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Chapter 13
Rates of
Reactions
Contents and Concepts
Reaction Rates
1. Definition of Reaction Rate
2. Experimental Determination of Rate
3. Dependence of Rate on Concentration
4. Change of Concentration with Time
5. Temperature and Rate; Collision and
Transition-State Theories
6. Arrhenius Equation
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Reaction Mechanisms
7. Elementary Reactions
8. The Rate Law and the Mechanism
9. Catalysis
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Learning Objectives
Reaction Rates
1. Definition of a Reaction Rate
a. Define reaction rate.
b. Explain instantaneous rate and average
rate of a reaction.
c. Explain how the different ways of
expressing reaction rates are related.
d. Calculate average reaction rate.
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2. Experimental Determination of Rate
a. Describe how reaction rates may be
experimentally determined.
3. Dependence of Rate on Concentration
a. Define and provide examples of a rate law,
rate constant, and reaction order.
b. Determine the order of reaction from the
rate law.
c. Determine the rate law from initial rates.
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4. Change of Concentration with Time
a. Learn the integrated rate laws for firstorder, second-order, and zero-order
reactions.
b. Use an integrated rate law.
c. Define half-life of a reaction.
d. Learn the half-life equations for first-order,
second-order, and zero-order reactions.
e. Relate the half-life of a reaction to the rate
constant.
f. Plot kinetic data to determine the order of a
reaction.
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5. Temperature and Rate; Collision and
Transition-State Theories
a. State the postulates of collision theory.
b. Explain activation energy (Ea).
c. Describe how temperature, activation
energy, and molecular orientation influence
reaction rates.
d. State the transition-state theory.
e. Define activated complex.
f. Describe and interpret potential-energy
curves for endothermic and exothermic
reactions.
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6. Arrhenius Equation
a. Use the Arrhenius equation.
Reaction Mechanisms
7. Elementary Reactions
a. Define elementary reaction, reaction
mechanism, and reaction intermediate.
Determine the rate law from initial rates.
b. Write the overall chemical equation from a
mechanism.
c. Define molecularity.
d. Give examples of unimolecular, bimolecular,
and termolecular reactions.
e. Determine the molecularity of an elementary
reaction.
f. Write the rate equation for an elementary
reaction.
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8. The Rate Law and the Mechanism
a. Explain the rate-determining step of a
mechanism.
b. Determine the rate law from a mechanism
with an initial slow step.
c. Determine the rate law from a mechanism
with an initial fast, equilibrium step.
9. Catalysis
a. Describe how a catalyst influences the rate
of a reaction.
b. Indicate how a catalyst changes the
potential-energy curve of a reaction.
c. Define homogeneous catalysis and
heterogeneous catalysis.
d. Explain enzyme catalysis.
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Chemical kinetics is the study of reaction rates,
including how reaction rates change with varying
conditions and which molecular events occur
during the overall reaction.
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The following questions are explored in this
chapter:
• How is the rate of a reaction measured?
• Which conditions will affect the rate of a
reaction?
• How do you express the relationship of rate to
the variables affecting the rate?
• What happens on a molecular level during a
chemical reaction?
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Which conditions will affect the rate of a reaction?
Four variables affect the rate of reaction:
1. The concentrations of the reactants
2. The concentration of the catalyst
3. The temperature at which the reaction
occurs
4. The surface area of the solid reactant or
catalyst
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Concentrations of Reactants
Often the rate of reaction increases when the
concentration of a reactant is increased. In some
reactions, however, the rate is unaffected by the
concentration of a particular reactant, as long as it
is present at some concentration.
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Concentration of Catalyst
Hydrogen peroxide, H2O2, decomposes rapidly in
the presence of HBr, giving oxygen and water.
Some of the HBr is oxidized to give the orange Br2,
as can be seen above.
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Temperature at Which Reaction Occurs
Usually reactions speed up when the temperature
increases. It takes less time to boil an egg at sea
level than on a mountaintop, where water boils at a
lower temperature. Reactions during cooking go
faster at higher temperature.
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Each test tube contains
potassium permanganate,
KMnO4, and oxalic acid,
H2C2O4, at the same
concentrations. Permanganate
ion oxidizes oxalic acid to
CO2 and H2O.
Top: One test tube was placed
in a beaker of warm water
(40°C); the other was kept at
room temperature (20°C).
Bottom: After 10 minutes, the
test tube at 40°C showed a
noticeable reaction, whereas
the other did not.
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Surface Area of a Solid
Reactant or Catalyst
If a reaction involves a solid with
a gas or liquid, then the surface
area of the solid affects the
reaction rate. Because the
reaction occurs at the surface of
the solid, the rate increases with
increasing surface area. For
example, a wood fire burns
faster if the logs are chopped
into smaller pieces. Similarly, the
surface area of a solid catalyst is
important to the rate of reaction.
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Reaction rate is the increase in molar
concentration of product of a reaction per unit time
or the decrease in molar concentration of reactant
per unit time. The unit is usually mol/(L • s) or M/s.
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The rate of this reaction can be found by
measuring the concentration of O2 at various
times. Alternatively, the concentration of NO2 could
be measured. Both of these concentrations
increase with time. The rate could also be
determined by measuring the concentration of
N2O5, which would decrease over time.
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2N2O5(g)  4NO2(g) + O2(g)
We can express these changes in concentration
as follows. Square brackets mean molarity.
Δ[O 2 ]
Rate of formation of O2 =
Δt
Δ[NO 2 ]
Rate of formation of NO2 =
Δt
Δ[N 2 O 5 ]
Rate of decomposition of N2O5 = 
Δt
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We can relate these expressions by taking into
account the reaction stoichiometry.
Δ[O 2 ]
1 Δ[NO 2 ]
1 Δ[N2 O 5 ]


2
Δt
4 Δt
Δt
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These equations give the average rate over the
time interval Dt. As Dt decreases and approaches
zero, the equations give the instantaneous rate.
The next slides illustrate this relationship
graphically for the increase in concentration of O2.
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The average rate
is the slope of
the hypotenuse
of the triangle
formed.
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As Dt gets smaller and
approaches zero, the
hypotenuse becomes
a tangent line at that
point.
The slope of the
tangent line equals
the rate at that point.
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?
Peroxydisulfate ion oxidizes iodide ion
to triiodide ion, I3−. The reaction is
S2O82−(aq) + 3I−(aq) 
2SO42−(aq) + I3−(aq)
How is the rate of reaction that is
expressed as the rate of formation of
I3− related to the rate of reaction of I−?
Δ[I3 ]
1 Δ[I ]

Δt
3 Δt
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?
Calculate the average rate of formation
of O2 in the following reaction during
the time interval from 1200 s to 1800 s.
At 1200 s, [O2] = 0.0036 M; at 1800 s,
[O2] = 0.0048 M.
Δ[O 2 ]  0.0048  0.0036  M

Δt
1800  1200  s
0.0012 M

600 s
6
 2.0  10 M/s
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Concept Check 13.1
Shown here is a plot of the
concentration of a reactant D
versus time.
a. How do the instantaneous rates
at points A and B compare?
b. Is the rate for this reaction
constant at all points in time?
a. The slope at point A is greater than the slope at
point B, so the instantaneous rate at point A is
greater than the instantaneous rate at point B.
b. No. If it were, the graph would be linear.
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Rates are determined experimentally in a variety of
ways.
For example, samples can be taken and analyzed
from the reaction at several different intervals.
Continuously following the reaction is more
convenient. This can be done by measuring
pressure, as shown on the next slide, or by
measuring light absorbance when a color change
is part of the reaction.
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The reaction rate usually depends on the
concentration of one or more reactant. This
relationship must be determined by experiment.
This information is captured in the rate law, an
equation that relates the rate of a reaction to the
concentration of a reactant (and catalyst) raised to
various powers. The proportionality constant, k, is
the rate constant.
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For the generic reaction
aA + bB +cC  dD + eE
the rate law is
Rate = k[A]m[B]n[C]p
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The reaction order with respect to a specific
reactant is the exponent of that species in the
experimentally determined rate law.
The overall order of a reaction is the sum of the
orders of the reactant species from the
experimentally determined rate law.
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For the reaction
the rate law is
Rate = k[NO2][F2]
In this case, the exponent for both NO2 and F2 is 1 (which
does not need to be written).
We say the reaction is first order in each. The reaction is
second order overall.
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?
Hydrogen peroxide oxidizes iodide ion
in acidic solution:
H2O2(aq) + 3I−(aq) + 2H+(aq) 
I3−(aq) + 2H2O(l)
The rate law for the reaction is
Rate = k[H2O2][I−]
a.
b.
What is the order of reaction with
respect to each reactant species?
What is the overall order?
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H2O2(aq) + 3I−(aq) + 2H+(aq)  I3−(aq) + 2H2O(l)
Given: Rate = k[H2O2][I−]
a.
The reaction is first order in H2O2.
The reaction is first order in I−.
The reaction is zero order in H+.
b.
The reaction is second order overall.
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Concept Check 13.2
Consider the reaction Q + R  S + T and the rate law for the
reaction:
Rate = k[Q]0[R]2
a. You run the reaction three times, each time starting with [R]
= 2.0 M. For each run you change the starting
concentration of [Q]: run 1, [Q] = 0.0 M; run 2, [Q] = 1.0 M;
run 3, [Q] = 2.0 M. Rank the rate of the three reactions
using each of these concentrations.
b. The way the rate law is written in this problem is not typical
for expressions containing reactants that are zero order in
the rate law. Write the rate law in the more typical fashion.
a. The concentration of Q does not affect the
rate. The rate will depend only on [R].
b. Rate = k[R]2
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The rate law for a reaction must be determined
experimentally.
We will study the initial rates method of
determining the rate law. This method measures
the initial rate of reaction using various starting
concentrations, all measured at the same
temperature.
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Returning to the decomposition of N2O5, we have
the following data:
2N2O5(g)  2N2(g) + 5O2(g)
Initial N2O5
Concentration
Initial Rate of
Disappearance of N2O5
Experiment 1
1.0 × 10−2 M
4.8 × 10−6 M/s
Experiment 2
2.0 × 10−2 M
9.6 × 10−6 M/s
Note that when the concentration doubles from
experiment 1 to 2, the rate doubles.
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Examining the effect of doubling the initial
concentration gives us the order in that reactant.
When rate is multiplied by
m is
½
−1
1
0
2
1
4
2
In this case, when the concentration doubles, the
rate doubles, so m = 1.
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?
2NO(g) + O2(g)  2NO2(g)
Obtain the rate law and the rate
constant using the following data for
the initial rates of disappearance of NO:
Initial Rate of
Exp. [NO]0, M [O2]0, M Reaction of NO, M/s
1 0.0125
0.0253
0.0281
2
0.0250
0.0253
0.112
3
0.0125
0.0506
0.0561
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To find the order in NO, we will first identify two
experiments in which the concentration of O2
remains constant. Then, we will divide the rate
laws for those two experiments. Finally, we will
solve the equation to find the order in NO.
To find the order in O2, we will repeat this
procedure, this time choosing experiments in
which the concentration of NO remains constant.
Once the order in NO and the order in O2 are
known, data from any experiment can be
substituted into the rate law. The rate law can then
be solved for the rate constant k.
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[O2] remains constant in experiments 1 and 2.
The generic rate law is rate = k[NO]m[O2]n. We will
put the result from experiment 2 in the numerator
because its rate is larger than that of experiment 1.
M
0.112
m
n
s  0.0250 M  0.0253 M 
M 0.0125 M m 0.0253 M n
0.0281
s
4  2m
m2
The reaction is second order in NO.
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[NO] remains constant in experiments 1 and 3.
The generic rate law is rate = k[NO]m[O2]n. We will
put the result from experiment 3 in the numerator
because its rate is larger than that of experiment 1.
M
0.0.0561
m
n
s  0.0125 M  0.0506 M 
m
n
M




0.0125
M
0.0253
M
0.0281
s
2  2n
n 1
The reaction is first order in O 2 .
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It is also possible to find the order by examining
the data qualitatively.
Again choosing experiments 1 and 2, we note
that the concentration of NO is doubled and the
rate is quadrupled. That means the reaction is
second order in NO.
Choosing experiments 1 and 3, we note that the
concentration of O2 is doubled and the rate is
doubled. That means the reaction is first order in
O2.
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The rate law is rate = k[NO]2[O2].
Solving for k , we obtain
k
rate
NO 2 O 2 
Using data from experiment 1:
M
0.0281
3
7.11

10
s
k

2
2
M
s
0.0125 M  0.0253 M 
Using data from experiment 2: k = 7.08 × 103/M2  s
Using data from experiment 3: k = 7.10 × 103/M2  s
These values are the same to 2 significant figures.
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?
Iron(II) is oxidized to iron(III) by chlorine
in an acidic solution:
2Fe2+(aq) + Cl2(aq) 
2Fe3+ (aq) + 2Cl−(aq)
What is the reaction order with respect
to Fe2+, Cl2, and H+? What is the rate
law and the relative rate constant?
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The following data were collected. The rates given
are relative, not actual.
Exp. [Fe2+], M
[Cl2], M [H+], M Rate, M/s
1
0.0020
0.0020
1.0
1.0 × 10−5
2
0.0040
0.0020
1.0
2.0 × 10−5
3
0.0020
0.0040
1.0
2.0 × 10−5
4
0.0040
0.0040
1.0
4.0 × 10−5
5
0.0020
0.0020
0.5
2.0 × 10−5
6
0.0020
0.0020
0.1
1.0 × 10−4
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The generic rate law is
Rate = k[Fe2+]m[Cl2]n[H+]p
To find the order in Fe2+, we choose two
experiments in which both [Cl2] and [H+] remain
constant: 1 and 2, or 3 and 4.
Exp [Fe2+]
[Cl2]
[H+] Initial Rate
1 0.0020 0.0020
1.0 1.0 × 10-5
2 0.0040 0.0020
1.0 2.0 × 10-5
3 0.0020 0.0040
1.0 2.0 × 10-5
4 0.0040 0.0040
1.0 4.0 × 10-5
In each case, the concentration of Fe2+ doubles
and the rate doubles.
The reaction is first order in Fe2+; m = 1.
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To find the order in Cl2, we choose two
experiments in which both [Fe2+] and [H+] remain
constant: 1 and 3, or 2 and 4.
Exp [Fe2+]
[Cl2]
[H+] Initial Rate
1 0.0020 0.0020
1.0 1.0 × 10-5
3 0.0020 0.0040
1.0 2.0 × 10-5
2 0.0040 0.0020
1.0 2.0 × 10-5
4 0.0040 0.0040
1.0 4.0 × 10-5
In each case, the concentration of Cl2 doubles and
the rate doubles.
The reaction is first order in Cl2; n = 1.
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To find the order in H+, we choose two
experiments in which both [Fe2+] and [Cl2] remain
constant: 1 and 5, 1 and 6, or 5 and 6.
1
0.0020
0.0020
1.0 1.0 × 10-5
5
0.0020
0.0020
0.5 2.0 × 10-5
From experiment 5 to experiment 1, the
concentration doubles while the rate is halved.
1
0.0020
0.0020
1.0 1.0 × 10-5
6
0.0020
0.0020
0.1 1.0 × 10-4
From experiment 6 to experiment 1, the
concentration is increased 10 times, the rate is cut
to one-tenth of the original rate.
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5
0.0020
0.0020
0.5 2.0 × 10−5
6
0.0020
0.0020
0.1 1.0 × 10−4
From experiment 6 to experiment 5, the
concentration is increased 5 times; the rate is cut
to one-fifth of the original rate.
Each of these describes the order in H+ as p = −1.
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The rate law is rate = k[Fe2+][Cl2][H+]−1 .
Another way to express this is
Rate 
k Fe2  Cl2 
H 
Using data from experiment 1:
M
1.0  10
1.0 M  2.50
rate H
s
k


2
Fe  Cl2   0.002 M  0.002 M  M  s
5
The same value for k is obtained using each
experiment.
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Note that the units on the rate constant are specific
to the overall order of the reaction.
• For a zero-order reaction, the unit is M/s or
mol/(L  s).
• For a first-order reaction, the unit is 1/s or s−1.
• For a second-order reaction, the unit is 1/(M  s)
or L/(mol  s).
If you know the rate constant, you can deduce the
overall rate of reaction.
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Concept Check 13.3
Rate laws are not restricted to chemical systems; they are used to help
describe many “everyday” events. For example, a rate law for tree growth
might look something like this:
Rate of growth = (soil type)w(temperature)x(light)y(fertilizer)z
In this equation, like chemical rate equations, the exponents need to be
determined by experiment. (Can you think of some other factors?)
a. Say you are a famous physician trying to determine the factors that
influence the rate of aging in humans. Develop a rate law, like the one
above, that would take into account at least four factors that affect the
rate of aging.
b. Explain what you would need to do in order to determine the exponents
in your rate law.
c. Consider smoking to be one of the factors in your rate law. You conduct
an experiment and fi nd that a person smoking two packs of cigarettes a
day quadruples (43) the rate of aging over that of a one-pack-a-day
smoker. Assuming that you could hold all other factors in your rate law
constant, what would be the exponent of the smoking term in your rate
law?
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a. These factors might include things such as
genetics, diet, exercise, weight, and stress.
b. You would need to run controlled studies
quantifying each of the factors and measures of
aging.
c. When smoking doubles, aging increases
fourfold. That is a second-order reaction, so
m = 2.
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The rate law tells us the relationship between the
rate and the concentrations of reactants and
catalysts.
To find concentrations at various times, we need to
use the integrated rate law. The form used
depends on the order of reaction in that reactant.
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?
Cyclopropane is used as an anesthetic.
The isomerization of cyclopropane to
propene is a first-order reaction with a
rate constant of 9.2/s. If an initial
sample of cyclopropane has a
concentration of 6.00 M, what will the
cyclopropane concentration be after
1.00 s?
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The rate law is
Rate = k[cyclopropane]
k = 9.2/s
[A]0 = 6.00 M
t = 1.00 s
[A]t = ?
The reaction is first order, so the integrated rate law
is
A
 t
ln
 A 0
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  kt
13 | 59
 A t
ln
 A 0
9.2

 1.00 s   9.2
s
A t  e 9.2  1.01  10 4
A 0
4
At  1.0110 6.00 M 
A t  6.110
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4
M
13 | 60
The half-life of a reaction, t½, is the time it takes for
the reactant concentration to decrease to one-half
of its initial value.
By substituting ½[A]0 for [A]t, we solve the
integrated rate law for the special case of t = t½.
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?
Ammonium nitrite is unstable because
ammonium ion reacts with nitrite ion to
produce nitrogen:
NH4+(aq) + NO2−(aq)  N2(g) + 2H2O(l)
In a solution that is 10.0 M in NH4+, the
reaction is first order in nitrite ion (at low
concentrations), and the rate constant at
25°C is 3.0 × 103/s.
What is the half-life of the reaction?
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The rate is first order in nitrite, NO2−:
k = 3.0 × 10−3/s
For first-order reactions:
kt½ = 0.693
0.693
0.693
t1/2 

3
3.0  10
k
s
t½ = 2.3 × 102 s = 3.9 min
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Let’s look again at the integrated rate laws:
y = mx + b
y = mx + b
y = mx + b
In each case, the rate law is in the form of y = mx +
b, allowing us to use the slope and intercept to find
the values.
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Zero Order : [A]t   kt  [A]0
For a zero-order reaction, a plot of [A]t versus t is
linear. The y-intercept is [A]0.
First Order : ln [A]t   kt  ln [A]0
For a first-order reaction, a plot of ln[A]t versus t is
linear. The graph crosses the origin (b = 0).
1
1
Second Order :
 kt 
[A]t
[A]0
For a second-order reaction, a plot of 1/[A]t versus t
is linear. The y-intercept is 1/[A]0.
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13 | 66
Graphs of concentration and time can also be
used to determine the order of reaction in that
reactant.
•
•
•
For zero-order reactions, [A] versus t is
linear.
For first-order reactions, ln[A] versus t is
linear.
For second order reactions, 1/[A] versus t is
linear.
This is illustrated on the next slide for the
decomposition of NO2.
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13 | 67
Left: The plot of ln[NO2] versus t is not linear, so
the reaction is not first order.
Right: The plot of 1/[NO2] versus t is linear, so the
reaction is second order in NO2.
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13 | 68
Concept Check 13.4
A reaction believed to be either fi rst or second order
has a half-life of 20 s at the beginning of the reaction
but a half-life of 40 s sometime later. What is the
order of the reaction?
[A]0
Zero Order : t1/2 
2k
0.693
First Order : t1/2 
k
1
Second Order : t1/2 
k [A]t
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The initial concentration
decreases in each time
interval. The only equation
that results in a larger value
for t½ is the second-order
equation.
The reaction is second
order.
13 | 69
Rate Constant and Temperature
The rate constant depends strongly on
temperature. How can we explain this
relationship?
Collision theory assumes that reactant molecules
must collide with an energy greater than some
minimum value and with the proper orientation.
The minimum energy is called the activation
energy, Ea.
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13 | 70
We will now explore the effect of a temperature
increase on each of the three requirements for a
reaction to occur.
The rate constant can be given by the equation
k = Zfp
where
Z = collision frequency
f = fraction of collisions with the minimum energy
p = orientation factor
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13 | 71
Certainly, Z will increase with temperature, as the
average velocity of the molecules increases with
temperature.
However, this factor alone cannot explain the
dramatic effect of temperature changes.
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13 | 72
The fraction of molecular collisions having the
minimum energy required is given by f:
f e

Ea
RT
1

e
Ea
RT
Now we can explain the dramatic impact of
temperature. The fraction of collisions having the
minimum energy increases exponentially with
temperature.
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13 | 73
The reaction rate also depends on p, the proper
orientation for the collision. This factor is
independent of temperature.
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13 | 74
Importance of Molecular Orientation in
the Reaction of NO and Cl2
(A) NO approaches with its N atom toward Cl2,
and an NOCl bond forms. Also, the angle of
approach is close to that in the product NOCl.
(B) NO approaches with its O atom toward Cl2. No
NOCl bond can form, so NO and Cl2 collide and
then fly apart.
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13 | 75
Transition-state theory explains the reaction
resulting from the collision of two molecules in
terms of an activated complex (transition state),
an unstable grouping of atoms that can break up to
form products.
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13 | 76
The potential energy diagram for a reaction
visually illustrates the changes in energy that
occur.
The next slide shows the diagram for the reaction
NO + Cl2  NOCl2‡  NOCl + Cl
where NOCl2‡ indicates the activated complex.
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13 | 78
The reaction of NO with Cl2 is an endothermic
reaction.
The next slide shows the curve for a generic
exothermic reaction.
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13 | 79
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13 | 80
?
Sketch a potential energy diagram for
the decomposition of nitrous oxide.
N2O(g)  N2(g) + O(g)
The activation energy for the forward
reaction is 251 kJ; DHo = +167 kJ.
Label your diagram appropriately.
What is the reverse activation energy?
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13 | 81
E
n
e
r
g
y
Ea(reverse)
= 84 kJ
Ea= 251 kJ
N2 + O
Products
p
e
r
m
o
l
DH = 167 kJ
N2O
Reactants
Progress of reaction
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13 | 82
Rate constants for most chemical reactions closely
follow an equation in the form
k  Ae

Ea
RT
This is called the Arrhenius equation. A is the
frequency factor, a constant.
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13 | 83
A more useful form of the Arrhenius equation is
 k2
ln 
 k1
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 Ea  1
1
 
 

 R  T1 T2 
13 | 84
?
In a series of experiments on the
decomposition of dinitrogen pentoxide,
N2O5, rate constants were determined
at two different temperatures:
•
At 35°C, the rate constant was
1.4 × 10−4/s.
•
At 45°C, the rate constant was
5.0 × 10−4/s.
What is the activation energy?
What is the value of the rate constant
at 55°C?
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13 | 85
This is actually two problems.
First, we will use the Arrhenius equation to find Ea.
Then, we will use the Arrhenius equation with Ea to
find the rate constant at a new temperature.
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13 | 86
T1 = 35°C = 308 K
k1 = 1.4 × 10−4/s
 k2
ln 
 k1
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T2 = 45°C = 318 K
k2 = 5.0 × 10−4/s
 Ea  1
1
 
 

 R  T1 T2 
13 | 87
 5.0  104 /s 
Ea
1 
 1
ln 




4
J
1.4

10
/s
308
K
318
K



 8.315
mol  K
We will solve the left side and rearrange the right side.




10 K


1.273  E a


J 
(318 K)(308 K) 
  8.315
mol  K 


J 

1.273 8.315
(318 K)(308 K)
mol  K 

Ea 
10 K
Ea  1.0  105 J/mol
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13 | 88
T1 = 35°C = 308 K
k1 = 1.4 × 10−4/s
Ea = 1.04 × 105 J/mol
 k2
ln 
 k1
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T2 = 55°C = 328 K
k2 = ?
 Ea  1
1
 
 

 R  T1 T2 
13 | 89
J
1.04  10
k2
1 


 1
mol
ln 





4
J
 1.4  10 /s  8.315
 308 K 328 K 
mol  K
J
5
1.04  10
k
20 K




2
mol
ln 




4
J
 1.4  10 /s  8.315
 308 K  328 K 
mol  K
k2


ln 
 2.476

4
 1.4  10 /s 
k2
 11.90
4
1.4  10 /s
5
k2  1.7  103 /s
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13 | 90
Reaction Mechanism
A balanced chemical equation is a description of
the overall result of a chemical reaction. However,
what actually happens on a molecular level may
be more involved than what is represented by this
single equation. For example, the reaction may
take place in several steps. That set of steps is
called the reaction mechanism.
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13 | 91
Each step in the reaction mechanism is called an
elementary reaction and is a single molecular
event.
The set of elementary reactions, which when
added give the balanced chemical equation, is
called the reaction mechanism.
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13 | 92
Because an elementary reaction is an actual
molecular event, the rate of an elementary reaction
is proportional to the concentration of each
reactant molecule. This means we can write the
rate law directly from an elementary reaction.
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13 | 93
The slowest step in the reaction mechanism is
called the rate-determining step (RDS). The rate
law for the RDS is the rate law for the overall
reaction.
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13 | 94
Rate Law and Reaction Mechanism
A reaction mechanism cannot be directly
observed.
We can, however, determine the rate law by
experiment and decide if the reaction mechanism
is consistent with that rate law.
The rate of a reaction is determined completely by
the slowest step, the rate-determining step.
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13 | 95
For example, the reaction of NO2 with F2 is
believed to occur in the following elementary
steps:
NO2 + F2  NO2F + F
(slow step)
F + NO2  NO2F
(fast step)
The overall reaction is
2NO2 + F2  2NO2F
What is the rate law for this mechanism?
Rate = k[NO2][F2]
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13 | 96
A reaction intermediate is a species produced
during a reaction that does not appear in the net
equation because it reacts in a subsequent step in
the mechanism.
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13 | 97
For example, we look again at the mechanism for
the reaction of NO2 with F2:
NO2 + F2  NO2F + F
(Slow step)
F + NO2  NO2F
(Fast step)
The overall reaction is
2NO2 + F2  2NO2F
F does not appear in the overall reaction equation.
It is produced in the first step and used in the
second step. Thus F is a reaction intermediate.
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13 | 98
Catalysis
Catalysis is an increase in the rate of reaction that
results from the addition of a catalyst.
Enzymes are biological catalysts.
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13 | 99
A catalyst is not consumed in a reaction. Rather, it
is present in the beginning, is used in one step,
and is produced again in a subsequent step. As a
result, the catalyst does not appear in the overall
reaction equation.
A catalyst increases the reaction rate by providing
an alternative reaction path with a lower activation
energy. When Ea is reduced, k increases
exponentially.
This relationship is illustrated on the potential
energy diagram for the decomposition of ozone.
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13 | 100
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13 | 101
Consider the following reaction:
NO2(g) + CO(g)  NO(g) + CO2(g)
At 500 K, the reaction mechanism is believed to
involve two elementary reactions:
NO2(g) + NO2(g)  NO3(g) + NO(g) (Slow step)
NO3(g) + CO(g)  NO2(g) + CO2(g)
The overall reaction equation is
NO2(g) + CO(g)  NO(g) + CO2(g)
NO3 is a reaction intermediate: It is produced in
the first step and used in the second step. There is
no catalyst.
Rate = k[NO2][NO2] = k[NO2]2
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?
Write the rate law for the first step in
the mechanism for the decomposition
of ozone.
Cl(g) + O3(g)  ClO(g) + O2(g)
ClO(g) + O(g)  Cl(g) + O2(g)
Rate = k[Cl][O3]
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13 | 103
?
Chlorofluorocarbons, such as CCl2F2,
decompose in the stratosphere as a
result of their irradiation with the shortwavelength ultraviolet light present at
those altitudes. The decomposition
yields chlorine atoms that catalyze the
decomposition of ozone in the
presence of oxygen atoms (present in
the stratosphere) to give oxygen
molecules.
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13 | 104
The mechanism of the decomposition is
Cl(g) + O3(g)  ClO(g) + O2(g) (Slow step)
ClO(g) + O(g)  Cl(g) + O2(g)
What is the overall reaction equation?
O3(g) + O(g)  2O2(g)
Is there a reaction intermediate? If so, what is it?
Yes. ClO.
Is there a catalyst? If so, what is it?
Yes. Cl.
What is the rate law for the overall reaction?
Rate = k[Cl][O3]
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?
The decomposition of hydrogen
peroxide is catalyzed by iodide ion. The
reaction mechanism is thought to be
H2O2(aq) + I−(aq)  H2O(l) + IO−(aq)
IO−(aq) + H2O2(aq)  H2O(l) + O2(g) + I−(aq)
At 25°C, the first step is slow relative to
the second step. What is the rate law
predicted by the reaction mechanism? Is
there a reaction intermediate?
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H2O2(aq) + I−(aq)  H2O(l) + IO−(aq)
IO−(aq) + H2O2(aq)  H2O(l) + O2(g) + I−(aq)
Overall reaction equation:
2H2O2(aq)  2H2O(l) + O2(g)
I− was present in the beginning, used in the first
step, and produced in the second step. It is a
catalyst.
IO− was produced in the first step and used in the
second step. It is an intermediate.
Rate = k[H2O2][I−]
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IO− is the intermediate.
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