TOPIC 16
CHEMICAL KINETICS
16.1
Rate Expression and
Reaction Mechanism
ESSENTIAL IDEA
Rate expressions can only be determined empirically
and these limit possible reaction mechanisms. In
particular cases, such as a linear chain of elementary
reactions, no equilibria and only one significant
activation barrier, the rate equation is equivalent to the
slowest step of the reaction.
NATURE OF SCIENCE (2.7)
Principle of Occam’s razor – newer theories need to remain
as simple as possible while maximizing explanatory power.
The low probability of three molecule collisions means
stepwise reaction mechanisms are more likely.
INTERNATIONAL-MINDEDNESS
The first catalyst used in industry was for the
production of sulfuric acid. Sulfuric acid
production closely mirrored a country’s
economic health for a long time. What are some
current indicators of a country’s economic
health?
THEORY OF KNOWLEDGE
Reaction mechanism can be supported by
indirect evidence. What is the role of
empirical evidence in scientific theories? Can
we ever be certain in science?
UNDERSTANDING/KEY IDEA
16.1.A
The order of a reaction can be
either integer or fractional in
nature. The order of a reaction can
describe, with respect to a reactant,
the number of particles taking part
in the rate-determining step.
 Distinguish
between the terms rate
constant, overall order of reaction
and order of reaction with respect
to a particular reactant.
DEFINITIONS
The order of reaction with respect to a
particular reactant is the power to which its
concentration is raised in the rate equation.
 The overall order for the reaction is the sum
of all the individual orders for all reactants.
 The rate constant (k) is a constant for a
particular reaction at a specified
temperature.

RATE LAW EXPRESSION
A+B
PRODUCTS
The rate law can be written as follows:
R = k [A]m[B]n
where R is the rate
k is the rate constant
A and B are reactants
m and n are the experimentally found
orders of the reaction with respect to
each reactant
2H2 + 2NO
2H2O + N2
R = k [NO]2[H2]
 Notice that the coefficients in the equation
are not the exponents in the rate expression.
Please do not confuse this with equilibrium.
 This is a 2nd order rxn with respect to NO
and a 1st order rxn with respect to H2.
 The overall order is found by adding the
individual orders 2 + 1 = 3.

UNDERSTANDING/KEY IDEA
16.1.B
The value of the rate constant (k) is
affected by temperature and its
units are determined from the
overall order of the reaction.
UNITS OF THE RATE
CONSTANT (k)




Zero order
Rate = k
k = units of rate
 = mol dm-3 s-1
First order
Rate = k[A] k = units of rate/units of conc
 = (mol dm-3 s-1)/(mol dm-3) = s-1
Second order
Rate = k[A]2 k = units of rate/(units of conc)2
 = (mol dm-3 s-1)/(mol dm-3)2 = mol-1dm3 s-1
Third order
Rate = k[A]3 k = units of rate/(units of conc)3
 = (mol dm-3 s-1)/(mol dm-3)3 = mol-2 dm6 s-1
UNDERSTANDING/KEY IDEA
16.1.C
Rate equations can only be
determined experimentally.
APPLICATION/SKILLS
Be able to deduce the rate
expression for an equation from
experimental data and solve
problems involving the rate
expression.
DETERMINING THE EXPONENTS IN A
RATE LAW
This must be done experimentally. NEVER assume
the exponents equal the coefficients in the balanced
equation.
 Factor by which
Factor by which
Order of
conc. is changed
rate changes
Reaction

2
3
2
3
4
2
3
4
2
3
4
No change
No change
2 = 21
3 = 31
4 = 41
4 = 22
9 = 32
16 = 42
8 = 23
27 = 33
64 = 43
0
0
1
1
1
2
2
2
3
3
3
INITIAL RATES METHOD
Problem - Write the rate law, determine the value of
the rate constant, k, and the overall order for the
following reaction:
2 NO(g) + Cl2(g)  2 NOCl(g)
Experiment
[NO]
(mol/dm3)
[Cl2]
(mol/dm3)
Rate
mol/dm3·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
11.4 x 10-6
OBSERVATIONS
When given initial concentration data,
you want to look at each reactant
separately.
 Try to find 2 reactions where your
identified reactant is changing and the
other reactant(s) are constant.
 For example in rxns 1 and 2, [NO] is
changing and [Cl2] is constant.

Part 1 – Determine the values for the exponents in the rate law:
R = k[NO]x[Cl2]y
Experiment
[NO]
(mol/dm3)
[Cl2]
(mol/dm3)
Rate
mol/dm3·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
1.14 x 10-5
In experiment 1 and 2, [Cl2] is constant while [NO] doubles.
The rate quadruples, so the reaction is second order with respect to [NO]
 R = k[NO]2[Cl2]y
Part 1 – Determine the values for the exponents in the rate law:
R = k[NO]2[Cl2]y
Experiment
[NO]
(mol/dm3)
[Cl2]
(mol/dm3)
Rate
mol/dm3·s
1
0.250
0.250
1.43 x 10-6
2
0.500
0.250
5.72 x 10-6
3
0.250
0.500
2.86 x 10-6
4
0.500
0.500
1.14 x 10-5
In experiment 2 and 4, [NO] is constant while [Cl2] doubles.
The rate doubles, so the reaction is first order with respect to [Cl2].
 R = k[NO]2[Cl2]
Part 2 – Determine the value for k, the rate constant, by using any set of
experimental data:
R = k[NO]2[Cl2]
Experiment
1
[NO]
(mol/dm3)
[Cl2]
(mol/dm3)
Rate
mol/dm3·s
0.250
0.250
1.43 x 10-6
2
mol
mol  
mol 

1.43 x 10
 k  0.250
  0.250

Ls
L  
L 

6
2
 1.43 x106   mol   L3 
L
5
k 

9.15
x
10


3
3 
2
0.250
L

s
mol
mol
s




APPLICATION/SKILLS
Be able to sketch, identify and
analyze graphical representations
for zero, first and second order
reactions.
GUIDANCE
Be familiar with concentration vs
time and rate vs concentration
graphs.
rate
concentration
ZERO-ORDER REACTION
time
concentration
Rate = k[A]0
rate
concentration
FIRST-ORDER REACTION
time
concentration
Rate = k[A]
HALF-LIFE
If a reactant has a constant half-life,
then the reaction must be first order
with respect to that reactant.
 This is only true for 1st order reactions.
 The shorter the half-life, the faster the
reaction.
 Half-life (t1/2) is the time it takes for
half the concentration to decrease.

rate
concentration
SECOND-ORDER REACTION
time
Rate =
k[A]2
concentration
The concentration – time graph starts out steeper than the first order
and levels off more than first order.
rate
concentration
SUMMARY
time
concentration
UNDERSTANDING/KEY IDEA
16.1.D
Reactions may occur by more than
one step and the slowest step
determines the rate of reaction.
(rate determining step/RDS)
APPLICATION/SKILLS
Evaluate proposed reaction
mechanisms to be consistent with
kinetic and stoichiometric data.
Reaction Mechanism
The reaction mechanism is the series of
elementary steps by which a chemical reaction
occurs. It is a theory about the sequence of
events in the progression of reactants to
products.
The sum of the elementary steps must give
the overall balanced equation for the reaction
Rate-Determining Step
In a multi-step reaction, the slowest step
is the rate-determining step. It therefore
determines the rate of the reaction.
The experimental rate law must agree
with the rate-determining step.
Identifying the Rate-Determining
Step
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
The experimental rate law is:
R = k[NO]2[H2]
Which step in the reaction mechanism is the rate-determining (slowest) step?
Step #1
H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2
N2O(g) + H2(g)  N2(g) + H2O(g)
Step #1 agrees with the experimental rate law
Identifying Intermediates
For the reaction:
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
Which species in the reaction mechanism are intermediates (do not
show up in the final, balanced equation?)
Step #1
H2(g) + 2NO(g)  N2O(g) + H2O(g)
Step #2
N2O(g) + H2(g)  N2(g) + H2O(g)
2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
 N2O(g) is an intermediate
GUIDANCE
Be able to use potential energy
level profiles to illustrate multi-step
reactions; showing the higher Ea in
the rate-determining step in the
profile.
www.chemguide.co.uk
The activation energy for the overall reaction is equal to
the activation energy of the rate determining step.
GUIDANCE
You should be familiar with
reactions where the ratedetermining step is not always the
first step.
www.chemistry.msu.edu
Notice the second step is the rate determining
step since it has a higher activation energy.
When the rate determining step is not the first
step in the mechanism, it is a bit more
complicated.
 The reactant concentrations depend upon
earlier steps so these earlier steps must be
taken into account.

2NO(g) + O2(g) 2NO2(g)
Step #1
Step #2
Overall
NO(g) + NO(g)  N2O2(g)
N2O2(g) + O2(g)  2NO2(g)
2NO(g) + O2(g) 2NO2(g)
fast
RDS (slow)
So the rate equation is taken from step 2
Rate = k[N2O2][O2] but N2O2 is an intermediate whose
concentration is based on step 1 which is NO(g) + NO(g).
We then substitute back into the rate equation getting
Rate = k[NO]2[O2] which matches the coefficients in the
overall balanced equation thus meeting the requirements
for the mechanism.
UNDERSTANDING/KEY IDEA
16.1.E
The molecularity of an elementary
step is the number of reactant
particles taking part in that step.
The term molecularity is used in reference to an
elementary step to indicate the number of
reactant species involved.
 Unimolecular – one reactant species
 Bimolecular – two reactant species
 Termolecular – three reactant species


The probability of more than two particles
colliding at the same time with sufficient energy
and correct orientation is extremely low.
UNDERSTANDING/KEY IDEA
16.1.F
Catalysts alter a reaction
mechanism, introducing a step with
lower activation energy.
GUIDANCE
Know that catalysts are involved in
the rate-determining step.

Catalysts alter the reaction
mechanism by providing an alternate
pathway for the rate-determining step
with a lower activation energy.
Duch.sd57.bc.ca
Citations
International Baccalaureate Organization. Chemistry Guide, First
assessment 2016. Updated 2015.
Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed.
N.p.: Pearson Baccalaureate, 2014. Print.
Most of the information found in this power point comes directly
from this textbook.
The power point has been made to directly complement the Higher
Level Chemistry textbook by Catrin and Brown and is used for direct
instructional purposes only.