Kinetics
What do we mean by kinetics?
Kinetics refers to
 the rate at which chemical
reactions occur.
 The reaction mechanism or
pathway through which a reaction
proceeds.
2.
Rates of reaction
 In
any reaction the concentration of reactants decreases
and concentration of products increases over time until
reaction reaches equilibrium or is complete.
 Rate
[mol/dm3s]
Factors That Affect Reaction Rates
 The
Nature of the Reactants
 Concentration of Reactants
 Temperature
 Catalysts.
4.
Measuring Reaction Rates
The rate of a chemical reaction can be determined by
monitoring the change of any property that changes
with time. For example: concentration of either
reactants or the products as a function of time.
[A] vs t
5.
1
3
4
colorimeter
2
5
pH meter
Rate = loss of mass
time
6
In a reaction a gas may be given off
and the system loses mass.
8
Rate = change in concentration
time
10
Monitor the concentration of a reactant
over time by taking samples at fixed time
intervals. You quench the sample to stop
the reaction. It’s a slow process, so not
7
suitable for fast reactions.
Rate = change in absorbance
time
9
An acid and a carbonate react together,
this is a suitable method for following the
rate of this reaction.
11
If one of the reactants or products is an
acid or a base, then you can use one of
these connected to a data logger.
Rate = change in concentration of H+ ions
time
13
12
Rate = volume of gas produced
time
14
If there is a colour change in the
reaction. You can measure the
absorbance of the reaction mixture.
15
Br2(aq) + HCOOH(aq) -------------- 2H+(aq) + 2Br-(aq) + CO2(g)
16
CH3COCH3(aq) + I2 -------------- CH3COCH2I(aq) + H+(aq) + I-(aq)
17
Mg(s) + HCl(aq)
--------- MgCl(aq) + H2(g)
18
2HCl(aq) + Na2 CO3(aq) -------------- 2NaClaq) + H2O (l) + CO2(g)
19
Measuring Reaction Rates
 Other



examples:
Mass or volume for gaseous reactions
Change in pH, for acid-base reactions
Use of colorimeter for color change reactions
9.
Example 1: Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)
[C4H9Cl] M
In this reaction, the
concentration of
chlorobutane, C4H9Cl,
was measured at
various times, t.
Rate =
[C4H9Cl]
t
10.
Reaction Rates Calculation
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl (aq)
Average Rate, M/s
The average rate of the
reaction over each
interval is the change in
concentration divided by
the change in time:
11.
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)


A plot of concentration vs.
time for this reaction yields
a curve like this.
The slope of a line tangent
to the curve at any point is
the instantaneous rate at
that time.
12.
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)

The reaction slows
down with time
because the
concentration of the
reactants decreases.
13.
The Collision Model

In a chemical reaction, bonds are
broken and new bonds are formed.
 Molecules can only react if they collide
with each other.
 These collisions must occur with
sufficient energy and at the appropriate
orientation.
14.
Activation Energy


There is a minimum amount of energy required for
reaction: the activation energy, Ea.
A reaction cannot occur unless the molecules
possess sufficient energy to get over the activation
energy barrier.
15.
Maxwell–Boltzmann Distributions

Temperature is
defined as a measure
of the average kinetic
energy of the
molecules in a
sample.
• At any temperature there is a wide
distribution of kinetic energies.
16.
Maxwell–Boltzmann Distributions

As the temperature
increases, the curve
flattens and broadens.
 Thus at higher
temperatures, a larger
population of
molecules has higher
energy.
17.
Factors that increase rate of reaction
 Increasing
concentration of reactants
or P of gaseous reactions => more
collisions
 Increasing the T of reaction => more
energy => go faster => more energy
 can posess activation energy
Factors that increase rate of reaction
 Tred
> Tblue
 With T red more
particles posees
enough Ea.
Factors that increase rate of reaction
 Adding




a suitable catalyst
Catalysts increase the rate of a reaction without being
chemically changed at the end of the reaction
They work by decreasing the activation energy of the reaction.
Catalysts change the mechanism by which the process occurs.
Some catalysts also make atoms line up in the correct
orientation so as to enhance the reaction rate
Catalysts
Catalysts may be
either homogeneous or
heterogeneous
A homogeneous
catalyst is in the same
phase as the
substances reacting.
A heterogeneous
catalyst is in a different
phase
21.
Heterogeneous catalysts
There are two types of catalysts: heterogeneous and
homogeneous.
Heterogeneous catalysts
are in a different phase to
the reactants. The catalyst
is usually a solid and the
reactants are liquids or
gases (e.g. solid catalysts
for gas reactions in catalytic
converters).
Industrial examples of heterogeneous catalysis include
the iron catalyst used in ammonia production and the
Ziegler–Natta catalyst used in poly(e)thene production.
Catalysts
One way a catalyst
can speed up a
reaction is by
holding the
reactants together
and helping bonds
to break.
Heterogeneous
catalysts often act
in this way
23.
How do catalysts work?
Catalysts increase the rate of reactions without being
used up during the reaction.
One way in which this occurs is for the catalyst to be
changed during the reaction, then changed back in a
second reaction with one of the reactants or products.
This is an alternative reaction pathway.
An example is the
oxidation of sulfur dioxide:
This is catalyzed by
vanadium(V) oxide:
SO2(g) + ½O2(g)  SO3(g)
SO2(g) + V2O5(s)  SO3(g) + V2O4(s)
The catalyst is re-formed
by reacting with oxygen:
V2O4(s) + ½O2(g)  V2O5(s)
Homogeneous catalysts
Homogeneous
catalysts are in the
same phase as the
reactants. The
catalyst and the
reactants are
usually liquids,
such as the
hardener added to
fibreglass resin.
Another example of homogeneous catalysis is the
destruction of atmospheric ozone catalyzed by chlorine
free radicals. In this reaction the catalyst and reactants
are in the gas phase.
Catalysts
Some catalysts
help to lower the
energy for
formation for the
activated complex
or provide a new
activated complex
with a lower
activation energy
AlCl3 + Cl2  Cl+ + AlCl4Cl+ + C6H6  C6H5Cl + H+
H+ + AlCl4-  HCl + AlCl3
Overall reaction
C6H6 + Cl2  C6H5Cl + HCl
26.
Enzymes

Enzymes are
catalysts in
biological systems.
 The substrate fits
into the active site of
the enzyme (induced
fit model).
 Lock and key model
27.
Orders of reaction
A(aq) + B(aq)
C(aq)
An investigation was done to see how the initial rate of the reaction
changed as the concentrations of A and B were changed.
Experiment
[A] (mol dm-3)
[B] (mol dm-3)
Initial rate (mol dm-3 s-1)
1
1
1
2
2
2
1
1
2
2
4
3
What happens when:
i) The concentration of A doubled?
ii) The concentration of B doubled?
Orders of reaction
D(aq) + E(aq)
F(aq)
An investigation was done to see how the initial rate of the reaction
changed as the concentrations of D and E were changed.
Experiment
[D] (mol dm-3)
[E] (mol dm-3)
Initial rate (mol dm-3 s-1)
1
5
3
6
2
10
5
3
6
24
6
3
What happens when:
i) The concentration of D doubled?
ii) The concentration of E doubled?
Orders of reaction - Rules
i)
If you double a reactants concentration and the rate stays the same, the order
with respect to the reactant is 0
ii)
If you double a reactants concentration and the rate doubles, the order with
respect to the reactant is 1
iii)
If you double a reactants concentration and the rate quadruples, the order with
respect to the reactant is 2
These are written as [X]0 or [Y]1 or [Z]2
Orders of reaction
G(aq) + H(aq)
I(aq)
An investigation was done to see how the initial rate of the reaction
changed as the concentrations of G and H were changed.
Experiment
[G] (mol dm-3)
[H] (mol dm-3)
Initial rate (mol dm-3 s-1)
1
0.5
0.75
1.33
2
1.0
0.5
0.75
1.50
2.66
1.33
3
i) What is the order with respect to G?
ii) What is the order with respect to H?
New rule
The overall order of the reaction is the sum of
the reactants orders.
[X]m [Y]n
Overall order = m + n
For example: [X]2 [Y]1 The overall order is 3
[X]1 [Y]1 The overall order is 2
[X]0 [Y]1 The overall order is 1
[X]0 [Y]2 The overall order is 2
Orders of reaction
J(aq) + K(aq)
L(aq)
An investigation was done to see how the initial rate of the reaction
changed as the concentrations of G and H were changed.
Experiment
[J] (mol dm-3)
[K] (mol dm-3)
Initial rate (mol dm-3 s-1)
1
0.70
0.125
0.07
2
1.4
0.7
0.125
0.250
0.14
0.14
3
i) What is the order with respect to J?
ii) What is the order with respect to K?
iii)What is the overall order of the reaction?
Orders of reaction
M(aq) + N(aq)
O(aq)
An investigation was done to see how the initial rate of the reaction
changed as the concentrations of M and N were changed.
Experiment
[M] (mol dm-3)
[N] (mol dm-3)
Initial rate (mol dm-3 s-1)
1
0.56
0.67
0.04
2
0.56
1.12
1.34
0.67
0.16
0.08
3
i) What is the order with respect to M?
ii) What is the order with respect to N?
iii)What is the overall order of the reaction?
The rate equation
rate =
m
n
k[A] [B]
This is the rate equation.
We use this equation to calculate the rate of a
reaction.
These are the orders of the
reaction
rate = k[A]
m[B] n
This is the rate constant.
These are the concentration of
The bigger it is, the faster the
the reactants
reaction.
It increases with temperature.
Orders of reaction
P(aq) + Q(aq)
R(aq)
An investigation was done to see how the initial rate of the reaction
changed as the concentrations of P and Q were changed.
Experiment
[P] (mol dm-3)
[Q] (mol dm-3)
Initial rate (mol dm-3 s-1)
1
0.34
0.15
0.13
2
0.34
0.68
0.30
0.15
0.26
0.26
3
i) What is the order with respect to P?
ii) What is the order with respect to Q?
iii) What is the overall order of the reaction?
iv) Can you construct a rate equation.
Orders of reaction
S(aq) + T(aq)
U(aq)
An investigation was done to see how the initial rate of the reaction
changed as the concentrations of S and T were changed.
Experiment
[S] (mol dm-3)
[T] (mol dm-3)
Initial rate (mol dm-3 s-1)
1
0.78
1.40
0.09
2
1.56
0.78
1.40
2.80
0.09
0.36
3
i) What is the order with respect to S?
ii) What is the order with respect to T?
iii) What is the overall order of the reaction?
iv) Can you construct a rate equation.
Orders of reaction
NO(g) + CO(g) + O2(g)
NO2(g) + CO2(g)
Experiment
[NO(g)]
(mol dm-3)
[CO(g)]
(mol dm-3)
[O2(g)]
(mol dm-3)
Initial rate (mol dm-3 s-1)
1
2.0 x 10-2
4.0 x 10-2
2.0 x 10-2
1.0 x 10-2
1.0 x 10-2
2.0 x 10-2
1.0 x 10-2
1.0 x 10-2
1.0 x 10-2
0.176
0.704
0.176
2.0 x 10-2
1.0 x 10-2
2.0 x 10-2
0.176
2
3
4
i) Find the order with respect to each reactant
ii) Construct a rate equation
Rate Law
Each reaction has its own equation that gives its rate as
a function of reactant concentrations.
This is called its Rate Law
The general form of the rate law for A + B  products, is
Rate = k[A]x[B]y
Where
k is the rate constant at specified T,
[A] and [B] are the concentrations of the reactants.
X and y are exponents known as rate orders that must be
determined experimentally
x+ y = overall order of reaction
40.
Rate Expression
 Value

First order: x+y=1



Rate = k [A] or rate = k [B]
Units: [k] = 1/seconds
Second order: x+y=2



of k’s units depend on overall rate of reaction
Rate = k [A]2 or rate= k[A][B] or rate = k [B] 2
Units: [k] = L/mol.s = dm3/mol.s
Third order: Units: [k] = L2/mol2.s = dm6/mol2.s
Determining the Rate constant and
Order
Rate equation can only be determined
experimentally
The following data was collected for the reaction of substances
A and B to produce products C and D.
Deduce the order of this reaction with respect to A and to B.
Write an expression for the rate law in this reaction and
calculate the value of the rate constant.
[NO] mol dm-3
0.40
0.40
0.20
[O2] mol dm-3
0.50
0.25
0.25
Rate mol dm-3 s-1
1.6 x 10-3
8.0 x 10-4
2.0 x 10-4
42.
Resolution
 Order
𝑅𝑎𝑡𝑒1

𝑅𝑎𝑡𝑒2
with respect to [H2]: 1
=
0.00600

0.144
𝑘 𝑁𝑂 𝑥 𝐻2 𝑦
𝑘 𝑁𝑂 𝑥 𝐻2 𝑦
=
 0.04166
 0.125
𝑘 0.0100 𝑥 0.0100
𝑘 0.0200 𝑥 0.0300
=
0.0100 𝑥 0.33333
0.0200 𝑥
𝑥
= 0.5
 ln 0.125 = 𝑥 ln 0.5
 X= 3
Determining the Rate Order

Order 1
Rate1=k[A]
Rate2=k[2A]=2k[A]=2Rate1

Order 2
Rate1=k[A]2
Rate2=k[2A]2 = 4k[A]2 = 4Rate1

Order 3
Rate1=k[A]3
Rate2=k[2A]3 = 8k[A]3 = 8Rate1
Determining the Rate Order
Concentration
Rate
Order
Doubles
Stays the same
0
Doubles
Doubles
1
Doubles
Multiplies by 4
2
Doubles
Multiplies by 8
3
Graphical representations
 Remember
Rate = k[A]x[B]y
 Can
you deduce the graphical representation of rate vs
concentration with respect to reactant A when:



The order is 0
The order is 1
The order is 2?
Graphical representations
 Remember
rate = difference in concentration over
difference in time
 Can you deduce the graphical representation of
concentration vs time with respect to reactant A when:



The order is 0
The order is 1
The order is 2?
Graphical representations
 Zero
order: rate is independent of
concentration of that reactant.

Concentration of reactant will decrease
uniformly over time
Graphical representations
 First

order:
rate is proportional to concentration of that
reactant.
[A] = [A]0 e- k t
Integrated Rate
law equation
deduction
1st order reactions
Half-Life of a Reaction

Half-life is defined as the
time required for one-half
of a reactant to react.
 Because [A] at t1/2 is onehalf of the original [A],
[A]t = 0.5 [A]0
 In particular 1st order
reactions have a constant
half life
 t1/2
= (ln 2)/k
= 0.693/k
53.
Half life of 1st order reactions

In first order reactions half life is constant. It is
independent of initial concentration
= [A]0 e- k t
 [A]0 /2= [A]0 e- k t
 [A]
1/2= ln e- k t
 ln 1/2 = -kt lne = -kt
 t= ln2 / k = 0.69 /k
 ln
Orders of reaction from experimental results
A(aq) + B(aq)
C(g)
The concentrations of A and B are measured over time
[A] (mol dm-3)
Time (s)
[B] (mol dm-3)
Time (s)
1
0.5
0.25
0.125
0
17
35
51
1
0.5
0.25
0.125
0
21
58
117
Plot 2 graphs.
i)
[A] against time
ii)
[B] against time
iii) Use the graphs to calculate the first three half-lives (t1/2) for each
reactant. (This is the same method as radioactive half-lives)
iv) Construct a rate equation for the reaction.
Graphical representations
 Second
order:
rate
[A]2
Parabolic as it depends
on square concentration
1st order and 2nd order graphs are both curved.
1st
order
graphs, the half-life is constant.
2nd order graphs, the half life increases.
Graphical representations
 Summary
Temperature and Rate

the reaction rate increases
as the temperature
increases.
 K is constant at fixed T
 But k is temperature
dependent.
 As T increases, k
increases exponentially
 As a rule of thumb a
reaction rate increases
about 10 fold for each 10oC
rise in temperature
59.
Arrhenius Equation
Arrhenius developed a mathematical relationship
between k and T and Ea:
where A is Arrhenius constant, a number that represents
the likelihood that collisions would occur with the
proper orientation for reaction.
Ea is the activation energy
T is the Kelvin temperature
R is the universal thermodynamics (gas) constant.
R = 8.314 J mol-1 K-1 or 8.314 x 10-3 kJ mol-1 K-1
60.
Arrhenius Equation
Ln A
Slope = -Ea/R
Taking the natural
logarithm of both
sides, the equation
becomes
1
RT
y = mx + c
When k is determined experimentally at
several temperatures, Ea can be calculated
from the slope of a plot of ln k vs. 1/T.
61.
 Activation
energy can be calculated if only two data points
given for k and T
lnk1 = -Ea/R.T1 + lnA
lnk2 = -Ea/R.T2 + lnA
ln(k1/k2) = (Ea/R) (1/T2 – 1/T1)
(equation in data booklet)
Worked example
 The
following data were collected
for a reaction
Rate constant (1/s)
Temperature (C)
2.88 x10-4
320
4.87 x10-4
340
7.96 x10-4
360
1.26 x10-3
380
1.94x10-3
400
 Find
Ea in kJ/mol
Reaction Mechanisms
The sequence of events that describes
the actual process by which reactants
become products is called the
reaction mechanism.
64.
Reaction Mechanisms

Reactions may occur all at once or through
several discrete steps.

Each of these processes is known as an
elementary reaction or elementary process.

Overall rate will depend on the rate of slowest
step: Rate determining step.

Principle of Occam’s razor: newer theories need to
remain as simple as possible while maximizing
explanatory power. The low probability of three
molecules colliding means reaction mechanism is
more likely.
65.
Reaction Mechanisms

The [ ] of each reactant in rate determining step must appear
in rate law for that step, raised to the power of its coefficients
in equation
Equation rate elementary step
Rate Law
A  products
Rate = k [A]
2A  products
Rate = k [A]2
A + B  products
Rate = k [A] [B]
66.
Slow Initial Step
NO2 (g) + CO (g)  NO (g) + CO2 (g)

The rate law for this reaction is found experimentally
to be
Rate = k [NO2]2


CO is necessary for this reaction to occur, but the
rate of the reaction does not depend on its
concentration.
This suggests the reaction occurs in two steps.
67.
Slow Initial Step

A proposed mechanism for this reaction is
Step 1: NO2 + NO2  NO3 + NO (slow)
Step 2: NO3 + CO  NO2 + CO2 (fast)

The NO3 intermediate is consumed in the second step.

As CO is not involved in the slow, rate-determining step, it
does not appear in the rate law.
68.
Fast Initial Step
Example:
2NO + O2  2NO2
Because termolecular (= trimolecular) processes are
rare a two-step mechanism is proposed


Step 1: NO + NO  N2O2 fast
Step 2: N2O2 + O2  2NO2 slow
Rate= k [N2O2] [O2]
But N2O2 depends on step 1
Rate= k [NO]2 [O2]
69.
Example
2NO + 2 H2  N2 + 2 H2O
 Rate = k [H2][NO]2


Possible mechanism
1.
2.
3.
2NO  N2O2
N2O2 + H2  N2O + H2O
N2O + H2  N2 + H2O
Which step is the slow one to match experimental rate?
Same reaction - another possible
mechanism


2NO + 2 H2  N2 + 2 H2O
Rate = k [H2][NO]2

Possible mechanism
1.
2.
3.
H2 + NO  H2O + N
N + NO N2O
N2O + H2  N2 + H2O
fast
slow
fast
Is it consistent with experimental rate?
Which is the most likely mechanism? Why?