Chapter 16
Chemical Kinetics
化學動力學
1
Outline
1. The Rate of a Reaction
2. Nature of the Reactants
3. Concentrations of the Reactants: The RateLaw Expression
4. Concentration Versus Time: The Integrated
Rate Equation
5. Collision Theory of Reaction Rates
6. Transition State Theory
7. Reaction Mechanisms and the Rate-Law
Expression
8. Temperature: The Arrhenius Equation
9. Catalysts
2
The Rate of a Reaction
• Kinetics is the study of rates of chemical
reactions and the mechanisms by which they
occur.
• The reaction rate is the increase in
concentration of a product per unit time or
decrease in concentration of a reactant per unit
time.
• A reaction mechanism is the series of molecular
steps by which a reaction occurs.
3
The Rate of a Reaction
• Thermodynamics (Chapter 15) determines if a
reaction can occur.
• Kinetics (Chapter 16) determines how quickly a
reaction occurs.
– Some reactions that are thermodynamically
feasible are kinetically so slow as to be
imperceptible.
C diamond  O 2 g   CO 2 g  G
o
298
 396 kJ
VERY SLOW
H +aq  + OH-aq   H 2 O l  G o298 = -79 kJ
INSTANTANE OUS
4
The Rate of Reaction
• Consider the hypothetical reaction,
aA(g) + bB(g)  cC(g) + dD(g)
• equimolar amounts of reactants, A and B, will be
consumed while products, C and D, will be formed
as indicated in this graph:
5
The Rate of Reaction
Concentrations of
Reactants & Products
1.2
1
0.8
[A] & [B]
[C] & [D]
0.6
0.4
0.2
0
35
0
30
0
25
0
20
0
15
0
10
50
0
0
Time
• [A] is the symbol for the concentration of A in M
( mol/L).
• Note that the reaction does not go entirely to
6
The Rate of Reaction
• Reaction rates are the rates at which
reactants disappear or products appear.
• This movie is an illustration of a reaction rate.
7
The Rate of Reaction
• Mathematically, the rate of a reaction can be
written as:
Rate =
-  A 
a t

-  B
b t

+  C
c t

+  D
d t
8
The Rate of Reaction
• The rate of a simple one-step reaction is directly
proportional to the concentration of the reacting
substance.
A (g)  B(g) + C(g)
Rate  A or Rate = kA
• [A] is the concentration of A in molarity
or moles/L.
• k is the specific rate constant.
– k is an important quantity in this chapter.
9
The Rate of Reaction
•
For a simple expression like Rate = k[A]
– If the initial concentration of A is doubled, the
initial rate of reaction is doubled.
• If the reaction is proceeding twice as fast, the
amount of time required for the reaction to reach
equilibrium would be:
A.The same as the initial time.
B.Twice the initial time.
C.Half the initial time.
•
If the initial concentration of A is halved the initial
rate of reaction is cut in half.
10
The Rate of Reaction
• If more than one reactant molecule
appears in the equation for a one-step
reaction, we can experimentally
determine that the reaction rate is
proportional to the molar concentration
of the reactant raised to the power of the
number of molecules involved in the
reaction.
2 X g   Yg  + Zg 
Rate  X or Rate = kX
2
2
11
The Rate of Reaction
•
Rate Law Expressions must be determined
experimentally.
– The rate law cannot be determined from the
balanced chemical equation.
– This is a trap for new students of kinetics.
• The balanced reactions will not work because
most chemical reactions are not one-step
reactions.
•
Other names for rate law expressions are:
1. rate laws
2. rate equations or rate expressions
12
The Rate of Reaction
•
•
Important terminology for kinetics.
The order of a reaction can be expressed
in terms of either:
1 each reactant in the reaction or
2 the overall reaction.
 Order for the overall reaction is the sum of the
orders for each reactant in the reaction.
•
For example:
2 N 2 O 5g   4 NO2 g  + O 2 g 
Rate = kN 2 O 5 
This reaction is first order in N 2 O 5
and first order overall.
13
The Rate of Reaction
• A second example is:
CH3 3 CBraq   OH-aq   CH3 3 COHaq   Br-aq 
Rate = k[CH3 3 CBr ]
This reaction is first order in CH3 3 CBr,
zero order in OH- , and first order overall.
14
The Rate of Reaction
• A final example of the order of a reaction is:
2 NO   + O    2 NO  
g
2 g
Rate = k[NO] O
2 g
2
2

This reaction is second order in NO,
first order in O , and third order overall
2
REMEMBER, ALL RATE EXPRESSION S ARE
DETERMINED EXPERIMENTALLY
15
The Rate of Reaction
• Given the following one step reaction and
its rate-law expression.
– Remember, the rate expression would have to
be experimentally determined.
2 A  g   B g   C  g 
Rate = kA
2
• Because it is a second order rate-law
expression:
– If the [A] is doubled the rate of the reaction will
increase by a factor of 4. 22 = 4
– If the [A] is halved the rate of the reaction will
decrease by a factor of 4. (1/2)2 = 1/4
16
Factors That Affect Reaction
Rates
•
1.
2.
3.
4.
•
There are several factors that can influence the
rate of a reaction:
The nature of the reactants.
The concentration of the reactants.
The temperature of the reaction.
The presence of a catalyst.
We will look at each factor individually.
17
Nature of Reactants
• This is a very broad category that
encompasses the different reacting
properties of substances.
• For example sodium reacts with water
explosively at room temperature to liberate
hydrogen
sodium
hydroxide.
2 Na s   and
2 H 2form
O   
2 NaOH
aq   H 2 g 
This is a violent and rapid reaction.
The H 2 ignites and burns.
18
Nature of Reactants
• Calcium reacts with water only slowly at room
temperature to liberate hydrogen and form calcium
hydroxide.
Ca s   2 H 2O   Ca OH2 aq   H 2g 
This is a rather slow reaction.
19
Nature of Reactants
• The reaction of magnesium with water at room
temperature is so slow that that the evolution of
hydrogen is not perceptible to the human eye.
Mg s   H 2O   No reaction
20
Nature of Reactants
• However, Mg reacts with steam rapidly
to liberate H2 and form magnesium
oxide.
Mg s   H 2O( g )  MgO s   H 2g 
100o C
• The differences in the rate of these three
reactions can be attributed to the
changing “nature of the reactants”.
21
Concentrations of Reactants:
The Rate-Law Expression
• This movie illustrates how changing the
concentration of reactants affects the rate.
22
Concentrations of Reactants:
The Rate-Law Expression
• This is a simplified representation of the effect of
different numbers of molecules in the same volume.
– The increase in the molecule numbers is
indicative of an increase in concentration.
A(g) + B (g)  Products
A
B
A B
4 different possible
A-B collisions
A B
B
A B
6 different possible
A-B collisions
A B
A B
A B
9 different possible 23
A-B collisions
Concentrations of Reactants:
The Rate-Law Expression
Example 16-1: The following rate data were
obtained at 25oC for the following reaction.
What are the rate-law expression and the
specific rate-constant for this reaction?
2 A(g) + B(g)  3 C(g)
Experiment
Number
1
Initial [A]
(M)
0.10
Initial [B]
(M)
0.10
Initial rate of
formation of C
(M/s)
2.0 x 10-4
2
0.20
0.30
4.0 x 10-4
3
0.10
0.20
2.0 x 10-4
24
Concentrations of Reactants:
The Rate-Law Expression
Experiment
Number
1
Initial [A]
(M)
0.10
Initial [B]
(M)
0.10
Initial rate of
formation of C
(M/s)
2.0 x 10-4
2
0.20
0.30
4.0 x 10-4
3
0.10
0.20
2.0 x 10-4
25
Concentrations of Reactants:
The Rate-Law Expression
Example 16-2: The following data were
obtained for the following reaction at 25oC.
What are the rate-law expression and the
specific rate constant for the reaction?
2 A(g) + B(g) + 2 C(g)  3 D(g) + 2 E(g)
26
Concentrations of Reactants:
The Rate-Law Expression
Initial [C] Initial rate
of formation
(M)
of D (M/s)
Experiment
Initial [A]
(M)
Initial [B]
(M)
1
2
0.20
0.20
0.10
0.30
0.10
0.20
2.0 x 10-4
6.0 x 10-4
3
4
0.20
0.60
0.10
0.30
0.30
0.40
2.0 x 10-4
1.8 x 10-3
27
Concentrations of Reactants:
The Rate-Law Expression
Example 16-3: consider a chemical reaction
between compounds A and B that is first order
with respect to A, first order with respect to B,
and second order overall. From the
information given below, fill in the blanks.
You do it!
28
Concentrations of Reactants:
The Rate-Law Expression
Experiment
1
Initial Rate
(M/s)
4.0 x 10-3
Initial [A]
(M)
0.20
Initial [B]
(M)
0.050
2
1.6 x 10-2
?
0.050
3
3.2 x 10-2
0.40
?
29
Concentration vs. Time: The
Integrated Rate Equation
• The integrated rate equation relates time and
concentration for chemical and nuclear reactions.
– From the integrated rate equation we can
predict the amount of product that is produced in
a given amount of time.
• Initially we will look at the integrated rate equation
for first order reactions.
These reactions are 1st order in the reactant and 1st
order overall.
30
Concentration vs. Time: The
Integrated Rate Equation
• An example of a reaction that is 1st order in the
reactant and 1st order overall is:
a A  products
This is a common reaction type for many chemical
reactions and all simple radioactive decays.
• Two examples of this type are:
2 N2O5(g)  2 N2O4(g) + O2(g)
238U  234Th + 4He
31
Concentration vs. Time: The
Integrated Rate Equation
• The integrated rate equation for first order
reactions is:

A0
ln
akt
A
where:
[A]0= mol/L of A at time t=0. [A] = mol/L of A at time t.
k = specific rate constant. t = time elapsed since
beginning of reaction.
a = stoichiometric coefficient of A in balanced overall32
equation.
Concentration vs. Time: The
Integrated Rate Equation
• Solve the first order integrated rate equation
for t.
1  A0 
t
ln 

a k  A 
• Define the half-life, t1/2, of a reactant as the
time required for half of the reactant to be
consumed, or the time at which [A]=1/2[A]0.
33
Concentration vs. Time: The
Integrated Rate Equation
• At time t = t1/2, the expression becomes:
t 1/2
t 1/2
t 1/2
 A 0
1

ln 
 1/2A 
ak
0

1
ln 2 

ak
0.693

ak




34
Concentration vs. Time: The
Integrated Rate Equation
Example 16-4: Cyclopropane, an anesthetic,
decomposes to propene according to the
following equation.
CH2
CH
H3C
CH2
H C CH
2
2 (g)
(g)
The reaction is first order in cyclopropane with k
= 9.2 s-1 at 10000C. Calculate the half life of
cyclopropane at 10000C.
35
Concentration vs. Time: The
Integrated Rate Equation
Example 16-5: Refer to Example 16-4. How
much of a 3.0 g sample of cyclopropane
remains after 0.50 seconds?
– The integrated rate laws can be used for any unit
that represents moles or concentration.
– In this example we will use grams rather than mol/L.
36
Concentration vs. Time: The
Integrated Rate Equation
Example 16-6: The half-life for the following first
order reaction is 688 hours at 10000C.
Calculate the specific rate constant, k, at
10000C and the amount of a 3.0 g sample of
CS2 that remains after 48 hours.
CS2(g)  CS(g) + S(g)
You do it!
37
Concentration vs. Time: The
Integrated Rate Equation

A ln A 0  k t  ln A   ln A   k t
ln
 k t  ln A   ln0A   k t
A A
-1
ln(3.0) - ln A  (0.00101
ln(3.0)
00101hr
hr )(
)(48
48 hr)
hr)
1.10 - ln A   0.048
ln A   -(0.048 - 1.10)  1.052
0
0
-1
A  e
1.052
 2.86 g  2.9 g or 97% unreacted
38
Concentration vs. Time: The
Integrated Rate Equation
• For reactions that are second order with
respect to a particular reactant and
second order overall, the rate equation is:
1
1

akt
A  A 0
• Where:
[A]0= mol/L of A at time t=0.
k = specific rate constant.
[A] = mol/L of A at time t.
t = time elapsed since
beginning of reaction.
a = stoichiometric coefficient of A in balanced overall
39
equation.
Concentration vs. Time: The
Integrated Rate Equation
• Second order reactions also have a half-life.
– Using the second order integrated rate-law as a
starting point.
• At the half-life, t1/2 [A] = 1/2[A]0.
2 1 1  1 akt


a
k
t
or
1/2
1/2
A01 / 2AA00 A0
which has a common denominato r of A 0
1
 a k t1/2
A0
40
Concentration vs. Time: The
Integrated Rate Equation
• If we solve for t1/2:
1
t1/2 
a k A 0
• Note that the half-life of a second
order reaction depends on the initial
concentration of A.
41
Concentration vs. Time: The
Integrated Rate Equation
Example 16-7: Acetaldehyde, CH3CHO,
undergoes gas phase thermal decomposition to
methane and carbon monoxide.

CH 3CHO g  
 CH 4 g  + CO g 
The rate-law expression is Rate = k[CH3CHO]2,
and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is
the half-life of CH3CHO if 0.10 mole is injected
into a 1.0 L vessel at 527oC?
42
Concentration vs. Time: The
Integrated Rate Equation
Example 16-7:

CH 3CHO g  
 CH 4 g  + CO g 
The rate-law expression is Rate = k[CH3CHO]2,
and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is
the half-life of CH3CHO if 0.10 mole is injected
into a 1.0 L vessel at 527oC?
43
Concentration vs. Time: The
Integrated Rate Equation
(b) How many moles of CH3CHO remain after
200 hours?
44
Concentration vs. Time: The
Integrated Rate Equation
(c) What percent of the CH3CHO remains after 200
hours?
45
Concentration vs. Time: The
Integrated Rate Equation
Example 16-8: Refer to Example 16-7. (a)
What is the half-life of CH3CHO if 0.10 mole is
injected into a 10.0 L vessel at 527oC?
– Note that the vessel size is increased by a factor of
10 which decreases the concentration by a factor of
10!
You do it!
46
Concentration vs. Time: The
Integrated Rate Equation
• (b) How many moles of CH3CHO remain after 200
hours?
You do it!
47
Concentration vs. Time: The
Integrated Rate Equation
• (c) What percent of the CH3CHO remains after 200
hours?
You do it!
48
Concentration vs. Time: The
Integrated Rate Equation
• Let us now summarize the results from the
last two examples.
Moles of
%
Initial
CH3CH CH3CHO
[CH3CHO]
[CH3CHO] O at 200 remainin
Moles
0
CH3CHO
(M)
(M)
hr.
g
Ex. 167
Ex. 168
0.10
0.10
0.071
0.071
71%
0.010
0.010
0.0096
0.096
96%
49
Enrichment - Derivation of
Integrated Rate Equations
• For the first order reaction
a A  products
the rate can be written as:
1  A  
Rate = - 

a  t 
50
Enrichment - Derivation of
Integrated Rate Equations
• For a first-order reaction, the rate is
proportional to the first power of [A].
1  A 
- 
  kA 
a  t 
51
Enrichment - Derivation of
Integrated Rate Equations
• In calculus, the rate is defined as the
infinitesimal change of concentration d[A] in
an infinitesimally short time dt as the
derivative of [A] with respect to time.
1  d A 
- 
  kA 
a  dt 
52
Enrichment - Derivation of
Integrated Rate Equations
• Rearrange the equation so that all of the [A]
terms are on the left and all of the t terms
are on the right.
d A 
 a k  dt
A 
53
Enrichment - Derivation of
Integrated Rate Equations
• Express the equation in integral form.
d A 
- 
 a k   dt
A


A
0
 0
A
t
54
Enrichment - Derivation of
Integrated Rate Equations
• This equation can be evaluated as:
-lnA   a k t or
t
0
t
0
-lnA t  lnA 0  a k  t - a k  0
which becomes
-lnA t  lnA 0  a k t
55
Enrichment - Derivation of
Integrated Rate Equations
• Which rearranges to the integrated first order
rate equation.

A 0
ln
akt
At
56
Enrichment - Derivation of
Integrated Rate Equations
• Derive the rate equation for a reaction that is
second order in reactant A and second
order overall.
• The rate equation is:
d A 
2

 kA 
adt
57
Enrichment - Derivation of
Integrated Rate Equations
• Separate the variables so that the A terms
are on the left and the t terms on the right.
d A 


k
d
t
2
a A 
58
Enrichment - Derivation of
Integrated Rate Equations
• Then integrate the equation over the limits as
for the first order reaction.
d A


a
k
d
t
A  A2 0
0
A 
t
59
Enrichment - Derivation of
Integrated Rate Equations
• Which integrates the second order
integrated rate equation.
1
1

akt
A A0
60
Enrichment - Derivation of
Integrated Rate Equations
• For a zero order reaction the rate expression
is:
d A 

k
adt
61
Enrichment - Derivation of
Integrated Rate Equations
• Which rearranges to:
 d A  a k d t
62
Enrichment - Derivation of
Integrated Rate Equations
• Then we integrate as for the other two cases:
A 
t
A 0
0



d
A

a
k
d
t


63
Enrichment - Derivation of
Integrated Rate Equations
• Which gives the zeroeth order integrated
rate equation.
A  A0  -a k t
or
A  A0 - a k t
64
Enrichment - Rate Equations
to Determine Reaction Order
• Plots of the integrated rate equations can
help us determine the order of a reaction.
• If the first-order integrated rate equation is
rearranged.
– This law of logarithms, ln (x/y) = ln x - ln y, was
applied to the first-order integrated rate-equation.
ln A0  ln A  a k t
or
ln A   a k t  ln A 0
65
Enrichment - Rate Equations
to Determine Reaction Order
• The equation for a straight line is:
y  mx  b
• Compare this equation to the rearranged
first order rate-law.
66
Enrichment - Rate Equations
to Determine Reaction Order
y  mx  b
ln A  a k t  ln A0
• Now we can interpret the parts of the
equation as follows:
 y can be identified with ln[A] and plotted on the
y-axis.
 m can be identified with –ak and is the slope of
the line.
 x can be identified with t and plotted on the x67
axis.
Enrichment - Rate Equations
to Determine Reaction Order
• Example 16-9: Concentration-versus-time
data for the thermal decomposition of ethyl
bromide are given in the table below. Use
the following graphs of the data to
determine the rate of the reaction and the

value of the rate
constant.
C2 H5Brg  
 C2H4g   HBrg  at 700K
68
Enrichment - Rate Equations
to Determine Reaction Order
Time
(min)
[C2H5Br]
0
1.00
1
0.82
2
0.67
3
0.55
4
0.45
5
0.37
ln [C2H5Br] 0.00 -0.20 -0.40 -0.60 -0.80 -0.99
1/[C2H5Br]
1.0
1.2
1.5
1.8
2.2
2.7
69
Enrichment - Rate Equations
to Determine Reaction Order
• We will make three different graphs of the data.
1 Plot the [C2H5Br] (y-axis) vs. time (x-axis)
– If the plot is linear then the reaction is zero order
with respect to [C2H5Br].
2 Plot the ln [C2H5Br] (y-axis) vs. time (x-axis)
– If the plot is linear then the reaction is first order with
respect to [C2H5Br].
3 Plot 1/ [C2H5Br] (y-axis) vs. time (x-axis)
– If the plot is linear then the reaction is second order
with respect to [C2H5Br].
70
Enrichment - Rate Equations
to Determine Reaction Order
• Plot of [C2H5Br] versus time.
– Is it linear or not?
[C2H5Br]
[C2H5Br] vs. time
1.2
1
0.8
0.6
0.4
0.2
0
0
1
2
3
Time (min)
4
5
71
Enrichment - Rate Equations
to Determine Reaction Order
• Plot of ln [C2H5Br] versus time.
– Is it linear or not?
ln [C2H5Br] vs. time
0
ln [C2H5Br]
-0.2
0
1
2
3
4
5
-0.4
-0.6
-0.8
-1
-1.2
Time (min)
72
Enrichment - Rate Equations
to Determine Reaction Order
• Plot of 1/[C2H5Br] versus time.
– Is it linear or not?
1/[C2H5Br] vs. time
1/[C2H5Br]
3
2
1
0
0
1
2
3
4
5
Time (min)
73
Enrichment - Rate Equations
to Determine Reaction Order
• Note that the only graph which is linear is the plot of
ln[C2H5Br] vs. time.
– Thus this is a first order reaction with respect to
[C2H5Br].
• Next, we will determine the value of the rate
constant from the slope of the line on the graph of
ln[C2H5Br] vs. time.
– Remember slope = (y2-y1)/(x2-x1).
y 2 - y1  0.80  (0.20)
slope 

x 2 - x1
4  1 min
 0.60
slope 
 0.20 min -1
3 min
74
Enrichment - Rate Equations
to Determine Reaction Order
• From the equation for a first order reaction
we know that the slope = -a k.
– In this reaction a = 1.
slope  -0.20  -k
Thus the rate constant k  0.20 min .
-1
75
Enrichment - Rate Equations
to Determine Reaction Order
• The integrated rate equation for a reaction
that is second order in reactant A and
second order overall.
1
1

akt
A A0
• This equation can be rearranged to:
1
1
akt
A
A0
76
Enrichment - Rate Equations
to Determine Reaction Order
y  mx  b
• Compare the equation for a straight line and
the second1order rate-law expression.
1
A
akt
A0
• Now we can interpret the parts of the
equation as follows:
 y can be identified with 1/[A] and plotted on the
y-axis.
 m can be identified with a k and is the slope of
the line.
 x can be identified with t and plotted on the x- 77
axis
Enrichment - Rate Equations
to Determine Reaction Order
Example 16-10: Concentration-versus-time
data for the decomposition of nitrogen dioxide
are given in the table below. Use the graphs
to determine the rate of the reaction and the
value of the rate constant

2 NO2 g  
 2 NOg   O2g  at 500K
78
Enrichment - Rate Equations
to Determine Reaction Order
Time
(min)
[NO2]
0
1.0
1
0.53
2
0.36
3
0.27
4
0.22
5
0.18
ln [NO2]
0.0
-0.63
-1.0
-1.3
-1.5
-1.7
1/[NO2]
1.0
1.9
2.8
3.7
4.6
5.5
79
Enrichment - Rate Equations
to Determine Reaction Order
• Once again, we will make three different
graphs of the data.
1. Plot [NO2] (y-axis) vs. time (x-axis).
– If the plot is linear then the reaction is zero order
with respect to NO2.
2.
Plot ln [NO2] (y-axis) vs. time (x-axis).
• If the plot is linear then the reaction is first order
with respect to NO2.
3.
Plot 1/ [NO2] (y-axis) vs. time (x-axis).
– If the plot is linear then the reaction is second
order with respect to NO2.
80
Enrichment - Rate Equations
to Determine Reaction Order
• Plot of [NO2] versus time.
– Is it linear or not?
[NO2]
[NO2] vs. time
1.2
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
Time (min)
81
Enrichment -Rate Equations
to Determine Reaction Order
• Plot of ln [NO2] versus time.
– Is it linear or not?
ln [NO2] vs. time
0
0
1
2
3
4
5
ln [NO2]
-0.5
-1
-1.5
-2
Time (min)
82
Enrichment - Rate Equations
to Determine Reaction Order
• Plot of 1/[NO2] versus time.
– Is it linear or not?
1/[NO2]
1/[NO2] vs.time
6
5
4
3
2
1
0
0
1
2
3
Time (min)
4
5
83
Enrichment - Rate Equations
to Determine Reaction Order
• Note that the only graph which is linear is the plot of
1/[NO2] vs. time.
• Thus this is a second order reaction with respect to
[NO2].
• Next, we will determine the value of the rate
constant from the slope of the line on the graph of
1/[NO2] vs. time.
84
Enrichment - Rate Equations
to Determine Reaction Order
y 2 -y1
slope 

x 2 -x1
1
(
(5.50  1.90)
(5  1)
M
min
)
3.60 1 M
slope 
 0.90 1 M min
4 min
• From the equation for a second order
reaction we know that the slope = a k
– In this reaction a = 2.
slope  0.90  2 k
Thus the rate constant k  0.45 M
1
min
-1
85
Collision Theory of
Reaction Rates
•
Three basic events must occur for a reaction to
occur the atoms, molecules or ions must:
1. Collide.
2. Collide with enough energy to break and form
bonds.
3. Collide with the proper orientation for a reaction
to occur.
86
Collision Theory of
Reaction Rates
• One method to increase the number of
collisions and the energy necessary to break
and reform bonds is to heat the molecules.
• As an example, look at the reaction of
methane and oxygen:
CH 4(g)  O 2(g)  CO2(g)  H 2 O (g)  891 kJ
• We must start the reaction with a match.
– This provides the initial energy necessary to
break the first few bonds.
– Afterwards the reaction is self-sustaining.
87
Collision Theory of
Reaction Rates
• Illustrate the proper orientation of molecules that is
necessary for this reaction.
X2(g) + Y2(g) 2 XY(g)
• Some possible ineffective collisions are :
X
Y Y
X
Y
X
X
X X
Y Y
Y
88
Collision Theory of
Reaction Rates
• An example of an effective collision is:
X
Y
X
Y
X
Y
+
X
Y
X
Y
X
Y
89
Collision Theory of
Reaction Rates
• This picture
illustrates effective
and ineffective
molecular collisions.
90
Transition State Theory
• Transition state theory postulates that reactants form a
high energy intermediate, the transition state, which
then falls apart into the products.
• For a reaction to occur, the reactants must acquire
sufficient energy to form the transition state.
– This energy is called the activation energy or Ea.
• Look at a mechanical analog for activation energy
91
Transition State Theory
Boulder
Eactivation
Epot=mgh2
Epot = mgh
h2
Height
Cross section
of mountain
h
h1
Epot=mgh1
92
Transition State Theory
Representation of a chemical reaction.
Eactivation - a kinetic quantity
Potential
Energy
E H
a thermodynamic
quantity
X2 + Y2
2 XY
Reaction Coordinate
93
Transition State Theory
94
Transition State Theory
• The relationship between the activation energy for
forward and reverse reactions is
– Forward reaction = Ea
– Reverse reaction = Ea + E
– difference = E
95
Transition State Theory
• The distribution of molecules possessing
different energies at a given temperature is
represented in this figure.
96
Reaction Mechanisms and
the Rate-Law Expression
• Use the experimental rate-law to postulate a
molecular mechanism.
• The slowest step in a reaction mechanism is
the rate determining step.
97
Reaction Mechanisms and
the Rate-Law Expression
• Use the experimental rate-law to postulate a
mechanism.
• The slowest step in a reaction mechanism is the rate
determining step.
• Consider the iodide ion catalyzed decomposition of
hydrogen peroxide to water and oxygen.
2 H 2O 2   2 H 2O  + O 2g 
I-
98
Reaction Mechanisms and
the Rate-Law Expression
• This reaction is known to be first order in
H2O2 , first order in I- , and second order
overall.
• The mechanism for this reaction
- is thought to
Slow step
H 2 O 2 + I  IO + H 2 O
be:
Fast step
IO - + H 2 O 2  H 2 O + O 2 + I -
Overall reaction 2 H 2 O 2  2 H 2 O + O 2
Experiment al rate law

R = kH 2 O 2  I
-
99
Reaction Mechanisms and
the Rate-Law Expression
•
1.
2.
3.
Important notes about this reaction:
One hydrogen peroxide molecule and one iodide
ion are involved in the rate determining step.
The iodide ion catalyst is consumed in step 1 and
produced in step 2 in equal amounts.
Hypoiodite ion has been detected in the reaction
mixture as a short-lived reaction intermediate.
100
Reaction Mechanisms and
the Rate-Law Expression
• Ozone, O3, reacts very rapidly with nitrogen
oxide, NO, in a reaction that is first order in
each reactant and second order overall.
O3g  + NOg   NO2g  + O 2g 
Experiment al rate - law is Rate = kO3 NO
101
Reaction Mechanisms and
the Rate-Law Expression
• One possible mechanism is:
Slow step
O3 + NO  NO3 + O
Fast step
O + NO3  NO2 + O 2
Overall reaction O3 + NO  NO2 + O 2
102
Reaction Mechanisms and
the Rate-Law Expression
• A mechanism that is inconsistent with
the rate-law expression is:
Slow step O3  O 2 + O
Fast step
O + NO  NO2
Overall reaction O3 + NO  NO2  O 2
The rate - law from this mechanism is Rate = kO3 
which proves this mechanism cannot be correct.
103
Reaction Mechanisms and
the Rate-Law Expression
•
1.
2.
Experimentally determined reaction orders
indicate the number of molecules involved in:
the slow step only, or
the slow step and the equilibrium steps preceding
the slow step.
104
Temperature:
The Arrhenius Equation
• Svante Arrhenius developed this relationship among
(1) the temperature (T), (2) the activation energy
(Ea), and (3) the specific rate constant (k).
k = Ae
or
-
Ea
RT
Ea
ln k = ln A RT
105
Temperature:
The Arrhenius Equation
• This movie illustrates the effect of
temperature on a reaction.
106
Temperature:
The Arrhenius Equation
• If the Arrhenius equation is written for two
temperatures, T2 and T1 with T2 >T1.
Ea
ln k1  ln A RT1
and
Ea
ln k 2  ln A RT2
107
Temperature:
The Arrhenius Equation
1. Subtract one equation from the other.
Ea  Ea 
ln k 2  ln k1  ln A - ln A  

RT2  RT1 
Ea Ea
ln k 2  ln k1 
RT1 RT2
108
Temperature:
The Arrhenius Equation
2. Rearrange and solve for ln k2/k1.
k 2 Ea  1 1 
ln
   
k1 R  T1 T2 
or
k 2 E a  T2 - T1 
ln
 

k1 R  T2T1 
109
Temperature:
The Arrhenius Equation
• Consider the rate of a reaction for which
Ea=50 kJ/mol, at 20oC (293 K) and at 30oC
(303 K).
– How much
 T two
k do
E the
- T rates differ?
ln
2
k1



R  T2T1 
a
2
1
k 2 50,000 J mol  303  293 
ln 

K
J
k1 8.314 K mol  303293
k2
ln  0.677
k1
k2
 e0.677  197
. 2
k1
110
Temperature:
The Arrhenius Equation
• For reactions that have an Ea50 kJ/mol, the
rate approximately doubles for a 100C rise in
temperature, near room temperature.
• Consider:
2 ICl(g) + H2(g)  I2(g) + 2 HCl(g)
• The rate-law expression is known to be
R=k[ICl][H2].
0
-1 -1
At 230 C, k = 0.163 M s
0
-1 -1
At 240 C, k = 0.348 M s
k approximately doubles
111
Catalysts
• Catalysts change reaction rates by
providing an alternative reaction pathway
with a different activation energy.
112
Catalysts
• Homogeneous catalysts exist in same phase
as the reactants.
• Heterogeneous catalysts exist in different
phases than the reactants.
– Catalysts are often solids.
113
Catalysts
• Examples of commercial catalyst systems
include:
2 C8 H18 g  +25 O 2 g  16 CO 2 g   18 H 2O g 
NiO and Pt
NiO and Pt
2 CO g  +O2 g  
 2 CO2 g 
NiO and Pt
2 NO g  
 N 2 g   O 2 g 
Automobile catalytic converter system
114
Catalysts
• This movie shows catalytic converter
chemistry on the Molecular Scale
115
Catalysts
• A second example of a catalytic system is:
2 SO 2g   O 2g   2 SO3g 
V2 O5 or NiO/Pt
Sulfuric acid preparatio n
116
Catalysts
• A third examples of a catalytic system is:
N 2g   3 H 2g    2 NH3g 
Fe or Fe 2 O 3
Haber Process
117
Catalysts
• Look at the catalytic oxidation of CO to
CO2
• Overall reaction
2 CO(g)+ O2(g) 2CO2(g)
• Absorption
CO(g) CO(surface) + O2(g)
O2(g) O2(surface)
• Activation
O2(surface)  O(surface)
• Reaction
CO(surface) +O(surface)  CO2(surface)
• Desorption
118
Catalysts
119