Chapter
12
Chemical Kinetics
Chemistry 4th Edition
McMurry/Fay
Dr. Paul Charlesworth
Michigan Technological University
Reaction Rates
•
01
Reaction Rate: The change in the concentration of
a reactant or a product with time (M/s).
Reactant  Products
aA  bB
[A]
Rate  
t
Prentice Hall ©2004
[B]
Rate 
t
Chapter 12
Slide 2
Reaction Rates
•
02
Consider the decomposition of N2O5 to give NO2
and O2:
2 N2O5(g)
4 NO2(g) + O2(g)
Prentice Hall ©2004
Chapter 12
Slide 3
Reaction Rates
Prentice Hall ©2004
Chapter 12
03
Slide 4
Rate Law & Reaction Order
01
•
Rate Law: Shows the relationship of the rate of a
reaction to the rate constant and the concentration
of the reactants raised to some powers.
•
For the general reaction: aA + bB  cC + dD
rate = k[A]x[B]y
x and y are NOT the stoichiometric coefficients.
• k = the rate constant
•
Prentice Hall ©2004
Chapter 12
Slide 5
Rate Law & Reaction Order
02
•
Reaction Order: The sum of the powers to which
all reactant concentrations appearing in the rate
law are raised.
•
Reaction order is determined experimentally:
1.
By inspection.
2.
From the slope of a log(rate) vs. log[A] plot.
Prentice Hall ©2004
Chapter 12
Slide 6
Rate Law & Reaction Order
• Determination
03
by inspection:
aA + bB  cC + dD
Rate = R = k[A]x[B]y
Use initial rates (t = 0)
 [ A]2 
R2 k [ A] [ B ]


 
R1 k [ A] [ B ]
 [ A]1 
x
2
x
1
R2  [ A]2 

 
R1  [ A]1 
Prentice Hall ©2004
y
2
y
1
x
 [ B ]2 


 [ B ]1 
y
x
if [B] 2  [B]1
Chapter 12
Slide 7
Rate Law & Reaction Order
•
04
Determination by plot of a log(rate) vs. log[A]:
aA + bB  cC + dD
Rate = R = k[A]x[B]y
Log(R) = log(k) + x·log[A] + y·log[B]
= const + x·log[A] if [B] held constant
Prentice Hall ©2004
Chapter 12
Slide 8
Rate Law & Reaction Order
05
•
The reaction of nitric oxide with hydrogen at
1280°C is: 2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)
•
From the following data determine the rate law and
rate constant.
Experiment
1
2
3
Prentice Hall ©2004
[NO]
5.0 x 10–3
10.0 x 10–3
10.0 x 10–3
[H2]
2.0 x10–3
2.0 x 10–3
4.0 x 10–3
Chapter 12
Initial Rate (M/s)
1.3 x 10–5
5.0 x 10–5
10.0 x 10–5
Slide 9
Rate Law & Reaction Order
•
06
The reaction of peroxydisulfate ion (S2O82-) with
iodide ion (I-) is:
S2O82-(aq) + 3 I-(aq)  2 SO42-(aq) + I3-(aq)
•
From the following data, determine the rate law and
rate constant.
Experiment
1
2
3
Prentice Hall ©2004
[S2O82-]
0.080
0.080
0.16
[I-]
0.034
0.017
0.017
Chapter 12
Initi al Rate (M/s)
2.2 x 10-4
1.1 x 10-4
2.2 x 10-4
Slide 10
Rate Law & Reaction Order
•
07
Rate Constant: A constant of proportionality
between the reaction rate and the concentration of
reactants.
rate  [Br2]
rate = k[Br2]
Prentice Hall ©2004
Chapter 12
Slide 11
First-Order Reactions
•
01
First Order: Reaction rate depends on the reactant
concentration
raised to first power.
Rate = k[A]
A
Rate = t
Prentice Hall ©2004
Chapter 12
Slide 12

First-Order Reactions
•
Using calculus we obtain the integrated rate equation:
[A] 
ln t   kt
[A]0 
•
02
or
ln[A]t  ln[A]o   kt
Plotting ln[A]t against t gives a straight line of slope –k.
An alternate expression is:
kt
[A]t  [A]0 e
Prentice Hall ©2004
exponential decay law
Chapter 12
Slide 13
First-Order Reactions
•
03
Identifying First-Order Reactions:
Prentice Hall ©2004
Chapter 12
Slide 14
First-Order Reactions
•
04
Show that the decomposition of N2O5 is first order
and calculate the rate constant.
Prentice Hall ©2004
Chapter 12
Slide 15
First-Order Reactions
•
06
Half-Life:
Time for reactant
concentration to
decrease by half
its original value.
ln2
t1 
k
2
Prentice Hall ©2004
Chapter 12
Slide 16
Second-Order Reactions
•Second-Order
Reaction:
A  Products
A + B  Products
Rate = k[A]2
•These
01
Rate = k[A][B]
can then be integrated to give:
1
1
 kt 
[A]t
[A]0
Prentice Hall ©2004
Chapter 12
Slide 17
Second-Order Reactions
•
02
Half-Life:
Time for reactant
concentration to
decrease by half
its original value.
1
t1 
k[A] 0
2
Prentice Hall ©2004
Chapter 12
Slide 18
Second-Order Reactions
•
03
Iodine atoms combine to form molecular iodine in the gas
phase.
I(g) + I(g)  I2(g)
•
This reaction follows second-order kinetics and
k = 7.0 x 10–1 M–1s–1 at 23°C. (a) If the initial concentration
of I was 0.086 M, calculate the concentration after 2.0 min.
(b) Calculate the half-life of the reaction if the initial
concentration of I is 0.60 M and if it is 0.42 M.
Prentice Hall ©2004
Chapter 12
Slide 19
Reaction Mechanisms
•
01
A reaction mechanism
is a sequence of
molecular events, or
reaction steps, that
defines the pathway
from reactants to
products.
Prentice Hall ©2004
Chapter 12
Slide 20
Reaction Mechanisms
02
•
Single steps in a mechanism are called elementary
steps (reactions).
•
An elementary step describes the behavior of
individual molecules.
•
An overall reaction describes the reaction
stoichiometry.
Prentice Hall ©2004
Chapter 12
Slide 21
Reaction Mechanisms
03
•
NO2(g) + CO(g)  NO(g) + CO2(g)
Overall
•
NO2(g) + NO2(g)  NO(g) + NO3(g)
Elementary
•
NO3(g) + CO(g)  NO2(g) + CO2(g)
Elementary
•
The chemical equation for an elementary reaction
is a description of an individual molecular event
that involves the breaking and/or making of
chemical bonds.
Prentice Hall ©2004
Chapter 12
Slide 22
Reaction Mechanisms
•
Molecularity: is the number of molecules (or
atoms) on the reactant side of the chemical
equation.
•
Unimolecular: Single reactant molecule.
Prentice Hall ©2004
Chapter 12
04
Slide 23
Reaction Mechanisms
•
Bimolecular: Two reactant molecules.
•
Termolecular: Three reactant molecules.
Prentice Hall ©2004
Chapter 12
05
Slide 24
Reaction Mechanisms
•
06
Determine the overall reaction, the reaction
intermediates, and the molecularity of each
individual elementary step.
Prentice Hall ©2004
Chapter 12
Slide 25
Rate Laws and Reaction
Mechanisms
01
Rate law for an overall reaction must be
determined experimentally.
• Rate law for elementary step follows from its
molecularity.
•
Prentice Hall ©2004
Chapter 12
Slide 26
Rate Laws and Reaction
Mechanisms
02
•
The rate law of each elementary step follows its
molecularity.
•
The overall reaction is a sequence of elementary
steps called the reaction mechanism.
•
Therefore, the experimentally observed rate law for
an overall reaction must depend on the reaction
mechanism.
Prentice Hall ©2004
Chapter 12
Slide 27
Rate Laws and Reaction
Mechanisms
03
•
The slowest elementary step in a multistep reaction
is called the rate-determining step.
•
The overall reaction cannot occur faster than the
speed of the rate-determining step.
•
The rate of the overall reaction is therefore
determined by the rate of the rate-determining step.
Prentice Hall ©2004
Chapter 12
Slide 28
Rate Laws and Reaction
Mechanisms
Prentice Hall ©2004
Chapter 12
04
Slide 29
Rate Laws and Reaction
Mechanisms
•
05
The following reaction has a second-order rate law:
H2(g) + 2 ICl(g)  I2(g) + 2 HCl(g)
Rate = k[H2][ICl]
•
Devise a possible mechanism.
•
The following substitution reaction has a first-order
rate law:
Co(CN)5(H2O)2–(aq) + I–  Co(CN)5I3–(aq) + H2O(l)
Rate = k[Co(CN)5(H2O)2–]
•
Suggest a mechanism in accord with the rate law.
Prentice Hall ©2004
Chapter 12
Slide 30
The Arrhenius Equation
01
•
Collision Theory: A bimolecular reaction occurs
when two correctly oriented molecules collide with
sufficient energy.
•
Activation Energy (Ea): The potential energy
barrier that must be surmounted before
reactants can be converted to products.
Prentice Hall ©2004
Chapter 12
Slide 31
The Arrhenius Equation
Prentice Hall ©2004
Chapter 12
02
Slide 32
The Arrhenius Equation
Prentice Hall ©2004
Chapter 12
03
Slide 33
The Arrhenius Equation
•
04
This relationship is summarized by the Arrhenius
equation.
E

 a RT


k  Ae
•
Taking logs and rearranging, we get:
 Ea 1 
ln k  
ln A


 
R T
Prentice Hall ©2004
Chapter 12
Slide 34
The Arrhenius Equation
05
Temp
k
(°C) (M-1 s-1)
283 3.52e-7
356
393
3.02e5
2.19e-4
427
1.16e-3
508
3.95e-2
Prentice Hall ©2004
Chapter 12
Slide 35
The Arrhenius Equation
07
The
The second-order
second-order rate
rate constant
constant for
for the
the
decomposition
decomposition of
of nitrous
nitrous oxide
oxide (N
(N22O)
O) into
into nitrogen
nitrogen
molecule
molecule and
and oxygen
oxygen atom
atom has
has been
been measured
measured at
at
different
different temperatures:
temperatures:
-1
-1 -1
-1
Determine
Determine graphically
graphically
the
the activation
activation energy
energy
for
for the
the reaction.
reaction.
Prentice Hall ©2004
kk (M
(M ss ))
-3
-3
1.87x10
1.87x10
0.0113
0.0113
0.0569
0.0569
0.244
0.244
Chapter 12
tt (°C)
(°C)
600
600
650
650
700
700
750
750
Slide 36
The Arrhenius Equation
•
09
A simpler way to use this is by comparing the rate
constant at just two temperatures:
Ea  1 1 
k2
ln  
  
k1
R T2 T1 
•
If the rate of a reaction doubles by increasing the
temperature by 10°C from 298.2 K to 308.2 K, what
is the activation energy of the reaction?

Prentice Hall ©2004
Chapter 12
Slide 37
Catalysis
Prentice Hall ©2004
01
Chapter 12
Slide 38
Catalysis
•
01
A catalyst is a substance that increases the rate of
a reaction without being consumed in the reaction.
Prentice Hall ©2004
Chapter 12
Slide 39
Catalysis
•
02
The relative rates of the reaction A + B  AB in vessels a–d
are 1:2:1:2. Red = A, blue = B, yellow = third substance C.
(a) What is the order of reaction in A, B, and C?
(b) Write the rate law.
(c) Write a mechanism that agrees with the rate law.
(d) Why doesn’t C appear in the overall reaction?
Prentice Hall ©2004
Chapter 12
Slide 40
Catalysis
03
Homogeneous Catalyst: Exists in the same phase
as the reactants.
• Heterogeneous Catalyst: Exists in different phase
to the reactants.
•
Prentice Hall ©2004
Chapter 12
Slide 41
Catalysis
•
04
Catalytic Hydrogenation:
Prentice Hall ©2004
Chapter 12
Slide 42
Catalysis
Prentice Hall ©2004
05
Chapter 12
Slide 43