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Chapter Outline
•
•
•
•
13.1 Cars, Trucks, and Air Quality
13.2 Reaction Rates
13.3 Effect of Concentration on Reaction Rates
13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
© 2014 W. W. Norton Co., Inc.
Cars, Trucks, and Air Quality
• Photochemical Smog: Mixture of gases
formed when sunlight interacts with compounds
produced in internal combustion engines
• Depends on chemical kinetics
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Important Reactions in Smog
1. N2(g) + O2(g) → 2 NO(g)
ΔH= 180.6 kJ
2. 2 NO(g) + O2(g) → 2 NO2(g) ΔH= -114.2 kJ
sunlight

3. NO2(g) 
NO(g) + O(g)
4. O2(g) + O(g) → O3(g)
5. O(g) + H2O(g) → 2 OH(g)
Note: Products of some reactions are reactants in
other reactions.
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Variations of Smog Components
Photodecomposition
of NO2 leads to high
levels of O3 in the
afternoon.
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Why Study Kinetics?
• We can control the reaction conditions to
obtain product as quickly and economically
as possible
• To understand reaction mechanisms – a
study of kinetics sheds light on how a
reaction proceeds
© 2014 W. W. Norton Co., Inc.
Factors that control reaction kinetics
• Chemical makeup of reactants and
products
• Concentration of reactants
• Temperature
• Catalysts
• Surface area
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Chapter Outline
•
•
•
•
13.1 Cars, Trucks, and Air Quality
13.2 Reaction Rates
13.3 Effect of Concentration on Reaction Rates
13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
© 2014 W. W. Norton Co., Inc.
Rates
Distance, miles 
If driving a car, rate =
change in position
∆dist d −d
= slope= ∆t = 2 1
t2 −t1
change in time
(t 2 , d2 )
rise =
𝑑2 − 𝑑1
(t 1 , d1 )
run = 𝑡2 − 𝑡1
Time, hours 
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Reaction Rates
Concentration, M

∆[A] [A] −[A]
change in conc.
rate = change in time = slope = ∆t = t2 −t 1
2 1
(t 2 , [A]2 )
rise =
[𝐴]2 −[𝐴]1
(t 1 , [A]1 )
run = 𝑡2 − 𝑡1
Time (s, min, hr) 
© 2014 W. W. Norton Co., Inc.
Reaction Rates:
rate of loss of reactant = negative slope
Concentration, M

∆[A] [A] −[A]
change in conc.
rate = change in time = slope = ∆t = t2 −t 1
2 1
run = 𝑡2 − 𝑡1
(t 1 , [A]1 )
rise =
[𝐴]2 −[𝐴]1
Here the slope is negative, so the
rate is multiplied by -1 to make
the rate positive (convention)
Rate = -
∆[A]
∆t
(t 2 , [A]2 )
Time (s, min, hr) 
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For the reaction A  B
Concentration, M
1.2
[B]t
1
0.8
Note how the reaction
rate decreases over time
0.6
0.4
Here the slope is
negative
0.2
[A]t
0
0
1
2
3
4
5
6
7
8
9
10
Time, min
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Reaction Rates and
Stoichiometry
2A
B
As an example, two moles of A disappear for each
mole of B that is formed, so the rate of disappearance
of A is twice the rate of appearance of B.
Mathematically:
−
∆[𝐴]
∆𝑡
=2
1 ∆[𝐴]
∆𝑡
−2
=
∆[𝐵]
∆𝑡
or
∆[𝐵]
∆𝑡
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Reaction Rates and
Stoichiometry (cont’d)
2A
3B
2
A
3
B
A trickier example: two moles of A
disappear and three moles of B are
formed. First divide both sides by 3 -So now the rate of disappearance of A is
2/3 times the rate of appearance of B.
Mathematically:
−
∆[A] 2 ∆[B]
=
∆t
3 ∆t
−
1 ∆[A] 1 ∆[B]
=
2 ∆t
3 ∆t
or
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In general aA + bB
-
1 ∆[𝐴]
𝑎 ∆𝑡
= -
1 ∆[𝐵]
𝑏 ∆𝑡
Reactants have
the minus sign
cC + dD
=
1 ∆[𝐶]
𝑐 ∆𝑡
=
1 ∆[𝐷]
𝑑 ∆𝑡
products
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Actual Reaction Rates
Smog
reaction #1
N2(g) + O2(g)  2 NO(g)
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Average Rates
N2(g) + O2(g)  2 NO(g)
Average Rate: Change in concentration of
reactant or product over a specific time interval
average rate =
=
∆[NO]
∆t
([NO]2 −[NO]1 )
(t2 −t1)
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Practice: Reaction Rates
• Calculate the average reaction rate for the
formation of NO between 5 s and 10 s, and
then between 25 s and 30 s, using the data
below: N2(g) + O2(g) → 2 NO(g)
(a)
(b)
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Average Rate =
Δ NO
Δt
(a)
(b)
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Instantaneous Reaction Rates
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Instantaneous Reaction Rates
• Instantaneous Rate:
• Reaction rate at a particular instant
• Determined graphically as tangential slope
of concentration vs. time plot
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Chapter Outline
•
•
•
•
13.1 Cars, trucks, and Air Quality
13.2 Reaction Rates
13.3 Effect of Concentration on Reaction Rates
13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
© 2014 W. W. Norton Co., Inc.
Effect of Concentration on
Reaction Rates
A typical plot of reactant concentration versus time. The
rate depends on the number of molecular collisions
which depends on the concentration of reactants.
a) Initial rate (will be used in the “Method of Initial
Rates”)
b) midpoint
c) end
The rate approaches zero
as the number of reactant
molecules decreases =
concentration effect
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Reaction Order and Rate Constants
We observe in chemical reactions that rates increase as
the reactant concentrations increase. The rate is found to
be proportional to the reactant concentrations raised to
some experimentally determined power -
General Rate Law:
Rate = k [A]m [B]n
o k = rate constant
o m and n are the reaction orders with
respect to reactants A and B
o (m + n ) = overall order
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Overall Order
The sum of the exponents for each reactant in the Rate Law.
rate = k[A]
1st order in A, 1st order overall
rate = k[A][B]
1st order in A and B, 2nd order overall
rate = k[A]2[B]
2nd order in A, 1st order in B, 3rd order overall
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The Method of Initial Rates
An experimental procedure for determining the
overall reaction order by systematically varying
the concentrations of each reactant
Rate = k [A]m [B]n
Example experimental setup:
Exp. #
[A], M
[B], M
Rate,
Ms-1
1.
0.10
0.10
1 x 10-5
2.
0.10
0.20
2 x 10-5
3.
0.20
0.20
8 x 10-5
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2 NO(g) + O2(g)  2 NO2(g)
rate = k [NO]m [O2]n
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Systematically varying the concentrations
of each reactant, e.g., the exponent n Hold the concentration of one reactant while doubling,
tripling, etc the other concentration of the other reactant:
rate = k [NO]m [O2]n
n: [NO] is held constant. [O2] doubles and the
rate doubles. Therefore n = 1
© 2014 W. W. Norton Co., Inc.
Systematically varying the concentrations
of each reactant, e.g., the exponent m rate = k [NO]m [O2]n
m: [O2] is held constant. [NO] doubles and
the rate quadruples. Therefore m = 2
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2 NO(g) + O2(g)  2 NO2(g)
rate = k [NO]2 [O2]
Solve for the rate constant k by using the data from any of
the experiments, e.g. experiment #1 k=
rate
2
[NO] [O2]
k=
1 x 10−6 M s−1
0.010 𝑀 2 [0.010 M]
k = 1.0 M-2 s-1
© 2014 W. W. Norton Co., Inc.
Characteristics of 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, e.g.
F2 (g) + 2 ClO2 (g)
2FClO2 (g)
Experimentally the rate = k [F2][ClO2] 1
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Equations for determining the exponents:
e.g. rate = k [A]m [B]n
Experiment
[A], M
[B], M
rate
1.
[A]1
[B]1
rate1
2.
[A]1
[B]2
rate2
[B]1
rate1
=
rate2
[B]2
n
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Want to solve for n, so take the log of both sides first -
[B]1
rate1
=
rate2
[B]2
rate1
log
rate2
rate1
log
rate2
n
[B]1
= log
[B]2
[B]1
= n log
[B]2
n
and since
log Xn = n log X
log
n=
log
rate1
rate2
[𝐵]1
[𝐵]2
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Sample Exercise 13.3 - Deriving a Rate
Law From Initial Reaction Rate Data
N2(g) + O2(g) → 2 NO(g)
rate = k [N2]m [O2]n
n
m
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N2(g) + O2(g) → 2 NO(g) rate = k [N2]m [O2]n
Calculating n, the exponent on O2:
log
n=
rate1
rate2
[O2 ]
1
log
[O2 ]2
707
500
=
0.020
log
0.010
log
=
0.15
0.30
= 1/2
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N2(g) + O2(g) → 2 NO(g) rate = k [N2]m [O2]n
Calculating m, the exponent on N2:
log
m=
rate2
rate3
[N2 ]2
log
[N2 ]
3
500
125
=
0.040
log
0.010
log
=
0.60
0.60
= 1
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N2(g) + O2(g) → 2 NO(g) rate = k [N2]m [O2]n
rate = k [N2][O2]1/2
Solve for the rate constant k by using the data
from any of the experiments, e.g. experiment #1 k=
k=
rate
[N2][O2]1/2
707 M s−1
0.040 [0.020]1/2
k = 1.25 x 105 M-1/2 s-1
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Chapter Outline
• 13.1 Cars, trucks, and Air Quality
• 13.2 Reaction Rates
• 13.3 Effect of Concentration on Reaction Rates
o Integrated Rate Laws: First-Order Reactions
o Reaction half-lives
o Integrated Rate Laws: Second-Order Reactions
• 13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
© 2014 W. W. Norton Co., Inc.
Relation Between Reactant Concentration
and Time: Integrated Rate Laws
rate = -
∆[A]t
= k1[A]
∆t
II. Second Order Kinetics:
rate = -
∆[A]t
= k1[A]2
∆t
[A]t 
I. First Order Kinetics:
0.120
0.100
0.080
0.060
0.040
0.020
0.000
2nd order
1st order
0
10
20
30
40
50
time 
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First Order Integrated Rate Equation
A (+ other reactants)  products
rate = -
∆[A]t
= k1[A] the solution to this equation is:
∆t
[A]t = concentration of A at some time = t
[A]o = concentration of A at time t = 0 (initial concentration)
k1 = first order rate constant
(units = s-1, min-1, h-1, etc)
© 2014 W. W. Norton Co., Inc.
Some mathematics: ln
Solving for [A]t
exp ln
[A]t
[A]o
= exp −k1 t
[A]t
= exp −k1 t
[A]o
[A]
= −k1 t
[A]o
Solving for time, t
ln
[A]
= −k1 t
[A]o
t=−
1
[A]
ln
k1 [A]o
[A]t = Ao exp −k1 t
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1st Order Kinetics Example – HW # 69
The following reaction is known to be 1st order with
a rate constant k = 6.32 x 10-4 s-1. How much N2O5
remains in solution after 1 hr if [N2O5]o = 0.50 M?
2 N2O5  4 NO2 + O2
[N2O5]t = [N2O5]o exp −k1 t
[N2O5]t = (0.50 M) exp[-(6.32 x 10-4 s-1)(3600 s)]
[N2O5]t = (0.50 M)(0.103)
[A]t = 0.052 M
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Linearizing the 1st Order Equation
Linearization of an equation is a common method
used in science where useful information can be
obtained from the slope and intercept (e.g. measuring
the rate constant).
a
[A]t
And since ln
= ln (a) − ln (b)
ln
= −k1 t
b
[A]o
ln [A]t − ln [A]o = −k1 t
ln [A]t
y
= −k1 t + ln [A]o
m = -𝑘1
=
b = ln [A]o
mx +
b
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Example Linearization
(spreadsheet on the webpage)
Example Data
0.012
0.01
[A]t →
1st order data
0.01
0.007408182
0.005488116
0.004065697
0.003011942
0.002231302
0.001652989
0.001224564
0.00090718
0.000672055
0.000497871
0.000368832
0.000273237
0.000202419
0.000149956
0.00011109
8.22975E-05
6.09675E-05
4.51658E-05
3.34597E-05
2.47875E-05
0.008
0.006
0.004
0.002
0
0
20
40
60
80 100 120 140 160 180 200
time, s
1st Order Linearization
0
-2
[A]t →
time, s
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
0
20 40 60 80 100 120 140 160 180 200
-4
-6
-8
-10
-12
time, s
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Integrated Rate Law: 1st-Order
h
O3 
 O2 + O
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Plot data after
taking the log
Fit to a
straight line
y=mx + b
m = -6.93 x 10-3 s-1
m = -k1 = -6.93 x 10-3 s-1
k1 = 6.93 x 10-3 s-1
Half-Life: 1st Order Reactions
• Half-Life (t1/2):
• The time in the course of a chemical reaction
during which the concentration of a reactant
decreases by half
• From integrated rate law, when [A]t = ½ [A]o:
A
ln t = - 0.693 = - kt
A 0
• Rearrange: t1/2 =
0.693
k
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Half-Life: 1st-Order Reactions
t1/2 (1) = t1/2 (2) = t1/2 (3)
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Sample Exercise 13.5: Calculating the Half-Life
of a First-Order Reaction.
The rate constant for the decomposition of N2O5 at a particular
temperature is 7.8 x 10-3 s-1. What is the half-life of N2O5 at that
temperature?
t1/2 =
0.693
k
0.693
t1/2 =
7.8 x 10−3 s−1
t1/2 = 89 s
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Some example half lives for First Order
processes Process
4.51 X 109 yr
5.73 X 103 yr
Rate constant,
k (s-1)
4.87 X 10-18
1.21 X 10-4
8.4 h
2.3 X 10-6
8.9 X 10-7 s
7.8 X 105
Half life, t1/2
Radioactive decay of 238U
Radioactive decay of 14C
C12H22O11(aq) + H2O 
C6H12O6(aq) + C6H12O6(aq)
(glucose)
(fructose)
HC2H3O2 + H2O  H3O+ +
C2H3O2-
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Second Order Kinetics
A (+ other reactants)  products
rate = -
∆[A]t
= k2 [A]2 the solution to this equation is:
∆t
1
1
= k2 t +
[A]t
[A]o
[A]t = concentration of A at some time = t
[A]o = concentration of A at time t = 0 (initial concentration)
k2 = second order rate constant (units = M-1 s-1, M-1 min-1, etc)
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1
1
Some mathematics: [A] = k2 t + [A]
o
t
Solving for [A]t
Solving for time, t
1
1
= k2 t +
[A]t
[A]o
k2 t =
[A]o k2 t + 1
1
=
[A]t
[A]o
k2 t =
[A]o
[A]t =
[A]o k2 t+1
t=
1
1
[A]t [A]o
[A]o − [A]t
[A]t [A]o
1 [A]o − [A]t
k2 [At ][A]o
© 2014 W. W. Norton Co., Inc.
Linearizing the 2nd Order Equation
1
1
= k2 t +
[A]t
[A]o
y = mx + b
m = k2
b=
1
[A]o
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How to graphically distinguish between 1st and 2nd order kinetics
ln [NO2] plot: curved
1/[NO2] plot: straight line
slope = k2; intercept = 1/[NO2]o
Sample Exercise 13.6: Distinguishing
between First and Second order reactions
Chlorine monoxide accumulates in the stratosphere above
Antarctica each winter and plays a key role in the formation of
the ozone hole above the South pole each spring. Eventually
ClO decomposes according to the reaction:
2 ClO(g)  Cl2(g) + O2(g)
The kinetics of this reaction were studied in a laboratory
experiment at 298 K, and the data are shown in Table
13.10(b). Determine the order of the reaction, the rate law,
and the value of the rate constant k.
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2 ClO(g)  Cl2(g) + O2(g)
Results:
y = 7.24E+06x + 6.63E+07
• because the plot of 1/[ClO]
is linear, the reaction is 2nd
order
• So the rate law = k2[ClO]2
• from the straight-line fit:
m = 7.24 x 106
m = k2 = 7.24 x 106 M-1s-1
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Half Life for Second Order Reactions
1
1
= k2 t +
[A]t
[A]o
1
1
k2 t =
[A]t [A]o
When t = t½ ,
then [A]t1/2 = [A]o/2
1
1
[A]o /2 [A]o
2
1
k2 t1/2 =
[A]o [A]o
1
k2 t1/2 =
[A]o
1
t1/2 =
k2 [A]o
k2 t1/2 =
© 2014 W. W. Norton Co., Inc.
Sample Exercise 13.7: Calculating the
half-life of Second order reactions
Calculate the half-life of the second-order decomposition of
NO2(g) if the rate constant is 0.544 M-1 s-1 at a particular
temperature; the initial concentration of NO2 = 0.0100 M.
t1/2 =
1
k2 [A]o
=
1
(0.544 𝑀−1 𝑠 −1 )(1.00 x 10−2 𝑀)
t1/2 = 184 s
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Zero Order Reactions
A (+ other reactants)  products
rate = -
∆[A]t
∆t
= k0 [A]0 = k0
And the integrated rate law is [A]t = −ko t + [A]o
y = mx+ b
[A]t
In zero-order reactions, the
concentration of the
reactant has no effect on
the rate. One important
application is measuring
enzyme kinetics.
m = -ko
© 2014 W. W. Norton Co., Inc.
Half Life for Zero Order Reactions
[A]t = −ko t + [A]o
When t = t½ ,
then [A]t1/2 = [A]o/2
[A]o
= −ko t½ + [A]o
2
[A]o
− [A]o = −ko t½
2
−
[A]o
= −ko t½
2
[A]o
= t½
2ko
© 2014 W. W. Norton Co., Inc.
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Glucose bound to hexokinase
http://www.rcsb.org/pdb/explore/jmol.do?structureId=1BDG&bionumber=1
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Rate of product formation vs. substrate
concentration in an enzyme-catalyzed reaction.
© 2014 W. W. Norton Co., Inc.
Summary of the Different Rate Laws and
How to Experimentally Distinguish Them
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Chapter Outline
•
•
•
•
13.1 Cars, Trucks, and Air Quality
13.2 Reaction Rates
13.3 Effect of Concentration on Reaction Rates
13.4 Reaction Rates, Temperature, and the Arrhenius
Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
© 2014 W. W. Norton Co., Inc.
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Factors Affecting Rate
• Temperature:
o Increased temperature increases kinetic
energy of molecules, molecular collisions
• Proper Molecular Orientation
• Activation Energy (Ea):
o Minimum energy of molecular collisions
required to break bonds in reactants,
leading to formation of products
• These factors are incorporated into the
Arrhenius equation
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Effect of Temperature
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Molecular Orientation
O3 + NO → O2 + NO2
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Reaction Energy Profile
High-energy transition state (“activated complex”)
Ea =
Activation
energy
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The Transition State or “Activated Complex”
H2(g) + I2(g)  2 HI(g)
I
I
H
H
H = -13 kJ
Transition state or
activated complex
H2 + I2
2 HI
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Transition state - bonds between reactants are
breaking at the same time that bonds are
forming between the products
Bonds forming
I-I
I
I

H-H

H
H
I
I
H H
Bonds breaking
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Arrhenius Equation
All of these controlling factors are combined
together into the Arrhenius Equation…..
k = Ae−Ea /RT
• k = rate constant
• A is a “collisional frequency factor” , includes the
probability of colliding with the correct orientation
• Ea = activation energy
• R = Ideal Gas Law constant = 8.314 J/mol K
• T = temperature (in kelvin)
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Arrhenius Equation
Ea and A can be measured graphically using
linearization again….
k = Ae-Ea/RT
ln k = ln (A▪e-Ea/RT)
and since ln (a▪b) = ln a + ln b
ln k = ln A + ln (e-Ea/RT)
E 1
ln k = − a
+ ln A
R T
y = m x + b
ln k = ln A + - Ea/RT
m = -Ea/R
b = ln A
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Graphical Determination of Ea
ln k 
-E a  1 
   ln A
R T 
m = -Ea/R
b = ln A
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y = -1580x + 27.99
E
m=− a
R
Ea = −Rm
b = ln A
A = eb
Ea = −(8.314 J/mol K)(−1580 K) = 13,140 J = 13.1 kJ
A = eb = e27.99 = 1.45 x 1012 s-1
Measuring Ea using data from two
different temperatures Another way to measure k if only two data points are available
ln k1 
-E a  1 
   ln A
R  T1 
ln k1 − ln k2 =
ln k2 
-E a  1 
   ln A
R  T2 
−Ea 1
−Ea 1
+ ln A −
+ ln A
R T1
R T2
k
−Ea 1
1
ln 1 =
−
k2
R T1 T2
k
Ea 1
1
ln 1 =
−
k2
R T2 T1
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Back to Exercise
13.9 where the first
two experiments
have been chosen.
k
Ea 1
1
ln 1 =
−
k2
R T 2 T1
Chapter Outline
•
•
•
•
13.1 Cars, trucks, and Air Quality
13.2 Reaction Rates
13.3 Effect of Concentration on Reaction Rates
13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
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Method of Initial
Rates: collect kinetic
data in the lab for
each reactant
General Rate Law:
analyze the kinetic data in
order to determine the
exponents on the
reactants
Propose a reasonable
reaction mechanism
that is consistent with
the rate law
Reaction Mechanisms
• Reaction Mechanism: A set of steps that describe
how a reaction occurs at the molecular level; must be
consistent with the rate law for the reaction
• Elementary Step: Molecular-level view of a single
process taking place in a chemical reaction
• Intermediate: Species produced in one step of a
reaction and consumed in a subsequent step
• Molecularity: The number of ions, atoms, or
molecules involved in an elementary step in a reaction
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Most reactions involve a
sequence of steps
Step 1
Br2(g) + NO(g)  Br2NO(g)
Step 2
Br2NO(g) + NO(g)  2 BrNO(g)
Overall
Reaction
Br2(g) + 2 NO(g)  2 BrNO(g)
“intermediate”
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Each step in the mechanism is called an “elementary
step” and has it’s own Ea and rate constant, k.
Ea1
Br2(g) + NO(g)
Ea2
Br2NO(g)
[+ NO(g)]
2 BrNO(g)
Reaction coordinate 
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Molecularity of Elementary Steps
1 molecule = unimolecular
2 molecules = bimolecular
3 molecules = termolecular
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1. termolecular (and higher) reactions are
improbable because it's difficult to have 3 or more
molecules colliding simultaneously with the
correct energy and orientation.
2. The exponents in the rate expressions for
elementary steps are the same as the
stoichiometric coefficients in each
elementary step. The reason why can be
explained by returning to Sec. 13-3…..
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Why the rate is proportional to the product of
concentrations (Sec.13-3 revisited)
NO(g) + O3(g)  NO2(g) + O2(g)
1x1=1
2x1=2
2x2=4
rate = k[NO][O3]
2x3=6
3x3=9
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What if the rate law contains exponents?
2 NO(g) + O2(g) → 2 NO2(g)
rate = k[NO]2[O2]
When holding O2 constant, doubling NO
should increase the rate by 22 = 4X;
tripling should increase by 32 = 9X, etc.
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Doubling Concentrations
equivalent
equivalent
Tripling Concentrations
equivalent
equivalent
equivalent
equivalent
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equivalent
equivalent
Reaction mechanisms have a RATE DETERMINING
STEP - the slowest step in the mechanism.
2 NH3 + OCl-  N2H4 + Cl- + H2O
Step 1
NH3 + OCl-  NH2Cl + OH-
fast
Step 2
NH2Cl + NH3  N2H5+ + Cl-
slow
Step 3
N2H5+ + OH-  N2H4 + H2O
fast
Overall
2 NH3 + OCl-  N2H4 + Cl- + H2O
Reaction
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The overall rate expression must be
consistent with the rate law for the slow step
2 NH3 + OCl-  N2H4 + Cl- + H2O
Step 1
NH3 + OCl-  NH2Cl + OH-
fast
Step 2
NH2Cl + NH3  N2H5+ + Cl-
slow
Step 3
N2H5+ + OH-  N2H4 + H2O
fast
Overall
2 NH3 + OCl-  N2H4 + Cl- + H2O
Reaction
so the rate law = k[NH2Cl][NH3]
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Writing plausible reaction mechanisms:
•
The sum of the elementary steps must give the overall
balanced equation for the reaction.
•
The rate-determining step is the slowest step in the
mechanism
•
The rate law for the reaction is given by the slow step,
unless…
•
There is a fast equilibrium step prior to the slow step, and so
a calculation has to be made starting with ratef = rater
(coming up)
•
the detection of an intermediate supports a given
mechanism
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Example Reaction Mechanisms
2 NO2 (g) → 2 NO(g) + O2(g)
Proposed mechanism:
rate = k[NO2]2
Known from exp.
Step 1: 2 NO2(g) → NO(g) + NO3(g)
Step 2: NO3(g) → NO(g) + O2(g)
Since the observed rate is k[NO2]2, the first
step must be the RDS.
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2 NO2 (g) → 2 NO(g) + O2(g)
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Rate Determining Step for Mechanisms
With an Equilibrium Elementary Step
2 NO(g) + O2(g)  2NO2(g)
rate = k[NO]2[O2]
Proposed mechanism:
Step 1: NO + O2
NO3
rate1 = k1[NO][O2] fast & rev.
Step 2: NO3 + NO → 2NO2
rate2 = k2[NO3][NO]
2 NO(g) + O2(g)  2NO2(g)
Is this mechanism consistent
with the observed rate law?
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2 NO(g) + O2(g)  2 NO2(g)
Step 1: NO + O2
ratef = kf [NO][O2]
rater = kr [NO3]
by rearrangement:
kf
kr
NO3 at equil. ratef = rater
so
kf [NO][O2] = kr [NO3]
[NO3] =
kf
[NO][O2]
kr
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[NO3] =
kf
[NO][O2]
kr
Substituting into rate2 = k2[NO3][NO]
= k2
kf
[NO][O2] [NO]
kr
= k’[NO]2[O2]
• Which is consistent with
the observed rate law
• mechanism verified
2 NO(g) + O2(g)  2 NO2(g)
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HW# 107
At a given temperature, the rate law of the reaction between
NO and Cl2 is proportional to the product of concentrations of
the two gases: [NO][Cl2]. The following two-step mechanism
has been proposed for the reaction:
Step 1:
NO(g) + Cl2(g) →
Step 2:
NOCl2(g) + NO(g) → 2 NOCl(g)
Overall:
2 NO(g) + Cl2(g) → NOCl2(g)
NOCl2(g)
Which step must be the RDS if this mechanism is correct?
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HW# 108
Mechanism of Ozone Destruction:
Ozone decomposes thermally to oxygen in the following
mechanism:
Step 1:
O3(g) → O(g) + O2(g)
Step 2:
O(g) + O3(g) → 2 O2(g)
2 O3(g) → 3 O2(g)
The reaction is second order in ozone. What properties of the
two elementary steps (specifically, relative rate and
reversibility) are consistent with this mechanism?
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HW# 110
The rate laws for the thermal and photochemical decomposition
of NO are different. Which of the following mechanisms are
possible for the thermal decomposition of NO2, and which are
possible for the photochemical decomposition of NO2? For
thermal decomposition, Rate = kT[NO2]2, and for photochemical
decomposition, Rate = kP[NO2].
slow
2 NO2(g) → N2O4(g)
(a)
fast
N2O4(g) → N2O3(g) + O(g)
fast
N2O3(g) + O(g) → N2O2(g) + O2(g)
fast
N2O2(g) → 2 NO(g)
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HW# 110, cont’d
slow
(b)
2 NO2(g) → NO(g) + NO3(g)
fast
NO3(g) → NO(g) + O2(g)
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HW# 110, cont’d
(c)
NO2(g)
slow
→ N(g) + O2(g)
N(g) + NO2(g)
fast
→ N2O2(g)
slow
N2O2(g) → 2 NO(g)
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Chapter Outline
• 13.1 Cars, trucks, and Air Quality
• 13.2 Reaction Rates
• 13.3 Effect of Concentration on Reaction
Rates
• 13.4 Reaction Rates, Temperature, and the
Arrhenius Equation
• 13.5 Reaction Mechanisms
• 13.6 Catalysts
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Catalysts and the Ozone Layer
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Natural Photodecomposition
of Ozone:
Mechanism:
Step 1:
sunlight
O3 (g) 
 O2 (g) + O(g)
Step 2:
O3 (g) + O(g) 
 2O2 (g)
Overall:
2 O3(g) → 3 O2(g)
slow
Ea for Step 2 = 17.7 kJ/mol (slow step)
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CFC Emissions and Ozone
Chlorofluorocarbons (CFC) e.g. CCl2F2, CCl3F, CClF3
Stable! Migrate into the stratosphere
hν
CCl3F(g) 
 CCl2F(g) + Cl(g)
Cl(g) + O3 (g)  ClO(g) + O2 (g)
ClO(g) + O3 (g)  Cl(g) + O2 (g)
Net:
2 O3(g) → 3 O2(g)
Ea for Cl-catalyzed reaction = 2.2 kJ/mol
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Energy Profiles for O3 Decomposition
hν
2 O3 
 3 O2
Cl atom:
• Not consumed
during the reaction
• Homogeneous
catalyst
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Types of Catalysts
Homogeneous
Catalyst:
• One in the same
phase as the
reactants
Heterogeneous
Catalyst:
• One in a different
phase from the
reactants
Platinum-rhodium gauze
catalyst used in the
production of HNO3
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Automobile Catalytic Converter
Catalyst = Pt-NiO, finely divided powder. Higher
surface area provides more sites for adsorption and
reaction
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Catalytic Converters
The heterogeneous catalyst provides a surface where
the reactants can be ADSORBED on an ACTIVE SITE -
Industrial production
of ammonia (Haber
Process)-
Fe
3 H2(g) + N2(g) 
2 NH3(g)
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