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Chapter 16
Kinetics: Rates and Mechanisms
of Chemical Reactions
16-1
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Kinetics: Rates and Mechanisms of Chemical Reactions
16.1 Factors That Influence Reaction Rates
16.2 Expressing the Reaction Rate
16.3 The Rate Law and Its Components
16.4 Integrated Rate Laws: Concentration Changes over Time
16.5 The Effect of Temperature on Reaction Rate
16.6 Explaining the Effects of Concentration and Temperature
16.7 Reaction Mechanisms: Steps in the Overall Reaction
16.8 Catalysis: Speeding Up a Chemical Reaction
16-2
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Figure 16.1
16-3
Reaction rate: the central focus of chemical kinetics.
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Factors That Influence Reaction Rate
Under a specific set of conditions, every reaction has its own
characteristic rate, which depends upon the chemical nature of
the reactants.
Four factors can be controlled during the reaction:
1.
2.
3.
4.
16-4
Concentration - molecules must collide to react;
Physical state - molecules must mix to collide;
Temperature - molecules must collide with enough energy to react;
The use of a catalyst.
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Figure 16.2
16-5
Collision energy and reaction rate.
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Expressing the Reaction Rate
reaction rate - changes in the concentrations of reactants or
products per unit time
reactant concentrations decrease while product concentrations
increase
A
for
rate of reaction = -
B
change in concentration of A
change in time
-
16-6
 (conc A)
t
=-
conc A2-conc A1
t2-t1
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Table 16.1 Concentration of O3 at Various Time in its
Reaction with C2H4 at 303K
C2H4(g) + O3(g)
Time (s)
-
16-7
 (conc A)
t
C2H4 O(g) + O2(g)
Concentration of O3 (mol/L)
0.0
3.20x10-5
10.0
2.42x10-5
20.0
1.95x10-5
30.0
1.63x10-5
40.0
1.40x10-5
50.0
1.23x10-5
60.0
1.10x10-5
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Figure 16.3
The concentrations of O3 vs. time during its reaction with C2H4
C2H4(g) + O3(g)
rate =
 [C2H4]
t
-
-
16-8
 [O3]
t
=
C2H4 O(g) + O2(g)
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Figure 16.4
16-9
Plots of [C2H4] and [O2] vs. time.
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In general, for the reaction
aA
rate =
-
+
1
[A]
a
t
bB
= -
cC
1
[B]
b
t
+
= +
dD
1
[C]
c
t
= +
1
[D]
d
t
The numerical value of the rate depends upon the substance that
serves as the reference. The rest is relative to the balanced
chemical equation.
16-10
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Sample Problem 16.1
PROBLEM:
Expressing Rate in Terms of Changes in
Concentration with Time
Because it has a nonpolluting product (water vapor), hydrogen
gas is used for fuel aboard the space shuttle and may be used
by Earth-bound engines in the near future.
2H2(g) + O2(g)
2H2O(g)
(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.
(b) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H2O]
increasing?
PLAN: Choose [O2] as a point of reference since its coefficient is 1. For every
molecule of O2 which disappears, 2 molecules of H2 disappear and 2
molecules of H2O appear, so [O2] is disappearing at half the rate of
change of H2 and H2O.
SOLUTION:
(a)
(b) -
16-11
[O2]
t
rate = -
1
2
= - 0.23mol/L*s
[H2]
t
=-
=+ 1
2
[O2]
t
[H2O]
t
1
=+
2
;
[H2O]
t
[H2O]
t
= 0.46mol/L*s
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Sample Problem 16.2 Determining Reaction Order from Rate Laws
PROBLEM:
For each of the following reactions, determine the reaction order
with respect to each reactant and the overall order from the
given rate law.
(a) 2NO(g) + O2(g)
(b) CH3CHO(g)
2NO2(g); rate = k[NO]2[O2]
CH4(g) + CO(g); rate = k[CH3CHO]3/2
(c) H2O2(aq) + 3I-(aq) + 2H+(aq)
I3-(aq) + 2H2O(l); rate = k[H2O2][I-]
PLAN: Look at the rate law and not the coefficients of the chemical reaction.
SOLUTION:
(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.
(b) The reaction is 3/2 order in CH3CHO and 3/2 order overall.
(c) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+,
while being 2nd order overall.
16-12
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Table 16.2 Initial Rates for a Series of Experiments in the
Reaction Between O2 and NO
2NO(g) + O2(g)
Experiment
16-13
2NO2(g)
Initial Reactant
Concentrations (mol/L)
O2
NO
Initial Rate
(mol/L*s)
1
1.10x10-2
1.30x10-2
3.21x10-3
2
2.20x10-2
1.30x10-2
6.40x10-3
3
1.10x10-2
2.60x10-2
12.8x10-3
4
3.30x10-2
1.30x10-2
9.60x10-3
5
1.10x10-2
3.90x10-2
28.8x10-3
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Determining Reaction Orders
Using initial rates Run a series of experiments, each of which starts with a different
set of reactant concentrations, and from each obtain an initial rate.
See Table 16.2 for data on the reaction
O2(g) + 2NO(g)
rate = k [O2]m[NO]n
2NO2(g)
Compare 2 experiments in which the concentration of one reactant
varies and the concentration of the other reactant(s) remains constant.
k [O2]2m[NO]2n
rate2
=
rate1
=
k
[O2]1m[NO]1n
6.40x10-3mol/L*s
[O2]2m
2.20x10-2mol/L
=
3.21x10-3mol/L*s
=
[O2]1m
[O2]2
[O2]1
m
;
2 = 2m
1.10x10-2mol/L
Do a similar calculation for the other reactant(s).
16-14
m
m=1
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Sample Problem 16.3 Determining Reaction Order from Initial Rate Data
PROBLEM:
Many gaseous reactions occur in a car engine and exhaust
system. One of these is
NO2(g) + CO(g)
NO(g) + CO2(g)
rate = k[NO2]m[CO]n
Use the following data to determine the individual and overall reaction orders.
Experiment
1
Initial Rate(mol/L*s) Initial [NO2] (mol/L) Initial [CO] (mol/L)
0.10
0.40
0.10
2
0.0050
0.080
3
0.0050
0.10
0.20
PLAN:
0.10
Solve for each reactant using the general rate law using the
method described previously.
SOLUTION:
rate = k [NO2]m[CO]n
First, choose two experiments in which [CO] remains
constant and the [NO2] varies.
16-15
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Sample Problem 16.3 Determining Reaction Order from Initial Rate Data
continued
rate 2
=
rate 1
0.080
0.0050
=
rate 3
rate 1
=
0.0050
0.0050
k [NO2]m2[CO]n2
k [NO2
]m
0.40
m
=
1
The reaction is
2nd order in NO2.
[NO2] 1
16 = 4m and m = 2
;
k [NO2]m3[CO]n3
k [NO2]m1 [CO]n1
=
[CO] 3
n
[CO] 1
n
;
1 = 2n and n = 0
0.10
rate = k [NO2]2[CO]0 = k [NO2]2
16-16
m
0.10
0.20
=
1
[CO]n
[NO2] 2
The reaction is
zero order in CO.
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Integrated Rate Laws
[A]
rate = -
t
first order rate equation
= k [A]
ln
[A]0
= - kt
[A]t
ln [A]0 = -kt + ln [A]t
[A]
rate = -
t
= k [A]2
1
[A]t
second order rate equation
-
1
[A]0
= kt
1
[A]t
[A]
rate = -
t
= k [A]0
zero order rate equation
[A]t - [A]0 = - kt
16-17
= kt
+
1
[A]0
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Table 16.3 Units of the Rate Constant k for Several Overall
Reaction Orders
Overall Reaction Order
16-18
Units of k (t in seconds)
0
mol/L*s (or mol L-1 s-1)
1
1/s (or s-1)
2
L/mol*s (or L mol -1 s-1)
3
L2 / mol2 *s (or L2 mol-2 s-1)
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Sample Problem 16.4 Determining Reaction Concentration at a Given Time
PROBLEM: At 10000C, cyclobutane (C4H8) decomposes in a first-order
reaction, with the very high rate constant of 87s-1, to two
molecules of ethylene (C2H4).
(a) If the initial C4H8 concentration is 2.00M, what is the
concentration after 0.010 s?
(b) What fraction of C4H8 has decomposed in this time?
PLAN:
Find the [C4H8] at time, t, using the integrated rate law for a 1st order
reaction. Once that value is found, divide the amount decomposed by
the initial concentration.
SOLUTION:
ln
(a)
[C4H8]0
[C4H8]t
= kt
; ln
2.00
= (87s-1)(0.010s)
[C4H8]
[C4H8] = 0.83mol/L
(b)
[C4H8]0 - [C4H8]t
[C4H8]0
16-19
2.00M - 0.87M
=
2.00M
= 0.58
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Figure 16.5
Integrated rate laws and reaction order.
1/[A]t = kt + 1/[A]0
ln[A]t = -kt + ln[A]0
16-20
[A]t = -kt + [A]0
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Figure 16.8
16-21
Graphical determination of the reaction order for the
decomposition of N2O5.
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Figure 16.9
A plot of [N2O5] vs. time for three half-lives.
for a first-order process
t1/2 =
16-22
ln 2
k
=
0.693
k
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Sample Problem 16.5 Determining the Half-Life of a First-Order Reaction
PROBLEM:
Cyclopropane is the smallest cyclic hydrocarbon. Because its
600 bond angles allow poor orbital overlap, its bonds are weak.
As a result, it is thermally unstable and rearranges to propene at
10000C via the following first-order reaction:
CH2

H3C CH CH2 (g)
H2C CH2 (g)
The rate constant is 9.2s-1, (a) What is the half-life of the reaction? (b) How
long does it take for the concentration of cyclopropane to reach one-quarter of
the initial value?
0.693
PLAN:
Use the half-life equation, t1/2 =
, to find the half-life.
k
One-quarter of the initial value means two half-lives have passed.
SOLUTION:
(a)
t1/2 = 0.693/9.2s-1 = 0.075s
16-23
(b)
2 t1/2 = 2(0.075s) = 0.150s
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Table 16.4 An Overview of Zero-Order, First-Order, and
Simple Second-Order Reactions
Zero Order
First Order
Second Order
Rate law
rate = k
rate = k [A]
rate = k [A]2
Units for k
mol/L*s
1/s
L/mol*s
Integrated rate law in
straight-line form
[A]t =
k t + [A]0
ln[A]t =
-k t + ln[A]0
1/[A]t =
k t + 1/[A]0
Plot for straight line
[A]t vs. t
ln[A]t vs. t
1/[A]t = t
Slope, y-intercept
k, [A]0
-k, ln[A]0
k, 1/[A]0
Half-life
[A]0/2k
ln 2/k
1/k [A]0
16-24
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Figure 16.10
16-25
Dependence of the rate constant on temperature
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The Effect of Temperature on Reaction Rate
The Arrhenius Equation
k  Ae 
Ea
RT
where k is the kinetic rate constant at T
Ea is the activation energy
R is the energy gas constant
ln k = ln A - Ea/RT
ln
16-26
k2
k1
Ea
= RT
T is the Kelvin temperature
A is the collision frequency factor
1
T2
1
T1
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Figure 16.11 Graphical determination of the activation energy
ln k = -Ea/R (1/T) + ln A
16-27
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Sample Problem 16.6 Determining the Energy of Activation
PROBLEM:
The decomposition of hydrogen iodide,
2HI(g)
H2(g) + I2(g)
has rate constants of 9.51x10-9L/mol*s at 500. K and 1.10x10-5
L/mol*s at 600. K. Find Ea.
PLAN:
Use the modification of the Arrhenius equation
to find Ea.
SOLUTION:
ln
k2
k1
= -
Ea
1
R
T2
Ea = - (8.314J/mol*K) ln
-
1
T1
1.10x10-5L/mol*s
9..51x10-9L/mol*s
Ea = 1.76x105 J/mol = 176 kJ/mol
16-28
Ea = - R ln
1
600K
-
k2
1
k1
T2
1
500K
-
1
T1
-1
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Figure 16.12
Information sequence to determine the kinetic parameters of a reaction.
Series of plots
of concentration vs. time
Initial
rates
Determine slope
of tangent at t0 for
each plot
Plots of
concentration
vs. time
16-29
Reaction
Rate constant
orders
(k) and actual
Compare initial
rate law
rates when [A]
Substitute initial rates,
changes and [B] is
orders, and concentrations
Find k at
held constant and
into general rate law:
varied T
m
n
vice versa
rate = k [A] [B]
Integrated
rate law
(half-life,
t1/2)
Use direct, ln or
inverse plot to
find order
Rate constant
and reaction
order
Rearrange to
linear form and
graph
Activation
energy, Ea
Find k at
varied T
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Figure 16.13
The dependence of possible collisions on the product
of reactant concentrations.
A
B
4 collisions
A
B
A
Add another
molecule of A
B
6 collisions
A
B
A
Add another
molecule of B
16-30
A
B
A
B
A
B
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Table 16.5 The Effect of Ea and T on the Fraction (f) of Collisions
with Sufficient Energy to Allow Reaction
Ea (kJ/mol)
50
1.70x10-9
75
7.03x10-14
100
2.90x10-18
T
16-31
f (at T = 298 K)
f (at Ea = 50 kJ/mol)
250C(298K)
1.70x10-9
350C(308K)
3.29x10-9
450C(318K)
6.12x10-9
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Figure 16.14
The effect of temperature on the distribution of collision energies
16-32
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Figure 16.15
Energy-level diagram for a reaction
Ea (forward)
Ea (reverse)
REACTANTS
PRODUCTS
The forward reaction is exothermic because the
reactants have more energy than the products.
16-33
Collision Energy
Collision Energy
ACTIVATED STATE
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Figure 16.16
16-34
An energy-level diagram of the fraction of collisions
exceeding Ea.
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Figure 16.17
The importance of molecular orientation to an effective collision.
NO + NO3
2 NO2
A is the frequency factor
A = pZ where
16-35
Z is the collision frequency
p is the orientation probability factor
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Figure 16.18
Nature of the transition state in the reaction between CH3Br and OH-.
CH3Br + OH-
CH3OH + Br -
transition state or activated complex
16-36
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Figure 16.19 Reaction energy diagram for the reaction of CH3Br and OH-.
16-37
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Figure 16.20
Reaction energy diagrams and possible transition states.
16-38
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Sample Problem 16.7 Drawing Reaction Energy Diagrams and
Transition States
PROBLEM:
A key reaction in the upper atmosphere is
O3(g) + O(g)
2O2(g)
The Ea(fwd) is 19 kJ, and the Hrxn for the reaction is -392 kJ. Draw a
reaction energy diagram for this reaction, postulate a transition state, and
calculate Ea(rev).
Consider the relationships among the reactants, products and
transition state. The reactants are at a higher energy level than the
products and the transition state is slightly higher than the
reactants.
SOLUTION:
Potential Energy
PLAN:
Ea= 19kJ
O3+O
transition state Ea(rev)= (392 + 19)kJ =
Hrxn = -392kJ
411kJ
2O2
16-39
Reaction progress
O
breaking
bond
O
O
forming
bond
O
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REACTION MECHANISMS
Table 16.6 Rate Laws for General Elementary Steps
Elementary Step
Molecularity
Rate Law
product
Unimolecular
Rate = k [A]
2A
product
Bimolecular
Rate = k [A]2
A+B
product
Bimolecular
Rate = k [A][B]
2A + B
product
Termolecular
Rate = k [A]2[B]
A
16-40
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Sample Problem 16.8 Determining Molecularity and Rate Laws for
Elementary Steps
PROBLEM:
(1)
The following two reactions are proposed as elementary steps in
the mechanism of an overall reaction:
NO2Cl(g)
NO2(g) + Cl (g)
(2)
NO2Cl(g) + Cl (g)
NO2(g) + Cl2(g)
(a) Write the overall balanced equation.
(b) Determine the molecularity of each step.
(c) Write the rate law for each step.
PLAN: (a) The overall equation is the sum of the steps.
(b) The molecularity is the sum of the reactant particles in the step.
SOLUTION:
NO2Cl(g)
NO2(g) + Cl (g)
(2) NO2Cl(g) + Cl (g)
NO2(g) + Cl2(g)
(a) (1)
2NO2Cl(g)
16-41
2NO2(g) + Cl2(g)
(b) Step(1) is unimolecular.
Step(2) is bimolecular.
(c) rate1 = k1 [NO2Cl]
rate2 = k2 [NO2Cl][Cl]
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The Rate-Determining Step of a Reaction Mechanism
The overall rate of a reaction is related to the rate of the slowest, or
rate-determining step.
Correlating the Mechanism with the Rate Law
The elementary steps must add up to the overall equation.
The elementary steps must be physically reasonable.
The mechanism must correlated with the rate law.
16-42
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Figure 16.21
Reaction energy diagram for the two-step reaction of NO2 and F2.
16-43
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CATALYSTS
•Each catalyst has its own specific way of functioning.
•In general a catalyst lowers the energy of activation.
•Lowering the Ea increases the rate constant, k, and
thereby increases the rate of the reaction
•A catalyst increases the rate of the forward AND the
reverse reactions.
•A catalyzed reaction yields the products more quickly,
but does not yield more product than the uncatalyzed
reaction.
•A catalyst lowers Ea by providing a different mechanism,
for the reaction through a new, lower energy pathway.
16-44
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Figure 16.22
Reaction energy diagram of a catalyzed and an uncatalyzed process.
16-45
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Figure 16.23
Mechanism for the catalyzed hydrolysis of an organic ester.
O
H
+
+
H O
fast
R C
R C
O
O
R'
R'
H O
resonance forms
H O
R C
R C
O
H O
R'
R C
O
R'
O
R'
resonance hybrid
H O
R C
O
R'
16-46
H O
O H
H
R C
slow, ratedetermining
step
O
R'
H+ O
O H
H
all fast
R C
OH
R'
O H
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16-47
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Figure 16.24
The metal-catalyzed hydrogenation of ethylene
H2C CH2 (g) + H2 (g)
16-48
H3C CH3 (g)
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Figure 16.24
16-49
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Tools of the Laboratory
Figure B16.1
16-50
Spectrophotometric monitoring of a reaction.
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Tools of the Laboratory
Figure B16.2
Conductometric monitoring
of a reaction
Figure B16.3
Manometric monitoring of a
reaction
16-51
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Figure B16.4
The widely separated amino acid groups that form an active site.
16-52
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Figure B16.5
16-53
Two models of enzyme action
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Figure B16.6
The increasing size of the Antarctic ozone hole
O2
16-54
2O
O + O2
O3 (ozone formation)
O + O3
2O2 (ozone breakdown)