Transcript Document

Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 14
Chemical Kinetics
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
14
14.1
14.2
14.3
14.4
14.5
14.6
Chemical Kinetics
Reaction Rates
Collision Theory of Chemical Reactions
Measuring Reaction Progress and Expressing Reaction
Rate
Average Reaction Rate
Instantaneous Rate
Stoichiometry and Reaction Rate
Dependence of Reaction Rate on Reactant Concentration
The Rate Law
Experimental Determination of the Rate Law
Dependence of Reactant Concentration on Time
First-Order Reactions
Second-Order Reactions
Dependence of Reaction Rate on Temperature
The Arrhenius Equation
14
14.7
14.6
Chemical Kinetics
Reaction Mechanisms
Elementary Reactions
Rate-Determining Step
Experimental Support for Reaction Mechanisms
Catalysis
Heterogeneous Catalysis
Homogeneous Catalysis
Enzymes: Biological Catalysts
14.1
Reaction Rates
Chemical kinetics is the study of how fast reactions take place.
Some happen almost instantaneously, while others can take millions
of years.
Increasing the rate of a reaction is important to many industrial
processes.
14.2
Collision Theory of Chemical Reactions
Most reactions happen faster at higher temperature.
Chemical reactions generally occur as a result of collisions between
reacting molecules.
According to collision theory of chemical kinetics, the reaction rate
is directly proportional to the number of molecular collisions per
second:
number of collisions
rate 
s
Collision Theory of Chemical Reactions
Collisions that result in a chemical reaction are called effective
collisions.
The activation energy (Ea ) is the minimum amount of energy
required to initiate a chemical reaction.
Molecules must also be oriented in a way that favors reaction.
Cl + NOCl → Cl2 + NO
Correct orientation to
facilitate reaction
An effective collision
results in reaction.
Collision Theory of Chemical Reactions
Collisions that result in a chemical reaction are called effective
collisions.
The activation energy (Ea ) is the minimum amount of energy
required to initiate a chemical reaction.
Molecules must also be oriented in a way that favors reaction.
Cl + NOCl → Cl2 + NO
Incorrect orientation does
not favor reaction
An ineffective collision
results in no reaction.
Collision Theory of Chemical Reactions
When molecules collide in an effective collision, they form an
activated complex (also called the transition state).
14.3
Measuring Reaction Progress and Expressing
Reaction Rate
Chemical kinetics is the study of how fast reactions take place.
A→B
rate = 
[A]
t
[A] decreases
rate =
[B]
t
Measuring Reaction Progress and Expressing Reaction
Rate
A→B
Measuring Reaction Progress and Expressing Reaction
Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
average rate = 
[Br ]  [Br2 ]initial
[B]
  2 final
t
tfinal  tinitial
Measuring Reaction Progress and Expressing
Reaction Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
Measuring Reaction Progress and Expressing Reaction
Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
Measuring Reaction Progress and Expressing Reaction
Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
First 50
seconds:
First 100
seconds:
0.0101 0.0120  M

[B]
average rate = 

 3.80×10-5M / s
t
50.0  0.0  s
average rate = 
0.00846  0.0120 M  3.54×10-5M / s
[B]

t
100.0  0.0  s
Measuring Reaction Progress and Expressing Reaction
Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
Measuring Reaction Progress and Expressing Reaction
Rate
The instantaneous rate is the rate for a specific instant in time.
Measuring Reaction Progress and Expressing Reaction
Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
Time (s)
[Br2](M)
Rate (M/s)
0.0
0.0120
4.20 x 10–5
50.0
0.0101
3.52 x 10–5
250.0
0.00500
1.75 x 10–5
Br2 50 s
Br2 250 s
2
rate  [Br2 ]
k is called the rate constant.
rate at 50.0 s 3.52 10 5

2
5
rate at 250.0 s 1.75 10
rate  k[Br2 ]
Measuring Reaction Progress and Expressing Reaction
Rate
Br2(aq) + HCOOH(aq) → 2Br–(aq) + 2H+(aq) + CO2(g)
Time (s)
[Br2](M)
Rate (M/s)
0.0
0.0120
4.20 x 10–5
50.0
0.0101
3.52 x 10–5
250.0
0.00500
1.75 x 10–5
rate  k[Br2 ]
rate
k 
[Br2 ]
3.52  10 5 M / s
k 
 3.49  10 3 s 1
0.0101 M
at t = 50.0 s
Measuring Reaction Progress and Expressing Reaction
Rate
2H2O2(aq) → 2H2O(l) + O2(g)
[O2 ] 1 PO 2
rate =

t
RT t
Measuring Reaction Progress and Expressing Reaction
Rate
aA + bB → cC + dD
rate = 
1 [A]
1 [B] 1 [C] 1 [D]



a t
b t
c t
d t
Measuring Reaction Progress and Expressing Reaction
Rate
Write the rate expressions for the following reaction:
CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)
Solution:
Use the equation below to write the rate expressions.
rate = 
rate = 
1 [A]
1 [B] 1 [C] 1 [D]



a t
b t
c t
d t
[CO 2 ]
1 [H2O] [CH4 ] 1 [O 2 ]



t
2 t
t
2 t
Worked Example 14.1
Write the rate expressions for each of the following reactions:
(a) I-(aq) + OCl-(aq) → Cl-(aq) + OI-(aq)
(b) 2O3(g) → 3O2(g)
(c) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
Strategy For reactions containing gaseous species, progress is generally
monitored by measuring pressure. Pressures are converted to molar
concentrations using the ideal gas equation, and rate expressions are written in
terms of molar concentrations.
Solution (a) All the coefficients in this equation are 1. Therefore,
Δ[I-]
Δ[OCl-]
Δ[Cl-]
Δ[OI-]
rate = −
=−
=
=
Δt
Δt
Δt
Δt
Worked Example 14.1 (cont.)
Solution
(b) rate = −
(c) rate = −
Δ[O2]
Δt
1
2
Δ[O3]
=
Δt
1
4
Δ[NH3]
=−
Δt
1
3
1
5
Δ[O2]
=
Δt
1
4
Δ[NO]
=
Δt
1
6
Δ[H2O]
Δt
Think About It Make sure that the change in concentration of each species is
divided by the corresponding coefficient in the balanced equation. Also make sure
that the rate expressions written in terms of reactant concentrations have a
negative sign in order to make the resulting rate positive.
Worked Example 14.2
Consider the reaction
4NO2(g) + O2(g) → 2N2O5(g)
At a particular time during the reaction, nitrogen dioxide is being consumed at the
rate 0.00130 M/s. (a) At what rate is molecular oxygen being consumed? (b) At
what rate is dinitrogen pentoxide being produced?
Strategy Determine the rate of reaction and, using the stoichiometry of the
reaction, convert to rates of change for the specified individual species.
Solution
rate = −
1
4
We are given
Δ[NO2]
Δ[O2]
=−
=
Δt
Δt
1
2
Δ[N2O5]
Δt
Δ[NO2]
= −0.00130 M/s
Δt
where the minus sign indicates that the concentration of NO2 is decreasing the
time.
Worked Example 14.2 (cont.)
Solution The rate of reaction, therefore, is
rate = −
1
4
Δ[NO2]
1
= − 4 (−0.00130 M/s)
Δt
= 3.25×10-4 M/s
Δ[O2]
(a) 3.25×10-4 M/s = −
Δt
Δ[O2]
-4 M/s
Think About
It Remember
that the negative sign in a rate
= −3.25×10
Δt
expression indicates that a species is being consumed rather than
-4 M/s.
produced.
are always
expressed
quantities.
Molecular
oxygenRates
is being
consumed
at a rateasofpositive
3.25×10
1 Δ[N2O5]
-4
(b) 3.25×10 M/s = 2
Δt
Δ[N2O5]
-4
2(3.25×10 M/s) =
Δt
Δ[N2O5]
= 6.50×10-4 M/s
Δt
Dinitrogen pentoxide is being produced at a rate of 6.50×10-4 M/s.
14.4
Dependence of Reaction Rate on Reactant
Concentration
The rate law is an equation that relates the rate of reaction to the
concentrations of reactants.
aA + bB → cC + dD
rate = k[A]x[B]y
k: the rate constant
x: order with respect to A
y: order with respect to B
determined experimentally
x + y: represents the overall reaction order.
Dependence of Reaction Rate on Reactant Concentration
F2(g) + 2ClO2(g) → 2FClO2(g)
The initial rate is the rate at the beginning of the reaction.
Initial Rate Data for the Reaction between F2 and ClO2
Experiment
[F2](M)
[ClO2](M)
Initial Rate (M/s)
1
0.10
0.010
1.2 x 10–3
2
0.10
0.040
4.8 x 10–3
3
0.20
0.010
2.4 x 10–3
Dependence of Reaction Rate on Reactant Concentration
F2(g) + 2ClO2(g) → 2FClO2(g)
rate = k[F2]x[ClO2]y
Initial Rate Data for the Reaction between F2 and ClO2
Experiment
[F2](M)
[ClO2](M)
Initial Rate (M/s)
1
0.10
0.010
1.2 x 10–3
Rate
0.040
doubles
4.8 x 10–3
0.010
2.4 x 10–3
2
[F2]
doubles
3
F2 3
F2 1
0.10
[ClO2]
constant
0.20
0.20 M

2
0.10 M
The reaction is first order in F2; x = 1
rate3 2.4  103 M / s

2
3
rate1 1.2  10 M / s
rate = k[F2][ClO2]y
Dependence of Reaction Rate on Reactant Concentration
F2(g) + 2ClO2(g) → 2FClO2(g)
rate = k[F2][ClO2]y
Initial Rate Data for the Reaction between F2 and ClO2
Experiment
[F2](M)
[ClO2](M)
Initial Rate (M/s)
1
0.10
0.010
1.2 x 10–3
2
3
ClO2 2
ClO2 1
[F2]
constant
0.10
[ClO2]
x4
0.20
0.040 M

4
0.010 M
The reaction is first order in ClO2; y = 1
0.040
0.010
Rate
x4
4.8 x 10–3
2.4 x 10–3
rate2 4.8  103 M / s

4
3
rate1 1.2  10 M / s
rate = k[F2][ClO2]
Dependence of Reaction Rate on Reactant Concentration
aA + bB → cC + dD
rate = k[A]x[B]y
Initial Rate Data for the Reaction between A and B
Experiment
[A](M)
[B](M)
Initial Rate (M/s)
1
0.10
0.015
2.1 x 10–4
2
0.20
0.015
4.2 x 10–4
3
0.10
0.030
8.4 x 10–4
Dependence of Reaction Rate on Reactant Concentration
aA + bB → cC + dD
rate = k[A]x[B]y
Initial Rate Data for the Reaction between A and B
Experiment
[A](M)
[B](M)
Initial Rate (M/s)
1
0.10
0.015
2.1 x 10–4
2
[A]
x2
3
 A 1
 A 2
0.20
[B]
constant
0.10
0.10 M

 0.5
0.20 M
The reaction is first order in A; x = 1
0.015
0.030
Rate
x2
4.2 x 10–4
8.4 x 10–4
rate1 2.1 104 M / s

 0.5
4
rate2 4.2  10 M / s
rate = k[A][B]y
Dependence of Reaction Rate on Reactant Concentration
aA + bB → cC + dD
rate = k[A][B]y
Initial Rate Data for the Reaction between A and B
Experiment
[A](M)
[B](M)
Initial Rate (M/s)
1
0.10
0.015
2.1 x 10–4
2
[A]
constant
3
B3
B1
0.20
[B]
x2
0.10
0.030 M

2
0.015 M
The reaction is second order in B; y = 2
0.015
0.030
Rate
x4
4.2 x 10–4
8.4 x 10–4
rate3 8.4  104 M / s

4
4
rate1 2.1 10 M / s
rate = k[A][B]2
Dependence of Reaction Rate on Reactant Concentration
Three important things to remember about the rate law:
1) The exponents in a rate law must be determined from a table of
experimental data.
2) Comparing changes in individual reactant concentrations with
changes in rate shows how the rate depends on each reactant
concentration.
3) Reaction order is always defined in terms of reactant
concentrations, never product concentrations.
Dependence of Reaction Rate on Reactant Concentration
The reaction of peroxydisulfate ion (S2O82–) with iodide ion (I–) is:
S2O82–(aq) + 3I–(aq) → 2SO42–(aq) + I3–(aq)
Determine the rate law and calculate the rate constant, including its units.
Initial Rate Data for the Reaction between S2O82– and I–
Experiment
[S2O82–](M)
[I–](M)
Initial Rate (M/s)
1
0.080
0.034
2.2 x 10–4
2
0.080
0.017
1.1 x 10–4
3
0.16
0.017
2.2 x 10–4
Dependence of Reaction Rate on Reactant Concentration
Solution:
Step 1:In experiments 1 and 2, [S2O82–] is constant. The [I–] is
doubled, and rate doubles.
Initial Rate Data for the Reaction between S2O82– and I–
Experiment
[S2O82–](M)
[I–](M)
Initial Rate (M/s)
1
0.080
0.034
2.2 x 10–4
2
0.080
0.017
1.1 x 10–4
3
0.16
0.017
2.2 x 10–4
I 
0.034 M
1

2

0.017 M
I 
2
rate1 2.2  104 M / s

2
4
rate2 1.1 10 M / s
The reaction is first
order in I–
Dependence of Reaction Rate on Reactant Concentration
Solution:
Step 2:In experiments 2 and 3, [S2O82–] is doubled, [I–] is constant,
and rate doubles.
Initial Rate Data for the Reaction between S2O82– and I–
Experiment
[S2O82–](M)
[I–](M)
Initial Rate (M/s)
1
0.080
0.034
2.2 x 10–4
2
0.080
0.017
1.1 x 10–4
3
0.16
0.017
2.2 x 10–4
4
S2O82 
rate
0.16
M
2.2

10
M/s
3
3
The reaction is first


2


2

4
2–
0.080 M
rate2 1.1 10 M / s
S2O82 
order
in
S
O
2
2 8
Dependence of Reaction Rate on Reactant Concentration
Solution:
Step 2:In experiments 2 and 3, [S2O82–] is doubled, [I–] is constant,
and rate doubles.
Initial Rate Data for the Reaction between S2O82– and I–
Experiment
[S2O82–](M)
[I–](M)
Initial Rate (M/s)
1
0.080
0.034
2.2 x 10–4
2
0.080
0.017
1.1 x 10–4
3
0.16
0.017
2.2 x 10–4
The rate law is: rate = k [S2O82–] [I–]
Dependence of Reaction Rate on Reactant Concentration
Solution:
Step 3:Use the data from any experiment to calculate k.
Initial Rate Data for the Reaction between S2O82– and I–
Experiment
[S2O82–](M)
[I–](M)
Initial Rate (M/s)
1
0.080
0.034
2.2 x 10–4
2
0.080
0.017
1.1 x 10–4
3
0.16
0.017
2.2 x 10–4
rate3
2.2  104 M / s
1
1
k


0.081
M

s
(0.16 M )(0.017 M )
S2O82  I 
3
3
Worked Example 14.3
The gas-phase reaction of nitric oxide with hydrogen at 1280°C is
2NO(g) + 2H2(g) → N2(g) +2H2O(g)
From the following data collected at 1280°C, determine (a) the rate law, (b) the
rate constant, including units, and (c) the rate of the reaction when [NO] =
4.8×10-3 M and [H2] = 6.2×10-3 M.
Experiment
[NO] (M)
[H2] (M)
Initial rate (M/s)
1
5.0×10-3
2.0×10-3
1.3×10-5
2
1.0×10-2
2.0×10-3
5.0×10-5
3
1.0×10-2
4.0×10-3
1.0×10-4
Strategy Compare two experiments at a time to determine how the rate depends
on the concentration of each reactant.
The rate law is rate = k[NO]x[H2]y.
Worked Example 14.3 (cont.)
Solution The rate of reaction, therefore, is
rate2
5.0×10-5 M/s
k(1.0×10-2 M)x(2.0×10-3 M)y
=
≈4=
1.3×10-5 M/s
k(5.0×10-3 M)x(2.0×10-3 M)y
rate1
Canceling identical terms in the numerator and denominator gives
(1.0×10-2 M)x
= 2x = 4
-3
x
(5.0×10 M)
Therefore, x = 2. The reaction is second order in NO.
Dividing the rate from experiment 3 by the rate from experiment 2, we get
rate3
1.0×10-4 M/s
k(1.0×10-2 M)x(4.0×10-3 M)y
=
=2=
rate2
5.0×10-5 M/s
k(1.0×10-2 M)x(2.0×10-3 M)y
Canceling identical terms in the numerator and denominator gives
(4.0×10-3 M)y
= 2y = 2
-3
y
(2.0×10 M)
Therefore, y = 1. The reaction is first order in H2. The overall rate law is
rate = k[NO]2[H2]
Worked Example 14.3 (cont.)
Solution We can use data from any of the experiments to calculate the value and
units of k. Using the data from experiment 1 gives
rate
1.3×10-5 M/s
= 2.6×102 M-2∙s-1
k=
=
2
-3
2
-3
[NO] [H2] (5.0×10 M) (2.0×10 M)
(c) Using the rate constant determined in part (b) and the concentrations of NO
and H2 given in the problem statement, we can determine the reaction rate as
follows:
rate = (2.6×102 M-2∙s-1)(4.8×10-3 M)2(6.2×10-3 M)
= 3.7×10-5 M∙s-1
Think About It The exponent for the concentration of H2 in the rate law is 1,
whereas the coefficient for H2 in the balanced equation is 2. It is a common error
to try to write a rate law using the stoichiometric coefficients as the exponents.
Remember that, in general, the exponents in the rate law are not related to the
coefficients in the balanced equation. Rate laws must be determined by
examining a table of experimental data.
14.5
Dependence of Reactant Concentration on Time
The rate law can be used to determine the rate of a reaction using the
rate constant and the reactant concentrations:
rate law
rate = k[A]x[B]y
rate
rate constant
A rate law can also be used to determine the concentration of a
reactant at a specific time during a reaction.
Dependence of Reactant Concentration on Time
A first-order reaction is a reaction whose rate depends on the
concentration of one of the reactants raised to the first power.
C2H6 → 2 ·CH3
rate = k[C2H6]
2N2O5(g) → 2NO2(g) + O2(g) rate = k[N2O5]
Dependence of Reactant Concentration on Time
In a first-order reaction of the type
A → products
The rate can be expressed as the rate of change in reactant
concentration,
A
rate = 
t
as well as in the form of the rate law:
rate = k[A]
Setting the two expressions equal to each other yields:

A
t
 k A
Dependence of Reactant Concentration on Time
Using calculus, it is possible to show that:
 At
ln
 A0
  kt
ln is the natural logarithm
[A]0 and [A]t refer to the concentration of A at times 0 and t
The equation above is sometimes called the integrated rate law for a
first order reaction.
Dependence of Reactant Concentration on Time
The rate constant for the reaction 2A → B is 7.5 x 10–3 s–1 at 110°C.
The reaction is first order in A. How long (in seconds) will it take for
[A] to decrease from 1.25 M to 0.71 M?
Solution:
Step 1:Use the equation below to calculate time in seconds.
 At
ln
 A0
0.71t
ln
1.250
   7.5  103 s1  t
t = 75 seconds
  kt
Worked Example 14.4
The decomposition of hydrogen peroxide is first order in H2O2.
2H2O2(aq) → 2H2O(l) + O2(g)
The rate constant for this reaction at 20°C is 1.8×10-5 s-1. If the start
concentration of H2O2 is 0.75 M, determine (a) the concentration of H2O2
remaining after 3 h and (b) how long it will take for the H2O2 concentration to
drop to 0.10 M.
Strategy Use ln ([A]t/[A]0) = –kt to find [H2O2]t where t = 3 h, and then solve
for t to determine how much time must pass for [H2O2]t to equal 0.10 M. [H2O2]0
= 0.75 M; time t for part (a) is (3 h)(60 min/h)(60 s/min) = 10,800 s.
Solution
[H O ]
(a)
ln 2 2 t = –kt
[H2O2]0
(b)
ln
[H2O2]t
= –(1.8×10-5 s-1)(10,800 s) = –0.1944
0.75 M
Worked Example 14.4 (cont.)
Solution Take the inverse natural logarithm of both sides of the equation to get
[H2O2]t
= e–0.1944 = 0.823
0.75 M
[H2O2]t = (0.823)(0.75 M) = 0.62 M
The concentration of H2O2 after 3 h is 0.62 M.
0.10 M
= –2.015 = –(1.8×10-5 s-1)t
(b) ln
0.75 M
2.015
= t = 1.12×105 s
-5
-1
1.8×10 s
The time required for the peroxide concentration to drop to 0.10 M is 1.1×105 s
or about 31 h.
Think About It Don’t forget the minus sign. If you calculate a concentration at
time t that is greater than the concentration at time 0 (or if you get a negative time
required for the concentration to drop to a specified level), check your solution
for this common error.
Dependence of Reactant Concentration on Time
Rearrangement of the first-order integrated rate law gives:
 At
ln
 A0
  kt
ln[A]t = –kt + ln[A]0
Rearrangement in this way has the form
of the linear equation y = mx + b.
ln[A]t = –kt + ln[A]0
Slope = –k
Intercept = ln[A]0
Dependence of Reactant Concentration on Time
The rate of decomposition of azomethane is studied by monitoring
the partial pressure of the reactant as a function of time.
CH3—N=N—CH3(g) → N2(g) + C2H6(g)
The data obtained at 300°C are listed in the following table:
Time (s)
Pazomethane (mmHg)
0
284
100
220
150
193
200
170
250
150
300
132
Dependence of Reactant Concentration on Time
Plotting the data gives a straight line, indicating the reaction is first
order.
ln[A]t = –kt + ln[A]0
5.70
5.60
5.50
Slope = –2.55 x 10–3 s–1
ln P
5.40
5.30
Intercept = 5.65
5.20
5.10
5.00
4.90
4.80
0
100
200
Time (s)
300
400
Dependence of Reactant Concentration on Time
Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas
phase as follows:
C2H5I(g) → C2H4(g) + HI(g)
Determine the rate of the reaction, after verifying that the reaction is
first order.
Time (s)
[C2H5I] (M)
0
0.36
15
0.30
30
0.35
48
0.19
75
0.13
Dependence of Reactant Concentration on Time
Solution:
Plot ln[C2H5I] vs time. If a straight line results, the reaction is first
order. The slope is equal to k.
[C2H5I] (M)
0
0.36
15
0.30
30
0.35
48
0.19
75
0.13
0.00
ln[C2H5I]
-1.02
-1.20
-1.39
-1.66
-2.04
0
20
40
-0.50
ln [C2H5I]
Time (s)
Slope = –1.3 x 10–2 s–1; k = 1.3 x 10–2 s–1
-1.00
-1.50
-2.00
-2.50
Time (s)
60
80
Worked Example 14.5
The rate of decomposition of azomethane is studied by monitoring the partial
pressure of the reactant as a function of time:
CH3−N=N−CH3(g) → N2(g) + C2H6(g)
The data obtained at 20°C are listed in the following table:
Time (s)
Pazomethane (mmHg)
0
284
100
220
150
193
200
170
250
150
300
132
Strategy We can use ln ([A]t/[A]0) = –kt only for first-order reactions, so we
must first determine if the decomposition of azomethane is first order. We do this
by plotting ln P against time. If the reaction is first order, we can use
ln ([A]t/[A]0) = –kt and the data at any two of the times in the table to
determine the rate constant.
Worked Example 14.5 (cont.)
Solution The table expressed in ln P is
Time (s)
ln P
0
5.649
100
5.394
150
5.263
200
5.136
250
5.011
300
4.883
Plotting these data gives a straight line, indicating that the reaction is indeed first
order. Thus, we can use ln ([A]t/[A]0) = –kt in terms of pressure.
Pt
ln P = –kt
0
Pt and P0 can be pressures at any two times during the experiment. P0 need not be
the pressure at 0 s–it need only be at the earlier of the two times.
Worked Example 14.5 (cont.)
Solution Using data from times 100 s and 250 s of the original table (Pazomethane
versus t), we get
150 mmHg
ln 220 mmHg = –kt
ln 0.682 = –k(150 s)
k = 2.55×10-3 s-1
Think About It We could equally well have determined the rate constant by
calculating the slope of the plot of ln P versus t. Using the two points labeled on
the plot, we get
5.011 – 5.394
slope = 250 – 100
= 2.55×10-3 s-1
Remember that slope = –k, so k = 2.55×10-3 s-1.
Dependence of Reactant Concentration on Time
The half-life (t1/2) is the
time required for the
reactant concentration
to drop to half its
original value.
t1/2
0.693

k
Worked Example 14.6
The decomposition of ethane (C2H6) to methyl radicals (CH3) is a first-order
reaction with a rate constant of 5.36×10-4 s-1 at 700°C.
C2H6 → CH3
Calculate the half-life of the reaction in minutes.
Strategy Use t½ = 0.693/k to calculate t½ in seconds, and then convert to
minutes.
0.693
0.693
Solution t½ =
=
= 1293 s
k
5.36×10-4 s-1
1 min
1293 s × 60 s = 21.5 min
The half-life of ethane decomposition at 700°C is 21.5 min.
Think About It Half-lives and rate constants can be expressed using any units
of time and reciprocal time, respectively. Track units carefully when you
convert from one unit of time to another.
Dependence of Reactant Concentration on Time
The half-life (t1/2) is the time required for the reactant concentration
to drop to half its original value.
 At
ln
 A0
  kt
rearranges
1  A 0
t  ln
k  A t
t = t1/2 when [A]t = ½[A]0.
t1/2
 A0
1
 ln
k 1 A
 0
2
t1/2 
simplifies
0.693
k
Dependence of Reactant Concentration on Time
Calculate the half-life of the decomposition of azomethane,
k = 2.55×10–3 s-1.
Solution:
Step 1:Use the equation below to calculate half-life:
t1/2 
t1/2 
0.693
k
0.693
 272 s
3
1
2.55 x 10 s
Dependence of Reactant Concentration on Time
A second-order reaction is a reaction whose rate depends on the
concentration of one reactant raised to the second power or on the
product of the concentrations of two different reactants (first order in
each).
Second-order integrated rate law:
1
1
 kt +
 A t
 A0
Second-order half-life:
t1/2 
1
k  A0
Worked Example 14.7
Iodine atoms combine to form molecular iodine in the gas phase:
I(g) + I(g) → I2(g)
This reaction is second order and has a rate constant of 7.0×109 M-1∙s-1 at 23°C.
(a) If the initial concentration of I is 0.086 M, calculate the concentration after
2.0 min. (b) Calculate the half-life of the reaction when the initial concentration
of I is 0.60 M and when the initial concentration of I is 0.42 M.
Strategy Use 1/[A]t = kt + 1/[A]0 to determine [I]t at t = 2.0 min; use
t½ = 1/k[A]0 to determine t½ when [I]0 = 0.60 M and when [I]0 = 0.42 M.
Solution t = (2.0 min)(60 s/min) = 120 s
(a)
1
1
= kt +
[A]t
[A]0
= (7.0×109 M-1∙s-1)(120 s) +
1
0.086 M
Worked Example 14.7 (cont.)
Solution
= 8.4×1011 M-1
[A]t =
1
= 1.2×10-12 M
11
-1
8.4×10 M
The concentration
of atomic
iodine like
afterthe
2 min
is halogens,
1.2×10-12exists
M. as
Think About
It (a) Iodine,
other
diatomic molecules at room temperature. It makes sense, therefore,
(b) When
M, would react quickly, and essentially completely, to
that[I]
atomic
iodine
0 = 0.60
form1I2 at room temperature.
The very low remaining concentration
1
-10 s
t½ of
= Ik[A]
= 2.4×10
after 2= min
makes
sense.
(b)
As
expected,
the
half-life of this
9
-1
-1
(7.0×10 M ∙s )(0.60 M)
0
second-order reaction is not constant. (A constant half-life is a
When characteristic
[I]0 = 0.42 M, of first-order reactions.)
1
1
t½ =
=
= 3.4×10-10 s
9
-1
-1
k[A]0 (7.0×10 M ∙s )(0.42 M)
Dependence of Reactant Concentration on Time
The rate of a zero-order reaction is a constant.
Third-order and higher are rare.
14.6
Dependence of Reaction Rate on Temperature
The dependence of the rate constant on temperature can be
expressed by the Arrhenius equation.
k  Ae
 Ea / RT
A represents the collision frequency and is called the frequency
factor.
Ea is the activation energy (in kJ/mol).
R is the gas constant (8.314 J/mol K).
T is the absolute temperature.
e is the base of the natural logarithm.
Dependence of Reaction Rate on Temperature
Taking the natural log of both sides, the Arrhenius equation may be
written as:
Ea
ln k  ln A 
RT
Rearrangement gives the linear form of the Arrhenius equation:
 Ea   1 
ln k   
   ln A

 R T 
Dependence of Reactant Rate on Temperature
Rate constants for the reaction
CO(g) + NO2(g) → CO2(g) + NO(g)
were measured at four different temperatures. The data are shown in
the table. Determine the activation energy for the reaction.
k (M–1·s–1)
0.0521
0.101
T (K)
288
298
0.184
0.332
308
318
Dependence of Reactant Rate on Temperature
Solution:
Plot ln k versus 1/T and determine the slope of the line;
slope = –Ea/R.
0.00
1/T
(K–1)
T (K)
ln k
0.0521
288
0.00347
-2.95
0.101
298
0.00336
-2.29
0.184
308
0.00325
-1.69
0.332
318
0.00314
-1.10
-0.50
-1.00
ln k
k
(M–1·s–1)
-1.50
-2.00
-2.50
-3.00
-3.50
slope = –5.6 x 103 K = –Ea/R
Ea = (5.6 x 103 K )(8.314 J/mol·K) = 46 kJ/mol
1/T (K–1)
Worked Example 14.8
Rate constants for the reaction
CO(g) + NO2(g) → CO2(g) + NO(g)
were measured at four different temperatures. The data are shown in the table.
Plot ln k versus 1/T, and determine the activation energy (in kJ/mol) for the
reaction.
k (M-1∙s-1)
T (K)
0.0521
288
0.101
298
0.184
308
0.332
318
Strategy Plot ln k versus 1/T, and determine the slope of the resulting line.
According to ln k = (–Ea/R)(1/T) + ln A, slope = –Ea/R. R = 8.314 J/mol∙K.
Worked Example 14.8 (cont.)
Solution Taking the natural log of each value of k and the inverse of each value
of T gives
ln k
1/T (K-1)
–2.95
3.47×10-3
–2.29
3.36×10-3
–1.69
3.25×10-3
–1.10
3.14×10-3
A plot of these data yields the following graph:
Worked Example 14.8 (cont.)
Solution The slope is determined using the x and y coordinates of any two
points on the line. Using the points that are labeled on the graph gives
–1.4 – (–2.5)
slope = 3.2×10-3 K-1 – 3.4×10-3 K-1 = –5.5×103 K
The value of the slope is –5.5×103 K. Because the slope = –Ea/R,
Ea = –(slope)(R)
= –(–5.5×103 K)(8.314 J/K∙mol)
= 4.6×104 J/mol or 46 kJ/mol
Think About It Note that while k has units M-1∙s-1, ln k has no units.
Dependence of Reactant Rate on Temperature
A two point form of the Arrhenius equation may be written:
k1 Ea  1 1 
ln 
  
k2 R  T2 T1 
If the rate constants at two different temperatures are known, it is
possible to calculate the activation energy.
If the activation energy and the rate constant at one temperature are
known, it is possible to determine the rate constant at any other
temperature.
Worked Example 14.9
The rate constant for a particular first-order reaction is given for three different
temperatures:
T (K)
k (s-1)
400
2.9×10-3
450
6.1×10-2
500
7.0×10-1
Using the data, calculate the activation energy of the reaction.
Strategy Rearrange and solve for Ea using the following
k1
ln
k2
Ea = R
1
1
–
T2
T2
Worked Example 14.9 (cont.)
Solution
ln
Ea = 8.314 J/K∙mol
2.9×103
1 6.1×10-1
–
450 K 2 400 K
= 91,173 J/mol = 91 kJ/mol
The activation energy of the reaction is 91 kJ/mol.
Think About It A good way to check your work is to use the value of Ea that
you calculated (and Equation 14.11) to determine the rate constant at 500 K.
Make sure it agrees with the value in the table.
Worked Example 14.10
A certain first-order reaction has an activation energy of 83 kJ/mol. If the rate
constant for this reaction is 2.1×10-2 s-1 at 150°C, what is the rate constant at
300°C?
Strategy Rearrange and solve for k2 using the following
k1
k =
Ea the1 rate constant
1
Think About It Make2 sure that
you calculate at a
–
R
T2
T2
higher temperature is in fact higher
than the
original rate constant.
e
According to the Arrhenius equation, the rate constant always
4 J/mol, T = 423 K, T = 573 K, R = 8.314 J/K∙mol, and
Ea = 8.3×10
1
2
increases
with increasing
temperature.
If you get a smaller k at a
-2
-1
k1 = 2.1×10
s .
higher temperature,
check your solution for mathematical errors.
Solution
2.1×10-2 s-1
k2 =
8.3×104
e
J/mol
8.314 J/K∙mol
The rate constant of 300°C is 10 s-1.
1
1 –
573 K
423 K
= 1.0×101 s-1
14.7
Reaction Mechanisms
A balanced chemical equation does not indicate how a reaction
actually takes place.
The sequence of steps that sum to give the overall reaction is called
the reaction mechanism.
Step 1:
A+B→C
Step 2:
C+B→D
Overall reaction: A + 2B → D
Reaction Mechanism
Chemical species that appear in the reaction mechanism, but not in
the overall chemical equation are called intermediates.
Step 1:
NO + NO
N2O2
Step 2:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
Reaction Mechanism
Each step in a reaction mechanism represents an elementary
reaction, one that occurs in a single collision of the reactant
molecules.
The molecularity of an elementary reaction is essentially the number
of reactant molecules involved in the collision.
unimolecular (one reactant molecule)
A → products
rate = k[A]
first order
bimolecular (two reactant molecules)
A + B → products
rate = k[A][B]
second order
A + A → products
rate = k[A]2
second order
termolecular (three reactant molecules)
Reaction Mechanism
In a reaction mechanism consisting of more than elementary step,
the rate law for the overall process is given by the rate-determining
step.
The rate determining step is the slowest step in the sequence.
A proposed mechanism must satisfy two requirements:
The sum of the elementary reaction must be the overall balanced
equation for the reaction.
The rate determining step must have the same rate law as that
determined from the experimental data.
Reaction Mechanism
The decomposition of hydrogen peroxide can be facilitated by iodide
ions:
Step 1: (slow)
H2 O2 + I–
H2O + IO–
Step 2:
Overall reaction:
H2O2 + IO–
2H2O2
H 2 O + O2 + I –
2H2O + O2
Rate = k1[H2O2][I–]
Worked Example 14.11
The gas-phase decomposition of nitrous oxide (N2O) is believed to occur in two
steps:
k1
Step 1: N2O →
N2 + O
k2
Step 2: N2O + O →
N2 + O2
Experimentally the rate law is found to be rate = k[N2O]. (a) Write the equation
for the overall reaction. (b) Identify the intermediate(s). (c) Identify the ratedetermining step.
Strategy Add the two equations, canceling identical terms on opposite sides of
the arrow, to obtain the overall reaction. The canceled terms will be the
intermediates if they were first generated and then consumed. Write rate laws for
each elementary step; the one that matches the experimental rate law will be the
rate-determining step.
Worked Example 14.11 (cont.)
Solution
k
1
Step 1: N2O →
N2 + O
k2
Step 2: N2O + O →
N2 + O 2
rate = k[N2O]
rate = k[N2O][O]
(a) 2N2O → 2N2 + O2
(b) O (atomic oxygen) is the intermediate.
(c) Step 1 is the rate-determining step because its rate law is the same as the
experimental law: rate = k[N2O].
Think About It A species that gets canceled when steps are added may be an
intermediate or a catalyst. In this case, the canceled species is an intermediate
because it was first generated and then consumed. A species that is first consumed
and then generated, but doesn’t appear in the overall equation, is a catalyst.
Worked Example 14.12
Consider the gas-phase reaction of nitric oxide and oxygen that was described at
the beginning of Section 14.5.
2NO(g) + O2(g) → 2NO2(g)
Show that the following mechanism is plausible. The experimentally determined
rate law is rate = k[N2O]2[O2].
k1
Step 1: NO(g) + NO(g) ⇌ N2O2(g)
(fast)
k-1
k
2
Step 2: N2O2(g) + O2(g) →
2NO2(g)
(slow)
Strategy To establish the plausibility of a mechanism, we must compare the rate
law of the rate-determining step to the experimentally determined rate law. In this
case, the rate-determining step has an intermediate (N2O2) as one of its reactants,
giving us a rate law of rate = k2[N2O2][O2]. Because we cannot compare this
directly to the experimental rate law, we must solve for the intermediate
concentration in terms of reactant concentrations.
Worked Example 14.12 (cont.)
Solution The first step is a rapidly established equilibrium. Both the forward
and reverse of step 1 are elementary processes, which enables us to write their
rate laws from the balanced equation.
rateforward = k1[NO]2
and
ratereverse = k-1[N2O2]
Because at equilibrium the forward and reverse processes are occurring at the
same rate,
we About
can set It
their
rates
equal to each
andrate-determining
solve for the intermediate
Think
Not
all reactions
haveother
a single
step. Analyzing the kinetics of reactions with two or more
concentration.
comparably slow steps isk1beyond
[NO]2 =the
k-1scope
[N2O2of
] this book.
k1[NO]2
[N2O2] =
k-1
Substituting the solution into the original rate law (rate = k[N2O2][O2]) gives
k1[NO]2
rate = k2
[O2] = k[NO]2[O2]
k-1
kk
where k = k2 1
-1
14.8
Catalysis
A catalyst is a substance that increases the rate of a chemical
reaction without itself being consumed.
Step 1: (slow)
Step 2:
Overall reaction:
H 2 O 2 + I–
H2O2 + IO–
2H2O2
H2O + IO–
H 2 O + O 2 + I–
2H2O + O2
A catalyst speeds up a reaction by providing a set of elementary
steps with more favorable kinetics than those that exist in its
absence.
A catalyst usually speeds up a reaction by lowering the activation
energy.
Catalysis
In the presence of a catalyst the rate constant is kc, called the
catalytic rate constant.
uncatalyzed
catalyzed
ratecatalyzed > rateuncatalyzed
Catalysis
In heterogeneous catalysis, the reactants and the catalysts are in
different phases.
Catalysis
In homogeneous catalysis, the reactants and catalyst are dispersed in
a single phase, usually liquid.
Advantages:
Reactions can be carried out under atmospheric conditions
Can be designed to function selectively
Are generally cheaper
Catalysis
Enzymes are biological catalysts.
Catalysis
The mathematical treatment of enzyme kinetics is complex, but can
be simplified:
k
E+S
ES
uncatalyzed
1




k1
ES
 E + P
k2
catalyzed
Catalysis
Generally, the rate of an enzyme catalyzed reaction is given by the
equation:
rate =
 P
t
= k ES
14
Key Points
Average Reaction Rates
Instantaneous Rate
Stoichiometry and Reaction Rate
The Rate Law
Experimental Determination of the Rate Law
First-Order Reactions
Second-Order Reactions
Collision Theory
Elementary Reactions
Rate-Determining Step
Experimental Support for Reaction Mechanisms
Heterogeneous Catalysis
Homogeneous Catalysis
Enzymes: Biological Catalysts