Chemistry 100
Chapter 14 - Chemical Kinetics
The Connection Between Chemical
Reactions and Time

Not all chemical reaction proceed
instantaneously!!!
2AB

combination reaction
Can we quantify the length of time it
takes for this (or any) chemical
reaction to occur?
Practical examples of how
long!
H2(g) + ½ O2 (g)  H2O (l)  Very Slow
N2O(g)  N2 (g) + ½ O2 (g)  SLOW
 combustion reactions  fast process
 TNT exploding  very fast reaction
 Food spoilage
 Drug decomposition
Chemical Kinetics


Chemical kinetics is concerned with
determining the speed or rate at which
a reaction occurs.
How is the reaction rate affected by





temperature?
states of reactants?
amount of reactants?
catalyst?
surface area of the reacting species?
Example
Br2 (aq) + HCOOH (aq)  2 Br - (aq) + 2
H+ (aq) + CO2 (g)
 Define the average rate

[ Br 2 ] final[ Br 2 ] initial 
 [ Br 2 ]


t
t final  tinitial
A Sample

Reaction
Br2 (aq) + HCOOH (aq)  2 Br – (aq) + CO2 (g) +
2 H+ (aq)

Note – orange colour fades as reaction
proceeds.
t/s
[Br2] / M
0
0.0120
50
0.0101
100
0.00846
150
0.00710
200
0.00596
250
0.00500
300
0.00420
350
0.00353
400
0.00296
Average Rate Data
Avg. Rate
(150 s)
3.27 x 10-5
2.76 x 10-5
2.31 x 10-5
1.93 x 10-5
1.62 x 10-5
Avg. Rate
(100 s)
3.54 x 10-5
3.00 x 10-5
2.50 x 10-5
2.10 x 10-5
1.76 x 10-5
1.47 x 10-5
Avg. Rate
(50 s)
3.80 x 10-5
3.28 x 10-5
2.72 x 10-5
2.28 x 10-5
1.92 x 10-5
1.60 x 10-5
1.34 x 10-5
Instantaneous Rates
Plot of the [Br2 ] vs. time
It’s best to define
an instantaneous
‘speed of
reaction’
0.014
0.012
[Bromine] / M

0.010
0.008
d[ Br 2 ]
0.006 dt

0.004
0.002
0.000
0
100
200
300
time / s
400
500
Instantaneous Rate Data
Time (s)
0
50
100
150
Rate (M/s)
4.19 x 10-5
3.52 x 10-5
2.95 x 10-5
2.48 x 10-5
200
2.08x 10-5
250
1.75 x 10-5
300
1.47 x 10-5
400
1.03 x 10-5
Plot of the ln [Br2] vs. time
-10.000
-10.200
0
100
200
300
[Bromine] / M
-10.400
-10.600
-10.800
-11.000
-11.200
-11.400
-11.600
time / s
400
500
The Rate Law


Relates rate of the reaction to the
reactant concentrations and rate
constant
For a general reaction
aA+bB+cC  dD+eE
rate = k[A]x[B]y[C]z
The exponents (x,y, and z) are called
the reactant orders.
The Rate Constant



The rate constant relates the ‘speed’ of the
chemical reaction to the instantaneous reactant
concentration.
k = constant for constant temperature
The rate of the reaction is dependent on
reactant concentration
RATE CONSTANT IS INDEPENDENT OF
THE REACTANT CONCENTRATION.
The Reaction Orders

Determine the superscripts (x, y, and
z) for a non-elementary chemical
reaction by experimentation.
S x + y + z = reaction order
e.g. x = 1; y = 1; z = 0
2nd order reaction (x + y + z = 2)
x = 0; y = 0; z = 1 (1st order reaction)
x = 2; y = 0; z = 0 (2nd order)
Reaction Rates and Reaction
Stoichiometry

Look at the reaction
O3(g) + NO(g)  NO2(g) + O2(g)
O3  [NO]
 [ NO2 ]
 [ O2 ]
rate = ==+
=+
t
t
t
t
Another Example
2 NOCl (g)  2 NO + 1 Cl2 (g)
1 NOCl 1 NO  1 Cl2 
rate  

2
t
2 t
1 t
WHY? 2 moles of NOCl disappear
for every 1 mole Cl2 formed.
The General Case
aA+bBcC+dD
rate = -1 [A] = -1 [B] = +1 [C] = +1 [D]

a t
b t
c t
d t
Why do we define our rate in this way?
 removes ambiguity

obtain a single rate for the entire equation,
not just for the change in a single reactant or
product.
The Isolation Method of Obtaining Rate
Laws
aA+bB cC+dD
rate = k[A]x[B]y


Fix the concentration of one reactant (say
reactant A).
We then perform a series of experiments to
examine how changing the [B] affects the initial
reaction rate?
rate = (constant) [B]y
Isolation Method (cont’d)


Now we fix the concentration of reactant B.
We then perform another series of experiments
to examine how changing the [A] affects the
initial reaction rate?
rate = (constant) [A]x
Types of Reactions


The rate law gives us information
about how the concentration of the
reactant varies with time
How much reactant remains after
specified period of time?
First Order Reaction

A  product
rate = -[A]/ t = k[A]
How does the concentration of the
reactant depend on time?
A
ln
 Ao

  kt

k has units of s-1
The Half-Life of a First Order
Reaction

For a first order reaction, the half-life
t1/2 is calculated as follows.
t1 /2
0.693

k
Radioactive Decay



Radioactive Samples decay according
to first order kinetics.
This is the half-life of samples
containing e.g. 14C , 239Pu, 99Tc.
Example 14
14
0
C  N  1 
Second Order Reaction
A + B  products
A  products


Rate = k[A][B]
Rate = k[A]2
Reaction 1 is 1st order in A and B and
2nd order overall
Reaction 2 is 2nd order in A
The Dependence of Concentration
on Time

For a second order process where rate
= k[A]2
1
1

 kt
A  A o
Half-life for a Second Order
Reaction.

[A] at t = t½ = ½ [A]0
1
1
=
+ kt 1 / 2
[A ]0 /2 [A ]o
1
or t 1 / 2 =
k [A ]0
A Pseudo-First Order Reaction

Example hydration of methyl iodide
CH3I(aq) + H2O(l)  CH3OH(aq) + H+(aq) + I-(aq)
Rate = k [CH3I] [H2O]

Since we carry out the reaction in aqueous
solution

[H2O] >>>> [CH3I] / [H2O] doesn’t change by a lot

Since the concentration of H2O is
essentially constant
rate = k [CH3I] [constant]
= k` [CH3I] where k` = k [H2O]

Pseudo first order since it appears to
be first order, but it is actually a
second order process.
Collision Theory of Kinetics

With few exceptions, the reaction rate
increases with increasing temperature.

Chemical reactions take place due to
collisions between reactant molecules
i.e. rate  number of collisions / unit time
A2 + B2  product rate = k[A2][B2]
The Reaction Profile

How does the
energy of the
reactants vary during
the reaction
sequence?
The Activation Energy

The minimum amount of
energy need for
initiation of a chemical
reaction is the activation
energy (Ea).

Colliding reactant
molecules possess
kinetic energy > the
activation energy or Ea.
The Activated Complex


The species temporarily formed by the
reactant molecules – the activated
complex.
A small fraction of molecules usually
have the required kinetic energy to get
to the transition state

The concentration of the activated
complex is extremely small.
The Arrhenius Equation
•
Arrhenius showed
how the rate
constant depended
on temperature.
Ea is the activation energy
ln k  ln A - Ea /RT
A is called the frequency factor –
an estimate of the number of
reactive collisions in the system
Activation Energies and the
Arrhenius Equation

Reaction possesses
a large activation
energy  small rate
constant


Slow reaction!!
Measure k at
several different
temperatures
 k2 
Ea  1 1 

ln   - 
R  T2 T1 
 k1 
R = 8.314 J/(K mole)
T in Kelvin units!!!
Arrhenius Equation (cont’d)

The Arrhenius equation is best suited
for studying reactions between simple
species (atoms, diatomic molecules).

The orientation of the reactants (how
they collide) becomes very important
when the species get bigger.
Catalysts

So far, we have considered one way of
speeding up a reaction

increasing T usually increases k.

Another way is by the use of a
catalyst.

A catalyst - a substance that speeds
up the rate of the reaction without
being consumed in the overall
reaction.

look at the following two reactions
A+B  C
rate constant k
A+B  C rate constant with catalyst is kc
 NOTE: RATE WITH CATALYST >
RATE WITHOUT CATALYST
Types of Catalyst

We will briefly discuss three types of
catalysts. The type of catalyst
depends on the phase of the catalyst
and the reacting species.



Homogeneous
Heterogeneous
Enzyme
Homogeneous Catalysis


The catalyst and the reactants are in
the same phase
e.g. Oxidation of SO2 (g) to SO3 (g)
2 SO2(g) + O2(g)  2 SO3 (g)

SLOW
Presence of NO (g), the following
occurs.
NO (g) + O2 (g)  NO2 (g)
NO2 (g) + 2 SO2 (g)  2 SO3 (g) + NO (g)
FAST



SO3 (g) is a potent acid rain gas
H2O (l) + SO3 (g)  H2SO4 (aq)
Note the rate of NO2(g) oxidizing
SO2(g) to SO3(g) is faster than the
direct oxidation.
NOx(g) are produced from burning
fossil fuels such as gasoline, coal, oil!!
Heterogeneous Catalysis

The catalyst and the reactants are in
different phases



adsorption the binding of molecules to
the surface to a surface.
Adsorption on the surface occurs on
active sites.
An active site is a place where reacting
molecules are adsorbed and physically
bond to the metal surface.

The hydrogenation of ethene (C2H4
(g)) to ethane
C2H4 (g) + H2(g)  C2H6 (g)

Reaction is energetically favourable


rH = -136.98 kJ/mole of ethane.
With a finely divided metal such as Ni
(s), Pt (s), or Pd(s), the reaction goes
very quickly .
Common Heterogeneous
Catalysts

Two other important heterogeneous
catalysis processes


petroleum cracking (refining crude oil)
catalytic converters  very efficient in
reducing exhaust emission when hot;
cold is another story!
Enzyme Catalysis



Enzymes - proteins (M > 10000 g/mol)
High degree of specificity (i.e., they will
react with one substance and one
substance primarily
Living cell > 3000 different enzymes
The Lock and Key Hypothesis



Enzymes are folded into
fixed configurations.
According to Fischer,
active site is rigid.
The substrate’s
molecular structure
exactly fits the “lock”
(hence, the “key”).
Simplified Model for Enzyme
Catalysis


E  enzyme; S  substrate; P 
product
E + S  ES
ES  P + E
rate = k [ES]
The reaction rate depends directly on
the concentration of the substrate.
Rate Laws for Multistep
Processes

Chemical reactions generally proceed
via a large number of elementary steps
- the reaction mechanism

The experimentally established rate
law must reflect the reaction rate of the
slowest elementary step  the rate
determining step (rds)
What do we mean by an rds?

A commuter goes through a two step
process to get to work in Halifax.


(1) highway  MacKay Bridge Toll
booth
(2)
toll booth  downtown
MacKay Bridge Toll Booth

Overall reaction
highway  downtown


Situation 1.  highway clogged, toll
booth is fast.
Situation 2.  fast highway, clogged
toll booth.
The Rate Determining Step
(rds)



Situation 1 - clogged highway is the slowest
step in the commuting process (rds).
Situation 2 - the clogged toll-booth is the
slowest step in the commuting process (the
rds).
Speed of overall process (highway 
downtown) depends which step is slowest!

Which is the rate-determining step.
Elementary steps and the
Molecularity


Any chemical reaction occurs via a
sequence of elementary steps.
Kinetics of the elementary step only
depends on the number of reactant
molecules in that step!

Molecularity  the number of reactant
molecules that participate in elementary
steps
The Kinetics of Elementary
Steps

Classes of elementary steps
A  products
rate  k A
A unimolecular step
A  B  products
rate  k AB 
A bimolecular step

For the step
2 A  B  products
rate  k A B 
A termolecular (three molecule) step.
2