Chapter 14
Chemical
Kinetics
14.1 Reaction Rates
• Kinetics: the study of how fast
reactions take place
• Some reactions are fast
(photosynthesis)
• Some reactions are slow (conversion of
diamond to graphite)
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Importance of Studying
Reaction Rates
• Speed up desirable reactions
• Minimize damage by undesirable
reactions
• Useful in drug design, pollution control,
and food processing
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Rate of Reaction
• Expressed as either:
– Rate of disappearance of reactants
(decrease or negative)
OR
– Rate of appearance of products (increase
or positive)
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Average Reaction Rate
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Average Reaction Rate
• Equation
A B
• rate =   [A] or  [B]
t
t
• Why the negative on [A]?
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Average Reaction Rate
Br2(aq) + HCOOH(aq)2Br(aq) + 2H+(aq) + CO2(g)
Note: Br2 disappears over time
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Average Reaction Rate
Br2(aq) + HCOOH(aq) 2Br(aq) + 2H+(aq) + CO2(g)
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Calculate Average Rate
• Avg. rate =
• Avg. rate =
 [Br2 ]
initial
t
 t
final
initial
[Br2 ]
final
0.0101 M  0.0120 M
= 3.80 M
50 s  0 s
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/s
Average Rate
• Average rate depends on time interval
• Plot of [Br2] vs time = curve
• Plot of Rate vs [Br2] = straight line
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Instantaneous Rate
• Instantaneous: rate at a specific
instance in time (slope of a tangent to
the curve)
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Rate Constant
• Using data from Table 14.1 - what can
you conclude?
Time (s)
50
250
[Br2]
0.0101
0.00596
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Rate (M/s)
3.52 x 105
1.75 x 105
Rate Constant
Answer:
• When the [Br2] is halved; the rate is
halved
• Rate is directly proportional to [Br2]
• rate = k [Br2]
• k = proportionality constant and is
constant as long as temp remains
constant
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Rate Constant
• Calculate the value of the rate constant for
any set of data and get basically the same
answer!
• k = rate / [Br2]
3.52 x 105 M/s
k 
 3.5x 103 /s
0.0101 M
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Stoichiometry and Reaction
Rate
• When stoichiometric ratios are not 1:1
rate of reaction is expressed as follows
General equation:
aA + bB  cC + dD
rate = 
1  [A]
1  [B] 1  [C] 1  [D]



a t
b t
c t
d t
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Stoichiometry and Reaction
Rate
• Write the rate expression for the
following reaction
2NO(g) + O2(g)  2NO2(g)
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Stoichiometry and Reaction
Rate
4PH3(g)  P4(g) + 6H2(g)
If molecular hydrogen is formed at a rate of
0.168 M/s, at what rate is P4 being
produced?
1  [PH3] 1  [P4] 1  [H2]



4 t
1 t
6 t
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Stoichiometry and Reaction
Rate
4PH3(g)  P4(g) + 6H2(g)
1  [H2]
1

(0.168 M /s)  0.028 M /s
6 t
6
1  [P4]
 0.028 M /s
1 t
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14.2 Dependence of Reaction
Rate on Reactant
Concentration
• Rate law expression
For the general equation:
aA + bB  cC + dD
rate law = k[A]x[B]y
k = proportionality constant
x and y = the order of the reaction with
respect to each reactant
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Order
• Exponents represent order
• Only determined via experimental data
– 1st order - rate directly proportional to
concentration
– 2nd order - exponential relationship
– 0 order - no relationship
• Sum of exponents (order) indicates
overall reaction order
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Experimental Determination of
Rate Law
• Method of initial rates - examine
instantaneous rate data at beginning of
reaction
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rate = k[F2]x [ClO2]y
Find order (exponents) by comparing data
Exp. 1 and 3: [ClO2] is held constant
[F2]
3  0.20 M  2
[F2]
0.10 M
1
[rate]
3 M /s
2.4

10
3 
2
[rate]
1.2  103 M /s
1
1st order with respect to [F2] (rate and M directly related)
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rate = k[F2]1[ClO2]y
Find order (exponents) by comparing data
Exp. 1 and 2: [F2] is held constant
[ClO2]
2  0.040 M  4
[ClO2]
0.010 M
1
[rate]
3 M /s
4.8

10
2 
4
[rate]
1.2  103 M /s
1
1st order with respect to [ClO2] (rate and M directly related)
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rate = k[F2]1[ClO2]1 overall order = 2
Find k (use any set of data)
k
[rate]
[A] [B]2

2.1  104 M /s
(0.10 M ) (0.015 M ) 2
 9.3 M 2  s1
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Determining Rate Law
Exp.
[A] (M)
[B] (M)
1
0.10
0.015
Initial
Rate
(M/s)
2.1 x 104
2
0.20
0.015
4.2 x 104
3
0.10
0.030
8.4 x 102
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What is Different?
• In experiment 1 and 2; [B] is constant;
[A] doubles and rate doubles - the
reaction is 1st order with respect to [A]
• In experiment 1 and 3; [A] is constant;
[B] doubles but the rate quadruples!
This means that the reaction is 2nd
order with respect to [B]
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Calculate the Rate Constant
•
•
•
•
Rate = k[A] [B]2
The rxn is 1st order w/ respect to [A]
The rxn is 2nd order w/ respect to [B]
The rxn is 3rd order overall (1 + 2)
k
[rate]
[A] [B]2

2.1  104 M /s
(0.10 M) (0.015 M )2
 9.3 M 2  s1
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14.3 Dependence of Reactant
Concentration on Time
• First-Order reactions may be
expressed in several ways
• Example: A  products
rate = k[A]
 [A]
rate  
t
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Integrated Rate Law
(First order)
When the two expressions are set equal to
each other,we get an expression that can be
rearranged in the form of a straight line.
ln  A  kt  ln  A0
y
=
mx
+
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b
Graphical Methods
• Given concentration and time data,
graphing can determine order
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Integrated Rate Law
(First order)
• For a 1st order reaction, a plot of ln [A]
vs time yields a straight line
• The slope = k (the rate constant)
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Try Graphing
Time (s)
P (mmHg)
0
284
100
220
150
193
200
170
250
150
300
132
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Graphing
• Plot ln [Pressure] on y-axis and time on
x-axis
• If the plot is a straight line, then the
integrated rate law equation can be
used to find the rate constant, k, or the
slope of the line can be calculated for
the rate constant.
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Integrated Rate Law
• The rate constant for the reaction
2A  B is 7.5 x 103 s1 at 110C. The
reaction is 1st order in A. How long (in
seconds) will it take for [A] to decrease
from 1.25 M to 0.71 M?
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• Another form of the integrated rate law
[A]
t   kt
ln
[A]
0
ln
(0.71 M)
  7.5  103 s1 (t )
(1.25 M)
t  75 s
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Your Turn!
• Consider the same first order reaction
2A  B, for which k = 7.5 x 103 s1 at
110C. With a starting concentration of
[A] = 2.25 M, what will [A] be after 2.0
minutes?
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Half-Life (1st order)
• Half-life: the time
that it takes for the
reactant
concentration to
drop to half of its
original value.
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Calculating First Order
Half-life
• Half-life is the time that it takes for the
reactant concentration to drop to half of
its original value.
• The expression for half-life is simplified
as
t1/ 2 
0.693
k
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Half-Life
• The decomposition of ethane (C2H6) to
methyl radicals (CH3) is a first order
reaction with a rate constant of
5.36 x 104 s1 at 700 C.
C2H6  2CH3
Calculate the half-life in minutes.
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Half-Life
t1/2
0.693

 1293 s
4
1
5.36  10 s
1 min
1293 s 
 21.5 min
60 s
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Dependence of Reactant
Concentration on Time
Second-order reactions may be expressed in
several ways
• Example: A  product
rate = k[A]2
rate  
 [A]
t
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Integrated Rate Law for
Second Order Reactions
Again, the relationships can be combined to
yield the following relationship in the form of a
straight line
1
1
 kt 
 A
 A 0
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Integrated Rate Law for
Second Order Reactions
• For a 2nd order reaction, a plot of 1/[A]
vs time yields a straight line
• The slope = k (the rate constant)
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Calculating Second-Order
Half-life
The expression for half-life is simplified as
t1/2
1

k [A]0
Note: half-life for 2nd order is inversely
proportional to the initial reaction
concentration
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Calculating Second Order
Half-life
I(g) + I(g)  I2(g)
The reaction is second order and has a rate
constant of 7.0 x 109 M1 s1 at 23C.
a) If the initial [I] is 0.086 M, calculate the
concentration after 2.0 min.
b) Calculate the half-life of the reaction
when the initial [I] is 0.60 M and when the
[I] is 0.42 M.
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a) Use integrated rate equation for 2nd
order
1
1
9
1 1
 (7.0  10 M s ) 120s 
A
0.086 M 
1
 8.4  10 M
1
12
[A] 
 1.2  10 M
11
1
8.4  10 M
11
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b) Use the half-life formula for 2nd order
(note: half-life does not remain constant for a
2nd-order reaction!)
t1/2 
t1/2 
1
(7.0  10 M
9
1
1
s ) (0.60 M )
1
(7.0  10 M
9
1
1
s ) (0.42 M )
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 2.4  10
 3.4  10
10
10
s
s
Zero Order
Zero Order reactions may exist but are
relatively rare
• Example: A  product
rate = k[A]0 = k
• Thus, a plot of [A] vs time yields a
straight line.
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Summary of Orders
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14.4 Dependence of Reaction
Rate on Temperature
• Most reactions occur faster at a higher
temperature.
• How does temperature alter rate?
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Collision Theory
• Particles must collide in order to react
• The greater frequency of collisions, the
higher the reaction rate
• Only two particles may react at one time
• Many factors must be met:
– Orientation
– Energy needed to break bonds (activation
energy)
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Collision Theory
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Collision Theory
• Though it seems simple, not all
collisions are effective collisions
• Effective collisions: a collision that
does result in a reaction
• An activated complex (transition state)
forms in an effective collision
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Activation Energy
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The Arrhenius Equation
• The dependence of the rate constant of a
reaction on temperature can be expressed
k  AeEa / RT
Ea = activation energy
R = universal gas constant
A = frequency factor T = Kelvin temp
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Arrhenius Equation
Ea
ln  k   
R
1
   ln  A 
t 
In the form of a straight line…what plot will give a straight line?
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Arrhenius Equation
• A plot of ln k vs 1/T will give a straight
line
• The slope of line will equal Ea/R
• The activation energy may be found by
multiplying the slope by “R”
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Graphing with Arrhenius
• Rate constants for the reaction
CO(g) + NO2(g)  CO2(g) + NO(g)
Were measured at four different
temperatures. Plot the data to
determine activation energy in kJ/mol.
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Graphing
k (M1 s1)
T (Kelvin)
0.0521
288
0.101
298
0.184
308
0.332
318
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Graphing
Steps:
• Make a column of ln k data
• Make a column of inverse temp (1/T)
• Plot ln k vs 1/T
• Calculate the slope
• Multiply slope by R
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Arrhenius Another Way!
• Another useful arrangement of the
Arrhenius equation enables calculation
of: 1) Ea (with k at two different temps)
2) the rate constant at a different
temperature (with Ea, k and temps)
1 
k1
Ea  1
ln



k2
R  T 2 T 1 
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Arrhenius Another Way!
• Use the data to calculate activation
energy of the reaction mathematically
T (Kelvin)
k (s1)
400
2.9 x 103
450
6.1 x 102
500
7.0 x 101
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Arrhenius
k1

ln

k2
Ea  R 
1
1


T1
 T2






2.9  103
 ln

6.1  102
Ea  8.3145J/Kmol 
1
1


400
 450

 91 kJ/mol
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






14.5 Reaction Mechanisms
• Most reactions occur in a series of steps
• The balanced equation does not tell us
how the reaction occurs!
• There are often a series of steps which
add together to give the overall reaction
• The series of steps is the reaction
mechanism
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Reaction Mechanisms
• Most chemical reactions occur in a
series of steps
• Energy of activation must be overcome
to form intermediates
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Reaction Mechanism
Consider:
2NO(g) + O2(g)  2NO2(g)
The reaction cannot occur in a single
step.
One proposed mechanism:
Step 1: NO + NO  N2O2
Step 2: N2O2 + O2  2NO2
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Reaction Mechanism
N2O2 is an intermediate in the reaction
mechanism
Intermediate: a substance that is
produced in an early step and
consumed in a later step
Elementary reaction: one that occurs in
a single collision of the reactant
molecules
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Reaction Mechanism
• Molecularity: the number of reactant
molecules involved in the collision
• Unimolecular: one reactant molecule
• Bimolecular: two reactant molecules
• Termolecular: three reactant molecules
(fairly rare)
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Rate Determining Step
• If the elementary reactions are known,
the order can be written from the
stoichiometric coefficients of the ratedetermining step
• Rate-determining step: the slowest
step in the mechanism
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Rate-Determining Step
• Steps of a mechanism must satisfy two
requirements
– Sum of elementary steps must equal the
overall balanced equation
– The rate law must have same rate law as
determined from experimental data
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• The decomposition of hydrogen
peroxide (2H2O2  2H2O + O2 )
may occur in the following two steps
k

¾
Step 1: H2O2 + I ¾ H2O + IO
k2

¾
¾ H O + O + I
Step 2: H2O2 + IO
2
2
1
If step 1 is the rate-determining step, then
the rate law is
rate = k1[H2O2] [I]
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• I does not appear in the overall
balanced equation
• I serves as a catalyst in the reaction it is present at the start of the reaction
and is present at the end
• IO is an intermediate
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Potential Energy Diagram
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Reaction Mechanism
• Given overall equation:
H2(g) + I2(g)  2HI(g)
k
¾¾
¾¾

Step 1: I2 ¬
2I (fast)
k
Step 2: H2 + 2I ¾k2¾ 2HI (slow)
rate = k2[H2] [I]2
This rate expression does not meet the
requirement..I is an intermediate and
should not appear in the rate expression
1
1
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• Consider the first equilibrium step: the
forward rate is equal to the reverse rate
k1[I2] = k-1 [I]2 k1/k-1 [I2] = [I]2
If we substitute for [I]2 , the rate law becomes
rate = k[H2] [I2]
This now matches the overall balanced
equation!
(When step 2 is rate-determining this
substitution is always possible)
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14.6 Catalysis
• Catalyst - a substance that increases
the rate of a chemical reaction without
being used up itself
• Provides a set of elementary steps with
more favorable kinetics than those that
exist in its absence
• Many times a catalyst lowers the
activation energy
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Reaction Pathway with
Catalyst
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Types of Catalysts
• Heterogeneous catalysts - reactants
and catalyst are in different phases
• Homogeneous catalysts - reactants
and catalysts are dispersed in single
phase
• Enzyme catalysts - biological catalysts
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Heterogeneous Catalysts
• Most important in industrial chemistry
• Used in catalytic converters in
automobiles
– Efficient catalytic converter serves two
purposes; oxidizes CO and unburned
hydrocarbons into CO2 and H2O; converts
NO and NO2 into N2 and O2
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Catalytic Converter
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Homogeneous Catalysts
• Usually dispersed in liquid phase
• Acid and base catalyses are the most
important types of homogeneous
catalysis in liquid solution
• Advantages of homogeneous catalysts
– Reactions performed at room conditions
– Less expensive
– Can be designed to function selectively
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Biological Catalysts
• Enzymes: large protein molecule that
contains one or more active sites where
interactions with substrates occur
• Enzymes are highly specific (lock and
key)
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Reaction Pathway without and
with Enzyme-Substrate
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Enzyme-Substrate Complex
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Biological Molecules
(binding of glucose to
hexokinase)
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Key Points
• Rate of reaction can be determined in
several ways
– Instantaneous rate
– Average rate
– Graphing using integrated rate laws
– Mechanisms
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Key Points
•
•
•
•
•
Write rate law expressions
Calculate rate constant with proper units
Distinguish orders: 1st, 2nd, 0 order
Calculate half-life
Collision theory and relationship to
Arrhenius equation
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Key Points
• Calculate activation energy graphically
and mathematically
• Reaction mechanisms
– Elementary reactions
– Molecularity
– Rate law from slow step
– Intermediates and catalysts
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Key Points
• Catalysis
– Heterogeneous
– Homogeneous
– Enzymes
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