Chemistry 142
Chapter 13: Chemical Kinetics
Outline
I. Kinetics
II. Rate
A. Integrated Rate Law
B. Reaction Rate and Temperature
III. Reaction Mechanisms
IV. Catalysis
Chapter 13 Examples – Rate
at t = 0
[X] = 8
[Y] = 8
[Z] = 0
at t = 0
[A] = 8
[B] = 8
[C] = 0
at t = 16
[X] = 7
[Y] = 7
[Z] = 1
at t = 16
[A] = 4
[B] = 4
[C] = 4
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2
Concentration as a function of time
2 N2O5 (g)  4 NO2 (g) + O2 (g)
Chapter 13 – Examples: Rate
[N2O5] (M)
0
0.0200
100
0.0169
200
0.0142
300
0.0120
400
0.0101
500
0.0086
600
0.0072
700
0.0061
[N2O5] vs. time (s)
0.025
0.02
[N2O5] (M)
2 N2O5 (g)  4 NO2 (g) + O2 (g)
a) Calculate the average
rate of decomposition of
dinitrogen
pentaoxide
during the time interval
200-300 s.
b) Calculate the average
rate of formation of
oxygen gas during the
same interval.
c) What
is
the
instantaneous rate of
nitrogen
dioxide
formation at 250 s?
Time (s)
0.015
0.01
0.005
0
0
100
200
300
400
time (s)
500
600
700
800
Continuous Monitoring
Sampling
Tro, Chemistry: A Molecular Approach
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Tro, Chemistry: A Molecular
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8
Energy Profile for the
Isomerization of Methyl Isonitrile
9
Effective Collisions
Orientation Effect
Tro, Chemistry: A Molecular
Approach
10
Rate Laws of Elementary Steps
Tro, Chemistry: A Molecular
Approach
11
Chapter 13 – Examples
Mechanism Validation
The balanced equation for the reaction of nitrogen dioxide
and fluorine gases is:
2 NO2 (g) + F2 (g)  2 NO2F (g)
the experimentally determined rate law is
rate = k[NO2][F2]
A suggested mechanism is
k1
NO2 (g) + F2 (g)  NO2F (g) + F (g)
slow
k2
F (g) + NO2 (g)  NO2F (g)
Is this an acceptable mechanism?
fast
Chapter 13 – Examples
Mechanism Validation
The balanced equation for the reaction of nitrogen dioxide
and fluorine gases is:
2 NO2 (g) + F2 (g)  2 NO2F (g)
the experimentally determined rate law is
rate = k[NO2][F2]
Another suggested mechanism is
NO2 (g) + F2 (g)  NOF2 (g) + O (g)
slow
NO2 (g) + O (g)  NO3 (g)
fast
NOF2 (g) + NO2 (g)  NO2F (g) + NOF (g)
fast
NO3 (g) + NOF (g)  NO2F (g) + NO2 (g)
fast
Is this an acceptable mechanism?
Chapter 13 – Examples
Mechanism Validation
The balanced equation for the reaction of nitrogen monoxide and
chlorine gases is:
2 NO (g) + Cl2 (g)  2 NOCl (g)
the experimentally determined rate law is
rate = k[NO]2[Cl2]
A suggested mechanism is
k1
NO (g) + Cl2 (g)  NOCl2 (g)
fast
k-1
k2
NOCl2 (g) + NO (g)  2 NOCl (g)
Is this an acceptable mechanism?
slow
Temperature’s Effect on Reaction Rates
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17
Arrhenius Plot
Chapter 13 – Examples
Arrhenius Plot
For the reaction: 2 N2O5 (g)  4 NO2 (g) + O2 (g)
Several values of k were obtained at different temperatures.
Calculate the activation energy for the reaction.
T (°C)
k (s-1)
20
2.0 x 10-5
30
7.3 x 10-5
40
2.7 x 10-4
50
9.1 x 10-4
60
2.9 x 10-3
Chapter 13 – Examples
Arrhenius Plot
2 N2O5 (g)  4 NO2 (g) + O2 (g)
T (°C)
T (K)
1/T (1/K)
k (s-1)
ln k
20
293
3.41 x 10-3
2.0 x 10-5
-10.82
30
303
3.30 x 10-3
7.3 x 10-5
-9.53
40
313
3.19 x 10-3
2.7 x 10-4
-8.22
50
323
3.10 x 10-3
9.1 x 10-4
-7.00
60
333
3.00 x10-3
2.9 x 10-3
-5.84
0
2.90E-03
-2
Calculate the activation energy for
the reaction.
Arrhenius Plot
ln k versus 1/T
3.00E-03
3.10E-03
3.20E-03
3.30E-03
3.40E-03
3.50E-03
ln (k)
-4
-6
ln(k) = (-12174 K)/T + 30.694
R² = 0.9998
-8
-10
-12
1/T (1/K)
Chapter 13 – Examples
Arrhenius Equation
For the gas phase reaction:
CH4 (g) + 2 S2 (g)  CS2 (g) + 2 H2S (g)
The rate constant at 550 °C is 1.1 L/mol s and at 625 °C the
value of k is 6.4 L/mol s. Calculate the activation energy for
the reaction.
Energetics in the Presence of a Catalyst
h
2 O 3 
 3 O
2
The catalyst, a chlorine
atom, is not consumed
during the chemical
reaction.
Catalysis
Enzyme-Substrate Binding
Lock and Key Mechanism
Tro, Chemistry: A Molecular
Approach
24
Enzymatic Hydrolysis of Sucrose
Tro, Chemistry: A Molecular
Approach
25
Types of Catalysts
Tro, Chemistry: A Molecular
Approach
26
Catalytic Hydrogenation
H2C=CH2 + H2 → CH3CH3
Tro, Chemistry: A Molecular
Approach
27
Molecular View of
Catalytic
Converters
2 NO (g)  N2 (g) + O2 (g)
First Order Reactions
ln[A]0
ln[A]
time
Tro, Chemistry: A Molecular
Approach
31
Rate Data for
C4H9Cl
(aq) +
H2O (l)  C4H9OH (aq) + HCl (aq)
Time (sec) [C4H9Cl], M
0.0
0.1000
Tro, Chemistry: A Molecular
Approach
50.0
100.0
150.0
0.0905
0.0820
0.0741
200.0
300.0
400.0
500.0
0.0671
0.0549
0.0448
0.0368
800.0
10000.0
0.0200
0.0000
32
C4H9Cl (aq) + H2O (l)  C4H9OH (aq) + 2 HCl (aq)
Concentration vs. Time for the Hydrolysis of C 4H9Cl
0.12
concentration, (M)
0.1
0.08
0.06
0.04
0.02
0
0
200
400
600
time, (s)
Tro, Chemistry: A Molecular
Approach
33
800
1000
C4H9Cl (aq) + H2O (l)  C4H9OH (aq) + 2 HCl (aq)
Rate vs. Time for Hydrolysis of C 4H9Cl
2.5E-04
Rate, (M/s)
2.0E-04
1.5E-04
1.0E-04
5.0E-05
0.0E+00
0
Tro, Chemistry: A Molecular
Approach
100
200
300
400
time, (s)
34
500
600
700
800
C4H9Cl (aq) + H2O (l)  C4H9OH (aq) + 2 HCl (aq)
LN([C4H9Cl]) vs. Time for Hydrolysis of C 4H9Cl
0
slope =
-2.01 x 10-3
k=
2.01 x 10-3 s-1
-0.5
LN(concentration)
-1
-1.5
-2
-2.5
-3
t1 
y = -2.01E-03x - 2.30E+00
2
-3.5

0.693
k
0.693
2.0110 3 s -1
 345 s
-4
-4.5
0
100
200
300
400
500
time, (s)
Tro, Chemistry: A Molecular
Approach
35
600
700
800
Chapter 13 – Examples
Reaction Order
Using the data below verify that the
decomposition
of
dinitrogen
pentaoxide is first order and
calculate the value of the rate
constant, where rate = -Δ[N2O5]/Δt
2 N2O5 (g)  4 NO2 (g) + O2 (g)
Time (s)
[N2O5] (mol/L)
0
0.100
50
0.0707
100
0.0500
200
0.0250
300
0.0125
400
0.00625
Chapter 13 – Examples
Reaction Order
Time (s)
[N2O5] (mol/L)
0
0.100
50
0.0707
100
0.0500
Concentration [N2O5] (mol/L)
Using the data below verify that the
decomposition of dinitrogen pentaoxide is first
order and calculate the value of the rate 0.120
constant, where rate = -Δ[N2O5]/Δt
2 N2O5 (g)  4 NO2 (g) + O2 (g)
0.100
First Order Reaction
Concentration [N2O5] versus Time
0.080
0.060
0.040
0.020
200
0.0250
300
0.0125
400
0.00625
0.000
0
100
200
300
time (s)
400
500
Chapter 13 – Examples
Reaction Order
Using the data below verify that the
decomposition of dinitrogen pentaoxide is first
order and calculate the value of the rate
constant, where rate = -Δ[N2O5]/Δt
2 N2O5 (g)  4 NO2 (g) + O2 (g)
First Order Reaction
ln[N2O5] versus Time
0.000
0
100
200
300
-1.000
Time (s)
[N2O5] (mol/L)
ln[N2O5]
0
0.100
-2.303
50
0.0707
-2.649
100
0.0500
-2.996
200
0.0250
-3.689
300
0.0125
-4.382
400
0.00625
-5.075
ln[N2O5]
-2.000
ln[N2O5]
= -0.0069(t) - 2.3026
-3.000
-4.000
-5.000
-6.000
time (s)
400
500
Half-Life of a First-Order Reaction Is Constant
Tro, Chemistry: A Molecular
Approach
39
Chapter 13 – Examples
Half-life
A certain first order reaction has a half-life of
20.0 minutes.
a. Calculate the rate constant.
b. How much time is required for this reaction to be
75% complete?
Second Order Reactions
1/[A]
l/[A]0
time
Tro, Chemistry: A Molecular Approach
41
Chapter 13 – Examples
Rate Laws
Butadiene reacts to form its
dimer according to:
2 C4H6 (g)  C8H12 (g)
a) Is this reaction first order or
second order?
b) What is the value of the rate
constant function of the
reaction?
c) What is the half-life for the
reaction under the conditions
of this experiment?
time (s)
0
1000
1800
2800
3600
4400
5200
6200
[C4H6] (mol/L)
0.0100
0.00625
0.00476
0.00370
0.00313
0.00270
0.00241
0.00208
Chapter 13 – Examples
Rate Laws
Butadiene reacts to form its
dimer according to:
2 C4H6 (g)  C8H12 (g)
a) Is this reaction first order or
second order?
b) What is the value of the rate
constant function of the
reaction?
c) What is the half-life for the
reaction
under
the
conditions
of
this
experiment?
time (s)
[C4H6] (mol/L)
ln[C4H6]
1/[C4H6] (L/mol)
0
0.0100
-4.605
100
1000
0.00625
-5.075
160
1800
0.00476
-5.348
210
2800
0.00370
-5.599
270
3600
0.00313
-5.767
319
4400
0.00270
-5.915
370
5200
0.00241
-6.028
415
6200
0.00208
-6.175
481
Second Order Reaction Plot
1/[C4H6] versus time
First Order Reaction Plot
ln[C4H6] versus time
600
0
1000
2000
3000
4000
ln[C4H6]
-5.000
-5.500
-6.000
5000
6000
7000
1/[C4H6] (L/mol)
-4.500
500
400
300
200
1/[C4H6] = (0.0612L/mol s)t + 99.36 L/mol
100
0
0
-6.500
time (s)
1000
2000
3000
4000
time (s)
5000
6000
7000
Half-life for a second order reaction
Zero Order Reactions
[A]0
[A]
time
Tro, Chemistry: A Molecular Approach
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Tro, Chemistry: A Molecular
Approach
46
Chapter 13 – Examples: Rate
a)
b)
c)
d)
2 N2O5 (g)  4 NO2 (g) + O2 (g)
Calculate the average rate
of
decomposition
of
dinitrogen
pentaoxide
during the time interval
200-300 s.
Calculate the average rate
of formation of oxygen gas
during the same interval.
What is the instantaneous
rate of nitrogen dioxide
formation at 250 s?
Is this reaction first order,
second order, or zero
order?
Time (s)
[N2O5] (M)
0
0.0200
100
0.0169
200
0.0142
300
0.0120
400
0.0101
500
0.0086
600
0.0072
700
0.0061
Chapter 13 – Examples: Rate
2 N2O5 (g)  4 NO2 (g) + O2 (g)
d) Is this reaction first order,
second order, or zero
order?
Time (s)
0
100
200
300
400
500
600
700
[N2O5] vs. time (s)
0.025
[N2O5] (M)
0.02
0.015
0.01
0.005
[N2O5] (M) ln[N2O5] 1/[N2O5]
0.0200
-3.912
50.0
0.0169
-4.080
59.2
0.0142
-4.255
70.4
0.0120
-4.423
83.3
0.0101
-4.595
99.0
0.0086
-4.76
116
0.0072
-4.93
139
0.0061
-5.10
164
0
0
200
400
600
800
1/[N2O5] vs. time (s)
time (s)
ln[N2O5] vs. time (s)
200
400
ln[N2O5]
-4.250
600
800
150.0
1/[N2O5]
-4.050 0
200.0
100.0
-4.450
50.0
-4.650
0.0
0
-4.850
-5.050
100
200
300
400
time (s)
time (s)
500
600
700
800
Chapter 13 – Examples
Initial rate method
Describe the general form of the rate law for
the reaction given the results of four
experiments below:
NH4+ (aq) + NO2- (aq)  N2 (g) + 2 H2O (l)
Experiment
[NH4+ ]0
[NO2- ]0
Initial rate (mol/L s)
1
0.100 M
0.0050 M
1.35 x 10-7
2
0.100 M
0.010 M
2.70 x 10-7
3
0.200 M
0.010 M
5.40 x 10-7
Chapter 13 – Examples
Initial rate method
Describe the general form of the rate law for
the reaction given the results of three
experiments below:
CH3COOCH3 (aq) + OH- (aq)  CH3COO- (aq) + CH3OH (aq)
Experiment
[CH3COOCH3 ]0
[OH- ]0
Initial rate (mol/L s)
1
0.040 M
0.040 M
0.00022
2
0.040 M
0.080 M
0.00045
3
0.080 M
0.080 M
0.00090
Chapter 13 – Examples
Initial rate method
Describe the general form of the rate law for
the reaction given the results of three
experiments below:
CH3COOH (aq) + OH- (aq)  CH3COO- (aq) + CH3OH (aq)
Experiment
[CH3COOH ]0
[OH- ]0
Initial rate (mol/L s)
k (L/mol s)
1
0.040 M
0.040 M
0.00022
0.0138
2
0.040 M
0.080 M
0.00045
0.0141
3
0.080 M
0.080 M
0.00090
0.0141
Chapter 13 – Examples
Initial Rate Method
The reaction between bromate ions and bromide ions in an
acidic solution is given by the equation:
BrO3- (aq) + 5 Br- (aq) + 6 H+ (aq)  3 Br2 (l) + 3 H2O (l)
Describe the general form of the rate law for this reaction
given the results of four experiments below.
Experiment
[BrO3- ]0
[Br- ]0
[H+ ]0
Initial rate (mol/L s)
1
0.10 M
0.10 M
0.10 M
8.0 x 10-4
2
0.20 M
0.10 M
0.10 M
1.6 x 10-3
3
0.20 M
0.20 M
0.10 M
3.2 x 10-3
4
0.10 M
0.10 M
0.20 M
3.2 x 10-3