Chemical Kinetics
Chapter 13
1
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Contents and Concepts
Reaction Rates
• Definition of Reaction Rate
• Experimental Determination of Rate
• Dependence of Rate on Concentration
• Change of Concentration with Time
• Temperature and Rate; Collision and
Transition-State Theories
• Arrhenius Equation
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13 | 2
Reaction Mechanisms
7.Elementary Reactions
8.The Rate Law and the Mechanism
9.Catalysis
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Chemical kinetics is the study of reaction
rates, including how reaction rates change
with varying conditions and which molecular
events occur during the overall reaction.
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Which conditions will affect the rate of a
reaction?
Four variables affect the rate of reaction:
1.
2.
3.
4.
The concentrations of the reactants
The concentration of the catalyst
The temperature at which the reaction occurs
The surface area of the solid reactant or
catalyst
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Concentrations of Reactants
Often the rate of reaction increases when the
concentration of a reactant is increased. In
some reactions, however, the rate is
unaffected by the concentration of a particular
reactant, as long as it is present at some
concentration.
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Concentration of Catalyst
Hydrogen peroxide, H2O2, decomposes rapidly in the
presence of HBr, giving oxygen and water. Some of the
HBr is oxidized to give the orange Br2, as can be seen
above.
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Temperature at Which Reaction Occurs
Usually reactions speed up when the temperature
increases. It takes less time to boil an egg at sea
level than on a mountaintop, where water boils at
a lower temperature. Reactions during cooking
go faster at higher temperature.
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Each test tube contains
potassium permanganate,
KMnO4, and oxalic acid,
H2C2O4, at the same
concentrations. Permanganate
ion oxidizes oxalic acid to
CO2 and H2O.
Top: One test tube was placed
in a beaker of warm water
(40°C); the other was kept at
room temperature (20°C).
Bottom: After 10 minutes, the
test tube at 40°C showed a
noticeable reaction, whereas
the other did not.
13 | 9
Surface Area of a Solid Reactant
or Catalyst
If a reaction involves a solid with a
gas or liquid, then the surface area
of the solid affects the reaction rate.
Because the reaction occurs at the
surface of the solid, the rate
increases with increasing surface
area. For example, a wood fire
burns faster if the logs are chopped
into smaller pieces. Similarly, the
surface area of a solid catalyst is
important to the rate of reaction.
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Chemical Kinetics
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
A
B
D[A]
rate = Dt
D[A] = change in concentration of A over
time period Dt
D[B]
rate =
Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
11
Reaction rate is the increase in molar
concentration of product of a reaction per unit
time or the decrease in molar concentration of
reactant per unit time. The unit is usually mol/(L
• s) or M/s.
13 | 12
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A
B
D[A]
rate = Dt
D[B]
rate =
Dt
13
2N2O5(g)  4NO2(g) + O2(g)
We can express these changes in concentration as
follows. Square brackets mean molarity.
Δ[O 2 ]
Rate of formation of O2 =
Δt
Δ[NO 2 ]
Rate of formation of NO2 =
Δt
Δ[N 2 O 5 ]

Rate of decomposition of N2O5 =
Δt
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red-brown
2Br- (aq) + 2H+ (aq) + CO2 (g)
Br2 (aq) + HCOOH (aq)
time
t 1< t 2 < t 3
393 nm
light
Detector
D[Br2] a D Absorption
15
Br2 (aq) + HCOOH (aq)
2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
[Br2]final – [Br2]initial
D[Br2]
average rate = =Dt
tfinal - tinitial
instantaneous rate = rate for specific instance in time
16
rate a [Br2]
rate = k [Br2]
rate
= rate constant
k=
[Br2]
= 3.50 x 10-3 s-1
17
The average rate
is the slope of
the hypotenuse
of the triangle
formed.
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Calculate the average rate of formation of O2
in the following reaction during the time
interval from 1200 s to 1800 s. At 1200 s,
[O2] = 0.0036 M; at 1800 s, [O2] = 0.0048 M.
Δ[O 2 ]  0.0048  0.0036  M

Δt
1800  1200  s
0.0012 M

600 s
6
 2.0  10 M/s
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?
2H2O2 (aq)
2H2O (l) + O2 (g)
PV = nRT
n
P=
RT = [O2]RT
V
1
[O2] =
P
RT
D[O2]
1 DP
rate =
=
RT Dt
Dt
measure DP over time
20
21
Reaction Rates and Stoichiometry
2A
B
Two moles of A disappear for each mole of B that is formed.
1 D[A]
rate = 2 Dt
aA + bB
D[B]
rate =
Dt
cC + dD
1 D[A]
1 D[B]
1 D[C]
1 D[D]
rate = ==
=
a Dt
b Dt
c Dt
d Dt
22
Peroxydisulfate ion oxidizes iodide ion to
triiodide ion, I3−. The reaction is
S2O82−(aq) + 3I−(aq) 
2SO42−(aq) + I3−(aq)
How is the rate of reaction that is expressed
as the rate of formation of I3− related to the
rate of reaction of I−?
Δ[I3 ]
1 Δ[I ]

Δt
3 Δt
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Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (g)
D[CH4]
D[CO2]
1 D[O2]
1 D[H2O]
rate = =
==
Dt
Dt
Dt
2 Dt
2
24
Rates are determined experimentally in a variety of
ways.
For example, samples can be taken and analyzed
from the reaction at several different intervals.
Continuously following the reaction is more
convenient. This can be done by measuring
pressure, as shown on the next slide, or by
measuring light absorbance when a color change
is part of the reaction.
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The reaction rate usually depends on the
concentration of one or more reactant. This
relationship must be determined by experiment.
This information is captured in the rate law, an
equation that relates the rate of a reaction to the
concentration of a reactant (and catalyst) raised to
various powers. The proportionality constant, k, is
the rate constant.
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For the generic reaction
aA + bB +cC  dD + eE
the rate law is
Rate = k[A]m[B]n[C]p
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The Rate Law
The rate law expresses the relationship of the rate of a reaction
to the rate constant and the concentrations of the reactants
raised to some powers.
aA + bB
cC + dD
Rate = k [A]x[B]y
Reaction is xth order in A
Reaction is yth order in B
Reaction is (x +y)th order overall
29
The reaction order with respect to a specific
reactant is the exponent of that species in the
experimentally determined rate law.
The overall order of a reaction is the sum of the
orders of the reactant species from the
experimentally determined rate law.
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Rate Laws
•
Rate laws are always determined experimentally.
•
Reaction order is always defined in terms of reactant
(not product) concentrations.
•
The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the balanced
chemical equation.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2] 1
31
For the reaction
the rate law is
Rate = k[NO2][F2]
In this case, the exponent for both NO2 and F2 is 1 (which
does not need to be written).
We say the reaction is first order in each. The reaction is
second order overall.
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Hydrogen peroxide oxidizes iodide ion in
acidic solution:
H2O2(aq) + 3I−(aq) + 2H+(aq) 
I3−(aq) + 2H2O(l)
The rate law for the reaction is
Rate = k[H2O2][I−]
a.
b.
What is the order of reaction with
respect to each reactant species?
What is the overall order?
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H2O2(aq) + 3I−(aq) + 2H+(aq)  I3−(aq) + 2H2O(l)
Given: Rate = k[H2O2][I−]
a.
The reaction is first order in H2O2.
The reaction is first order in I−.
The reaction is zero order in H+.
b.
The reaction is second order overall.
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Determining rate Law by the Initial Rate Method
This method measures the initial rate of reaction using
various starting concentrations, all measured at the same
temperature.
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F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles
x=1
Quadruple [ClO2] with [F2] constant
rate = k [F2][ClO2]
Rate quadruples
y=1
36
The decomposition of N2O5 have the following
data:
2N2O5(g)  2N2(g) + 5O2(g)
Initial N2O5
Concentration
Initial Rate of
Disappearance of N2O5
Experiment 1
1.0 × 10−2 M
4.8 × 10−6 M/s
Experiment 2
2.0 × 10−2 M
9.6 × 10−6 M/s
Note that when the concentration doubles from
experiment 1 to 2, the rate doubles.
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Examining the effect of doubling the initial
concentration gives us the order in that reactant.
When rate is multiplied by
m is
½
−1
1
0
2
1
4
2
In this case, when the concentration doubles, the
rate doubles, so m = 1.
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?
2NO(g) + O2(g)  2NO2(g)
Obtain the rate law and the rate
constant using the following data for
the initial rates of disappearance of NO:
Initial Rate of
Exp. [NO]0, M [O2]0, M Reaction of NO, M/s
1 0.0125
0.0253
0.0281
2
0.0250
0.0253
0.112
3
0.0125
0.0506
0.0561
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To find the order in NO, we will first identify two
experiments in which the concentration of O2
remains constant. Then, we will divide the rate
laws for those two experiments. Finally, we will
solve the equation to find the order in NO.
To find the order in O2, we will repeat this
procedure, this time choosing experiments in
which the concentration of NO remains constant.
Once the order in NO and the order in O2 are
known, data from any experiment can be
substituted into the rate law. The rate law can then
be solved for the rate constant k.
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[O2] remains constant in experiments 1 and 2.
The generic rate law is rate = k[NO]m[O2]n. We will
put the result from experiment 2 in the numerator
because its rate is larger than that of experiment 1.
M
0.112
m
n
s  0.0250 M  0.0253 M 
M 0.0125 M m 0.0253 M n
0.0281
s
4  2m
m2
The reaction is second order in NO.
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[NO] remains constant in experiments 1 and 3.
The generic rate law is rate = k[NO]m[O2]n. We will
put the result from experiment 3 in the numerator
because its rate is larger than that of experiment 1.
M
0.0.0561
m
n
s  0.0125 M  0.0506 M 
m
n
M




0.0125
M
0.0253
M
0.0281
s
2  2n
n 1
The reaction is first order in O 2 .
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It is also possible to find the order by examining
the data qualitatively.
Again choosing experiments 1 and 2, we note
that the concentration of NO is doubled and the
rate is quadrupled. That means the reaction is
second order in NO.
Choosing experiments 1 and 3, we note that the
concentration of O2 is doubled and the rate is
doubled. That means the reaction is first order in
O2.
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The rate law is rate = k[NO]2[O2].
Solving for k , we obtain
k
rate
NO 2 O 2 
Using data from experiment 1:
M
0.0281
3
7.11

10
s
k

2
2
M
s
0.0125 M  0.0253 M 
Using data from experiment 2: k = 7.08 × 103/M2  s
Using data from experiment 3: k = 7.10 × 103/M2  s
These values are the same to 2 significant figures.
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?
Iron(II) is oxidized to iron(III) by chlorine
in an acidic solution:
2Fe2+(aq) + Cl2(aq) 
2Fe3+ (aq) + 2Cl−(aq)
What is the reaction order with respect
to Fe2+, Cl2, and H+? What is the rate
law and the relative rate constant?
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The following data were collected. The rates given
are relative, not actual.
Exp. [Fe2+], M
[Cl2], M [H+], M Rate, M/s
1
0.0020
0.0020
1.0
1.0 × 10−5
2
0.0040
0.0020
1.0
2.0 × 10−5
3
0.0020
0.0040
1.0
2.0 × 10−5
4
0.0040
0.0040
1.0
4.0 × 10−5
5
0.0020
0.0020
0.5
2.0 × 10−5
6
0.0020
0.0020
0.1
1.0 × 10−4
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The generic rate law is
Rate = k[Fe2+]m[Cl2]n[H+]p
To find the order in Fe2+, we choose two
experiments in which both [Cl2] and [H+] remain
constant: 1 and 2, or 3 and 4.
Exp [Fe2+]
[Cl2]
[H+] Initial Rate
1 0.0020 0.0020
1.0 1.0 × 10-5
2 0.0040 0.0020
1.0 2.0 × 10-5
3 0.0020 0.0040
1.0 2.0 × 10-5
4 0.0040 0.0040
1.0 4.0 × 10-5
In each case, the concentration of Fe2+ doubles
and the rate doubles.
The reaction is first order in Fe2+; m = 1.
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To find the order in Cl2, we choose two
experiments in which both [Fe2+] and [H+] remain
constant: 1 and 3, or 2 and 4.
Exp [Fe2+]
[Cl2]
[H+] Initial Rate
1 0.0020 0.0020
1.0 1.0 × 10-5
3 0.0020 0.0040
1.0 2.0 × 10-5
2 0.0040 0.0020
1.0 2.0 × 10-5
4 0.0040 0.0040
1.0 4.0 × 10-5
In each case, the concentration of Cl2 doubles and
the rate doubles.
The reaction is first order in Cl2; n = 1.
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To find the order in H+, we choose two
experiments in which both [Fe2+] and [Cl2] remain
constant: 1 and 5, 1 and 6, or 5 and 6.
1
0.0020
0.0020
1.0 1.0 × 10-5
5
0.0020
0.0020
0.5 2.0 × 10-5
From experiment 5 to experiment 1, the
concentration doubles while the rate is halved.
1
0.0020
0.0020
1.0 1.0 × 10-5
6
0.0020
0.0020
0.1 1.0 × 10-4
From experiment 6 to experiment 1, the
concentration is increased 10 times, the rate is cut
to one-tenth of the original rate.
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5
0.0020
0.0020
0.5 2.0 × 10−5
6
0.0020
0.0020
0.1 1.0 × 10−4
From experiment 6 to experiment 5, the
concentration is increased 5 times; the rate is cut
to one-fifth of the original rate.
Each of these describes the order in H+ as p = −1.
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The rate law is rate = k[Fe2+][Cl2][H+]−1 .
Another way to express this is
Rate 
k Fe2  Cl2 
H 
Using data from experiment 1:
M
1.0  10
1.0 M  2.50
rate H
s
k


2
Fe  Cl2   0.002 M  0.002 M  M  s
5
The same value for k is obtained using each
experiment.
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Determine the rate law and calculate the rate constant for the
following reaction from the following data:
S2O82- (aq) + 3I- (aq)
2SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate
(M/s)
1
0.08
0.034
2.2 x 10-4
2
0.08
0.017
1.1 x 10-4
3
0.16
0.017
2.2 x 10-4
rate = k [S2O82-]x[I-]y
y=1
x=1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
2.2 x 10-4 M/s
rate
k=
=
= 0.08/M•s
2[S2O8 ][I ] (0.08 M)(0.034 M)
52
The rate law tells us the relationship between the
rate and the concentrations of reactants and
catalysts.
To find concentrations at various times, we need to
use the integrated rate law. The form used
depends on the order of reaction in that reactant.
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First-Order Reactions
A
k=
product
D[A]
rate = Dt
rate
M/s
=
= 1/s or s-1
M
[A]
[A] = [A]0e−kt
rate = k [A]
D[A]
= k [A]
Dt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
ln[A] = ln[A]0 - kt
55
Graphical Determination of k
2N2O5
4NO2 (g) + O2 (g)
56
The reaction 2A
B is first order in A with a rate constant
of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease
from 0.88 M to 0.14 M ?
[A]0 = 0.88 M
ln[A] = ln[A]0 - kt
[A] = 0.14 M
kt = ln[A]0 – ln[A]
ln[A]0 – ln[A]
=
t=
k
ln
[A]0
[A]
k
ln
=
0.88 M
0.14 M
2.8 x
10-2
s-1
= 66 s
57
First-Order Reactions
The half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln
t½ =
[A]0
[A]0/2
k
ln 2
0.693
=
=
k
k
What is the half-life of N2O5 if it decomposes with a rate constant
of 5.7 x 10-4 s-1?
0.693
t½ = ln 2 =
= 1200 s = 20 minutes
-4
-1
k
5.7 x 10 s
How do you know decomposition is first order?
units of k (s-1)
58
First-order reaction
A
product
# of
half-lives [A] = [A]0/n
1
2
2
4
3
8
4
16
59
Cyclopropane is used as an anesthetic. The
isomerization of cyclopropane to propene
is a first-order reaction with a rate constant
of 9.2/s. If an initial sample of
cyclopropane has a concentration of 6.00
M, what will the cyclopropane
concentration be after 1.00 s?
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The rate law is
Rate = k[cyclopropane]
k = 9.2/s
[A]0 = 6.00 M
t = 1.00 s
[A]t = ?
The reaction is first order, so the integrated rate law
is
A
 t
ln
 A 0
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  kt
13 | 61
 A t
ln
 A 0
9.2

 1.00 s   9.2
s
A t  e 9.2  1.01  10 4
A 0
4
At  1.0110 6.00 M 
A t  6.110
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4
M
Second-Order Reactions
A
product
D[A]
rate = Dt
rate
M/s
=
= 1/M•s
k=
2
2
M
[A]
1
1
=
+ kt
[A]
[A]0
rate = k [A]2
D[A]
= k [A]2
Dt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
1
t½ =
k[A]0
63
Zero-Order Reactions
A
product
D[A]
rate = Dt
D[A]
=k
Dt
rate
= M/s
k=
0
[A]
[A] = [A]0 - kt
rate = k [A]0 = k
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
[A]0
t½ =
2k
64
Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
0
Rate Law
rate = k
1
rate = k [A]
2
[A]2
rate = k
Concentration-Time
Equation
[A] = [A]0 - kt
ln[A] = ln[A]0 - kt
1
1
=
+ kt
[A]
[A]0
Half-Life
t½ =
[A]0
2k
t½ = ln 2
k
1
t½ =
k[A]0
65
Let’s look again at the integrated rate laws:
y = mx + b
y = mx + b
y = mx + b
In each case, the rate law is in the form of y = mx +
b, allowing us to use the slope and intercept to find
the values.
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Zero Order : [A]t   kt  [A]0
For a zero-order reaction, a plot of [A]t versus t is
linear. The y-intercept is [A]0.
First Order : ln [A]t   kt  ln [A]0
For a first-order reaction, a plot of ln[A]t versus t is
linear. The graph crosses the origin (b = 0).
1
1
Second Order :
 kt 
[A]t
[A]0
For a second-order reaction, a plot of 1/[A]t versus t
is linear. The y-intercept is 1/[A]0.
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13 | 67
Graphs of concentration and time can also be
used to determine the order of reaction in that
reactant.
•
•
•
For zero-order reactions, [A] versus t is
linear.
For first-order reactions, ln[A] versus t is
linear.
For second order reactions, 1/[A] versus t is
linear.
This is illustrated on the next slide for the
decomposition of NO2.
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13 | 68
Left: The plot of ln[NO2] versus t is not linear, so
the reaction is not first order.
Right: The plot of 1/[NO2] versus t is linear, so the
reaction is second order in NO2.
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13 | 69
Rate Constant and Temperature
The rate constant depends strongly on
temperature. How can we explain this
relationship?
Collision theory assumes that reactant molecules
must collide with an energy greater than some
minimum value and with the proper orientation.
The minimum energy is called the activation
energy, Ea.
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13 | 70
We will now explore the effect of a temperature
increase on each of the three requirements for a
reaction to occur.
The rate constant can be given by the equation
k = Zfp
where
Z = collision frequency
f = fraction of collisions with the minimum energy
p = orientation factor
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13 | 71
Certainly, Z will increase with temperature, as the
average velocity of the molecules increases with
temperature.
However, this factor alone cannot explain the
dramatic effect of temperature changes.
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13 | 72
The fraction of molecular collisions having the
minimum energy required is given by f:
f e

Ea
RT
1

e
Ea
RT
Now we can explain the dramatic impact of
temperature. The fraction of collisions having the
minimum energy increases exponentially with
temperature.
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13 | 73
The reaction rate also depends on p, the proper
orientation for the collision. This factor is
independent of temperature.
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13 | 74
Importance of Molecular Orientation in
the Reaction of NO and Cl2
(A) NO approaches with its N atom toward Cl2,
and an NOCl bond forms. Also, the angle of
approach is close to that in the product NOCl.
(B) NO approaches with its O atom toward Cl2. No
NOCl bond can form, so NO and Cl2 collide and
then fly apart.
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13 | 75
Transition-state theory explains the reaction
resulting from the collision of two molecules in
terms of an activated complex (transition state),
an unstable grouping of atoms that can break up to
form products.
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13 | 76
The potential energy diagram for a reaction
visually illustrates the changes in energy that
occur.
The next slide shows the diagram for the reaction
NO + Cl2  NOCl2‡  NOCl + Cl
where NOCl2‡ indicates the activated complex.
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13 | 77
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13 | 78
The reaction of NO with Cl2 is an endothermic
reaction.
The next slide shows the curve for a generic
exothermic reaction.
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13 | 79
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13 | 80
A+B
Exothermic Reaction
+
AB+
C+D
Endothermic Reaction
The activation energy (Ea ) is the minimum amount of
energy required to initiate a chemical reaction.
81
?
Sketch a potential energy diagram for
the decomposition of nitrous oxide.
N2O(g)  N2(g) + O(g)
The activation energy for the forward
reaction is 251 kJ; DHo = +167 kJ.
Label your diagram appropriately.
What is the reverse activation energy?
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13 | 82
E
n
e
r
g
y
Ea(reverse)
= 84 kJ
Ea= 251 kJ
N2 + O
Products
p
e
r
m
o
l
DH = 167 kJ
N2O
Reactants
Progress of reaction
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13 | 83
Temperature Dependence of the Rate Constant
k  A  e ( Ea / RT )
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor
Alternate format:
Ea 1
ln k = + lnA
R T
84
Alternate Form of the Arrhenius Equation
At two temperatures, T1 and T2
or
85
Importance of Molecular Orientation
effective collision
ineffective collision
86
?
In a series of experiments on the
decomposition of dinitrogen pentoxide,
N2O5, rate constants were determined
at two different temperatures:
•
At 35°C, the rate constant was
1.4 × 10−4/s.
•
At 45°C, the rate constant was
5.0 × 10−4/s.
What is the activation energy?
What is the value of the rate constant
at 55°C?
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13 | 87
This is actually two problems.
First, we will use the Arrhenius equation to find Ea.
Then, we will use the Arrhenius equation with Ea to
find the rate constant at a new temperature.
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13 | 88
T1 = 35°C = 308 K
k1 = 1.4 × 10−4/s
 k2
ln 
 k1
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T2 = 45°C = 318 K
k2 = 5.0 × 10−4/s
 Ea  1
1
 
 

 R  T1 T2 
13 | 89
 5.0  104 /s 
Ea
1 
 1
ln 




4
J
1.4

10
/s
308
K
318
K



 8.315
mol  K
We will solve the left side and rearrange the right side.




10 K


1.273  E a


J 
(318 K)(308 K) 
  8.315
mol  K 


J 

1.273 8.315
(318 K)(308 K)
mol  K 

Ea 
10 K
Ea  1.0  105 J/mol
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13 | 90
T1 = 35°C = 308 K
k1 = 1.4 × 10−4/s
Ea = 1.04 × 105 J/mol
 k2
ln 
 k1
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T2 = 55°C = 328 K
k2 = ?
 Ea  1
1
 
 

 R  T1 T2 
13 | 91
J
1.04  10
k2
1 


 1
mol
ln 





4
J
 1.4  10 /s  8.315
 308 K 328 K 
mol  K
J
5
1.04  10
k
20 K




2
mol
ln 




4
J
 1.4  10 /s  8.315
 308 K  328 K 
mol  K
k2


ln 
 2.476

4
 1.4  10 /s 
k2
 11.90
4
1.4  10 /s
5
k2  1.7  103 /s
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13 | 92
Reaction Mechanisms
The overall progress of a chemical reaction can be represented
at the molecular level by a series of simple elementary steps
or elementary reactions.
The sequence of elementary steps that leads to product
formation is the reaction mechanism.
2NO (g) + O2 (g)
2NO2 (g)
N2O2 is detected during the reaction!
Elementary step:
NO + NO
N 2O 2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
93
2NO (g) + O2 (g)
2NO2 (g)
Mechanism:
94
Intermediates are species that appear in a reaction
mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step
and consumed in a later elementary step.
Elementary step:
NO + NO
N 2O 2
+ Elementary step:
N2O2 + O2
2NO2
Overall reaction:
2NO + O2
2NO2
The molecularity of a reaction is the number of molecules
reacting in an elementary step.
•
Unimolecular reaction – elementary step with 1 molecule
•
Bimolecular reaction – elementary step with 2 molecules
•
Termolecular reaction – elementary step with 3 molecules
95
Rate Laws and Elementary Steps
Unimolecular reaction
A
products
rate = k [A]
Bimolecular reaction
A+B
products
rate = k [A][B]
Bimolecular reaction
A+A
products
rate = k [A]2
Writing plausible reaction mechanisms:
•
The sum of the elementary steps must give the overall
balanced equation for the reaction.
•
The rate-determining step should predict the same rate
law that is determined experimentally.
The rate-determining step is the slowest step in the
sequence of steps leading to product formation.
96
Sequence of Steps in Studying a Reaction Mechanism
97
The experimental rate law for the reaction between NO2 and CO
to produce NO and CO2 is rate = k[NO2]2. The reaction is
believed to occur via two steps:
Step 1:
NO2 + NO2
NO + NO3
Step 2:
NO3 + CO
NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO
NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2
98
Chemistry In Action: Femtochemistry
CH2
CH2
CH2
CH2
2 CH2
CH2
CH2
CH2
•
CH2
CH2
CH2
CH2
CH2
2 CH2
CH2
•
CH2
99
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13 | 100
In heterogeneous catalysis, the reactants and the catalysts
are in different phases.
•
Haber synthesis of ammonia
•
Ostwald process for the production of nitric acid
•
Catalytic converters
In homogeneous catalysis, the reactants and the catalysts
are dispersed in a single phase, usually liquid.
•
Acid catalysis
•
Base catalysis
101
Haber Process
N2 (g) + 3H2 (g)
Fe/Al2O3/K2O
catalyst
2NH3 (g)
102
Ostwald Process
4NH3 (g) + 5O2 (g)
Pt catalyst
2NO (g) + O2 (g)
2NO2 (g) + H2O (l)
4NO (g) + 6H2O (g)
2NO2 (g)
HNO2 (aq) + HNO3 (aq)
Pt-Rh catalysts used
in Ostwald process
103
Catalytic Converters
catalytic
CO + Unburned Hydrocarbons + O2 converter
CO2 + H2O
catalytic
2NO + 2NO2 converter
2N2 + 3O2
104
Enzyme Catalysis
105
Binding of Glucose to Hexokinase
106
Enzyme Kinetics
D[P]
rate =
Dt
rate = k [ES]
107