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1. Chapter 12
Chemical Kinetics

2. Section 12.1
Reaction Rates
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Reaction Rate
Change in concentration of a reactant or product
per unit time.
[A] means concentration of A in mol/L; A is the
reactant or product being considered.
21
21
concentration of A at time concentration of A at time
Rate =

A
=




tt
tt
t

3. Section 12.1
Reaction Rates
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The Decomposition of Nitrogen Dioxide

4. Section 12.1
Reaction Rates
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The Decomposition of Nitrogen Dioxide

5. Section 12.1
Reaction Rates
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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.

6. Section 12.2
Rate Laws: An Introduction
Rate Law
Shows how the rate depends on the concentrations of
reactants.
For the decomposition of nitrogen dioxide:
2NO
2(g) → 2NO(g) + O
2(g)
Rate = k[NO
2]
n
:
k= rate constant
n= order of the reactant
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7. Section 12.2
Rate Laws: An Introduction
Rate Law
Rate = k[NO
2]
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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8. Section 12.2
Rate Laws: An Introduction
Rate Law
Rate = k[NO
2]
n
The value of the exponent nmust be determined by
experiment; it cannot be written from the balanced
equation.
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9. Section 12.2
Rate Laws: 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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10. Section 12.2
Rate Laws: 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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11. Section 12.2
Rate Laws: 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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12. Section 12.3
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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13. Section 12.3
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 tas 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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14. Section 12.3
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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15. Section 12.3
Determining the Form of the Rate Law
How do exponents (orders) in rate laws compare to
coefficients in balanced equations?
Why?
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CONCEPT CHECK!

16. 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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17. Section 12.4
The Integrated Rate Law
Plot of ln[N
2O
5] vs Time
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18. Section 12.4
The Integrated Rate Law
First-Order
Time required for a reactant to reach half its original
concentration
Half–Life:
k= rate constant
Half–life does not depend on the concentration of
reactants.
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2
0.693
= t
k

19. Section 12.4
The Integrated Rate Law
A first order reaction is 35% complete at the end of
55 minutes. What is the value of k?
k = 7.8 ×10
–3
min
–1
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20. Section 12.4
The Integrated Rate Law
Second-Order
Rate = k[A]
2
Integrated:
[A] = concentration of A at time t
k= rate constant
t= time
[A]
o= initial concentration of A
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0
11
= +
AA
kt

21. Section 12.4
The Integrated Rate Law
Plot of ln[C
4H
6] vs Time and Plot of 1/[C
4H
6] vs Time

22. Section 12.4
The Integrated Rate Law
Second-Order
Half–Life:
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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1
2
0
1
=
A
t
k

23. Section 12.4
The Integrated Rate Law
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 ×10
-3
M
–1
min
–1
c)Calculate [A] at t= 525 minutes.
[A] = 0.23 M
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EXERCISE!

24. 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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25. Section 12.4
The Integrated Rate Law
Plot of [A] vs Time
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26. Section 12.4
The Integrated Rate Law
Zero-Order
Half–Life:
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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0
1
2
A
=
2
t
k

27. Section 12.4
The Integrated Rate Law
How can you tell the difference among 0
th
, 1
st
, and 2
nd
order rate laws from their graphs?
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CONCEPT CHECK!

28. Section 12.4
The Integrated Rate Law
Rate Laws
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29. Section 12.4
The Integrated Rate Law
Summary of the Rate Laws
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30. Section 12.4
The Integrated Rate Law
Consider the reaction aA Products.
[A]
0= 5.0 Mand k= 1.0 ×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
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4.7 M
3.7 M
2.0 M
EXERCISE!

31. 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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32. Section 12.5
Reaction Mechanisms
A Molecular Representation of the Elementary Steps in the Reaction
of NO
2and CO
NO
2(g) + CO(g) → NO(g) + CO
2(g)
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33. 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. Very rare.
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34. 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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35. 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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36. Section 12.5
Reaction Mechanisms
Decomposition of N
2O
5
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37. Section 12.5
Reaction Mechanisms
Decomposition of N
2O
5
2N
2O
5(g) 4NO
2(g) + O
2(g)
Step 1:N
2O
5 NO
2+ NO
3 (fast)
Step 2:NO
2+ NO
3→ NO + O
2+ NO
2(slow)
Step 3:NO
3+ NO → 2NO
2 (fast)
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2( )

38. Section 12.5
Reaction Mechanisms
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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CONCEPT CHECK!

39. Section 12.6
A Model for Chemical Kinetics
Collision Model
Molecules must collide to react.
Main Factors:
Activation energy, E
a
Temperature
Molecular orientations
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40. Section 12.6
A Model for Chemical Kinetics
Activation Energy, E
a
Energy that must be overcome to produce a chemical
reaction.
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41. Section 12.6
A Model for Chemical Kinetics
Transition States and Activation Energy
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42. Section 12.6
A Model for Chemical Kinetics
Change in Potential Energy
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43. 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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44. Section 12.6
A Model for Chemical Kinetics
The Gas Phase Reaction of NO and Cl
2
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45. Section 12.6
A Model for Chemical Kinetics
Arrhenius Equation
A = frequency factor
E
a= activation energy
R = gas constant (8.3145 J/K·mol)
T = temperature (in K)
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=

a
E RT
k Ae

46. Section 12.6
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
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a
E1
ln( ) = ln
RT



k + A

47. Section 12.6
A Model for Chemical Kinetics
Linear Form of Arrhenius Equation
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48. Section 12.6
A Model for Chemical Kinetics
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 energybe for this
statement to be true for a temperature increase from
25°C to 35°C?
E
a= 53 kJ
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EXERCISE!

49. 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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50. Section 12.7
Catalysis
Energy Plots for a Catalyzed and an Uncatalyzed Pathway
for a Given Reaction
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51. Section 12.7
Catalysis
Effect of a Catalyst on the Number of Reaction-Producing
Collisions
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52. 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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53. Section 12.7
Catalysis
Heterogeneous Catalysis
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54. 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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55. Section 12.7
Catalysis
Homogeneous Catalyst
Exists in the same phase as the reacting molecules.
Enzymes are nature’s catalysts.
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56. Section 12.7
Catalysis
Homogeneous Catalysis
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