Chemistry: The Molecular Nature of Matter, 6 th edition
By Jesperson, Brady, & Hyslop

CHAPTER 14 Chemical Kinetics
Learning Objectives:
 Factors Affecting Reaction Rate: o Concentration o State o Surface Area o Temperature o Catalyst
 Collision Theory of Reactions and Effective Collisions
 Determining Reaction Order and Rate Law from Data
 Integrated Rate Laws
 Rate Law

Concentration vs Rate
 Integrated Rate Law

Concentration vs Time
 Units of Rate Constant and Overall Reaction Order
 Half Life vs Rate Constant (1 st Order)
 Arrhenius Equation
 Mechanisms and Rate Laws
 Catalysts
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Lecture Road Map:
① Factors that affect reaction rates
② Measuring rates of reactions
③ Rate Laws
④ Collision Theory
⑤ Transition State Theory & Activation Energies
⑥ Mechanisms
⑦ Catalysts
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Integrated
Rate Laws
Concentration & Time
Rate law tells us how speed of reaction varies with concentrations.
Sometimes want to know o Concentrations of reactants and products at given time during reaction o How long for the concentration of reactants to drop below some minimum optimal value
 Need dependence of rate on time
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
5

Integrated
Rate Laws
First Order Integrated Rate Law
Rate
[ t
A
]
k
[
A
]
• Corresponding to reactions
– A
 products
• Integrating we get ln
[
[
A ]
0
A ] t
= kt
• Rearranging gives ln[ A ] t
= kt + ln[
• Equation of line y = mx + b
A ]
0
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
6

Integrated
Rate Laws
First Order Integrated Rate Law
A
t
  kt

A
0
Yields straight line o Indicative of first order kinetics o Slope = –k o Intercept = ln [A]
0 o If we don't know already
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Integrated
Rate Laws
2 nd Order Integrated Rate Law
Rate
= k [ B ]
2 = - D
[ B ]
D t
• Corresponding to special second order reaction
– 2B
 products
• Integrating we get
[
1
B ] t
• Rearranging gives
1
[ B ] t
=
-
[
1
B ] kt +
0
=
1 kt
[ B ]
0
• Equation of line y = mx + b
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
8

Integrated
Rate Laws
2 nd Order Integrated Rate Law
B
t
 kt

B
0
Yields straight line o Indicative of 2 nd order kinetics o Slope = +k o Intercept = 1/[B]
0
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
9

Integrated
Rate Laws
Graphically determining Order
Make two plots:
1. ln [ A ] vs. time
2. 1/[ A ] vs. time o If ln [ A ] is linear and 1/[ A ] is curved, then reaction is 1 st order in [ A ] o If 1/[A] plot is linear and ln [ A ] is curved, then reaction is 2 nd order in [ A ] o If both plots give horizontal lines, then 0 th order in [ A ]
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Integrated
Rate Laws
Graphically determining Order
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Integrated
Rate Laws
Example: SO
2
Cl
2

SO
2
+ Cl
2
Time, min [SO
2
Cl
2
], M ln[SO
2
Cl
2
]
0 0.1000
-2.3026
100
200
0.0876
0.0768
-2.4350
-2.5666
300
400
0.0673
0.0590
-2.6986
-2.8302
500
600
700
800
900
1000
1100
0.0517
0.0453
0.0397
0.0348
0.0305
0.0267
-2.9623
-3.0944
-3.2264
-3.3581
-3.4900
-3.6231
Molecular Nature of Matter, 6E
1/[SO
2
Cl
2
] (L/mol)
10.000
11.416
13.021
14.859
16.949
19.342
22.075
25.189
28.736
32.787
37.453
42.735
12

Integrated
Rate Laws
Example: SO
2
Cl
2

SO
2
+ Cl
2
-2.2
-2.4
-2.6
-2.8
-3.0
-3.2
-3.4
-3.6
-3.8
0
First Order Plot for SO
2
Cl
2
Decomposition
200 400 600 time (min)
800 1000 1200
45
Second order plot for SO
2
Cl
2
Decomposition
40
35
30
25
20
15
10
0 200 400 600 time (min)
800 1000 1200
st
Jesperson, Brady, Hyslop. Chemistry: The
2
2
Molecular Nature of Matter, 6E
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Integrated
Rate Laws
0
50
100
150
200
250
300
350
Example:
HI(g)  H
2
(g) + I
2
(g)
0.1000
0.0716
0.0558
0.0457
0.0387
0.0336
-2.3026
-2.6367
-2.8860
-3.0857
-3.2519
-3.3932
0.0296
-3.5200
0.0265
-3.6306
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
10.000
13.9665
17.9211
21.8818
25.840
29.7619
33.7838
37.7358
14

Integrated
Rate Laws
Example:
HI(g)  H
2
(g) + I
2
(g)
-2.2
First Order Plot for HI Decomposition at 508 o
C
-2.4
40
Second order plot for HI
Decomposition at 508 o
C
35
-2.6
-2.8
30
-3.0
-3.2
-3.4
25
20
-3.6
-3.8
0 50 100 150 200 250 300 350 time (s)
15
10
0 50 100 150 200 250 300 350 time (s)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
15

Group
Problem
A plot for a zeroth order reaction is shown. What is the proper label for the y-axis in the plot ?
A. Concentration Zeroth Order Plot
B. ln of Concentration
C. 1/Concentration
D. 1/ ln Concentration
0 200 400 600 time (min)
800 1000 1200
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
16

Integrated
Rate Laws
Half Life (t
1/2
) for first order reactions
Half-life = t
½
We often use the half life to describe how fast a reaction takes place
First Order Reactions o Set [ A ] t
=
1
2
[ A ]
0 o Substituting into ln
[
[
A ]
0
A ] t
= kt o Gives ln
1
2
[ A ]
0
[ A ]
0
= kt
1
2 o Canceling gives ln 2 = kt
½ o t
Rearranging gives 1
2
= ln k
2
=
0
1
Jesperson, Brady, Hyslop. Chemistry: The
.
k
693
1
Molecular Nature of Matter, 6E
17

Integrated
Rate Laws
Half Life (t
1/2
) for First Order Reactions
Observe:
1.
t
½ is independent of [A] o o For given reaction (and T ) o Takes same time for concentration to fall from o 2 M to 1 M as from o 5.0

10 –3 M to 2.5

10 –3 M
2.
k
1 has units (time) –1 , so t
½ has units (time) o t
½ called half-life o Time for ½ of sample to decay
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
18

Integrated
Rate Laws
Half Life (t
1/2
)
Does this mean that all of sample is gone in two half-lives (2 × t
½
)?
No! o In 1 st t
½
, it goes to ½[A] o o In 2 nd t
½
, it goes to ½(½[A] o
) = ¼[A] o o In 3 rd t
½
, it goes to ½(¼[A] o
) = ⅛[A] o o In n th t
½
, it goes to [A] o
/2 n
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
19

Integrated
Rate Laws
Half Life (t
1/2
)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
20

Integrated
Rate Laws
Half Life (t
1/2
): First Order Example
131 I is used as a metabolic tracer in hospitals. It has a half-life, t
½
= 8.07 days.
How long before the activity falls to 1% of the initial value?
N = N o e kt t ln
N
N o
= kt = t ln2 t
1
2
= t
1
2 ln
N ln
2
N o = -
(
8
.
07 days
)ln
æ
è
1
100
ö
ø ln
2
=
53
.
6 days
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s
–1
?
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
The half-life of I-132 is 2.295 h. What percentage remains after 24 hours?
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Integrated
Rate Laws
Half Life (t
1/2
): Carbon-14 Dating
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Integrated
Rate Laws
Half Life (t
1/2
): Second Order Reactions
How long before [A ] = ½[A] o
?
t
1/2
=
1 k [ A ]
0 o t
½
, depends on [A] o o t
½
, not useful quantity for a second order reaction
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
The rate constant for the second order reaction 2 A → B is 5.3
× 10
–5
M
–1 s
–1
. What is the original amount present if, after 2 hours, there is 0.35 M available?
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Collision
Theory
Reaction Rates
Collision Theory
As the concentration of reactants increase o The number of collisions increases o Reaction rate increases
As temperature increases o Molecular speed increases o Higher proportion of collisions with enough force (energy) o There are more collisions per second o Reaction rate increases
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Collision
Theory
Reaction Rates
Rate of reaction proportional to number of effective collisions/sec among reactant molecules
Effective collision o One that gives rise to product e.g. At room temperature and pressure o H
2 and I
2 molecules undergoing 10 o Yet reaction takes a long time
10 collisions/sec o Not all collisions lead to reaction
Only very small percentage of all collisions lead to net change
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Collision
Theory
Molecular Orientation
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Collision
Theory
Temperature
As T increases o More molecules have E a o So more molecules undergo reaction
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Collision
Theory
Activation Energy (E a
)
Molecules must possess certain amount of kinetic energy
(KE) in order to react
Activation Energy, E a reaction to occur
= Minimum KE needed for o Get energy from collision with other molecules o If molecules move too slowly, too little KE, they just bounce off each other o Without this minimum amount, reaction will not occur even when correctly oriented
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Transition
State
Molecular Basis of Transition State Theory
KE KE
KE decreasing as PE increases
Is the combined KE of both molecules enough to overcome
Activation Energy
KE
PE
KE
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
34

Transition
State
Molecular Basis of Transition State Theory
a
(

H)
= difference in PE between products and reactants

H reaction
= H products
– H reactants
Reaction Coordinate (progress of reaction)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
Products
35

Transition
State
Exothermic Reactions
Exothermic
Reaction

H = –
Products
Reaction Coordinate (progress of reaction)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
36

Transition
State
Exothermic Reactions o 
H reaction
< 0 (negative) o Decrease in PE of system o Appears as increase in KE o So the temperature of the system increases o Reaction gives off heat o Can ’t say anything about E a from size of could be low and reaction rapid

H o E a

H could be high and reaction slow even if rxn large and negative o E a
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
37

Transition
State
Endothermic Reactions

reaction
products
reactants
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
Endothermic
Reaction

H = +
38

Transition
State
Endothermic Reactions o 
H reaction
> 0 (positive) o Increase in PE o Appears as decrease in KE o So temperature of the system decreases o Have to add E to get reaction to go o E a
 
H rxn o If

H rxn as E a includes large and positive

H rxn o E a must be high o Reaction very slow
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
39

Transition
State
Activated Complex o Arrangement of atoms at top of activation barrier o Brief moment during successful collision when o bond to be broken is partially broken and o bond to be formed is partially formed
Example
H
3
C N C
N
H
3
C
C
Transition State
H
3
C C N
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
Draw the transition state complex, or the activated complex for the following reaction:
CH
3
CH
2
O + H
3
O +  CH
3
CH
2
OH + H
2
O
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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E a
Arrhenius Equation
The rate constant is dependent on Temperature, which allows us to calculate Activation Energy, E a
Arrhenius Equation:
Equation expressing temperature-dependence of k k = Ae E a
/ RT o A = Frequency factor has same units as k o R = gas constant in energy units
= 8.314 J mol –1 K –1 o E a
= Activation Energy o T = Temperature in K
—has units of J/mol
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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E a
Calculating Activation Energy
• Method 1. Graphically
• Take natural logarithm of both sides
• Rearranging ln k
ln A
E
R a
• Equation for a line
• y = b + mx
1
T
Arrhenius Plot
• Plot ln k ( y axis) vs. 1/T ( x axis)  yield a straight line
• Slope = -E a
/R
• Intercept = A
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
44

E a
Arrhenius Equation: Graphing Example
Given the following data, predict k at 75 ˚C using the graphical approach ln k = -
E a
R
´
1
T
+ ln A ln (k) = –36.025/T – 6.908
k ( M /s) T , ˚C T , K
0.000886 25 298
0.000894 50 348
0.000908 100 398
0.000918 150 448
?
75 348 ln (k) = –36.025/(348) – 6.908 = – 7.011
k = e -
7
.
012 =
9
.
01
´
10
-
4
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
45

E a
Arrhenius Equation: Graphing Example
-6,99
-7,00
-7,01
-7,02 y = –36.025x – 6.908
-7,03
0,0022 0,0024 0,0026 0,0028 0,003 0,0032 0,0034
1/ T (K –1 )
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
46

E a
Arrhenius Equation
Sometimes a graph is not needed o Only have two k s at two T s
Here use van't Hoff Equation derived from
Arrhenius equation: ln
k k
2
1
E a
R
T
1
2
1
T
1
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
47

E a
Arrhenius Equation: Ex Vant Hoff Equation
CH
4
+ 2 S
2

CS
2
+ 2 H
2
S
k (L/mol s) T ( ˚ C) T (K)
1.1 = k
1
6.4 = k
2
550
625
823 = T
1
898 = T
2 ln çç
æ
è
E a
6
.
4
1
.
1
=
ö
ø
÷÷ =
–
E a
8.3145 J/K mol
çç
æ
è
-
(
8.314 J/K mol
çç
æ
è
1
898 K
–
1
) ln
823 K
÷÷
ö
ø
çç
æ
è
6.4
1.1
898 K
÷÷
ö
ø
1
–
1
823 K
=
1
.
4
´
10 5
÷÷
ö
ø
J/mol
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
48

Group
Problem
Given that k at 25 ˚C is 4.61 × 10 –1 M /s and that at 50 ˚C it is
4.64 × 10
–1
M /s, what is the activation energy for the reaction?
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
49

Group
Problem
A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70˚C to 80 ˚C?
A. Rate increases approximately 1.5 times
B. Rate increases approximately 5000 times
C. Rate does not increase
D. Rate increases approximately 3 times
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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