRIUM at 0

at 10C : G = 6.02kJ

mol (283.15K)

(0.02204

kJ

molK

)= 0.2204 kJ

mol

SPONTANEOUS at 10

at -10C : G = 6.02kJ

mol (263.15K)

(0.02204

kJ

molK

)= 0.2204

kJ

mol

NONSPONTANEOUS at -10

2. 6CO2(g) +6H2O(l) C6H12O6(s) +6O2(g)Hf -393.5 -285.83 -1273.02 0 kJ/molS 213.6 69.91 212.1 205.0 J/molK

T = 298.15K

G = H TS= [nHf , products nHf , reactants] T [nS, products nS, reactants]

=

{[(1)

(1273.02 kJ

mol

)+ (6)

(0kJ

mol

)][(6)

(393.5 kJ

mol

)+ (6)

(285.83 kJ

mol

)]} (288.15 K)

{[(1)

(212.1

J

molK

)+ (6)

(205.0

J

molK

)][(6)

(213.6

J

molK

)+ (6)

(69.91

J

molK

)]}(

kJ

1000J

)= 2880.17 kJ, NONSPONTANEOUS

G = H - TS + Reaction is spontaneous at all T Reaction is spontaneous at low T+ + Reaction is spontaneous at high T+ Reaction is nonspontaneous at all T

Grxn = nGf , products nGf , reactants

Gf, elements = 0

Phase Changes (Constant T and P)

Find q, w, U, H, S, and G for 2mol H2O

H2O(l) H2O(g)Hvap = 40.67 kJ/mol


q = nHvap = (2 mol)

(40.67

kJ

mol

)= 81.34 kJ

w = (reversible, since slow process)

= PextV2

V1

dV = Pext(V2 V1)

= Pext(Vgas Vliq)Assume Vgas >>> Vliq; for liquid to solid reactions use density to convert

= PVgas = nRT

= (2 mol)(8.314

J

molK

)(373.15 K) = 6205 J

U = q + w

= 81.34 kJ 6.205 kJ = 75.135 kJH = qP = 81.34 kJ

Svap =nHvap

Tb=

(2 mol)(40.67 kJmol

)373.15 K

= 0.218kJ

K

G = H TS = 81.34 kJ (373.15 K)(0.218

kJ

K

)= 0

Third Law of Thermodynamics

Spure, crystalline, solid at 0K = 0

- you can calculate for the S and not only S

0K 298.15KS = S298.15K S0KS = S298.15K


q qP = nCPT qV = nCVT q = w 0

wirrev

PextdV 0

PextdV

PextdV

wrev PV = nRT 0 nRT lnV2V1

PgasdV

U nCVT nCVT 0 w = nCVT

H nCPT nCPT 0 nCPT

S nCP lnT2T1

nCV lnT2T1

nR lnV2V1

0

Chemical Equilibrium- the moment where there is no change in the concentration of reactants/products - the reaction is still occurring (dynamic state)

Approaches to EquilibriumKinetic Approach

1

2N2(g) +O2(g) NO2(g)


at equilibrium: Rateforward = Ratebackward

Kf [N2]12 [O2] = Kb[NO2]

KfKb

=[NO2]

[N2]12 [O2]

KfKb

= Keq/KC (equilibrium constant)

Thermodynamic Approach- natural direction of reactions is toward equilibrium - equilibrium: maximum entropy of the universe. lowest G of the system - when G = 0, it has achieved the above conditions

1

2N2(g) +O2(g) NO2(g)

G = G +RT lnQa

G = G +RT lnaNO2

aN212 aO2

Q: reaction quotient

Qa =aNO2

aN212 aO2

at equilibrium:

G = 0

0 = G +RT lnKaG = RT lnKalnKa = G

RT

Ka = eGRT (thermodynamic/true equilibrium constant)

Ka =

aNO2

aN212 aO2

activity = true pressure = P

= activity coefficient (correction factor to account for non-ideal pressure

Ka =PNO2

(PN2)12 PO2

=NO2

N212 O2

PNO2PN2

12PO2

Ka = K KP

If gases are ideal, = 1, a = P

Ka KP = PNO2PN2

12PO2

P =nRT

V=MRT = CRT

KP =CRTNO2

CRTN212CRTO2

=CNO2

CN212CO2

(RT ) 12

KP = KC(RT )ng

Example

Gf,NO2(g) = 51.84kJ

molT = 298.15K

1

2N2(g) +O2(g) NO2(g)


Grxn = nGf,p nGf,r

= 51.84kJ

mol 1 mol (0 + 0) = 51.84 kJ

Ka = eGRT

= e 51.84 kJ

(8.314103 kJ/molK)(298.15 K)

= 8.27 1010

since the gases are ideal,

Ka = KP = 8.27 1010

KP = KC(RT )ng

KC =KP

(RT )ng=

8.27 1010[(8.314 103 kJ/molK)(298.15 K)] 12

= 4.09 109

Manipulating Equations

1

2N2(g) +O2(g) NO2(g) KP = 8.27 1010

2

[1

2N2(g) +O2(g) NO2(g)

]KP = (8.27 1010)2

1

2

[1

2N2(g) +O2(g) NO2(g)

]KP = (8.27 1010) 12

1[1

2N2(g) +O2(g) NO2(g)

]KP = (8.27 1010)1

Adding/Subtracting

12N2(g) +O2(g) O2(g) KC1

NO(g) 12N2(g) +12O2(g) KC2

NO(g) +12O2(g) NO2(g) KC1 KC2

Quiz

2NO(g) N2(g) +O2(g) KP = 1 1030

3NO(g) +3

2Br2(g) 3NOBr(g) KP = 715.5

12[2NO(g) N2(g) +O2(g)] (KP = 1 1030) 12

1

3[3NO(g) +

3

2Br2(g) 3NOBr(g)] (KP = 715.5)

13

Overall:1

2N2(g) +

1

2O2(g) +

1

2Br2(g) NOBr(g) KP = 8.944 1015

G = RT lnKa Ka KP= (8.314 103kJ/molK)(298.15 K)(ln 8.944 105) = 91.05 kJ

mol

Molecular EquilibriumSolving Equilibrium Problems 1. Write the balanced equilibrium equation


2. ICE table 3. Relate equilibrium concentrations to K 4. Solve

Homogenous Equilibria

1. Given:

1 mol SO3(g) 2L vessel 303.15 K 12.5% decomposed at equilibrium

2SO3g 2SO2g O2gI 0.5 M 2 M 0 MC -0.0625 M +0.0625 M +(0.5)0.0625 ME 0.4375 M 0.0625 M 0.03125 M

KC =[SO2]

2[O2]

[SO3]2

KC =(0.0625M)2(0.03125M)

(0.4375M)2

KC = 6.38 104

KP = KC(RT )ng = 6.38 104

((0.08206

Latm

molK

)(303.15 K)

)KP = 0.0159

2. Given:

PNO = 300 mmHg PCl = 250 mmHg Peq = 420 mmHg T = 473.15 K

NO(g) + O2(g) NOCl(g)I 300 0 MC -x -(0.5)x +xE 300-x 250-0.5x x

300 x+ 250 0.5x+ x = 420 x = 260NO : 40mmHg Cl2 : 120mmHg NOCl : 260mmHg

KP =PNOCl

PNOP12

CL2

KP =260/760

(140/760)(120/760)12

KP = 16.36

KC =KP

(RT )ng=

16.36((0.08206LatmmolK

)(473.15 K)

) 12KC = 101.94

3. Given: Initial: 0.40 mol NH3 Equilibrium: 0.15 mol H2 at 2 atm

2NH3 N2 + 3H2I (mol) 0.4 0 0C (mol) -0.1 +0.05 +0.15E (mol) 0.30 0.05 0.15

mTOTAL = 0.5 mol


PT = 2 atm

PNH3 = NH3PT =0.3 mol

0.5mol 2 atm = 1.2 atm

PN2 = N2PT =0.05 mol


PH2 = H2PT =0.15 mol


KP =PN2P

3H2

P 2NH3=

(0.2)(0.6)3

(1.2)2= 0.030

QUIZ

Given: Initial: 0.112 mol O2 0.0400 mol N2O in 2.00L Equilibrium: 0.0400 mol NO2

2N2O(g) + 3O2(g) 4NO2(g)I 0.02 M 0.56 M 0C - 0.01 - 0.015 + 0.02E 0.01 M 0.041 M 0.02 M

KC =[NO2]

4

[N2O]2[O2]3=

(0.02)4

(0.01)2(0.041)3= 23.21

4. Given: Initial: 0.125 mol COCl2 in 2.00L KC = 8 at 298.15 K

COCl2(g) CO(g) + Cl2(g)I 0.125 M 0 0C - x + x + xE 0.125 - x M x x

KC =[CO][Cl2]

[COCl2]8 =

x2

0.125 xx = 0.123

At Equilibrium: [COCl2] = 0.002 M [CO] = 0.123 M [Cl2] = 0.123 M

Heterogenous Equilibria

NH4Cl(s) NH3(g) +HCl(g)Ka =

aNH3aHClaNH4Cl

aNH4Cl = 1 (a = 1 for pure solids and liquids)

= aNH3aHCl = (P )NH3(P )HCl

= NH3HCl PNH3PHCl= K KP KP (for ideal gases, = 1) = PNH3PHClKC = [NH3][HCl]

Fe(s) +H2O(l) FeO(s) +H2(g) KC = [H2][H2O]2Ag(s) +

12O2(g) Ag2O(s) KP =

1P 0.5O2

C(s) + CO2(g) 2CO(g) KC = [CO]2

[CO2]

1. Given: PT = 450 mmHg at 298.15 K

NH4Cl(s) NH3(g) + HCl(g)I - 0 0C - + x + xE - x x

PT = 450 mmHg = 2x

x = 225 mmHg

KP = PNH3PHCl =

(225

760

)2= 0.0876

KC =KP

(RT )ng=

0.0876

[(0.09206)(298.15)]2= 1.46 104


2. Given: 2.5 mol NH4HS in 2L KC = 1.2 104 at 303.15 K

NH4HS(s) NH3(g) + H2S(g)I - 0 0C - + y + yE - y y

KC = [NH3][H2S]

1.2 104 = y2y = 0.011 M

At Equilibrium: [NH3]=0.011 M [H2S]=0.011 M

3. Given: 0.30 mol CCl4, 0.010 mol Cl2 in 2L KC = 0.0132 at 700 K

CCl4(g) C(s) + 2Cl2(g)I 0.15 M - 0.005 MC - z - + 2zE 0.15 - z - 0.005 + 2z

KC =[Cl2]

2

[CCl4]0.0132 =

(0.005 + 2z)2

0.15 zz = 0.0183

At Equilibrium: [CCl4] = 0.1317 M [Cl2] = 0.0416 M

Factors Affecting EquilibriumLe Chatelier's Principle- when a stress factor is applied to a system in equilibrium, the system will shift to relieve the stress (relieve the balance) factors: changes in concentration, P, V, T, addition of catalyst

2A(g) +B(g) P(g) + heat

1. Concentration - changes in concentration of solids and pure liquids do not affect equilibrium

Addition of A: shift to the RIGHT Removal of A: shift to the LEFT Addition of B: shift to the RIGHT Removal of B: shift to the LEFT Addition of P: shift to the LEFT Removal of P: shift to the RIGHT

2. Pressure/Volume (for gases)P (resulting in V) shift to the RIGHT P (resulting in V) shift to the LEFT P, shift to the lesser number of moles) P, (shift to the higher number of moles)

addition of an inert gas at constant P: decreases the partial pressure addition of an inert gas at constant V: no effect, additional pressure

3. Temperature - alters the value of the rate constant

T: shift to the LEFT T shift to the RIGHT T favors endothermic processes T favors exothermic processes H>0 (heat in the reactants) H < 0 (heat in the products)

4. Catalyst - no shift in equilibrium, they affect both the forward and backward reactions

Quantitative Approach

G = G +RT lnQ G = RT lnKG = RT lnK +RT lnQG = RT ln

Q

K

Q < K,G < 0, forward shift

Q = K,G = 0, equilibrium

Q > K,G > 0, backward shift


Example: Given: 0.5 M COCl2 in 2L KC = 8

COCl2(g) CO(g) + Cl2(g)I 0.5 M 0 0C - a + a + aE 0.5 - a a a

KC =[CO][Cl2]

[COCl2]8 =

a2

0.5 aa = 0.472

At Equilibrium: [COCl2] = 0.028 M [CO] = 0.472 M [Cl2] = 0.472 M

COCl2(g) CO(g) + Cl2(g)I 0.028 M 0.472 0.472C + 0.5E 0.528

QC =[CO][Cl2]

[COCl2]=

(0.472)2

0.528= 0.422

KC > Qc (forward shift)

COCl2(g) CO(g) + Cl2(g)I 0.028 M 0.472 0.472C + 0.5E 0.972

QC =[CO][Cl2]

[COCl2]=

(0.472)(0.972)

0.028= 16.39

KC < Qc (backward shift)

COCl2(g) CO(g) + Cl2(g)I 0.028 M 0.472 M 0.972 MC + b - b - bE 0.028 + b 0.472 - b 0.972 - b

KC =[CO][Cl2]

[COCl2]

8 =(0.472 b)(0.972 b)

0.028 + b

b = 0.0249

At New Equilibrium: [COCl2] = 0.0529 M [CO] = 0.4471 M [Cl2] = 0.9471 M

Temperature: alters K

G = RT lnKalnKa = G

RT= H

TSRT

lnKa = H

R

(1

T

)+S

R(vant Hoff equation)

Plot lnKa vs1

Tslope: Endothermic: - Exothermic: +

y-int:S

R

Two-point form:

lnKa2Ka1

= H

R

(1

T2 1T1

)Ionic Equilibria Nonelectrolyte - does not conduct electricity - does not dissociate into ions - molecular Examples: organic molecules C6H12O6, C12H22O11, CO(NH)2 Electrolyte - solutions that conduct electricity (salts, acids, bases) - dissociates into ions Strong Electrolyte - strongly conducts electricity - completely dissociates into ions Examples: HCl (strong acid), NaOH (strong base), NaCl (salt) Weak Electrolyte - partially dissociates into ions


Examples: CH3COOH (weak acid), NH3 (weak base), AgCl (partially soluble)

Theories on Acids and Bases

Arrhenius Acid: H+ producers Bases: OH- producers

HCl(aq) +H2O(l) H3O+(aq) + Cl(aq)NH3(aq) +H2O(l) NH+4(aq) +OH

(aq)

HCl(aq) +NH3(aq) NH+4(aq) + Cl(aq) (limitation) - the substance must be in an aqueous solution

Bronsted-Lowry Acid: H+ donors Bases: H+ acceptors

HCl(aq) acid 1

+H2O(l) base 2

H3O+(aq) acid 2

+Cl(aq) base 1

NH3(aq) base 1

+H2O(l) acid 2

NH+4(aq) acid 1

+OH(aq) base 2

HCl(aq) acid 1

+NH3(aq) base 2

NH+4(aq) acid 2

+Cl(aq) base 1

- water can be an acid or a base (amphoteric or amphiprotic) - dependent on the partner of water

Lewis Acid: electron pair acceptors Bases: electron pair donors

- forms coordinate covalent bonds (only one atom donates the electron pair) - not all Lewis acids are Bronsted-Lowry acids - other species besides H+ can accept electron pairs Example: central atoms in complexes

Examples of Lewis Acids: electron deficient (partially positive)1. H+

2. metals3. molecules with incomplete octets Examples: BF3, AlCl34. CO2 / SO2 (with double bonds)

Lewis Base: electron rich (with lone pairs)

Acid Strength- capacity to donate the proton (based from the Bronsted-Lowry theory)

1. Polarity of the H-X bond - having the greatest proton-like characteristics (most positive) 2. Strength of the H-X bond - must be connected trough a weaker bond to be donated more easily 3. Stability of Conjugate - acids are stronger when their conjugate base is more stable - unstable conjugate bases tend to reclaim the donated proton

Acid Strength for Binary Acids (HnX)From LEFT TO RIGHT: Acidity INCREASES Example: Acidity of CH4 < NH3 < H2O < HF- due to the polarity of H-X bond / electronegativity - HF has the most proton-like H+ (most positive)

From TOP TO BOTTOM: Acidity INCREASES


Example: Acidity of HF < HCl < HBr < HI- bigger atomic size, longer bond length - it becomes easier to donate the H+

Priority: the ability to donate (atomic size) over positivity of H+ (electronegativity)

From left to right, atomic size increases negligibly, so polarity is more prioritized.

Acid Strength of OxyacidsMore O atoms, HIGHER Acidity Example: Acidity of HNO2 < HNO3- Inductuve Effect: oxygen atoms exert a pull on the electron cloud, causing the H+ to be more positive. O atoms, pull on electron cloud, positive H+

Stability of Conjugate BaseAssessment: looking at the negative charge- the more atoms the electron is distributed to (by having more resonance structures), the more spread the negativity is and therefore, the more stable the conjugate base is

HIGHER Electronegativity of Central Atom, HIGHER Acidity- the H is not attached to the central atom. Therefore, the central atom only can pull the electron cloud.

Stability of Conjugate Base- so that the effect of the negative can be lessened, it may be pulled by the central atom, so the higher the pull of the central atom, the more stable the conjugate base is.

Basicity Acidity, Basicity- the stronger the acid, the weaker is its conjugate base.Examples: Acidity: HF < HCl < HBr < HI Basicity: F- > Cl- > Br- > I- Acidity: HNO3 > HNO2 Basicity: NO3

- < NO2-

Acidity: NH3 < H2O < HF Basicity: N3- > O2- > F-

Ionization of Water- water is amphoteric, so it can autoionize (Endothermic Reaction)

H2O(l) +H2O(l) H3O+(aq) +OH(aq)

KW = [H3O+][OH]

at 298.15 K, pure water, KW = 1.0 1014[H3O

+] = 1.0 107M [OH] = 1.0 107M - the concentrations are so small that water is almost a nonelectrolyte

logKW = log[H3O+] + log[OH]

log[H3O+] log[OH] = logKwpH + pOH = pKw

at 298.15 K, pH + pOH = 14

at 298.15 K,

Acid : [H3O+] > 1.0 107 [OH] < 1.0 107

pH < 7 pOH > 7

Neutral : [H3O+] = [OH] = 1.0 107

pH = pOH = 7

Base : [H3O+] < 1.0 107 [OH] > 1.0 107

pH > 7 pOH < 7

Strong ElectrolytesStrong Acids Strong BasesHCl HBr HI NaOH KOH LiOHHNO3 HClO4 HClO3 RbOH CsOH Ba(OH)2H2SO4 (first ionization) Sr(OH)2


Levelling Effect: Strongest acid that can exist in a solvent is the solvent's conjugate acid.Strongest base that can exist in a solvent is the solvent's conjugate base.- the degree of ionization is dependent on the solvent Example: many acids that are weak in water fully ionize in liquid ammonia.

Weak Electrolytes

HA(aq) +H2O(l) H3O+(aq) +A(aq)

Ka - acid ionization/dissociation constant

Ka =[H3O

+][A][HA]

Examples: HF, organic acids (R-COOH)

B(aq) +H2O(l) BH+(aq) +OH(aq)

Kb - base ionization/dissociation constant

Kb =[BH+][OH]

[B]Examples: organic bases (R-NH2)

Polyprotic Acids

H3PO4(aq) +H2O(l) H3O+(aq) +H2PO4(aq) Ka1 = 7.5 103

H2PO4(aq) +H2O(l) H3O

+(aq) +HPO

24(aq) Ka2 = 6.2 108

HPO24(aq) +H2O(l) H3O+(aq) + PO

34(aq) Ka3 = 4.2 1013

- the Ka becomes smaller after each ionization - since the acid becomes more negative, removing the H+ becomes harder ionization Ka

Salts- product of the neutralization of an acid with a base- the cation is from the base and the anion is from the acid Example: NaCl Na from NaOH; Cl from HCl

Salts of Strong Acids and Strong Bases

KNO3(aq) K+(aq) + NO3(aq)from a strong base from a strong acid

no hydrolysis no hydrolysis

- both are conjugates of strong acids and bases so no hydrolysis occurs, pH 7

Salts of Weak Acids and Strong Bases

NaF(aq) Na+(aq) + F(aq)from a strong base from a weak acid

no hydrolysis F(aq) +H2O(l) HF(aq) +OH(aq) (Kb)

pH > 7

Salts of Strong Acids and Weak Bases

NH4Cl(aq) NH+4(aq) + Cl(aq)from a weak base from a strong acid

NH+4(aq) +H2O(l) H3O+(aq) +NH3(aq) (Ka) no hydrolysis

pH < 7

More Examples:


KI(aq) +H2O(l) KOH(aq) +HI(aq) neutral

NaNO3(aq) +H2O(l) NaOH(aq) +HNO3(aq) neutral

NaNO2(aq) +H2O(l) NaOH(aq) +HNO2(aq) basic

NH4Br(aq) +H2O(l) NH3(aq) +HBr(aq) acidic

NaCH3COO(aq) +H2O(l) NaOH(aq) + CH3COOH(aq) basic

CH3NH3Cl(aq) +H2O(l) CH3NH2(aq) +HCl(aq) acidic

Relating Ka and Kb of conjugate acid-base pairs

A(aq) +H2O(l) HA(aq) +OH(aq)

Kb =[HA][OH]

[A] [H3O

+]

[H3O+]

Kb =[HA]

[A][H3O+] [OH][H3O+]

Kb =1

KaKW

KaKb = Kw (for conjugate acid-base pairs)

- this also explains why when an acid is very strong, its conjugate base is weak

Salts of Weak Acids and Weak Bases

Example: NH4FNH+4(aq) F

(aq)

from a weak base from a weak acidNH+4(aq) +H2O(l) H3O

+(aq) +NH3(aq) (Ka) F

(aq) +H2O(l) HF(aq) +OH

(aq) (Kb)

Ka > Kb acidic saltKa = Kb neutral saltKa < Kb basic salt

Kb : NH3 = 1.8 105 Ka : HF = 6.8 104Ka : NH

+4 = 5.56 1010 > Kb : F = 1.47 1011

acidic salt

Amphoteric Compounds

H2CO3(aq) +H2O(l) H3O+(aq) +HCO3(aq) Ka1

HCO3(aq) +H2O(l) H3O+(aq) + CO

23(aq) Ka2

NaHCO3 Na+(aq) +HCO3(aq) HCO3 is amphotericHCO3(aq) +H2O(l) H3O

+(aq) + CO

23(aq) Ka2 = 5.6 1011

HCO3(aq) +H2O(l) H2CO3(aq) +OH(aq) Kb =

KWKa1

= 2.33 108Kb is larger, so HCO

3 is more basic than acidic

From H3PO4(aq) +H2O(l) H3O+(aq) +H2PO4(aq) Ka1 = 7.5 103

H2PO4(aq) +H2O(l) H3O

+(aq) +HPO

24(aq) Ka2 = 6.2 108

HPO24(aq) +H2O(l) H3O+(aq) + PO

34(aq) Ka3 = 4.2 1013

Find out if Na3PO4, Na2HPO4 and NaH2PO4 are basic or acidic.

Common Ion Effect


- the common ion suppresses the ionization of weak electrolytes.

% ionization pH

0.10 M HCl 100% 1.0

0.10 M HCl + 0.05 M NaCl 100% 1.0

0.10 M HOAc 1.34% 2.87

0.10 M HOAc + 0.05 M NaOAc 0.036% 4.44

Weak Bases lower ionization lower pH

Buffer Solutions- solutions that resist drastic changes in pH upon addition of small amount of acid or base - because buffers have acids and bases that can neutralize that addition- biological samples are very sensitive to pH so they need to have buffers- buffers have ranges in pH that they are effective at (1.0 pH)- composed of a weak acid/base and its conjugate

HA(aq) +H2O(l) H3O+(aq) +A(aq)

Ka =[H3O

+][A][HA]

logKa = log[H3O+] + log

[A][HA]

log[H3O+] = logKa + log [A]

[HA]

pH = pKa + log[A][HA]

(Henderson-Hasselbalch Equation)

pOH = pKb + log[BH+]

[B]

- the closer pH to the pKa, the better the buffer - pH = pKa 1.0 (buffer capacity)

Initial pH + 10mL 1M HCl + 10 mL 1M NaOH

Buffer 7.00 6.98 7.02

H2O 7.00 1.71 12.29

- you can form a buffer by neutralizing the acid alone, thus forming a conjugate base (like in titration)

Acid-Base Indicators- weak organic acids and weak organic bases that are sensitive to pH changes

HIn(aq) +H2O(l) H3O+(aq) + In(aq)

The acid formed has a different color than the base formed.

pH range[In][HIn] 10 base color predominates +1

pKa[HIn][In] 10 acid color predominates -1

- pH ranges are dependent on pKa Examples: phenolphthalein, bromthymol blue


Acid-Base Titration- technique used to determine the concentration of an acid or base, using a known concen-tration of a known base or acid

titrant: known base or acid analyte: unknown acid or base indicator: 2 to 3 drops are added - a small amount because indicators are also acids/bases and can affect titration

Complex Ion Equilibrium- involves the formation of a complex

Cu2+(aq) + 4NH3(aq) Cu(NH3)2+4(aq)

Kf : formation constant

Kf =[Cu(NH3)

2+4 ]

[Cu2+][NH3]4= 5 1013

Ag+(aq) + 2NH3(aq) Ag(NH3)+2(aq)

Kf : formation constant

Kf =[Ag(NH3)

2+2 ]

[Ag+][NH3]2= 1.7 107

- since formation constants are very large, complex ion formation is very favorable

Slightly Soluble Salts

PbI2(s) Pb2+(aq) + 2I(aq)

KSP = [Pb2+][I]2 = 1.4 108

Al(OH)3(s) Al3+(aq) + 3OH(aq)

KSP = [Al3+][OH]3 = 1.8 1033

Ksp = solubility product constant

- solubility product constants are very small, therefore, the salts are very sparingly soluble

- you cannot say which is more soluble by looking at the Ksp unless they have the same formula type

Factors Affecting Solubility1. Temperature - temperature alters Ksp but it rarely affects the solubility of slightly soluble salts - since Ksp are small, the change temperature makes is not large enough - there are exceptions like PbCl2 (Ksp = 1.7 x 10

-5) Example: PbCl2 dissolves in hot water while AgCl doesnt (Ksp = 1.8 x 10

-10)

2. Common Ion Effect - the presence of the common ion suppresses the dissolution of salts

3. Uncommon Ion EffectAgBr(s) Ag+(aq) +Br

(aq) in 0.10 M KNO3

- the ion products form ion pairs with the uncommon ions (Ag+ with NO3- and Br- with K+)

- the amount of free ions decrease, so the equilibrium will shift forward (solubility)

4. Presence of Complexing Agent - increases solubility since complex ion formation reduces the amount of an ion formed - equilibrium shifts forward

5. pH - only affects salts with basic anions (conjugate bases of weak acids) - larger amount of basic anions (higher pH) shifts equilibrium backwards - precipitate forms - larger amount of acidic anions, (lower pH) neutralizes basic anions - equilibrium shifts forward, precipitate dissolves


Mg(OH)2(s) Mg2+(aq) + 2OH(aq)

OH(aq) +H3O+(aq) 2H2O(l)

PbF2(s) Pb2+(aq) + 2F(aq)

F(aq) +H3O+(aq) HF(aq) +H2O(l)

Exceptions:CuS CdS PbS NiS FeS ZnS

Ksp 6 1037 8 1028 3 1028 3 1020 6 1019 2 1025- for CuS, CdS and PbS, an increase in pH doesn't increase their solubility -since their Ksp are low, the removal of the basic anion doesn't cause a large shift forward

Precipitation

PbI2(s) Pb2+(aq) + 2I(aq)

Ksp = [Pb2+][I]2

Qsp = [Pb2+]0[I

]20 Q: ion product

Q < Ksp no precipitate (unsaturated solution)Q = Ksp maximum amount dissolved (saturated solution)Q > Ksp precipitate forms

Fractional Precipitation- precipitation of different compounds in a specific order- you cannot compare by looking at the Ksp unless they have the same formula type and concentration. - if the above is true, then the compound with the lowest Ksp forms first

Dissolution1. Variation in Temperature - no significant effect on most salts because of their low Ksp 2. Variation in pH - only applicable to salts with basic anions

BaCO3(s) Ba2+(aq) + CO23(aq)

CO23(aq) +H3O+(aq) HCO

3(aq) +H2O(l)

HCO3(aq) +H3O+(aq) H2CO3(aq)

CO2+H2O

+H2O(l)

3. Presence of Complexing Agent4. Presence of Oxidizing Agent

3CuS(s) + 8HNO3(aq) 3Cu(NO3)2(aq) + 2NO(g) + 3S(s) + 4H2O(l)

Amphoteric Hydroxides- soluble on both acids and bases

Al(OH)3(s) + 3H3O+(aq) Al

3+(aq) + 6H2O(l)

Al(OH)3(s) +OH(aq) Al(OH)

4(aq)

Zn(OH)2(s) + 3H3O+(aq) Zn

2+(aq) + 4H2O(l)

Zn(OH)2(s) + 2OH(aq) Zn(OH)

24(aq)

Electrochemistry- deals with interconversion of electrical energy and chemical energy- deals with reduction-oxidation reactions


Electrochemical Cell- cells that harness electrical energy from chemical energy in solution Example: Daniell Cell

electrode - metal where the reduction/oxidation reaction occurs anode: (- electrode) oxidation (AN OX) cathode: (+ electrode) reduction (RED CAT)electron flow: from anode to cathodesalt bridge - saturated salt solution, usually KCl or KNO3 - cation and anion have equal mobilities anode: Zn2+ is formed, excess of positive charge, anion is attracted cathode: SO4

2- is formed, excess of negative charge, cation is attracted - retains neutrality of solution

Two Types of Electrochemical Cell1. Galvanic or Voltaic Cell: generates electricity - chemical energy to electrical energy - reaction is spontaneous - work is done by the system - anode: (-) electrode cathode: (+) electrode2. Electrolytic Cell: consumes electricity - electrical energy to chemical energy - reaction is nonspontaneous, electricity is needed to drive it - work is done by the surroundings - anode: (+) electrode cathode: (-) electrode

Cell Potential / Cell EMF- sum of the contribution of both half cell potentials

Electrode potential - cannot be exactly determined - measured using a reference electrode - Standard Hydrogen Electrode (SHE)

2H+(aq) + 2e H2(g) E = 0V

- electrode potential arbitrarily assigned as 0 - used platinum (an inert electrode, doesnt react) as the solid electrode


-covered with platinum to increase surface area for more adsorption of H+ - alternative inert electrode: C(s, graphite) - paired up different electrodes with the SHE to determine their respective potentials - convention: E must be written as the REDUCTION potential - oxidation potential = -E

E, chance to be reduced, strength as Oxidizing Agent Example: Cu2+ vs Zn2+ Cu2+ is reduced, Cu2+ is a stronger Oxidizing Agent

E, chance to be oxidized, strength as Reducing Agent

Example:Arrange the following in order of:1. Increasing strength as Oxidizing Agent: Cl2, Br2, Ce

4+ Br2 < Cl2 < Ce4+

2. Increasing strength as Reducing Agent: Ni, Mn, Cd Ni < Cd < Mn3. Decreasing strength as Oxidizing Agent in acidic medium: Cr2O7

2-, MnO4-, H2O2 H2O2 > MnO4

- > Cr2O72-

4. Decreasing strength as Reducing Agent: Co, Ag, Fe Fe > Co > Ag

Cell Representation: Anode | Anode solution || Cathode solution | Cathode Example: Zn | Zn2+ || Cu2+ | Cu

Example:Calculate the standard cell potential of the following:

1. Zn|Zn2+||Ni2+|NiAnode: [Zn2+ + 2e Zn(s)] E = 0.76V

Cathode: Ni2+ + 2e Ni(s) E = 0.25VCell Reaction: Zn(s) +Ni

2+ Ni(s) + Zn2+ Ecell = Ecathode EanodeEcell = 0.51VEcell > 0, Galvanic

If Ecell > 0, you may say that the reaction is spontaneous at standard state - the cell is GALVANIC If Ecell < 0, you may say that the reaction is nonspontaneous at standard state - the cell is ELECTROLYTIC

2. Ag|Ag+||Fe2+|FeAnode: 2[Ag+ + e Ag(s)] E = 0.80V

Cathode: Fe2+ + 2e Fe(s) E = 0.44VCell Reaction: 2Ag(s) + Fe

2+ Fe(s) + 2Ag+ Ecell = 1.24VEcell < 0, Electrolytic

Note: Changing the moles of the reaction doesn't change E, it its intrinsic. To drive the reaction in number 2, you must apply more than 1.24V electricity

3. Cr|Cr3+||Cl, Cl2(g)|PtAnode: 2[Cr3+ + 3e Cr(s)] E = 0.74V

Cathode: 3[Cl2(g) + 2e 2Cl] E = 1.36V

Cell Reaction: 3Cl2(g) + 2Cr(s) 6Cl(aq) + 2Cr3+aq Ecell = 2.10VEcell > 0, Galvanic

Thermodynamics of Electrochemical Cells

G =Welectrical

G = nFE

n: moles of electrons (Faraday)

F: Faradays Constant: 96,485 C/mol e-

E: electrode potential


E > 0, G < 0, spontaneous, GALVANIC

E = 0, G = 0, equilibrium, neither

E < 0, G > 0, nonspontaneous, ELECTROLYTIC

G = G +RT lnQ nFE = nFE +RT lnQ

E = E RTnF

lnQ Nernst Equation

at 25C, E = E 8.314J

molK 298.15 Kn 96485 Cmol

2.303 logQ

E = E 0.0592n

logQ (at 25C)

Determine the potential of the following cells:1. P b|Pb2+(0.010M)||Co3+(0.050M), Co2+(0.0040M)|Pt

Anode: [Pb2+ + 2e Pb(s)] E = 0.13VCathode: 2[Co3+ + e Co2+] E = 1.82V

Cell Reaction: Pb(s) + 2Co3+ Pb2+ + 2Co2+ Ecell = 1.95V

Ecell = Ecell

0.0592

nlog

[Pb2+][Co2+]2

[Co3+]2

Ecell = 1.95V 0.05922

log(0.010)(0.0040)2

(0.050)2= 2.07V

2. Cu|Cu2+(0.50M)||Cr2O27 (0.060M), H+(0.0040M), Cr3+(0.080M)|PtAnode: 3[Cu2+ + 2e Cu(s)] E = 0.34V

Cathode: Cr2O27 + 14H

+ + 6e 2Cr3+ + 7H2O E = 1.33VCell Reaction: 3Cu(s) + Cr2O

27 + 14H

+ 3Cu2+ + 2Cr3+ + 7H2O Ecell = 0.99V

Ecell = Ecell

0.0592

nlog

[Cu2+]3[Cr3+]2

[Cr2O27 ][H

+]14

Ecell = 0.99V 0.05926

log(0.50)3(0.080)2

(0.060)(0.0040)14= 0.677V

Applications1. Determination of Equilibrium Constant

G = RT lnK nFE = RT lnKlnK =

nFERT

K = enFERT K = 10

nE0.0592 at 25C

Example:1. P b(s) + 2CO

3+ Pb2+ + 2Co2+ Find KEcell = E

cathode Eanode = 1.82V (0.13V ) = 1.95V

K = 10nE0.0592 = 10

21.95V0.0592 = 7.56 1065


2. Mg(s) + 2HCl(aq) MgCl2(aq) +H2(g) Find KAnode: [Mg2+ + 2e Mg(s)] E = 2.37V

Cathode: 2H+ + 2e H2(g) E = 0VCell Reaction: Mg(s) + 2H+ Mg2+ +H2(g) Ecell = 2.37V

K = 10nE0.0592 = 10

22.37V0.0592 = 1.17 1080

since K is very very large, the reaction is complete

2. Calculation of Other Equilibrium Constants

1. Ag|Ag+||Cl, AgCl|AgAgCl(s) Ag+(aq) + Cl

(aq)

Anode: [Ag+ + e Ag(s)] E = 0.80VCathode: AgCl(s) + e

Ag(s) + Cl(aq) E = 0.22VCell Reaction: AgCl(s) Ag+(aq) + Cl

(aq) E

cell = 0.58V

Ksp = 10nE0.0592 = 10

10.58V0.0592 = 1.59 1010

Quiz

Pt|Fe3+(1.20M), F e2+(0.0075M)||MnO4 (0.0020M), H+, Mn2+(2.50M)|Pt

Anode: 5[Fe3+ + e Fe2+] E = 0.77VCathode: MnO4 + 8H

+5e Mn2 + 4H2O] E = 1.51VCell Reaction: MnO4 + 8H

+ + 5Fe2+ Mn2 + 4H2O + 5Fe3+ Ecell = 0.74V

Ecell = Ecell

0.0592

nlog

[Mn2+][Fe3+]5

[MnO4 ][H+]8[Fe2+]5> 0

0.74V 0.05925

log(2.50)(1.20)5

(0.0020)[H+]8(0.0075)5> 0[

0.05925

log1.31072 1014

[H+]8> 0.74

] 5

0.0592

log1.31072 1014

[H+]8< 62.5

log1.31072 10143.162 1062 < [H

+]8

[H+] > 8.96 107MpH < 6.05

Concentration Cellelectrode: determines concentration or activity reference electrode: potential does not vary with the concentration of the analyte - independent, fixed potential Example: SCE - standard calomel electrode (Hg2Cl2) indicator electrode: sensitive to concentrations of analyte - potential varies with concentration of analytepH meter: reference electrodes: SCE, Ag|Ag+

indicator electrode: - glass tipped with solution of known pH - E is generated from difference in potentials between electrode and surroundings - calibration: using buffers, constant pH of 4/7/10 - setting what concentration will be pH 4/7/10


- checking the proper slope - in order to find the relationship of potential and pH

For SHE: Ecell = 0.0592pH

Electrolytic CellsTypes of electrical Conduction1. Electronic Conduction (Metallic Conduction) - the metal is the conductor because of a large sea of electrons2, Electrolytic Conduction - electrolytes are conductors - presence of ions

Electrolysis - application of electricity to a solution of electrolyte to drive a reaction

Mechanism of Electrolysis1. Migration of Ions towards the electrodes - Cations to Cathode - Anions to Anode2. Reaction occurs at the electrodes - Cathode: Reduction - Anode: Oxidation

Factors that affect electrode reaction1. Nature of Electrode a. Active: May also oxidize if it is the anode electrode b. Inert: No reaction (Pt, Cgraphite)2. Nature of Electrolyte a. Molten: Pure liquid, contains only the cation and anion b. Aqueous: In water solvent, involves competition with water

Predict the products formed:1. Electrolysis of molten LiBr between Pt electrodes Electrode: Inert Electrolyte: Molten

ProductAnode: 2Br Br2 + 2e E = 1.07V liquid/gaseous Br2

Cathode: Li+ + e Li(s) E = 3.05V deposit of Li(s)


2. Electrolysis of aqueous LiBr in Pt electrodes Electrode: Inert Electrolyte: Aqueous

Cathode: Li+ + e Li(s) E = 3.05V2H2O + 2e

H2(g) + 2OH(aq) E = 0.83VAnode: 2Br Br2 + 2e E = 1.07V

2H2O O2(g) + 4H+ + 4e E = 1.23V

3. Electrolysis of molten NaCl between Pt electrodes Electrode: Inert Electrolyte: Molten

Cathode: Na+ + e Na(s)Anode: 2Cl Cl2(g) + 2e

4. Electrolysis of aqueous NaCl between Pt electrodes Electrode: Inert Electrolyte: Aqueous

Cathode: Na+ + e Na(s) E = 2.71V2H2O + 2e

H2(g) + 2OH(aq) E = 0.83VAnode: 2Cl Cl2 + 2e E = 1.36V

2H2O O2(g) + 4H+ + 4e E = 1.23V

Useful products: H2(g), Cl2(g), NaOH

EXCEPTION FOR Cl- vs H2O overpotential - when gases are involved, an overpotential happens - added potential - added for O2 is more than added for Cl2 - Efinal O2 > Efinal Cl2 - dependent on nature of gas - only applies between Cl- and H2O

5. Electrolysis of Aqueous AgNO3 with Ag electrodes Electrode: Active Electrolyte: Aqueous

Cathode: Ag+ + e Ag(s) E = 0.80V2H2O + 2e

H2(g) + 2OH(aq) E = 0.83VAnode: XNO3 most oxidized form of N

2H2O O2(g) + 4H+ + 4e E = 1.23VAg(s) Ag+ + e E = 0.80V

if Oxidation State = Valence electrons, it is the most oxidized form

Faraday's Law of Electrolysis- products formed n current

Q = It

Q: charge (coulomb)

I: current (ampere, C/s)

t: time

ne =Q

F

n: moles of electrons (Faraday)

F: Faradays constant (96485 or 96500 C/mol e-)

Example: electrolysis of molten CaCl2 between Pt electrodes using 10A current for 5 min Electrode: Inert Electrolyte: Aqueous

Cathode: Ca2+ + 2e Ca(s)Anode: 2Cl Cl2(g) + 2e


WCa = 300 s

(10 C

s

)(1 mol e96500 C

)(1 mol Ca

2 mol e)(

40.08 g Ca

1 mol Ca

)= 0.623 g

VCl2, STP = 300 s

(10 C

s

)(1 mol e96500 C

)(1 mol Cl22 mol e

)(22.4 L Cl21 mol Cl2

)= 0.348 L = 348 mL

Given: Coin, 2.5 cm diameter, 0.15 cm height, 0.0025 cm layer of Au

= 19.3 g/cc, I =0.100 A, 90% yield

Find mass of Au and the time needed to form the Au layer

VAu = V = (pir2h)f (pir2h)i

VAu = pi[(1.2525)2(0.155) (1.25)2(0.15)]

VAu = 0.0276 cc

VAu = SA thicknessSA = 2 pir2 + pidhSA = 2 pi(1.25 cm)2 + pi(2.5 cm)(0.155 cm) = 11.03 cm2VAu = 11.03 cm

2 0.0025 cm = 0.0276 cc

WAu = 0.0276 cc

(19.3 g

cc

)= 0.533 g

Au(CN)4 + 3e Au(s) + 4CN

0.533 g

(1 mol

196.97 g

)(3 mol e

1 mol

)(96485 C

mol e

)( s0.100 C

)(109

)= 8703 s = 145.05 min

Coordination Chemistry- study of coordination compounds

History- started with CoCl36NH3, fascinated scientists- Alfred Werner

Compounds have Two Valences 1. Principal/Primary Valence - oxidation number 2. Secondary/Subsidary Valence - coordination number (number of coordinate covalent bonds)

- changed CoCl36NH3 to [Co(NH3)6]Cl3 Primary Valence: +3 Secondary Valence: 6

Complex / Coordination Compound - central metal atom - attached ligands - formed by a Lewis acid-base reaction

H3N: Ag+ :NH3 Lewis Acid: Ag

+ Lewis Base: NH3 Donor Atom: N (forms the coordinate covalent bond)

Coordination Number: number of donor atoms that surround the central atom

First Coordination Sphere: composed of the central atom and its attached ligands Example: [Ag(NH3)2]

+

K4[Fe(CN)6](aq) 4K+

(aq) + Fe(CN)64-

(aq) (first coordination sphere stays intact)

Conductivity


- electrical conductivity depends on the number of ions that dissociate

Platinum(IV) Complex Molar Conductance (0.001 M)

[Pt(NH3)6]Cl4 523

[PtCl2(NH3)4]Cl2 228

[PtCl4(NH3)2] 0

Note: Only Complexes with Coordination Numbers 2, 4 and 6 will be studied for Chem 18

Shapes of Complexes

Coordination Number Shape

2 Linear

4 Tetrahedral or Square Planar

6 Octahedral Example: [PtCl5(NH3)] coordination number: 6 shape: octahedral

Ligands - denticity: number of donor atoms available monodentate: one donor atom Examples: H2O, NH3, Cl

-, Br-, I-, F-, CN-, CO (C donates), SCN- OH-

bidentate: 2 donor atoms

- glycinate polydentate: EDTA: 6 donors: 4 O, 2 N

- polydentate ligands are good chelating agents


Biological Complexes: hemoglobin: central atom: Fe ligands: heme, globin, oxygen

Complex Primary Valence Secondary Valence Denticity

[Ru(NH3)5(H2O)]Cl2 +2 6 ALL monodentate

[CoBr2(en)2]2SO4 +3 6Br-: monodentate

en: bidentate

[Fe(CO)5] 0 5 ALL monodentate

Mg[Cr(C2O4)2(H2O)2]2 +3 6C2O4

2-: bidentateH2O: monodentate

Nomenclature of Complexes1. Cation is named first before the anion2. In complex parts, ligands are named first before the metal a. alphabetical order (not including greek prefixes) b. Greek prefixes to specify number of a ligand - di, tri, tetra, penta, etc. - if the ligand already contains a greek prefix, use bis, tris, tetrakis, pentakis and enclose the ligand it describes in a parenthesis Ag(NH3)2

+ diamminesilver(I) [Fe(en)3]

3+ tris(ethylenediamine)iron(III) - if adding a Greek prefix can mean another thing for the ligand, use bis, tris etc. (

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Chemistry 18 Notes