Chapter 12
Chemical Kinetics
Chapter 12
Table of Contents
12.1
12.2
12.3
12.4
12.5
12.6
12.7
Reaction Rates
Rate Laws: An Introduction
Determining the Form of the Rate Law
The Integrated Rate Law
Reaction Mechanisms
A Model for Chemical Kinetics
Catalysis
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Section 12.1
Reaction Rates
Reaction Rate
• Change in concentration of a reactant or
product per unit time.
Rate =
=
concentration of A at time t2  concentration of A at time t1
t2  t1
 A
t
[A] means concentration of A in mol/L; A is the
reactant or product being considered.
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Section 12.1
Reaction Rates
The Decomposition of Nitrogen Dioxide
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Section 12.1
Reaction Rates
The Decomposition of Nitrogen Dioxide
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Section 12.1
Reaction Rates
Instantaneous Rate
• Value of the rate at a particular time.
• Can be obtained by computing the slope of a
line tangent to the curve at that point.
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Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
• Shows how the rate depends on the
concentrations of reactants.
• For the decomposition of nitrogen dioxide:
2NO2(g) → 2NO(g) + O2(g)
Rate = k[NO2]n:
 k = rate constant
 n = order of the reactant
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Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
Rate = k[NO2]n
• The concentrations of the products do not
appear in the rate law because the reaction rate
is being studied under conditions where the
reverse reaction does not contribute to the
overall rate.
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Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Law
Rate = k[NO2]n
• The value of the exponent n must be
determined by experiment; it cannot be written
from the balanced equation.
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Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Types of Rate Laws
• Differential Rate Law (rate law) – shows how
the rate of a reaction depends on
concentrations.
• Integrated Rate Law – shows how the
concentrations of species in the reaction
depend on time.
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Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Laws: A Summary
• Because we typically consider reactions only
under conditions where the reverse reaction is
unimportant, our rate laws will involve only
concentrations of reactants.
• Because the differential and integrated rate
laws for a given reaction are related in a well–
defined way, the experimental determination of
either of the rate laws is sufficient.
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Section 12.2
Atomic
Rate
Laws:
Masses
An Introduction
Rate Laws: A Summary
• Experimental convenience usually dictates
which type of rate law is determined
experimentally.
• Knowing the rate law for a reaction is important
mainly because we can usually infer the
individual steps involved in the reaction from the
specific form of the rate law.
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Section 12.3
The Mole
Determining
the Form of the Rate Law
• Determine experimentally the power to which
each reactant concentration must be raised in
the rate law.
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Section 12.3
The Mole
Determining
the Form of the Rate Law
Method of Initial Rates
• The value of the initial rate is determined for
each experiment at the same value of t as close
to t = 0 as possible.
• Several experiments are carried out using
different initial concentrations of each of the
reactants, and the initial rate is determined for
each run.
• The results are then compared to see how the
initial rate depends on the initial concentrations
of each of the reactants.
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Section 12.3
The Mole
Determining
the Form of the Rate Law
Overall Reaction Order
• The sum of the exponents in the reaction rate
equation.
Rate = k[A]n[B]m
Overall reaction order = n + m
k = rate constant
[A] = concentration of reactant A
[B] = concentration of reactant B
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Section 12.3
The Mole
Determining
the Form of the Rate Law
Concept Check
How do exponents (orders) in rate laws
compare to coefficients in balanced equations?
Why?
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Section 12.4
The Integrated Rate Law
Half-Life of Reactions
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Section 12.4
The Integrated Rate Law
First-Order
• Rate = k[A]
• Integrated:
ln[A] = –kt + ln[A]o
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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Section 12.4
The Integrated Rate Law
Plot of ln[N2O5] vs Time
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Section 12.4
The Integrated Rate Law
First-Order
• Half–Life:
t1
2
0.693
=
k
k = rate constant
• Half–life does not depend on the
concentration of reactants.
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Section 12.4
The Integrated Rate Law
Exercise
A first order reaction is 35% complete at the end
of 55 minutes. What is the value of k?
k = 7.8 x 10–3 min–1
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Section 12.4
The Integrated Rate Law
Second-Order
• Rate = k[A]2
• Integrated:
1
1
= kt +
A
 A 0
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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Section 12.4
The Integrated Rate Law
Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time
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Section 12.4
The Integrated Rate Law
Second-Order
• Half–Life:
t1
2
1
=
k  A 0
k = rate constant
[A]o = initial concentration of A
• Half–life gets longer as the reaction progresses
and the concentration of reactants decrease.
• Each successive half–life is double the
preceding one.
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Section 12.4
The Integrated Rate Law
Exercise
For a reaction aA  Products,
[A]0 = 5.0 M, and the first two half-lives are 25
and 50 minutes, respectively.
a) Write the rate law for this reaction.
rate = k[A]2
b) Calculate k.
k = 8.0 x 10-3 M–1min–1
c) Calculate [A] at t = 525 minutes.
[A] = 0.23 M
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Section 12.4
The Integrated Rate Law
Zero-Order
• Rate = k[A]0 = k
• Integrated:
[A] = –kt + [A]o
[A] = concentration of A at time t
k = rate constant
t = time
[A]o = initial concentration of A
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Section 12.4
The Integrated Rate Law
Plot of [A] vs Time
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Section 12.4
The Integrated Rate Law
Zero-Order
• Half–Life:
t1 =
2
 A 0
2k
k = rate constant
[A]o = initial concentration of A
• Half–life gets shorter as the reaction
progresses and the concentration of reactants
decrease.
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Section 12.4
The Integrated Rate Law
Concept Check
How can you tell the difference among 0th, 1st,
and 2nd order rate laws from their graphs?
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Section 12.4
The Integrated Rate Law
Rate Laws
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Section 12.4
The Integrated Rate Law
Summary of the Rate Laws
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Section 12.4
The Integrated Rate Law
Exercise
Consider the reaction aA  Products.
[A]0 = 5.0 M and k = 1.0 x 10–2 (assume the units
are appropriate for each case). Calculate [A] after
30.0 seconds have passed, assuming the
reaction is:
a) Zero order
b) First order
c) Second order
4.7 M
3.7 M
2.0 M
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Section 12.5
Reaction Mechanisms
Reaction Mechanism
• Most chemical reactions occur by a series
of elementary steps.
• An intermediate is formed in one step and
used up in a subsequent step and thus is
never seen as a product in the overall
balanced reaction.
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Section 12.5
Reaction Mechanisms
A Molecular Representation of the Elementary Steps in the
Reaction of NO2 and CO
NO2(g) + CO(g) → NO(g) + CO2(g)
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Section 12.5
Reaction Mechanisms
Elementary Steps (Molecularity)
• Unimolecular – reaction involving one
molecule; first order.
• Bimolecular – reaction involving the
collision of two species; second order.
• Termolecular – reaction involving the
collision of three species; third order.
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Section 12.5
Reaction Mechanisms
Rate-Determining Step
• A reaction is only as fast as its slowest
step.
• The rate-determining step (slowest step)
determines the rate law and the
molecularity of the overall reaction.
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Section 12.5
Reaction Mechanisms
Reaction Mechanism Requirements
• The sum of the elementary steps must give
the overall balanced equation for the
reaction.
• The mechanism must agree with the
experimentally determined rate law.
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Section 12.5
Reaction Mechanisms
Decomposition of N2O5
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Section 12.5
Reaction Mechanisms
Decomposition of N2O5
2N2O5(g)  4NO2(g) + O2(g)
Step 1: 2(N2O5
NO2 + NO3 )
(fast)
Step 2: NO2 + NO3 → NO + O2 + NO2 (slow)
Step 3: NO3 + NO → 2NO2
(fast)
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Section 12.5
Reaction Mechanisms
Concept Check
The reaction A + 2B  C has the following
proposed mechanism:
A+B
D
(fast equilibrium)
D+BC
(slow)
Write the rate law for this mechanism.
rate = k[A][B]2
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Section 12.6
A Model for Chemical Kinetics
Collision Model
• Molecules must collide to react.
• Main Factors:
 Activation energy, Ea
 Temperature
 Molecular orientations
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Section 12.6
A Model for Chemical Kinetics
Activation Energy, Ea
• Energy that must be overcome to produce
a chemical reaction.
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Section 12.6
A Model for Chemical Kinetics
Transition States and Activation Energy
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Section 12.6
A Model for Chemical Kinetics
Change in Potential Energy
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Section 12.6
A Model for Chemical Kinetics
For Reactants to Form Products
• Collision must involve enough energy to
produce the reaction (must equal or
exceed the activation energy).
• Relative orientation of the reactants must
allow formation of any new bonds
necessary to produce products.
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Section 12.6
A Model for Chemical Kinetics
The Gas Phase Reaction of NO and Cl2
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Section 12.6
A Model for Chemical Kinetics
Arrhenius Equation
k = Ae
 Ea / RT
A
Ea
R
T
=
=
=
=
frequency factor
activation energy
gas constant (8.3145 J/K·mol)
temperature (in K)
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Section 12.6
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
Ea  1 
ln(k ) =    + ln  A 
R T
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Section 12.6
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
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Section 12.6
A Model for Chemical Kinetics
Exercise
Chemists commonly use a rule of thumb that an
increase of 10 K in temperature doubles the
rate of a reaction. What must the activation
energy be for this statement to be true for a
temperature increase from 25°C to 35°C?
Ea = 53 kJ
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Section 12.7
Catalysis
Catalyst
• A substance that speeds up a reaction
without being consumed itself.
• Provides a new pathway for the reaction
with a lower activation energy.
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Section 12.7
Catalysis
Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a
Given Reaction
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Section 12.7
Catalysis
Effect of a Catalyst on the Number of Reaction-Producing
Collisions
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Section 12.7
Catalysis
Heterogeneous Catalyst
• Most often involves gaseous reactants
being adsorbed on the surface of a solid
catalyst.
• Adsorption – collection of one substance
on the surface of another substance.
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Section 12.7
Catalysis
Heterogeneous Catalysis
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Section 12.7
Catalysis
Heterogeneous Catalyst
1. Adsorption and activation of the reactants.
2. Migration of the adsorbed reactants on the
surface.
3. Reaction of the adsorbed substances.
4. Escape, or desorption, of the products.
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Section 12.7
Catalysis
Homogeneous Catalyst
• Exists in the same phase as the reacting
molecules.
• Enzymes are nature’s catalysts.
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Section 12.7
Catalysis
Homogeneous Catalysis
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