Chapter 13
Rates of Reaction
Dr. Peter Warburton
peterw@mun.ca
http://www.chem.mun.ca/zcourses/1011.php
Reaction Rates
Reaction rate is concentration change
divided by time change
Reaction rate = D[X] / Dt
D[X] = [X]final – [X]initial
Dt = tfinal – tinitial
We most often use molL-1 as units of concentration
This means that rate often has units molL-1s-1
2
Reaction Rate
The reaction rate is defined either as the
increase in the concentration of a product
over time, or the decrease in the
concentration of a reactant over time.
A+BC+D
rate = -D[A] / Dt = -D[B] / Dt = +D[C] / Dt = +D[D] / Dt
Rate is always positive, so we must put negative
signs in front of reactant concentration changes!
3
2 N2O5 (g)  4 NO2 (g) + O2 (g)
4
2 N2O5 (g)  4 NO2 (g) + O2 (g)
Between 300 and 400 seconds:
Rate of decomposition of N2O5 =
- D[N2O5] / Dt =
- (0.0101 molL-1 – 0.0120 molL-1) / (400 s – 300 s)
= 1.9 x 10-5 mol(L·s)-1
5
2 N2O5 (g)  4 NO2 (g) + O2 (g)
Between 300 and 400 seconds:
Rate of formation of NO2 =
+ D[NO2] / Dt =
+ (0.0197 molL-1 – 0.0160 molL-1) / (400 s – 300 s)
= 3.7 x 10-5 mol(L·s)-1
6
2 N2O5 (g)  4 NO2 (g) + O2 (g)
Between 300 and 400 seconds:
Rate of formation of O2 =
+ D[O2] / Dt =
+ (0.0049 molL-1 – 0.0040 molL-1) / (400 s – 300 s)
= 9 x 10-6 mol(L·s)-1
7
1) Average reaction rate
2) Slopes
3) Time = 0
8
2 N2O5 (g)  4 NO2 (g) + O2 (g)
The three values for rate that we
calculated are not the same!
Why?
We have different molar amounts.
But the relative rates ARE THE SAME!
9
2 N2O5 (g)  4 NO2 (g) + O2 (g)
The relative rate of formation of O2 is
(1/1) 9 x 10-6 mol(L·s)-1 = 9 x 10-6 mol(L·s)-1
The relative rate of formation of NO2 is
(1/4) 3.7 x 10-5 mol(L·s)-1 = 9.3 x 10-6 mol(L·s)-1
The relative rate of decomposition of
N2O5 is
(1/2) 1.9 x 10-5 mol(L·s)-1 = 9.5 x 10-6 mol(L·s)-1
10
Instantaneous Reaction Rates
What’s happening at “this instant in time”?
We can use
instantaneous
reaction rates.
The initial rate is
the instantaneous
reaction rate for a
reaction at time
zero.
11
Problem
Consider the following reaction
3 I- (aq) + H3AsO4 (aq) + 2 H+ (aq)
→ I3– (aq) + H3AsO3 (aq) + H2O (l)
a) If –D[I-]/Dt = 4.8 x 10-4 mol(L·s)-1, what is
the value of D[I3-]/Dt during the same time
interval?
b) What is the average rate of consumption
of H+ during the same time interval?
12
Rate Laws and Reaction Order
The rate of a chemical reaction
depends on the concentration of some
or all of the reactants.
A reactant might not affect
the rate, regardless of its
concentration.
13
Rate laws
The rate law for a reaction
is the equation showing the
dependence of the reaction rate on
the
concentrations of the
reactants.
14
aA + bB  products
Rate = k [A]m[B]n
k is a constant for the reaction at a given
temperature, and is called the rate
constant.
m does not have to equal a
n does not have to equal b
15
“Sensitivity” to concentration change
16
Reaction order
Reaction order
with respect to a given reactant
is the value of the exponent of the rate law
equation for the specific reactant only.
The overall reaction order is the
sum of the reaction orders for
all reactants.
17
Reaction order example
rate = k [A]2[B]
The reaction order with respect to A is 2
or the reaction is second order in A
The reaction order with respect to B is 1
or the reaction is first order in B
The overall reaction order is 3 (2 + 1 = 3)
or the reaction is third order overall
18
Problem
Consider three reactions with their
given rate laws below. What is the
order of each reaction in the various
reactants, and what is the overall
reaction order for each reaction?
19
Experimental Determination of a
Rate Law
Reaction rate laws can only be
determined experimentally!
We most commonly carry out a series of
experiments in which the
initial rate of the reaction is measured
as a function of
different initial concentrations of
reactants
20
Method of initial rates
If you see a table like this with chemical concentrations
or pressures and rate data, chances are good the
question is a method of initial rates problem.
21
Method of initial rates
IGNORE THE REACTION with this type
of problem. The chemicals in the TABLE
are the interesting ones.
You always require at least one more
experimental reaction than your number
of chemicals given in your table!
Sometimes we are given a table with an
extra experiment which we can use to
check if we’ve done everything correctly.
22
2 NO (g) + O2 (g)  NO2 (g)
Since rate laws are always expressed in
terms of reactants (and sometimes
catalysts – we’ll see these later), lets
create a general form of the rate law for
this reaction based on what chemicals the
TABLE tells us are involved in the rate of
the reaction.
23
2 NO (g) + O2 (g)  NO2 (g)
m
n
rate = k [NO] [O2]
24
Method of initial rates
For our initial reactant order determination
we need to choose a pair of reactions where
only one reactant concentration changes.
Experiments #1 and #2 fulfill this condition.
25
Method of initial rates
m
n
rate = k [NO] [O2]
Since k is a constant then
k for experiment 1
IS EQUAL TO
k for experiment 2!
m
n
k = rate / [NO] [O2]
rate1
rate 2
rate1 NO O

so

m
n
m
n
NO1 O2 1 NO2 O2 2 rate 2 NO O
m
1
m
2


n
2 1
n
2 2
26
Reaction order w.r.t. NO
rate1 NO1 O 2 1

rate 2 NOm2 O 2 n2
m
m
n
0.048 M  s -1 (0.015 M) (0.015 M)

-1
0.192 M  s
(0.030 M) m (0.015 M) n
0.25  (0.50) m

log 0.25  log (0.50) m
log 0.25  m log 0.50

n
log 0.25
log 0.50
 0.602
m
 0.301
m2
m
27
Reaction order w.r.t. O2
rate1 NO1 O2 1

rate 3 NO32 O2 3n
2
2
n
0.048 M  s -1 (0.015 M) (0.015 M)

-1
0.096 M  s
(0.015 M) 2 (0.030 M) n
0.50  (0.50) n

log 0.50  log (0.50) n
log 0.50  n log 0.50

n
log 0.50
log 0.50
 0.301
n
 0.301
n 1
n
28
Our rate law
2
1
rate = k [NO] [O2]
29
Rate constant using experiment 1
2
1
k = rate / [NO] [O2]
0.048 M  s 1
k


2
2
6
3
3.3
x
10
M
NO O 2  0.015 M  0.015 M 
8
rate
0.048 M  s 1
k  1.4 2 x 10 4 M -2  s 1
30
Rate constant using experiment 2
2
1
k = rate / [NO] [O2]
rate
0.192 M  s 1
0.192 M  s 1
k


2
2
NO O2  0.030 M  0.015 M  1.35 x 105 M3
k  1.42 x 104 M -2  s 1
The rate constant is the same, as it should be!
31
Check using extra experiment
rate = (1.42 x 10 M s ) [NO] [O2]
4

rate  1.4
rate  1.4
-2
-1
2
1

0.030 M  0.030 M 
2.7 x 10 M 
rate  1.4 2 x 10 4 M -2  s 1 NO O 2 
2
2
x 10 4 M -2  s 1
2
x 10 4 M -2  s 1
2
-5
3
rate  3.83 x 10 -1 M  s 1
The rate is the same as the experimentally observed
rate (within rounding errors). We MUST have done
everything right!
32
Units of rate constants
Rate always has the units
mol(L·s)-1
To ensure we get the right
units for rate means the rate
constant must have
different units depending
on the overall reaction
order.
33
Problem
H2O2 (aq) + 3 I- (aq) + 2 H+ (aq)
 I3- (aq) + 2 H2O (l)
D[I3-]/Dt can be determined by measuring
the rate of appearance of the colour.
34
Problem
a) What is the rate law for the formation of I3-?
b) What is the value for the rate constant?
c) What is the initial rate of formation of
triiodide when the concentrations are [H2O2] =
0.300 molL-1 and [I-] = 0.400 molL-1?
35
Reaction Rates and Temperature
Increasing the temperature
increases a chemical reactions
rate.
In general, reaction rates
approximately double if you increase
the temperature by 10 °C.
36
Bumper cars
A gasp of surprise
could be a
“reaction”
when riding in a
bumper car.
“Reactions” occur ONLY when the
bumps are “very hard” and occur
“from behind”.
37
Collision theory
A + BC  AB + C
If this reaction occurs in a single step, then
at some point in time, the B-C bond starts to
break, while the A-B bond starts to form.
At this point, all three nuclei are weakly
linked together.
38
Collision theory
Molecules tend to repel each other
when they get close.
We must insert energy to force the molecules
close together. This is like forcing together the
north poles of two magnets.
This inserted energy is the kinetic energy of the
molecules. It becomes potential energy as the
molecules get closer.
39
Collision theory
A---B---C‡
has a higher potential energy than either
A + B-C or A-B + C
A---B---C‡
is the transition state
or the activated complex
40
Figure 13.11
41
Figure 13.11
There are two
useful energy
differences in the
Figure.
The difference in energy between products
and reactants is DH
The difference in energy between the
transition state and the reactants is
Ea – the activation energy
42
Activation energy
The activation energy (Ea) of a reaction is
the will always be positive!
The energy of
collision between two
molecules must be
AT LEAST as big as
Ea otherwise we
cannot make it to the
transition state.
43
Collisions between molecules at
higher temperatures are more
likely to have collision energy
GREATER THAN the activation
energy.
Higher temperatures mean
higher rates of reaction!
44
Collisions
An individual molecule collides with other
molecules about once every billionth of a
second (one billion collisions per second).
If every collision was successful in creating
products, then every reaction would be almost
instantaneous. This is not the case.
Not every collision breaks the
activation energy barrier!
45
Collisions
The fraction of collisions that have enough energy
to break the activation barrier is given by
f = e-Ea/RT
e is approximately 2.7183,
Ea is the activation energy,
T is the temperature in Kelvin,
R is the gas law constant
(8.314 JK-1mol-1)
46
47
Bumper cars and energy
bumper car - a more
energetic collision is
more likely to make us
gasp (our “reaction”)
molecular collisions higher energy collisions
are more likely to lead
to reaction
(by overcoming the
activation energy)
48
Bumper cars and orientation
You are also more likely to gasp if you are
hit from behind by another bumper car.
The orientation of how the collision
occurs is also important to get a “reaction.”
The same is true for molecules where the
fraction of collisions that have the right
orientation is p. We call this fraction p the
steric factor.
49
Figure 13.9
Cl2 MUST collide
with the N side of
NO to form the
transition state
O=N--Cl--Cl‡.
50
Figure 13.9
If Cl2 hits the O
side, we get a
different transition
state that might
not give the same
products or has a
higher activation
energy. (molecules
“bounce off” each
other)
51
Steric factor
Our steric factor in this case would be p ~
0.5 since half the collisions lead to the
wrong transition state.
52
General reaction A + BC  AB + C
Collision rate = Z [A] [BC]
Z is a constant related to the collision
frequency.
Recall only a fraction (f) of the collisions
have a collision energy greater than or
equal to the activation energy.
Of those collisions, only a fraction (p)
have the correct orientation to proceed
through the transition state to the
products.
53
General reaction A + BC  AB + C
Reaction rate = p x f x Collision rate
Reaction rate = pfZ [A] [BC]
Since for our general reaction
Reaction rate = k [A] [BC]
k = pfZ = pZ e-Ea/RT = A e-Ea/RT
(where A = pZ)
(frequency factor)
54
Arrhenius Equation
pZ = A
As T increases
k increases
55
Problem
AB + CD  AC +BD
What is the value of the activation energy for
this reaction? Is the reaction endothermic or
exothermic?
Suggest a
plausible
structure for
the transition
state.
56
Using the Arrhenius Equation
If we know the rate constants for
a reaction at two different
temperatures, we can then
calculate the activation energy.
k = A e-Ea/RT
ln k = ln (A e-Ea/RT)
ln k = ln (A) + ln (e-Ea/RT)
57
ln k = ln (A) – (Ea/RT)
This is the equation for a straight line!
58
If we graph the natural logarithm of the
rate constant versus inverse
temperature
ln k (y axis) vs 1/T (x axis)
we get a straight line where the
slope = -Ea/R
So Ea = - slope x R
59
ln k vs 1/T
60
ln k2 – ln k1 = (–Ea/R) (1/T2 – 1/T1)
OR
D (ln k) = (–Ea/R) D(1/T)
61
Your textbook says
ln (k2/k1) = (Ea/R) (1/T1 – 1/T2)
This is absolutely correct as well!
Use whichever form of the relation that you
feel more comfortable with mathematically.
62
Problem
Rate constants for the decomposition of
gaseous dinitrogen pentaoxide are
4.8 x 10-4 s-1 at 45 °C
and 2.8 x 10-3 s-1 at 60 °C
2 N2O5 (g)  4 NO2 (g) + O2 (g)
What is the activation energy of this
reaction in kJmol-1?
What is the rate constant at 35C?
63
Reaction Mechanisms
A reaction mechanism is the sequence
of molecular events (elementary steps
or elementary reactions) that defines the
pathway from the reactants to the
products in the overall reaction.
The elementary reactions describe the
behaviour of individual molecules while
the overall reaction tells us stoichiometry.
64
NO2 (g) + CO (g)  NO (g) + CO2 (g)
(Overall Reaction)
The reaction actually takes place in
two elementary reactions!
 2 NO2  NO and NO3
NO3 + CO NO2 and CO2
65
NO2 (g) + CO (g)  NO (g) + CO2 (g)
(Overall Reaction)
Elementary reactions must add
together to give the overall
equation!
66
NO2 (g) + CO (g)  NO (g) + CO2 (g)
(Overall Reaction)
Some of the “crossed-out” chemicals are
neither reactants nor products in the overall
reaction.
For example, in the above reaction NO3 is
formed in one elementary step and
consumed in a later elementary step.
67
Reaction intermediate
A reaction intermediate is a species that
is formed in an elementary step reaction,
that is consumed in a later elementary
step reaction.
We never see reaction intermediates
in the overall reaction!
68
Molecularity
The molecularity of an elementary
reaction is the number of molecules
on the reactant side
of the elementary step reaction.
A one molecule elementary reaction is unimolecular.
A two molecule elementary reaction is bimolecular.
A three molecule elementary reaction is termolecular.
69
Molecularity
70
Chances for molecularity
The chances of a unimolecular reaction only depend
on the one molecule, and are good.
A bimolecular reaction requires that two molecules
collide with each other. This isn’t difficult and happens
quite often.
A termolecular reaction requires that three molecules
collide with each other at the same time. The chances
of this happening are not very good.
The chances of four or more molecules colliding at
the same time are almost impossible.
71
Bumper cars
Consider bumper cars. Very often, you will hit
one other bumper car. Every once and a while,
you and another car will hit a third car at the
same time. It is a
very rare occurrence
to have a “bumper
car” pile-up where
many cars hit a single
car at exactly the
same time.
72
Problem
A suggested mechanism for the reaction of
nitrogen dioxide and molecular fluorine is
73
Problem
a) Give the chemical equation for the overall
reaction, and identify any reaction intermediates
b) What is the molecularity of each of the
elementary reactions?
74
Rate Laws and Reaction
Mechanisms
Unlike an overall reaction the
rate law for an elementary reaction
follows DIRECTLY from the
molecularity of the step reaction!
For a general elementary step reaction
aA + bB  products
rate = k [A]a [B]b
75
Ozone
Unimolecular decomposition of ozone.
O3 (g)  O2 (g) + O (g)
The rate law will
respect to ozone
be first order with
rate = k [O3]
76
Bimolecular reaction
A + B  products
Reaction depends on collisions between
molecules A and B
Increase [A], you increase # collisions
Increase [B], you increase # collisions
rate = k [A] [B]
77
78
Elementary reaction rate laws
79
Mechanisms and overall rate law
The mechanism of the overall
reaction is predicted through the
elementary reactions therefore
the elementary reactions will
determine the rate law of the
overall reaction!
80
Mechanisms and overall rate law
If the overall reaction occurs in ONE elementary
step, then the elementary reaction and the overall
reaction ARE THE SAME.
The rate law for the overall reaction is given by the
rate law for the step reaction
rate = k [CH3Br] [OH-]
81
Rate-determining step
The rate-determining step
of an overall reaction with a
mechanism of two or more steps is the
elementary step reaction
which has the slowest rate.
The overall reaction can occur NO FASTER
than its SLOWEST elementary reaction.
82
NO2 (g) + CO(g)  NO (g) + CO2 (g)
The second step has to wait for the first step
to create the NO3, which is then used rapidly
for the second step reaction.
83
NO2 (g) + CO(g)  NO (g) + CO2 (g)
Is the proposed mechanism plausible?
The elementary steps MUST ADD UP to
give the overall reaction
AND
the mechanism rate law MUST BE
CONSISTENT with the observed rate law.
84
NO2 (g) + CO(g)  NO (g) + CO2 (g)
The elementary reactions DO add up to the
overall reaction.
The rate law of the rate-determining step is
rate = k1 [NO2]2
Since this is the same as the experimentally
observed rate law, so this mechanism is
plausible.
85
Just because a mechanism
is plausible doesn’t mean it
is right!
86
Problem
Write the rate law for each of the
elementary reactions:
O3 (g) + O (g)  2 O2 (g)
Br (g) + Br (g) + Ar (g)  Br2 (g) + Ar (g)
Co(CN)5(H2O)2- (aq)  Co(CN)52- (aq) + H2O (l)
87
Problem
The following substitution reaction has a
first order rate law:
Co(CN)5(H2O)2- (aq) + I- (aq)  Co(CN)5I3- (aq) + H2O (l)
rate = k [Co(CN)5(H2O)2-]
Suggest a possible reaction mechanism,
and show that your reaction mechanism
is in accord with the observed rate law.
88
Catalysis
Reaction rates are not just affected by
reactant concentrations and
temperatures.
A catalyst is a substance that
increases the rate of a reaction
without being consumed in the
reaction.
89
How does a catalyst work?
A catalyst makes available a
different reaction
mechanism that is more
efficient than the
uncatalyzed mechanism.
90
How does a catalyst work?
To get from one side of a mountain to the other
we have to climb up to the top (the activation
energy), and then down the other side of the
mountain.
If there is a mountain pass partway up the
mountain then we can climb up to the pass (a
lower activation energy) and then climb down
to the other side.
Going through the pass will be quicker
than going to the top!
91
2 H2O2 (aq)  2 H2O (l) + O2 (g)
Ea for this reaction is 76 kJmol-1
At room temperature, the reaction is slow.
In the presence of iodide ion the reaction is
faster because a new pathway with a lower
activation energy is made available.
92
This catalyzed overall reaction is faster than
the uncatalyzed reaction because it has a
lower activation energy (our “mountain pass”)
of 19 kJmol-1.
Because the activation energy is about 3.75
times lower than in the uncatalyzed reaction,
the catalyzed reaction rate will be about 40
times faster than the uncatalyzed rate
constant.
93
Figure 13.13 (O3 + O is catalyzed by Cl atoms)
94
Homogeneous and Heterogeneous
Catalysts
A homogeneous catalyst exists in the
same phase as the reactants.
A heterogeneous catalyst exists in a
different phase (usually solid) than the
reactants.
I- is a
homogeneous
catalyst here
Pt is a
heterogeneous
catalyst here
Pt
C2 H 4 ( g )  H 2 ( g ) solid

 C2 H 6 ( g )
95
Figure 13.15
96
Heterogeneous catalysts
Most catalysts used in industry
are heterogeneous
It is much easier to separate a solid from a
gas or liquid (for example) than two liquids
or gases).
97
Enzymes are catalysts
In living beings catalysts are
usually called enzymes
Carbonic anhydrase catalyzes the reaction
of carbon dioxide with water
CO2 (g) + H2O (l)  H+ (aq) + HCO3- (aq)
The enzyme increases the rate of this
reaction by a factor of 106. Equivalent to
about a 200 K increase …
98
Enzymes
Lockand-key
model
99