KINETICS Chapter 14 Slide 2 14.1 Factors that Affect Reaction
Rates Chemical kinetics is the study of how fast chemical reactions
occur. Generally, the more frequently collisions between reaction
particles occur, the faster the reaction. Slide 3 There are several
important factors which affect rates of reactions: Physical state
of the reactants Homogeneous mixtures react faster than
heterogeneous mixtures. When reactants are in different phases:
more surface area faster reaction Concentration of the reactants
more reactant molecules so they collide more frequently faster
reaction Temperature of the reaction higher temperature molecules
move faster and collide more frequently faster reaction Presence or
absence of a catalyst use a catalyst usually changes mechanism
faster reaction Slide 4 14.2 Reaction Rates The speed of a reaction
is defined as the change that occurs per unit time. It is often
determined by measuring the change in concentration of a reactant
or product with time [M/sec] (could also use change in moles but if
volume is constant [most reactions], then molarity and moles are
directly proportional concentration is easier to directly measure
than moles. Pressure is sometimes used for gases.) The speed of the
chemical reaction is its reaction rate. Slide 5 Suppose A reacts to
form B. A B Let us begin with [A] = 1.00 M At t = 0 (time zero)
there is 1.00 M A and no B present. At t = 20 sec, there is 0.54 M
A and 0.46 M B. At t = 40 sec, there is 0.30 M A and 0.70 M B. We
can uses this information to find the average rate of the reaction.
Slide 6 For the reaction A B, there are 2 ways of measuring rate:
The rate of appearance of product B (i.e., change in concentration
of B per unit time) as in the preceding example. The rate of
disappearance of reactant A (i.e., the change in concentration of A
per unit time): Note the negative sign! Reaction rates are always
positive! Slide 7 Change of Rate with Time The average rate
generally decreases with time The rate at any instant in time is
called the instantaneous rate. It is the slope of the straight line
tangent to the curve at that instant. Instantaneous rate is
different from average rate. We usually call the "instantaneous
rate" the rate. Slide 8 Example: Using the data in table above,
calculate the average rate of disappearance of C 4 H 9 Cl over the
time interval from 50.0 to 150.0 seconds. (b) Using the figure,
estimate the instantaneous rate of disappearance of C 4 H 9 Cl at t
= 0 (the initial rate). Slide 9 Slide 10 (b)the initial rate is
given by the slope of the dashed line ( y / x) pick two points on
the tangent line: (0,.100) and (200,.060) Slide 11 Reaction Rates
and Stoichiometry For the reaction: C 4 H 9 Cl(aq) + H 2 O(l) C 4 H
9 OH(aq) + HCl(aq) The rate of appearance of C 4 H 9 OH must equal
the rate of disappearance of C 4 H 9 Cl. Slide 12 What if the
stoichiometric relationships are not one- to-one? For the reaction:
2HI(g) H 2 (g) + I 2 (g) The rate may be expressed as: We can
generalize this equation a bit. For the reaction: aA + bB cC + dD
The rate may be expressed as: Slide 13 The isomerization of methyl
isonitrile CH 3 NC to acetonitrile, CH 3 CN, was studied in the gas
phase at 215 o C and the following data was obtained: Time (s)0
2000 5000 8000 1200015000 [CH 3 NC] (M)0.0165 0.01100.00591 0.00314
0.001370.00074 (a) Calculate the average rate of reaction in M/s,
for the time interval between each measurement. (a) TimeTime
Interval [CH 3 NC] M Rate (sec) (t) (sec) (M) (M/ t) 0 0.0165 2000
2000 0.0110- 0.0055 2.8 10 -6 5000 3000 0.00591- 0.0051 1.7 10 -6
8000 3000 0.00314- 0.00277.923 10 -6 12000 4000 0.00137-
0.00177.443 10 -6 15000 3000 0.00074- 0.00063.21 10 -6 - ( - 0.0055
M / 2000 s) Slide 14 Consider the hypothetical reaction: A(aq)
B(aq). A flask is charged with 0.065 mol of A and allowed to react
to form B in a total volume of 100.0 mL. The following data are
collected: Time (min)010203040 Mol of A0.0650.0510.0420.0360.031
(a) Calculate the number of moles of B at each time in the table,
assuming there are no molecules of B at time zero (b) Calculate the
average rate of disappearance of A for each 10 min interval, in
units of M/s. (c) Between t = 10 min and t = 30 min, what is the
average rate of appearance of B in units of M/s? Assume the volume
of the solution is constant. Time(min) Mol A (a) Mol B [A] [A] (b)
Rate -( [A]/time) 0 0.0650 0.65 10 0.0510.014 0.51 - 0.14 2.3 10 -4
20 0.0420.023 0.42 - 0.09 2 10 -4 30 0.0360.029 0.36 - 0.06 1 10 -4
40 0.0310.034 0.31 - 0.05 0.8 10 -4 (c) - ( - 0.14 M / 600 s) Slide
15 Slide 16 Consider the combustion of H 2 (g), 2 H 2 (g) + O 2 (g)
2 H 2 O(g). If hydrogen is burning at the rate of 0.85 mol/s, what
is the rate of consumption of oxygen? What is the rate of formation
of water vapor? Slide 17 The reaction 2 NO(g) + Cl 2 (g) 2 NOCl(g)
is carried out in a closed vessel. If the partial pressure of NO is
decreasing at the rate of 23 torr/min, what is the rate of change
of the total pressure in the vessel? The change in total pressure
is the sum of the changes of each partial pressure. NO and Cl 2 are
disappearing and NOCl is appearing. P NO /t = -23 torr/min (given)
P Cl /t = (P NO /t ) = (-23) = -11.5 torr/min P NOCl /t = - P NO /t
= -(-23) = +23 torr/min P TOT = -23 11.5 + 23 = -11.5 = -12
torr/min Pressure is decreasing at a rate of 12 torr/min Slide 18
14.3 Concentration and Rate In general, rates: Increase when
reactant concentration is increased. Decrease when reactant
concentration is decreased. We often examine the effect of
concentration on reaction rate by measuring the way in which
reaction rate at the beginning of a reaction depends on different
starting conditions. Slide 19 Consider the reaction: NH 4 + (aq) +
NO 2 (aq) N 2 (g) + 2H 2 O(l) We measure initial reaction rates.
The initial rate is the instantaneous rate at time t = 0. We find
this at various initial concentrations of each reactant. Slide 20
As [NH 4 + ] doubles with [NO 2 ] constant the rate doubles. We
conclude the rate is proportional to [NH 4 + ]. As [NO 2 ] doubles
with [NH 4 + ] constant the rate doubles We conclude that the rate
is proportional to [NO 2 ]. Slide 21 Rate Law The overall
concentration dependence of reaction rate is given in a rate law or
rate expression. For a general reaction the rate law is: Rate = k
[A] m [B] n The proportionality constant k is called the rate
constant. Once we have determined the rate law and the rate
constant, we can use them to calculate initial reaction rates under
any set of initial concentrations. Slide 22 Exponents in the Rate
Law The exponents m and n are called reaction orders The rate is m
order in A The rate is n order in B We commonly encounter reaction
orders of 0, 1 or 2. But, fractional or negative values are
possible. The overall reaction order is the sum of the reaction
orders. The overall order of a reaction is m + n + . Note that
reaction orders must be determined experimentally. They DO NOT
necessarily correspond to the stoichiometric coefficients in the
balanced chemical equation! (but they can and sometimes do) Slide
23 Using Initial Rates to Determine Rate Laws To determine the rate
law, we observe the effect of changing initial concentrations. A
reaction is nth order if multiplying the concentration by x causes
a x n increase in rate. x increase in concentration = x n increase
in rate (n is the order) For our example the concentration was
multiplied by 2 and the rate was also multiplied by 2 (2 1 ) so the
rate is 1 st order with respect to both NH 4 + (aq) and NO 2 (aq) 2
(increase in concentration) = 2 1 (increase in rate) (1 st order)
Slide 24 Most Common Scenarios: change in conc. of reactantchange
in rateorder doubledno change - rate is multiplied by 1 or 2 0
doubleddoubled - rate is multiplied by 2 or 2 1 doubledquadrupled -
rate is multiplied by 4 or 2 2 tripledno change - rate is
multiplied by 1 or 3 0 tripledtripled - rate is multiplied by 3 or
3 1 tripledrate is multiplied by 9 or 3 2 Zero order rxn: change in
concentration = no change in rate 1 st order rxn: change in
concentration = change in rate 2 nd order rxn: change in
concentration = change in rate is the square of change in
concentration Another way to look at order of reaction vs. change
in concentration: 0 1 2 0 1 2 Slide 25 For the reaction in our
example: NH 4 + (aq) + NO 2 (aq) N 2 (g) + 2H 2 O(l) The rate law
is: Rate = k [NH 4 + ] [ NO 2 ] The reaction is said to be first
order in [NH 4 + ], first order in [NO 2 ], and second order
overall. Slide 26 Units of Rate Constants Units of the rate
constant depend on the overall reaction order. Second order
overall: Rate = k [A] 2 First order overall: Rate = k [A] Slide 27
Note that the rate, not the rate constant, depends on
concentration. The rate constant IS affected by temperature and by
the presence of a catalyst (more on this later). Slide 28 Examples
of rate laws What are the individual and overall reaction orders
for the reactions described in the following equations: (1)2 N 2 O
5 (g) 4 NO 2 (g) + O 2 (g) Rate = k [N 2 O 5 ] (2)CHCl 3 (g) + Cl 2
(g) CCl 4 (g) + HCl (g) Rate = k [CHCl 3 ][Cl 2 ] 1/2 Reaction (1)
is first order in N 2 O 5 and first order overall. Reaction (2) is
first order in CHCl 3 and order in Cl 2 and 3/2 order overall (add
the exponents). Slide 29 A reaction A + B C, obeys the following
rate law: rate = k [B] 2. (a) if [A] is doubled, how will the rate
change? Will the rate constant change? (b) What are the reaction
orders for A and B? What is the overall reaction order? (c) What
are the units of the rate constant? (a) if [A] is doubled, there
will be no change in the rate or the rate constant (b)the reaction
is zero order in A, second order in B and second order overall (c)
Slide 30 The decomposition of N 2 O 5 in carbon tetrachloride
proceeds as follows: 2 N 2 O 5 2 NO 2 + O 2. The rate law is first
order in N 2 O 5. At 64 o C, the rate constant is 4.82 10 -3 s -1.
(a) Write the rate law for the reaction. (b) What is the rate of
reaction when [N 2 O 5 ] = 0.0240 M? (c) What happens to the rate
when the concentration of N 2 O 5 is doubled to 0.0480 M? (a)rate =
k [N 2 O 5 ] = (4.82 10 -3 s -1) [N 2 O 5 ] (b)rate = (4.82 10 -3 s
-1 ) (0.0240 M) = 1.16 10 -4 M/s (c)one way: rate = 4.82 10 -3 s -1
(0.0480 M) = 2.31 10 -4 M/s quicker way: We know that the rxn is 1
st order in N 2 O 5, so doubling the conc. will double the rate
Slide 31 Consider the following reaction: CH 3 Br + OH -1 CH 3 OH +
Br -1. The rate law for this reaction is first order in CH 3 Br and
first order in OH -1. When [CH 3 Br] is 0.0050 M and [OH -1 ] is
0.050 M, the reaction rate at 298 K is 0.0432 M/s. (a) What is the
value of the rate constant? (b) What are the units of the rate
constant? (c) What would happen to the rate if the concentration of
OH -1 were tripled? (a,b) rate = k [CH 3 Br] [OH -1 ] (c)Since the
rate law is first order in [OH -1 ], if [OH -1 ] is tripled, the
rate triples. Slide 32 The iodide ion reacts with hypochlorite ion
(the active ingredient in bleach) in the following way: OCl -1 + I
-1 OI -1 + Cl -1. This rapid reaction gives the following rate
data: Experiment[OCl -1 ] (M)[I -1 ] (M)Rate (M/s)
10.00150.00151.36 10 -4 20.00300.00152.72 10 -4 30.00150.00302.72
10 -4 (a) What is the rate law for the reaction? (b) What is the
value of the rate constant? (c) Calculate the rate when [OCl -1 ] =
0.0020 M and [I -1 ] = 0.00050 M (a)from exp. 1&2, doubling
[OCl -1 ] while keeping [I -1 ] constant doubles the rate rxn is 1
st order in OCl -1 from exp. 1&3, doubling [I -1 ] while
keeping [OCl -1 ] constant doubles the rate rxn is 1 st order in I
-1 rate = k [OCl -1 ][I -1 ] (b)solving the rate law for k and
using the data of experiment 1: (c)using the rate law (from a) and
the value of k (from b): Slide 33 Consider the gas-phase reaction
between nitric oxide and bromine at 273 o C: 2 NO + Br 2 2 NOBr.
The following data for the initial rate of NOBr were obtained:
Experiment[NO] (M)[Br 2 ] (M)Initial Rate (M/s) 10.100.20 24
20.250.20150 30.100.50 60 40.350.50735 (a)Determine the rate law.
(b) Calculate the value of the rate constant for the appearance of
NOBr. (c)How is the rate of appearance of NOBr related to the rate
of disappearance of Br 2 ? (d)What is the rate of disappearance of
Br 2 when [NO] = 0.075 M and [Br 2 ] = 0.25? (a)From exp. 1&2,
multiplying [NO] by 2.5 multiplies the rate by 2.5 2 rxn is 2 nd
order in NO From exp. 1&3, multiplying [Br 2 ] by 2.5
multiplies the rate by 2.5 rxn is 1 st order in Br 2 The rate law
for the appearance of NOBr is : rate = k [NO] 2 [Br 2 ] (b) Slide
34 Consider the gas-phase reaction between nitric oxide and bromine
at 273 o C: 2 NO + Br 2 2 NOBr. The following data for the initial
rate of NOBr were obtained: Experiment[NO] (M)[Br 2 ] (M)Initial
Rate (M/s) 10.100.20 24 20.250.20150 30.100.50 60 40.350.50735
(a)Determine the rate law. (b) Calculate the value of the rate
constant for the appearance of NOBr. (c)How is the rate of
appearance of NOBr related to the rate of disappearance of Br 2 ?
(d)What is the rate of disappearance of Br 2 when [NO] = 0.075 M
and [Br 2 ] = 0.25? (c) (d) The rate law for the appearance of NOBr
is : rate = k [NO] 2 [Br 2 ] The rate of disappearance of Br 2 is
the rate of appearance of NOBr. Slide 35 Consider the reaction of
peroxydisulfate ion, S 2 O 8 -2, with iodide ion, I -1, in aqueous
solution: S 2 O 8 -2 + 3 I -1 2 SO 4 -2 + I 3 -1 At a particular
temperature, the rate of disappearance of S 2 O 8 -2 varies with
reactant concentrations in the following manner: Experiment[S 2 O 8
-2 ] (M)[I -1 ] (M)Initial Rate (M/s) 10.0180.0362.6 10 -6
20.0270.0363.9 10 -6 30.0360.0547.8 10 -6 40.0500.0721.4 10 -5
(a)Determine the rate law for the reaction. (a) from experiments 1
and 2, increasing [S 2 O 8 -2 ] by 1.5 increases the rate by 1.5 so
[S 2 O 8 -2 ] is 1 st order Slide 36 Experiment[S 2 O 8 -2 ] (M)[I
-1 ] (M)Initial Rate (M/s) 10.0180.0362.6 10 -6 20.0270.0363.9 10
-6 30.0360.0547.8 10 -6 40.0500.0721.4 10 -5 ALTERNATE SOLUTION In
exp 1 and 2, keeping [I -1 ] constant and increasing [S 2 O 8 -2 ]
by 1.5 increases the rate by 1.5 meaning that the reaction is first
order in [S 2 O 8 -2 ] So we know that the rate law is: rate = k [S
2 O 8 -2 ] [I -1 ] x and we need to find x Looking at exp 1 and 3,
the rate inc by a factor of 3 or Rate exp 3 / Rate exp1 = 3 Rate 3
k [S 2 O 8 -2 ] [I -1 ] x k [.036] [.054] x 2 [.054] x -------- =
-------------------- = -------------------- = ------------- = 3
Rate 1 k [S 2 O 8 -2 ] [I -1 ] x k [.018] [.036] x [.036] x [.054]
x 3 -------- = ----- (or 1.5) [.036] x 2 (1.5) x = 1.5 x = 1, so
the reaction is 1 st order in [I -1 ] rate = k [S 2 O 8 -2 ] [I -1
] Slide 37 Kinetics Part 2 14.4 Slide 38 Goal: Convert the rate law
into a convenient equation that gives concentration as a function
of time. Slide 39 First-Order Reactions Rate must depend on only
one reactant and be raised to the 1st power. For a first-order
reaction, the rate doubles as the concentration of a reactant
doubles. Rate = k[A] After some calculus A plot of ln[A] t versus t
is a straight line with slope -k and intercept ln[A] 0 ln[A] t Time
(s) (slope= k) y = mx + b Slide 40 Second-Order Reactions A
second-order reaction is one whose rate depends on the reactant
concentration to the second power or on the concentration of two
reactants, each raised to the first power. For a 2nd order reaction
with just one reactant: calculus A plot of 1/[A]t versus t is a
straight line with slope k and intercept 1/[A] 0. y = mx + b 1/[A]
t Time (s) (slope= k) Slide 41 Half-life Half-life, t , is the time
required for the concentration of a reactant to decrease to half
its original value. That is, half life, t , is the time taken for
[A] 0 to reach [A] 0. Slide 42 Mathematically, the half life of a
first-order reaction is: Note that the half-life of a first-order
reaction is independent of the initial concentration of the
reactant. Radioactive decay is 1 st order. Slide 43 We can show
that the half-life of a second order reaction is: Note that the
half-life of a second-order reaction is dependent on the initial
concentration of reactant. Slide 44 Sample Exercise #1 The first
order rate constant for the decomposition of a certain insecticide
in water at 12 o C is 1.45 yr -1. A quantity of this insecticide is
washed into a lake on June 1, leading to a concentration of 5.0 x
10 -7 g/cm 3 of water. Assume that the effective temperature of the
lake is 12 o C. (a) What is the concentration of the insecticide on
June 1 of the following year? (b) How long for the conc. of the
insecticide to drop to 3.0 x 10 -7 g/cm 3 ? Slide 45 (a) ln [A] t =
- k t + ln [A] 0 ln [A] t = - (1.45 yr -1 )(1.00 yr) + ln(5.0 x 10
-7 ) e ln [A] t = e -15.96 [A] t = 1.2 x 10 -7 g/cm 3 (note: conc.
units for [A] and [A] 0 must be the same) Slide 46 (b) ln [A] t = -
k t + ln [A] 0 ln (3.0 10 -7 ) = - (1.45)(t) + ln (5.0 10 -7 ) t =
0.35 yr Slide 47 Using the figure to the right, estimate the
half-life of the reaction of C 4 H 9 Cl with water. The initial
value of [C 3 H 9 Cl] is 0.100 M. The half-life for this reaction
is the time for [C 3 H 9 Cl] to be 0.050 M. This point occurs at
approx. 340 s. At the end of the second half-life, about 680 s, the
concentration should decrease by another factor of 2 to about 0.025
M. The graph indicates that this is true. Slide 48 The reaction SO
2 Cl 2 SO 2 + Cl 2 is first order in SO 2 Cl 2. Using the following
kinetic data, determine the magnitude of the first order rate
constant: Time (s)Pressure SO 2 Cl 2 (atm)ln Pressure SO 2 Cl 2
01.0000 25000.947- 0.0545 50000.895- 0.111 75000.848- 0.165
100000.803- 0.219 Graph ln P vs. time (first order) The graph is
linear with slope of about - 2.19 10 -5 s -1 Slide 49 Sample
Exercise #4 The following data was obtained for the gas phase
decomposition of NO 2 (g) at 300 o C: 2 NO 2 (g) 2 NO(g) + O 2 (g)
Time (s)[NO 2 ] (M) 0.00.01000 50.00.00787 100.00.00649
200.00.00481 300.00.00380 What is the rate law for this reaction?
Slide 50 To test whether the reaction is first or second order, we
can construct plots of ln [NO 2 ] and 1 / [NO 2 ] against time.
Time (s) [NO 2 ]ln [NO 2 ] 1/[NO 2 ] 0.00.01000 50.00.00787
100.00.00649 200.00.00481 300.00.00380 -4.610 -4.845 -5.038 -5.337
-5.573 100 127 154 208 263 Slide 51 From the graphs above, only the
plot of 1/[NO 2 ] is a straight line. Thus, the reaction is second
order in NO 2. Rate = k [NO 2 ] 2 From the slope of the straight
line graph, we can get: k = 0.543 M -1 s -1 Slide 52 The
decomposition of sulfuryl chloride (SO 2 Cl 2 ) is a first-order
process. The rate constant for the decomposition at 660 K is 4.5 10
-2 s -1. (a) if we begin with an initial SO 2 Cl 2 pressure of 375
torr, what is the pressure of this substance after 65 s? (b) at
what time will the pressure of SO 2 Cl 2 decline to one-tenth its
initial value? (a)ln P t = - kt + ln P o ln P 65 = - 4.5 10 -2 s -1
(65) + ln(375) P 65 = 20 torr (b)P t = 0.10 P 0 = (.10)(375) = 37.5
torr ln P t = - kt + ln P o ln (37.5) = - (4.5 x 10 -2 s -1 )t + ln
(375) t = 51 s Slide 53 Consider the data presented (a)By using
appropriate graphs, determine whether the reaction is first or
second order. (b)What is the value of the rate constant for the
reaction? (c)What is the half life for the reaction? (a)make both
first and second order plots to see which is linear the plot of
1/[A] vs. time is linear, so the reaction is second order in [A]
(b)the slope of the line in the graph is the rate constant (c) use
the half-life formula for a 2 nd order reaction: t 1/2 = 1 / k[A] o
= 1 / [(0.043 M -1 min -1 ) (0.65 M)] = 36 min 1/[A] Slide 54 14.5
Temperature and Rate Most reactions speed up as temperature
increases The rate law has no temperature term in it, so the rate
constant (k) must depend on temperature. We see an approximate
doubling of the rate with each 10 o C increase in temperature.
Slide 55 The Collision Model Rates of reactions are affected by
concentration and temperature. An explanation is provided by the
collision model, based on kinetic-molecular theory. In order for
molecules to react they must collide. The greater the number of
collisions the faster the rate. - The more molecules present, the
greater the probability of collision faster rate - The higher the
temperature, the faster molecules move and more often they collide
faster rate However, not all collisions lead to products. In fact,
only a small fraction of collisions lead to products. In order for
a reaction to occur the reactant molecules must collide in the
correct orientation AND with enough energy to form products. Slide
56 The Orientation Factor The orientation of a molecule during
collision can have a profound effect on whether or not a reaction
occurs. Consider the reaction between Cl and NOCl: If Cl collides
with Cl of NOCl, the products are Cl 2 and NO. If the Cl collides
with the O of NOCl, no products are formed. Cl Cl-N=O O=N-Cl Slide
57 Activation Energy Arrhenius: molecules must posses a minimum
amount of energy to react. In order to form products, bonds must be
broken in the reactants. Bond breakage requires energy. (H rxn =
bonds broken bonds formed) Molecules moving too slowly, with too
little kinetic energy, dont react when they collide. Activation
energy, E a is the minimum energy required to initiate a chemical
reaction (varies with the reaction). Slide 58 Slide 59 Energy is
required to stretch and break the bond between the CH 3 group and
the N C group The species at the top of the barrier is called the
activated complex or transition state. Energy is released when the
C-C bond is formed. The change in energy for the reaction is the
difference in energy between CH 3 NC and CH 3 CN. The activation
energy is the difference in energy between reactants, (CH 3 NC) and
the transition state. The rate depends on the magnitude of the E a.
In general, the lower the E a, the faster the rate. Slide 60 Notice
that if a forward reaction is exothermic (CH 3 NC CH 3 CN), then
the reverse reaction is endothermic (CH 3 CN CH 3 NC). Slide 61 How
does this relate to temperature? At any particular temperature, the
molecules present have an average kinetic energy. Some molecules
have less energy than the average while others have more than the
average value. Molecules that have an energy equal to or greater
than E a have sufficient energy to react. The fraction of molecules
with an energy equal to or greater than E a is given by: (R is
8.314 J/mol-k, Temp in K, E a in J) Slide 62 At a higher
temperature, more molecules have the minimum E a needed to react
and the rate is faster. Slide 63 The Arrhenius Equation Arrhenius
discovered that most reaction-rate data obeyed an equation based on
three factors 1. The number of collisions per unit time. 2. The
fraction of collisions that occur with the correct orientation.
3.The fraction of the colliding molecules that have an energy equal
to or greater than E a. Arrhenius equation: Where k is the rate
constant, E a is the activation energy(J), R is the ideal-gas
constant (8.314 J/Kmol) and T is the temperature (K). A is the
frequency factor. It is related to the frequency of collisions and
the probability that a collision will have a favorable orientation.
Slide 64 More Useful Versions of the Arrhenius Equation y = mx + b
A graph of ln k vs 1/T will be a line with slope of E a /R and a
y-intercept of ln A. Useful when we have 2 different k values at 2
different temperatures slope = -E a / R ln k 1 / T ln A Slide 65
Solution The lower the activation energy, the faster the reaction.
The value of E does not affect the rate. Hence the order is (2)
< (3) < (1). SAMPLE 1: Consider a series of reactions having
the following energy profiles: Assuming that all three reactions
have nearly the same frequency factors, rank the reactions from
slowest to fastest. Slide 66 Calculate the fraction of atoms in a
sample of argon gas at 400 K that have an activation energy of 10.0
kJ or greater. Slide 67 The gas phase reaction Cl(g) + HBr(g)
HCl(g) + Br(g) has an overall enthalphy change of - 66 kJ. The
activation energy for the reaction is 7 kJ. (a) Sketch the energy
profile for the reaction and label E a and E. (b) What is the
activation energy for the reverse reaction? (b) the reverse
reaction has an E a value of 73 kJ Slide 68 A certain first order
reaction has a rate constant of 2.75 10 -2 s -1 at 20 o C. What is
the value of k at 60 o C if E a = 75.5 kJ/mol T 1 = 293 K T 2 = 333
K Slide 69 The rate of the reaction CH 3 COOC 2 H 5 + OH -1 CH 3
COO -1 + C 2 H 5 OH was measured at several temperatures and the
following data were collected. Using the data, construct a graph of
ln k versus 1/T and determine the value of E a the slope = -E a / R
E a = - (-5.6 10 3 )(8.314 J/mol) = 47000 J/mol or 47 kJ/mol Slide
70 The activation energy of a certain reaction is 65.7 kJ/mol. How
many times faster will the reaction occur at 50 o C rather then 0 o
C assuming equal initial concentrations? Since we are considering
one reaction and temp only affects k which directly affects ratethe
ratio of k values equals the ratio of rates at the two
temperatures. T 1 = 323 K ; T 2 = 273 K The reaction will occur
about 88 times faster at 50 o C Slide 71 14.6 Reaction Mechanisms A
chemical equation provides information about substances present
before and after a reaction. The reaction mechanism is the process
of how the reactants become products. Mechanisms provide a picture
of which bonds are broken and formed during the course of a
reaction. Most reactions occur as a series of small steps or
changes. Elementary steps are processes that occur in a single
step. The number of molecules present in an elementary step is the
molecularity of that elementary step. Unimolecular: one molecule in
the elementary step. Bimolecular: two molecules in the elementary
step. Termolecular: three molecules in the elementary step. Slide
72 Multistep Mechanisms consist of a sequence of elementary steps.
The elementary steps must add together to give the balanced
chemical equation. Intermediates - these are species that appear in
an elementary step but are neither a reactant nor product. They are
formed in one elementary step and consumed in another. They are NOT
found in the balanced equation for the overall reaction. Slide 73
Rate Laws for Elementary Steps The rate laws of the elementary
steps determine the overall rate law of the reaction. For
elementary processes, we DO NOT need to find the reaction order by
experimentation it matches the stoichiometry of the equation The
rate law of an elementary step is determined by its molecularity.
Unimolecular processes are first order. Bimolecular processes are
second order. Termolecular processes are third order. Slide 74
Slide 75 Rate Laws for Multistep Mechanisms When a reaction occurs
by mechanisms with more than one step, the slowest step limits the
overall reaction rate. This is called the rate-determining step of
the reaction. This step governs the overall rate law for the
overall reaction. Slide 76 Consider the reaction: NO 2 (g) + CO(g)
NO(g) + CO 2 (g) Here is a proposed a mechanism for the reaction:
Step 1: NO 2 (g) + NO 2 (g) NO 3 (g) + NO(g)slow step Step 2: NO 3
(g) + CO(g) NO 2 (g) + CO 2 (g) fast step (Note that NO 3 is an
intermediate.) The overall reaction rate will depend on the slowest
step Rate = k[NO 2 ] 2 If we do an experiment and find that the
experimentally derived rate law is: Rate = k[NO 2 ] 2 Then our
theoretical rate law is in agreement with the experimental rate
law. This supports (but does not prove) our mechanism. Slide 77 The
following mechanism has been proposed for the reaction of methane
gas with chlorine gas. All species are in the gas phase. Step 1 Cl
2 2 Cl fast equilibrium Step 2 CH 4 + Cl CH 3 + HCl slow Step 3 CH
3 + Cl 2 CH 3 Cl + Cl fast Step 4 CH 3 Cl + Cl CH 2 Cl 2 + H fast
Step 5 H + Cl HCl fast (e) Identify the order of the reaction with
respect to each of the following according to the mechanism. In
each case, justify your answer. (i) CH 4 (g)(ii) Cl 2 (g) Slide 78
Slide 79 14.7 Catalysis A catalyst is a substance that changes the
speed of a chemical reaction without any permanent change to itself
Works by providing an alternate mechanism for the reaction one that
has a lower activation energy Lowering the activation energy allows
for more molecules with E a large enough to react more collisions
that can make products higher rxn rate (without changing
temperature) Very common in nature Much research to find new
catalysts AND, much research to find inhibitors(substances that
slow down an undesired reaction) Example: corrosion Homogeneous
catalyst catalyst is in the same phase as reactants Heterogeneous
catalyst catalyst is in different phase than reactants Enzymes
catalysts in living organisms Slide 80 Effect of a catalyst on a
reaction A catalyst provides an alternate mechanism that has a
lower E a than the original mechanism this allows the reaction to
occur at a faster rate Slide 81 What is the molecularity of each of
the following elementary processes? Write the rate law for each.
(a)Cl 2 2 Cl (b)OCl -1 + H 2 O HOCl + OH -1 (c)NO + Cl 2 NOCl 2
(a)unimolecularrate = k [Cl 2 ] (b) bimolecular rate = k [OCl -1 ]
[H 2 O] (c)bimolecularrate = k [NO] [Cl 2 ] Slide 82 (a)Based on
the following reaction profile, how many intermediates are formed
in the reaction A D? (b)How many transition states are there?
(c)Which step is the fastest? (d)Is the reaction A D exothermic or
endothermic? This is a three-step mechanism, A B, B C, C D (a) 2
intermediates, B and C (A is reactant, D is product) (b) 3 energy
maxima, so there are 3 transition states (c)Step C D has the lowest
activation energy, so it is the fastest (d)The energy of D is
greater than the energy of A, so the overall reaction is
endothermic Slide 83 The following mechanism has been proposed for
the gas-phase reaction of H 2 with ICl: H 2 + ICl HI + HCl HI + ICl
I 2 + HCl (a)Write the balanced equation for the overall reaction
(b)Identify any intermediates (c)Write rate laws for each
elementary reaction in the mechanism (d)If the first step is slow
and the second one is fast, what rate law do you expect to be
observed for the overall reaction? (e) If the second step is the
rds, what rate law would you expect for the reaction (a)H 2 + ICl
HI + HCl HI + ICl I 2 + HCl H 2 + 2 ICl I 2 + 2 HCl (b)HI (c)first
step: rate = k 1 [H 2 ] [ICl] second step: rate = k 2 [HI] [ICl]
(d)rate = k [H 2 ] [ICl] (e)rate = k [H 2 ] [ICl] 2 Slide 84 The
reaction 2 NO + Cl 2 2 NOCl obeys the rate law: rate = k [NO] 2 [Cl
2 ]. The following mechanism has been proposed for this reaction:
NO + Cl 2 NOCl 2 NOCl 2 + NO 2 NOCl (a) What would the rate law be
if the first step were rate determining? (b) Based on the observed
rate law, what can we conclude about the relative rate of the the
two steps? (a) rate = k [NO] [Cl 2 ] (b) Assuming the proposed
mechanism is correct the 2 nd step is the r-d-s (slower step). (The
rate law based on this mechanism with the 2 nd step as the r-d- s
matches the experimentally determined rate law.) Slide 85 The
oxidation of SO 2 to SO 3 is catalyzed by NO 2. The reaction
proceeds as follows: NO 2 (g) + SO 2 (g) NO (g) + SO 3 (g) 2 NO (g)
+ O 2 (g) 2 NO 2 (g) (a)Show that the two reactions can be summed
to give the overall oxidation of SO 2 by O 2 to give SO 3. Hint:
the top reaction must be multiplied by a factor so the NO and NO 2
cancel out. (b) Why do we consider NO 2 a catalyst and not an
intermediate in this reaction? (c) Is this an example of
homogeneous catalysis or heterogeneous catalysis? (a) 2 [NO 2 + SO
2 NO + SO 3 ] = 2 NO 2 + 2 SO 2 2 NO + 2 SO 3 2 NO + O 2 2 NO 2 __
2 SO 2 + O 2 2 SO 3 (b)NO 2 is a catalyst because it is present at
the beginning of the reaction it is consumed and then reproduced in
the reaction sequence. (NO is an intermediate because it is
produced and then consumed.) (c)homogeneous catalysis
