CHEMICAL KINETICS.pptx

CHEMICAL KINETICS
INTRODUCTION

The branch of Physical chemistry which deals with the rate of reactions is called
Chemical Kinetics. In chemical kinetics we shall study:
• The rate of the reactions and rate laws.
• The factors as temperature, pressure, concentration and catalyst, that influence the
rate of a reaction.
• The mechanism or the sequence of steps by which a reaction occurs.
•
ENOCH DORMAN (dorman.enochkojo@gmail.com) 27/06/2023

REACTION RATE
 The rate of a reaction tells as to what speed the reaction occurs.
 Let us consider a simple reaction; A ⎯⎯→ B
 The concentration of the reactant ‘A’ decreases and that of B
increases as time passes. The rate of reactions is defined as the
change in concentration of any of reactant or products per unit
time.

For the given reaction the rate of reaction may be
equal to the rate of disappearance of A which is
equal to the rate of appearance of B.
Thus rate of reaction = rate of disappearance of A
= rate of appearance of B
or rate = –
𝑑⦋𝐴⦌
𝑑𝑡
= +
𝑑⦋𝐵⦌
𝑑𝑡

Where [ ] represents the concentration in moles per litre
whereas d’ represents infinitesimally small change in
concentration. Negative sign shows the concentration of
the reactant A’ decreases whereas the positive sign
indicates the increase in concentration of the product B.

UNITS OF RATE
 Reactions rate has the units of concentration divided by time. We express
concentrations in moles per litre (mol/litre) but time may be given in any
convenient unit such as seconds second (s), minutes (min), hours (h), days (d) or
possible years. Therefore, the units of reaction rates may be
 mol dm-3 s-1 , mol dm-3 min–1 , mol dm-3 h–1 and, so on

PHYSICAL CHANGES WHICH INDICATE PROGRESS OF REACTION
 Rates, can be conveniently studied by measuring the suitable physical property of the
system as a whole or of one of the constituent of the reaction. Some common
examples are given below:
 Disappearance of Reactants and Appearance of Products.
 Colorimetry or Colour Change. Here increase or decrease in the intensity of colour is
observed at regular intervals
 Volume Changes. Measure the volume of the gas given off

• pH Measurement.
• Pressure Changes
• Change in Mass.
• Change in Temperature.

Analysis and Interpretation of Simple Graphs on Rates of Reaction
ENOCH DORMAN (dorman.enochkojo@gmail.com)
Graph 1
It shows the decrease in the amount of a solid reactant with time
27/06/2023

Graph 2
It shows the increase in the amount of a solid product
with time. The graph tends towards a maximum
amount possible when all the solid reactant is used up
and the graph becomes horizontal. This means the
speed has become zero as the reaction has stopped.

Graph 3
It shows the decrease in reaction time with increase
in temperature as the reaction speeds up. The reaction
time can represent how long it takes to form a fixed
amount of gas in the first few minutes of a metal
carbonate-acid reaction.

Graph 4
It shows the increase in speed of a
reaction with increase in temperature as
the particles have more and more kinetic
energy.

Graph5
It shows the increase in the amount of a gas
formed in a reaction with time. Again, the
graph becomes horizontal as the reaction
stops when one of the reactants is all used
up!

Graph 6
It shows the effect of Increasing concentration,
which decreases the reaction time, as the speed
increases because the greater the concentration
the greater the chance of fruitful collision.

Graph 7
It shows the rate/speed of reaction is
often proportional to the concentration
of one particular reactant. This is due to
the chance of a fruitful collision
forming products being proportional to
the concentration.

Theories of Reaction Rate
 This is made up of the collision theory and the energy of
activation or the transition state theory. Collision theory,
was developed by Max Trautz and William Lewis in
1916-18 and, provides a greater insight into the energetic
and mechanistic aspects of reactions and their rates

The collision theory
 The collision theory combines the following reasoning;
 Reactions occur due to approach and collisions of reactant particles
(atoms, molecules or ions).
 An effective collision, the one required for changing reactants to
products, can occur between two molecules only if they possess a certain
minimum amount of energy in excess of the normal energy of molecules.

Only a small fraction of collisions is successful in producing
a reaction. These collisions are called effective collisions.
The minimum energy which molecules must possess before
collision should be equal to or greater than the activation
energy.
 The rate of reaction is proportional to the frequency of
effective collisions per second.

NB: It must be noted here that the collisions
between reactant molecules will not lead to
reaction even if the energy requirement is satisfied.
It is because the colliding molecules should also
have proper orientation

Proper Orientation Improper Orientation

The energy of activation or the transition state
theory
 ACTIVATION ENERGY THEORY: Colliding
molecules need minimum amount of energy before
they can equal or go over the intermediate transition
state in order to bring about reaction. The minimum
amount of energy is called activation energy

Energy profile diagrams for exothermic reactions

In short,
For fast reactions; activation energies are
low.
For slow reactions; activation energies
are high.

Factors affecting rate of reaction
 CONCENTRATION
If the concentration of any reactant in a solution is increase, the rate of
reaction is increase. This because increasing the concentration, increases the
number of molecules per unit volume of the reactant vessel. Thus, the
number of collisions as well as the number of effective collisions per unit
time increase and so the rate of reaction increase.

THE EFFECT OF TEMPERATURE
Increasing temperature increases the rate of
reaction.
Reason: increasing temperature increases the kinetic
energy of reactant molecules and so the number of
collisions as well as the number of effective collisions per
unit time increases. The greater the number of collisions
per unit time, the higher the rate of reaction

Again when temperature is increased, the K.Es
of reactant molecules are increased. The fraction
of the reactant molecules whose K. Es is greater
than the activation energy of the reaction is
increased and this increases the rate of reaction.

Nature of reactants/ surface area.
The greater the surface area of reactants, the
greater the number of collisions and effective
collisions and the higher the rate of reaction. The
surface area depends on the nature of reactants.
Thus surface area decreases in order; Gas ►
liquid ►solid

Pressure
Pressure affects reactions whose reactant are gaseous
species. If the pressure of reactant is increased, the rate
of reaction is increased. This is because the number of
collisions as well as the number of effective collisions
per unit time increase and so the rate of reaction
increase.

Catalyst
Catalyst provides new reaction pathway with lowered
activation energy such that more reactant molecules
have their K.E greater than the activation energy of the
reaction. The greater the fraction of reactant molecules
with K.E greater than the activation energy, the higher
the reaction

Reaction Rates and Stoichiometry
 During our discussion of the hypothetical reaction,
A → B, we saw that the stoichiometry requires
that the rate of disappearance of A’ equal the rate
of appearance of B.
 What happens when the stoichiometric
relationships are not one-to-one?

For example, consider the reaction 2HI (g) → H2 (g) + I2
(g)
To fix this problem, we need to take into account the
reaction stoichiometry.
In general, for the reaction, a A + b B → c C + d D, the
rate is given by:
Rate = -
1
𝑎
𝑑⦋𝐴⦌
𝑑𝑡
= -
1
𝑏
𝑑⦋𝐵⦌
𝑑𝑡
= -
1
𝑐
𝑑⦋𝐶⦌
𝑑𝑡
= -
1
𝑑
𝑑⦋𝐷⦌
𝑑𝑡

Rate = -
1
2
𝑑⦋𝐻𝐼⦌
𝑑𝑡
=
𝑑⦋𝐻2⦌
𝑑𝑡
=
𝑑⦋𝐼2⦌
𝑑𝑡

PRACTICE EXERCISE
1. Write the rate expressions for the following reactions in terms of the
disappearance of the reactants and the appearance of the products:
(a) I2 (aq) + OCl2 (aq) →Cl2 (aq) + OI2 (aq)
(b) 4NH3 (g) + 5O2 (g) →4NO (g) + 6H2O (g)
2. If the rate of decomposition of N2O5 in the reaction 2N2O5(g)→4
NO2(g) + O2(g) at a particular instant is 4.2 ×10-7 M/s, what is the rate of
appearance of (a) NO2 and (b) O2 at that instant?

3. Consider the reaction; 4PH3 (g) →P4(g) +
6H2(g)
Suppose that, at a particular moment during
the reaction, molecular hydrogen is being
formed at the rate of 0.078 M/s. (a) At what
rate is P4 being formed? (b) At what rate is
PH3 reacting?

RATE LAW
 At a fixed temperature the rate of a given reaction
depends on concentration of reactants. The exact
relation between concentration and rate is determined
by measuring the reaction rate with different initial
reactant concentrations

By a study of numerous reactions it is shown that: the rate of a
reaction is directly proportional to the reactant
concentrations, each concentration being raised to some
power.
Thus for a substance A undergoing reaction;
Rate ∝ [A]n
Or rate = k [A]n ….1
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For a reaction: 2A + B ⎯⎯→ products
The reaction rate with respect to A or B is determined by varying the
concentration of one reactant, keeping that of the other constant. Thus
the rate of reaction may be expressed as:
Rate = k [A]m [B]n ……..2
Expressions such as (1) and (2) tell the relation between the rate of a
reaction and reactant concentrations.
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An expression which shows how the reaction rate is
related to concentrations is called the rate law or rate
equation. The proportionality constant k is called the
specific rate constant for the reaction. The magnitude of k
changes with temperature and therefore determines how
temperature affects rate
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m and n are numbers that must be determined
experimentally. Note that, in general, m and n are not
equal to the stoichiometric coefficients. Added
together, they give us the overall reaction order,
defined as the sum of the powers to which all
reactant concentrations appearing in the rate law are
raised.
27/06/2023

ORDER OF A REACTION
A reaction has an individual order
“with respect to” or “in” each
reactant. For a simple reaction;
A→ Product:

 When ⦋A⦌ doubles and the rate doubles, the rate depends on ⦋A⦌1
and the reaction is first order with respect to A
 When [A] doubles and the rate quadruples, then the rate depends
on [A]2 and the reaction second order with respect to [ A].
 If the rate does not change when [A] is doubled, the rate does not
depend on [A]0 and the reaction is zero order with respect to [A]
27/06/2023

APPLICATION
 In a certain reaction, the rate determining step is A + 2B → C +D.
a. Write the expression for the rate equation.
b. Deduce what will be the effect on the rate of reaction if
i. The concentration of A is doubled
ii. The concentration of is B doubled

a.For R.D.S the stoichiometric moles of the reactants is
also their respective orders. Therefore, R = [A] [B] 2
b.i. Let R1 the Rate when [A] is doubled.
Then; R = K [A] [B] 2 …………1
And R1 = 2K [A] [B] 2 …………2
Divide equation 2 by 1
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𝐑𝟏
𝐑
=
= 𝟐𝐊 [𝐀] [𝐁]𝟐
𝐊 [𝐀] [𝐁]𝟐
𝐑𝟏
𝐑
= 2
R1 =2R
Hence doubling [A] doubles the reaction rate
ii. Let R2 the Rate when [B] is doubled
Then; R = K [A] [B] 2 …………1
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And R2= 4K [A] [B] 2 …………3
Divide equation 2 by 1
𝐑𝟐
𝐑
=
= 𝟒𝐊 [𝐀] [𝐁]𝟐
𝐊 [𝐀] [𝐁]𝟐
𝐑𝟐
𝐑
= 4
R2 =4R
Hence doubling [B] quadruples the reaction rate
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Example 2:
The rate law for the reaction 2A +3B+ C → 3D is given
as R = K [A] [C] 2.
Deduce what will be the effect on the rate of reaction if
a.The concentration of A is doubled
b.The concentration of is B doubled
The concentration of is C doubled
27/06/2023

Answer:
Students are to do it by
themselves
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The initial rate of a reaction was measured for several different starting
concentrations of A and B, and the results are as follows: A + B → C
Experiment
Number [A] (M) [B] (M) Initial Rate (M/s)
1 0.100 0.100 4.0 × 10-5
2 0.100 0.200 4.0×10-5
3 0.200 0.100 16.0 × 10-5
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Using these data, determine (a) the rate law for the
reaction, (b) the rate constant, (c) the rate of the
reaction when and [A] = 0.050 M [B] = 0.100M
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Solution
a)We assume that the rate law has the following form: Rate = k
[A]m[B]n
If we compare experiments 1 and 2, we see that [A] is held constant
and [B] is doubled. Thus, this pair of experiments shows how [B]
affects the rate, allowing us to deduce the order of the rate law with
respect to B.
Thus;
𝑹𝟐
𝑹𝟏
=
𝑲 𝑨 𝒎 𝑩 𝒏
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𝟒.𝟎 × 𝟏𝟎
−
𝟓
𝟒.𝟎 × 𝟏𝟎
−
𝟓 =
𝑲 𝟎.𝟏𝟎𝟎 𝐦 𝟎.𝟐𝟎𝟎 𝐧
𝑲( 𝟎.𝟏𝟎𝟎 𝐦 ( 𝟎.𝟏𝟎𝟎 𝐦
1 = 2n
n = 0
(Because the rate remains the same when [B] is doubled, the
concentration of B has no effect on the reaction rate. The rate
law is therefore zero order in B (that is, n = 0).
27/06/2023

Again; In experiments 1 and 3, [B] is held constant, so these data show how [A] affects rate.
Thus;
𝑹𝟑
𝑹𝟏
=
𝟏𝟔.𝟎 × 𝟏𝟎
−
𝟓
𝟒.𝟎 × 𝟏𝟎
−
𝟓 =
𝑲 𝟎.𝟐𝟎𝟎 𝐦 𝟎.𝟏𝟎𝟎 𝐧
𝑲( 𝟎.𝟏𝟎𝟎 𝐦 ( 𝟎.𝟏𝟎𝟎 𝐦
4 = 2m
22= 2n
n = 2
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(Holding [B] constant while doubling [A] increases the
rate fourfold. This result indicates that rate is proportional
to [A] 2, that is, the reaction is second order in A
Hence the rate expression is: Rate = k [A] 2[B] 0 = k [A] 2
27/06/2023

b. Using the rate law and the data from experiment
1, we have;
K =
𝑅
[𝐴]2
=
4.0 × 10−5 𝑀/s
(0.100 𝑀)2
= 4.0 × 10-3 M-1 s-1
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c. Using the rate law from part (a) and the rate constant
from part (b), we have;
Rate = k [A]2 = (4.0×10-3 M-1 s-1)(0.050 M)2 = 1.0 ×10-5
M/s
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Example 4
The following data were measured for the reaction of nitric oxide with hydrogen
2NO (g) + 2 H2 (g) →N2 (g) + 2 H2O (g)
Experiment
Number [NO] (M) [H2] (M) Initial Rate (M/s)
1 0.100 0.100 1.23 × 10-3
2 0.100 0.200 2.46×10-3
3 0.200 0.100 4.92 × 10-3
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(a)Determine the rate law for this reaction.
(b) Calculate the rate constant. (c)
Calculate the rate when [NO] = 0.050 M
and [H2] = 0.150 M
27/06/2023

Answer:
Students are to do it by themselves
27/06/2023

Integrated equation for first order reaction
 A first-order reaction is a reaction whose rate
depends on the reactant concentration raised to
the first power. In a first-order reaction of the type
A→ product

R=
−𝑑[𝐴]
𝑑𝑡
…………………………1
From the rate law we also know that
R = k [A]…………………………2
Combining the first two equations for the rate we get
−𝑑[𝐴]
𝑑𝑡
= k [A] ……………………….3
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Rearranging, we get
𝑑[𝐴]
[𝐴]
= -kdt
Integrating between t= 0 and t= t gives
𝑜
𝑡 𝑑[𝐴]
[𝐴]
= -k 0
𝑡
𝑑𝑡
In [A]t – In [A] 0 = -kt…………………4
In (
𝐴 𝑡
[𝐴]0
) = -kt………………………5
27/06/2023

Equation 4 can be rearranged as follows; ln [A]t = -kt + ln [A] 0 ……………..6
27/06/2023

Equation 6 has the form of the linear equation y = mx+ b, in
which m is the Slope of the line that is the graph of the equation.
The slope of first order reaction is –k. For a first-order reaction, if
we plot ln [A]t versus time (y versus x), we obtain a straight line
with a slope equal to -k and a y intercept equal to ln [A]0 [Figure
2 Thus, we can calculate the rate constant from the slope of this
plot.
27/06/2023

Fig 2
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Reaction Half-life
As a reaction proceeds, the concentration of the reactant(s)
decreases. Another measure of the rate of a reaction, relating
concentration to time, is the half-life, t12 which is the time required
for the concentration of a reactant to decrease to half of its initial
concentration. We can obtain an expression for t12 for a first-order
reaction as follows.
By the definition of half-life, when t = t12, [A]t = [A] 0 /2, so
substituting into equation 5 gives
27/06/2023

Equation 5 can be rearranged to gives;
In (
𝑨 𝟎
𝟐[𝑨]𝟎
) = -k t12
In (
𝟏
𝟐
) = -k t12
- 0.693= -k t
t12 =
𝟎.𝟔𝟗𝟑
𝒌
………………7
Equation 7 tells us that the half-life of a first-order reaction is
independent of the initial concentration of the reactant.
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TUTORIALS
27/06/2023

1. Derive an expression for the half-life period of the
following reaction:
A→B, rate α [A]
2. The reaction, A + B + C →Products, is found to obey
the rate law R= k [A]2 [B]3/ 2 [C]-1/ 2 What is the order of
overall reaction?.
27/06/2023

1.If the half-life of a first order reaction in A is 15
min., how long it will take for [A] to reach 10
percent of the initial concentration?
2. The rate constant for a first order reaction is
1.54 × 10–3 sec–1. Calculate its half-life period.
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1.If the half-life of a first order in A is 2
min, how long will it take A to reach 25%
of its initial concentration?
27/06/2023

1. Consider the reaction in which nitric oxide is oxidized to nitrogen
dioxide,
2NO (g) + O2(g) → 2NO2(g), for which the rate law is = k [NO]2 [O2].
If this reaction takes place in a sealed vessel and the partial pressure
of nitric oxide is doubled, what effect would this have on the rate of
reaction?
27/06/2023

1.For first-order reactions the rate constant, k, has the unit(s)
(a)l mol–1 (b) time–1 (c) (mol/l)–1 time–1 (d) time mol l–1
1. What are the units of the rate constant for a reaction in
solution that has an overall reaction order of two? (M is
molarity, s is seconds.)
(a)M–1 s–1 (b) M–1 (c) s–1 (d) M s–
27/06/2023

1. The first-order rate constant for the decomposition of
N2O5 to NO2 and O2 at 70°C is 6.82 × 10–3 s–1.
Suppose we start with 0.300 mol of N2O5(g) in a 0.500 L
container. How many moles of N2O5 will remain after
1.5 min?
(a)0.081 mol (b) 0.555 mol (c) 0.325 mol (d) 0.162
mol
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1.For a first–order reaction of the form A → P, t½ = 9
hours. If the concentration of A is 0.0013 M right now,
what is the best estimate of what it was THERE?
(a)0.0026 M (b) 0.0065 M (c) 0.0052 M (d) 0.042 M
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1. Consider the reaction 3A → 2B. The average rate of appearance
of B is given by [B]/t. How is the average rate of appearance of
B related to the average rate of disappearance of A?
(a)–2[A]/3t (b) [A]/t (c) –[A]/t (d) –3[A]/2t
27/06/2023

1. For the reaction 2NO2 + O3 → N2O5 + O2 the following observations
are made: Doubling the concentration of [NO2] doubles the rate, and
doubling the concentration of [O3] doubles the rate. What is the rate
law for the reaction?
(a)rate = k [NO2] (b) rate = k [NO2]2 [O3] (c) rate = k [NO2]2 [O3]2
(d) rate = k [NO2] [O3]
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1. The reaction, 2NO (g) →N2 (g) + O2 (g), proceeds in a single elementary
step. This reaction is thus
(a)the molecularity cannot be determined from the given
information
(b)termolecular
(c)bimolecular
(d)unimolecular
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1.For a certain reaction, the rate = k [NO]2 [O2], when the
initial concentration of NO is tripled, the initial rate
(a)decreases by a factor of nine
(b)increases by a factor of three
(c)increases by a factor of six
(d)increases by a factor of nine
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1. A first order reaction requires 8.96 months for the
concentration of reactant to be reduced to 25.0% of its
original value. What is the half life of the reaction?
(a)4.48 months (b) 2.24 months (c) 8.96 months (d) 17.9
months
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1. The half-life for a first order reaction is 2768 years. If the
concentration after 11,072 years is 0.0216 M, what was
the initial concentration?
(a)0.0690 M (b) 0.345 M (c) 0.173 M (d) 1.000 M
27/06/2023

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