Dufile Seed SS S6 Chemistry term I, 2024

CHEMICAL KINETICS
Chemical kinetics is the study of rates of chemical reactions and how the reactions depend on the
factors affecting them.
Rate of a chemical reaction
Rate of a chemical reaction is the decrease in concentration of reactants or the increase in
concentration of products formed per unit time.
Reaction rates cannot be predicted theoretically and so they are determined experimentally.
Consider the reaction:
𝑊 +𝑋 →𝑌+𝑍
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒
−𝑑[𝑊] −𝑑[𝑋]
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = or
𝑑𝑡 𝑑𝑡

The negative sign inserted indicates that the concentrations of reactants decrease with time.
In terms of products;
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 =
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒
𝑑[𝑌] 𝑑[𝑍]
𝑅𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = or =
𝑑𝑡 𝑑𝑡

The rate of the reaction is not the same when expressed in terms of 𝑊 or 𝑋 as it is in terms of 𝑌
or 𝑍. It is usually necessary to specify which species is used when expressing rate of a chemical
reaction.
Units of rate of reaction
Different units can be used but if concentration is given in 𝑚𝑜𝑙𝑑𝑚−3 and time in seconds, then;-
𝑢𝑛𝑖𝑡 𝑜𝑓 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑚𝑜𝑙𝑑𝑚−3
𝑈𝑛𝑖𝑡 𝑎𝑡𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = = = 𝑚𝑜𝑙𝑑𝑚−3 𝑠 −1
𝑢𝑛𝑖𝑡 𝑜𝑓 𝑡𝑖𝑚𝑒 𝑠
Graphically;

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Dufile Seed SS S6 Chemistry term I, 2024

The Rate law

states that the rate of a reaction is directly proportional to the molar concentrations of reactants
raised to their appropriate powers which are experimentally determined at a given temperature.
Consider a reaction 𝒙𝐴 + 𝒚𝐵 + 𝒛𝐶 → products
Rate of reaction ∝ [𝐴]𝑙 [𝐵]𝑚 [𝐶]𝑛
Rate of reaction = 𝑘[𝐴]𝑙 [𝐵]𝑚 [𝐶]𝑛
Where 𝑘 is the rate constant. 𝑙, 𝑚 and 𝑛 are experimentally determined values The equation
above is known as the rate law or rate equation.
The rate equation is an expression that relates the rate of reaction and the product of the molar
concentrations of reactants raised to their appropriate powers.
The rate law for any reaction cannot be from the balanced chemical equation using the
stoichiometric coefficients but it is determined experimentally.
The rate constant, 𝑘. Rate constant of a reaction is the proportionality constant in an
experimentally determined rate equation.
Or
Rate constant of a reaction is the ratio of rate of reaction to the product of the molar
concentrations of reactants raised to appropriate powers which are experimentally determined at
a given temperature.
Order of reaction. Order of a reaction is the sum of the powers to which the molar concentrations
of the reactants are raised in the experimental rate equation.
For the reaction discussed above with the rate equation;
Rate of reaction = 𝑘[𝐴]𝑙 [𝐵]𝑚 [𝐶]𝑛
Where 𝑙, 𝑚 and 𝑛 are orders of reaction with respect to reactants 𝐴, 𝐵 and 𝐶 respectively.
The overall order is (𝑙 + 𝑚 + 𝑛).
Note:
- Order of a reaction helps us classify reactions zero order, first order or third order
reactions.
- It can be a whole number (positive or negative) or a fraction.
- The order of a reaction with respect to a given reactant cannot be deduced using
stoichiometric coefficients in the equation.

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Molecularity.
Molecularity of a reaction is the total number of molecules, atoms or ions involved in the rate
determining step of a reaction.
The rate determining step is the slowest step in the reaction mechanism for a reaction that takes
place in more than one step.
Molecularity is always a whole number and never greater than three. Molecularity is a theoretical
postulate and not experimentally measured. It must be a whole number.

Difference between order and molecularity of a reaction

Molecularity of a reaction Order of a reaction
The number of molecules, atoms or ions Sum of the powers to which the molar
involved in the rate determining step of a concentrations of the reactants are raised in the
reaction experimental rate equation
Molecularity is a theoretical concept It is determined experimentally
Molecularity is always a whole number Order can be a zero or any whole number
Molecularity explains mechanism of a Order does not tell anything about mechanism
reaction of a reaction

Integrated rate equation for zero, first and second order of reaction.
A. Zero order
A zero order reaction is a reaction in which the rate of reaction is independent of the
concentration of the reactant.
Consider a reaction below where the rate of reaction do not depend on the concentration
of the reactants.
𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
At 𝑡 = 0, 𝑎0 0
At 𝑡 = 𝑡, 𝑎0 − 𝑥 𝑥
𝑑 [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠]
From 𝑅𝑎𝑡𝑒 = , but 𝑅𝑎𝑡𝑒 = 𝒌[𝐴]0 since order is zero
𝑑[𝑡𝑖𝑚𝑒]

𝑑𝑥
= 𝒌[𝐴]0
𝑑𝑡
𝑑𝑥
= 𝒌(𝑎0 − 𝑥 )0
𝑑𝑡
𝑑𝑥
=𝒌
𝑑𝑡
𝑑𝑥 = 𝒌𝑑𝑡
Integrating on both sides

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𝑥=𝒙 𝑡=𝒕

∫ 𝑑𝑥 = ∫ 𝒌𝑑𝑡
𝑥=𝟎 𝑡=𝟎

𝑥=𝒙 𝑡=𝒕

∫ 𝑑𝑥 = 𝒌 ∫ 𝑑𝑡
𝑥=𝟎 𝑡=𝟎

𝑥 = 𝒌𝑡

A graph of 𝒙 versus 𝒕

𝑆𝑙𝑜𝑝𝑒 = −𝑘

A graph of rate of reaction against concentration of reactant is a straight line with zero
gradient.
(𝑅𝑎𝑡𝑒 = 𝒌[𝐴]0 = 𝒌)

Examples of zero order reactions include:

i. Iodination of propanone.
𝐶𝐻3 𝐶𝑂𝐶𝐻3 (𝑎𝑞) + 𝐼2 (𝑎𝑞) → 𝐶𝐻3 𝐶𝑂𝐶𝐻2 𝐼 (𝑎𝑞) + 𝐻𝐼 (𝑎𝑞)
This reaction is zero order with respect to iodine. The rate of reaction does not change
if the concentration of iodine is changed.
ii. The decomposition of hydrogen iodide on a gold surface.
2𝐻𝐼 (𝑎𝑞) → 𝐻2 (𝑔) + 𝐼2 (𝑔)
iii. Decomposition of ammonia to nitrogen and hydrogen by a hot tungsten wire catalyst.

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B. First order
A first order reaction is a reaction in which the rate of reaction is proportional to the
concentration of the reactant raised to power one.
Consider a reaction.
𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
At 𝑡 = 0, 𝑎0 0
At 𝑡 = 𝑡, 𝑎0 − 𝑥 𝑥
𝑑 [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠]
Using 𝑟𝑎𝑡𝑒 = 𝑑[𝑡𝑖𝑚𝑒]
𝑑𝑥
= 𝒌[𝐴]
𝑑𝑡
𝑑𝑥
= 𝒌( 𝑎0 − 𝑥)
𝑑𝑡
𝑑𝑥 = 𝒌( 𝑎0 − 𝑥)𝑑𝑡
𝑑𝑥
= 𝒌𝑑𝑡
(𝑎0 −𝑥)
Integrating on both sides
𝑥=𝒙 𝑡=𝒕
𝑑𝑥
∫ = 𝒌 ∫ 𝑑𝑡
(𝑎0 − 𝑥)
𝑥=𝟎 𝑡=𝟎

ln|( 𝑎0 − 𝑥)|0𝑥 = −𝒌𝑡

ln(𝑎0 − 𝑥) − ln(𝑎0 ) = −𝒌𝑡
(𝑎0 −𝑥)
ln = −𝒌𝑡
(𝑎0 )
(𝑎0 −𝑥)
= 𝑒 −𝒌𝑡
(𝑎0 )
(𝑎0 − 𝑥) = 𝑎0 𝑒 −𝒌𝑡
Let (𝑎0 − 𝑥) be the concentration at any time, 𝑡 as 𝒂 and 𝒂𝟎 be the initial concentration
of the reactant, the

𝒂 = 𝒂𝟎 𝒆−𝒌𝒕

𝑎0
Or 2.303𝑙𝑜𝑔 ( 𝑎 ) = 𝒌𝑡
0 −𝑥

Examples of first order reactions

- All radioactive decays
- Hydrolysis of sucrose in presence of acid.
- Acid catalyzed hydrolysis of an ester.
- Radioactive decay.
- Decomposition of nitrogen pentaoxide, N2 O5

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- Decomposition of hydrogen peroxide to form water and oxygen Decomposition of

Benzene diazonium chloride.
Conditions for first order reaction
- One of the reactants is in large excess that only a small fraction of it will be used up
in the reaction.
- Use of acid as catalyst; where its concentration doesn’t change during the course
of the reaction e.g. acid catalyzed hydrolysis of esters and sucrose.

Half-life
This is the time taken for a radioactive material to disintegrate to half of its original amount,
it’s denoted as (𝑡1⁄ )
2
𝑎0
At half-life, 𝑎0 − 𝑥 = 2
𝑎0
Then 2.303𝑙𝑜𝑔 ( 𝑎0 ) = 𝒌𝑡1⁄
2
2
2.303𝑙𝑜𝑔(2) = 𝒌𝑡1⁄
2
2.303𝑙𝑜𝑔 (2)
𝑡1⁄ =
2 𝒌

0.693
𝑡1⁄ =
2 𝒌

A plot of 𝑎0 − 𝑥 against 𝑡 gives (A graph of concentration of reactant against time)

𝒕𝟎 𝒕𝟏 𝒕𝟐

𝒕
For a first order reaction, 𝒕𝟎 = 𝒕𝟏 = 𝒕𝟐 = 𝒕𝟏⁄
𝟐

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A 𝑙𝑜𝑔(𝑎0 − 𝑥) or 𝑙𝑜𝑔𝑎 against 𝑡, OR: If all values of (𝑎0 − 𝑥)are < 1

𝑙𝑜𝑔𝒂𝒐

−𝒌
𝑆𝑙𝑜𝑝𝑒 = 𝑘 =
𝟐.𝟑𝟎𝟑
𝒍𝒐𝒈𝒂

𝒕
Using
𝑎0
2.303𝑙𝑜𝑔 ( 𝑎 ) = 𝒌𝑡
0 −𝑥
𝑎0 𝒌
𝑙𝑜𝑔 ( )= ( )𝑡
𝑎0 −𝑥 𝟐.𝟑𝟎𝟑

𝑎0
A 𝑙𝑜𝑔 ( 𝑎 ) is plotted against time
0 −𝑥

𝑎0
𝑙𝑜𝑔 ( 𝑎 )
0 −𝑥

A graph of 𝐼𝑛( 𝑎0 − 𝑥) against time 𝑡. OR: If all values of (𝑎0 − 𝑥) are < 1

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𝑎0
A graph of 𝐼𝑛 ( 𝑎 ) against time 𝑡
0 −𝑥

Summary of graphs

Questions
1. The kinetics data for the reaction between P and sodium hydroxide is shown in the table
below.
Concentration 1.05 0.88 0.74 0.51 0.37 0.26 0.16 0.10
of P
Time 0.0 3.5 70 14.5 20.0 27.0 35.5 45.0
(minutes)
a) Plot a graph of concentration of P against time.
b) From the graph determine:
i) the half – life of P
ii) order of the reaction
iii) the rate constant for the reaction.

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2. Hydrogen peroxide decomposes in presence of iron(III) chloride catalyst. The order of the
reaction can be found by withdrawing aliquots of the solution, adding dilute sulphuric acid
and titrating with potassium manganite (VII) solution. The volume of manganite (VII)
solution is a measure of concentration of hydrogen peroxide remaining after a certain
time. The table below shows the volume of potassium manganite (VII) required at various
time intervals.
Time (minutes) 0 5 10 15 20 25 30
Volume of 30 23.4 18.3 14.7 14.2 8.7 6.8
3
𝐾𝑀𝑛𝑂4 (𝑐𝑚 )
a) Plot a graph of volume of potassium manganite (VII) solution against time.
b) From the graph determine the:
i) half-life of the reaction.
ii) order of the reaction. Give a reason for your answer.
c) Write down the rate equation for the decomposition of hydrogen peroxide
d) Calculate the rate constant for the reaction.

3. a) State what is meant by the term:

i) Order of reaction
ii) Half-life of reaction
b) The table below shows the kinetic data obtained for hydrolysis of methylethanoate in
acidic media.
𝐶𝐻3 𝐶𝑂𝑂𝐶𝐻3 (𝑚𝑜𝑙𝑑𝑚−3 ) 0.241 0.161 0.109 0.073 0.046 0.034
Time (minutes) 0 60 120 180 240 320
Plot a graph of concentration of methyl ethanoate against time
b) Using the graph in (b), determine the:
i) half-life of the reaction
ii) order of the reaction with respect to 𝐶𝐻3 𝐶𝑂𝑂𝐶𝐻3 . Give a reason for your answer
c) Calculate the rate constant and indicate its units.

4. a) Dinitrogen pentoxide decomposes according to the equation;

2𝑁2 𝑂5 (𝑔) → 4𝑁𝑂2 (𝑔) + 𝑂2 (𝑔)
The mechanism for the reaction is as shown below:
Step 1: 𝑁2 𝑂5 → 𝑁𝑂2 + 𝑁𝑂3
Step 2: 𝑁2 𝑂5 + 𝑁𝑂3 → 3𝑁𝑂2 + 𝑂2
i) Define the term molecularity
ii) Given that the above reaction is unimolecular. Identify the rate determining step
of the reaction. Give a reason for your answer.

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b) 0.2M benzene diazonium chloride decomposes in the presence of water according to

the equation:
𝐶6 𝐻5 𝑁2+ 𝐶𝑙 −(𝑎𝑞) + 𝐻2 𝑂2 (𝑙) → 𝐶6 𝐻5 𝑂𝐻 (𝑎𝑞) + 𝑁2 (𝑔)
The volume of nitrogen gas evolved was recorded at different time intervals shown
below.

Time (s) 0 10 20 30 40 50 60 70 80
Vol of 𝑁2 (𝑑𝑚3 ) 0.0 0.42 0.70 0.90 1.08 1.22 1.28 1.30 1.30
i. Plot a graph of volume of nitrogen gas against time. Use your graph to determine
the initial rate
ii. The experiment was repeated with 0.05M benzene diazonium chloride at 40℃
and the initial rate was found to be 0.0205𝑑𝑚3 𝑠 −1 . Calculate the order of
reaction with respect to benzene diazonium chloride.
c) Write the rate equation if the concentration of water remains constant during the
reaction
d) State the effect of increasing temperature on the value of the rate constant.

5. The table below shows the kinetic data that was obtained for the conversion of sucrose to
glucose in acid solution.
Conc. of sucrose (𝑚𝑜𝑙𝑑𝑚−3 ) 0.08 0.06 0.04 0.02 0.01
−3
Rate of reaction (𝑚𝑜𝑙𝑑𝑚 ) 0.004 0.003 0.02 0.01 0.05
a. Plot a graph of rate of reaction against the concentration of sucrose.
b. State the order of the reaction. Give a reason for your answer.
c. Determine the rate constant for the reaction and indicate its units.
d. Calculate the rate of the reaction when the concentration of sucrose was 0.12𝑚𝑜𝑙𝑑 −3 .

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Methods to calculate the rate constant of first order reaction (determining order of reactions)
1. Titration or titrimetric method:
In this method an unknown solution is titrated against a standard solution of known
concentration. The volume of standard solution required at different time intervals is
measured from which the concentration of unknown solution can be calculate.
Let 𝑉𝑜 and 𝑉𝑡 be the volume of standard solution used at time 𝑡𝑜 and 𝑡 respectively.
If 𝑉𝑡 is the volume of standard solution used at time, 𝑡 and 𝑉∞ is volume of standard
solution used at completion of titration i.e. at 𝑡∞
Then from
[𝑎0 ]
ln = 𝒌𝑡
[𝑎0 − 𝑥]

2.303 [𝑉∞ − 𝑉0 ]
𝑘= log
𝑡 [𝑉∞ − 𝑉𝑡 ]

2. Optical method (by measuring angle of rotation)

In this method the total angle of rotation of the sample is measured at different time
intervals from which we can calculate the amount present at any given time. Consider the
reaction below
𝐴 → 𝐵
Let 𝑟𝑜 and 𝑟𝑡 be the initial angle of rotation and angle of rotation at time 𝑡 respectively.
If 𝑟∞ is angle of rotation after completion of reaction
Initial amount is 𝑟𝑜 − 𝑟∞ and amount present at any time t is 𝑟∞− 𝑟𝑡
2.303 𝑟𝑜 − 𝑟∞
𝑘= log ( )
𝑡 𝑟∞ − 𝑟𝑡
3. Pressure measurement
Consider the reaction below
𝐴(𝑔) → 𝐵(𝑔) + 𝐶(𝑔)
At 𝑡 = 0 𝑃𝑜
At 𝑡 = 𝑡 𝑃𝑜 − 𝑥 𝑥 𝑥
Total pressure at time t: 𝑃𝑡 = 𝑃𝑜 − 𝑥 + 𝑥 + 𝑥
= 𝑃𝑜 + 𝑥
Hence from 𝑃𝑡 = 𝑃𝑜 + 𝑥,
𝑥 = 𝑃𝑡 − 𝑃𝑜
2.303 𝑃𝑜
𝑘= log
𝑡 2𝑃𝑜 − 𝑃𝑇
Where 𝑃𝑜 = Initial partial pressure of A
𝑃𝑇 = Total pressure of gaseous system at time 𝑡

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4. Calorimetric method
5. Conductimetric analysis
6. Conductivity measurement

C. Second order
A second order reaction is a reaction in which the rate of reaction is proportional to the
concentration of the reactant raised to the second power.
Or a reaction in which the rate of reaction is proportional to the product of the
concentration of two reactants each raised to power one.
Consider a reaction
2𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
At 𝑡 = 0, 𝑎0 0
At 𝑡 = 𝑡, 𝑎0 − 𝑥 𝑥
𝑑 [𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠]
Using 𝑟𝑎𝑡𝑒 =
𝑑[𝑡𝑖𝑚𝑒]
𝑑𝑥
= 𝒌[𝐴]2
𝑑𝑡
𝑑𝑥
= 𝒌(𝑎0 − 𝑥)2
𝑑𝑡
𝑑𝑥 = 𝒌(𝑎0 − 𝑥)2 𝑑𝑡
𝑑𝑥
= 𝒌𝑑𝑡
(𝑎0 −𝑥)2
Integrating on both sides
𝑥=𝒙 𝑡=𝒕
𝑑𝑥
∫ = 𝒌 ∫ 𝑑𝑡
(𝑎0 − 𝑥)2
𝑥=𝟎 𝑡=𝟎

𝑥
= 𝒌𝑡
𝑎0 ( 𝑎0 − 𝑥)
(1)
1
A plot of against 𝑡 gives
𝑎0 −𝑥

1
𝑠𝑙𝑜𝑝𝑒 𝑘
𝑎0 −𝑥

1
𝑎0

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Half - life
𝑎0 𝑎0
At 𝑡1⁄ , 𝑎0 − 𝑥 = also 𝑥 =
2 2 2
Implying that equation (𝑖) becomes;-
𝑎0
2 = 𝒌𝑡
𝑎
𝑎0 ( 20 )

1
𝑡1⁄ =
2 𝑎0 𝒌

Examples of second order reactions include:

i. Reaction between hydrogen and iodine to form hydrogen iodide.
𝐻2 (𝑔) + 𝐼2 (𝑔) → 2𝐻𝐼 (𝑎𝑞)
ii. Oxidation of iodide ions to iodine by peroxodisulphate ions.
𝑆2 𝑂82−(𝑎𝑞) + 2𝐼(𝑎𝑞)
−
→ 2𝑆𝑂42−(𝑎𝑞) + 𝐼2 (𝑎𝑞)
iii. Reaction of bromoethane with sodium hydroxide.
𝐶𝐻3 𝐶𝐻2 𝐵𝑟 + 𝑁𝑎𝑂𝐻 → 𝐶𝐻3 𝐶𝐻2 𝑂𝐻 + 𝑁𝑎𝐵𝑟

D. PSEUDO-ORDER REACTIONS.
Definition:
These are reactions whose order is different from that expected using the rate law.
Pseudo-first order reactions.
These are reaction which are practically first order but theoretically are of higher orders.
Examples of pseudo-order reactions includes; -
- Hydrolysis of sucrose in the presence of an acid.
- Hydrolysis of an ester in the presence of an acid.
Consider the hydrolysis of the ester below;
𝐶𝐻3 𝐶𝑂𝑂𝐶𝐻3 + 𝐻2 𝑂 ⇌ 𝐶𝐻3 𝐶𝑂𝑂𝐻 + 𝐶𝐻3 𝑂𝐻
Rate = [𝐸𝑠𝑡𝑒𝑟][𝐻2 𝑂][𝐻 +]
Practically, the order of reaction is one, but theoretically three hence a pseudo first order
of reaction.
Condition for pseudo-order reaction to occur
- When one of the reactant is in excess. The concentration of the excess reactant
remains practically unchanged thus the rate of reaction will not be affected by such
excess concentration but it will depend on the concentration of the reactant which
tend to remain minimum.
- When one of the reactant is a catalyst. A catalyst is continuously regenerated and
hence its concentration remains unchanged.

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Summary:

Determining order of reaction using the initial rate method T

he initial rate method can be used to find the order of reaction with respect to each reactant hence
the overall order. The rate equation / rate law can therefore be deduced hence the rate constant
calculated.
The orders with respect to all the reactants are then determined either by deduction or calculation.

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Worked examples
1. For the reaction 𝐴(𝑔) + 𝐵(𝑔) → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠. The following data was obtained.
Experiment Concentration of A Concentration of B Initial rate
−3
(𝑚𝑜𝑙𝑑𝑚 ) (𝑚𝑜𝑙𝑑𝑚−3 ) (𝑚𝑜𝑙𝑑𝑚−3 𝑠 −1 )
1 2.50 × 10−1 4.00 × 10−1 1.50 × 10−3
2 2.50 × 10−1 8.00 × 10−1 3.00 × 10−3
3 5.00 × 10−1 4.00 × 10−1 6.00 × 10−3
a) Determine the order of reaction with respect to:
i) A
ii) B
b) Write the rate equation
c) Calculate the rate constant and state its units.
Solution

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2. a) Explain what is meant by the term order of a reaction.

b) The following kinetics data was obtained for the reaction between an alkylhalide, 𝑄
and sodium hydroxide solution
Experiment Concentration of A Concentration of B Initial rate
(𝑚𝑜𝑙𝑑𝑚−3 ) (𝑚𝑜𝑙𝑑𝑚−3 ) (𝑚𝑜𝑙𝑑𝑚−3 𝑠 −1 )
1 0.100 0.500 2.0 × 10−3
2 0.050 0.250 1.0 × 10−3
3 0.100 0.250 2.0 × 10−3
4 0.075 0.250 1.5 × 10−3
Determine the order of the reaction with respect to 𝑄 and sodium hydroxide and give
a reason for your answer.
i) order with respect to 𝑄 and reason
ii) order with respect to sodium hydroxide and reason
c) i) Write the rate equation for the reaction
iii) Calculate the rate constant and give its units
iv) Write the general structure of 𝑄 and give a reason for your answer.

Solution

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3. a) What is meant by the term rate constant

b) The following results were obtained for two compounds 𝐴 and 𝐵 reacting to give
product 𝐶.
𝐴+𝐵 →𝐶
Experiments Concentration of A Concentration of B Initial rate
(𝑚𝑜𝑙𝑑𝑚−3 ) (𝑚𝑜𝑙𝑑𝑚−3 ) (𝑚𝑜𝑙𝑑𝑚−3 𝑠 −1 )
1 0.2 0.24 2.0 × 10−4
2 0.4 0.24 8.0 × 10−4
3 0.6 0.48 2.88 × 10−2
i. Deduce the rate equation
ii. Calculate the rate constant and give its units.

Solution

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4. a) i) Explain what is meant by the order of chemical reaction.

iii) Name two methods used to determine orders of reactions.
b) The results obtained for the kinetics of the decomposition of nitrogen (V) oxide are
given in the table below.
[𝑁2 𝑂5 ](𝑚𝑜𝑙𝑑𝑚−3 ) Initial rate (𝑚𝑜𝑙𝑑𝑚−3 𝑠 −1 )
1.6 × 10−3 0.12
2.4 × 10−3 0.18
−3 𝒙
3.2 × 10
Calculate the:
i) order of reaction
ii) value of 𝑥.
Solution

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5. a) Differentiate between order of reaction and molecularity

b) The rate equation for the reaction between substances 𝐴, 𝐵 and 𝐶 is given by;
𝑅𝑎𝑡𝑒 = 𝑘[𝐴]𝑥 [𝐵]𝑦 [𝐶] 𝑧 , where 𝑥 + 𝑦 + 𝑧 = 4 . The kinetics of the reaction is given
below.
Exp’t [A] (𝑚𝑜𝑙𝑑𝑚−3 ) [B] (𝑚𝑜𝑙𝑑𝑚 −3 ) [C] (𝑚𝑜𝑙𝑑𝑚−3 ) Initial rate
(𝑚𝑜𝑙𝑑𝑚−3 𝑠 −1 )
1 0.20 0.10 0.20 8.0 × 10−5
2 0.05 0.10 0.20 2.0 × 10−5
3 0.10 0.05 0.20 2.0 × 10−5
4 0.10 0.10 0.10 𝒏
i) Deduce the order of reaction with respect to 𝐴, 𝐵 and 𝐶 respectively.
ii) Determine the value of 𝒏.
iii) Calculate the value of the rate constant, and state its units.
iv) Find the initial rate of reaction when [A] = 0.15𝑚𝑜𝑙𝑑𝑚−3 ,
[B] = 0.25𝑚𝑜𝑙𝑑𝑚−3 and [C] = 0.30𝑚𝑜𝑙𝑑𝑚 −3
Solution

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Experimental determination of reaction rates and orders of a reaction

The rate of a chemical reaction can be obtained by following some property of the reaction e.g.
concentration of the reactant or product which changes with extent of this reaction, then
analyzing the reaction mixture at suitable intervals of time which makes it possible to determine
the concentration of both the reactants and products at different time hence obtain a suitable
measure of reaction rate.

Note: To determine rate of reaction with respect to 𝑋, 𝑌 is fixed.

- For zero order reaction, change in concentration has no effect on the rate.
- For first order, change in concentration changes rate the same i.e. if concentration of the
reactants doubles, the rate also doubles.
- For second order, change in concentration changes the rate twice. i.e. if concentration is
doubled, then rate increases 4 times.
- For 3rd order, change in concentration changes the rate 3 times.

TITRIMETRIC ANALYSIS.
Experiment to show that iodination of propanone is zero order with respect to iodine.
Procedure
- A solution of propanone, a solution of iodine in potassium iodide and a solution of sulphuric
acid, all of known concentration are prepared and put into a thermostat bath at the
required temperature.
- A known volume of the propanone solution is pipetted and put into a conical flask followed
by a known volume of the sulphuric acid to catalyze the reaction.
- A known volume of iodine in potassium iodide solution is added to the mixture and a stop
watch is started simultaneously.
- The mixture is shaken and allowed to settle at a constant temperature After a few minutes,
a fixed volume of the reacting mixture is pipetted and put into a container containing
sodium hydrogencarbonate solution to stop the reaction by neutralizing the acid.
- The time at which the reaction stops is recorded. The resultant solution is titrated against
a standard solution of sodium thiosulphate using starch indicator to determine the amount
of iodine remaining.
2𝑆2 𝑂32−(𝑎𝑞) + 𝐼2(𝑎𝑞) ⇌ 𝑆4 𝑂62−(𝑎𝑞) + 2𝐼 −(𝑎𝑞)
- The analysis of the reacting mixture is repeated several times by changing the time taken
for the reaction to take place before adding the reacting mixture to sodium hydrogen
carbonate.
- The original iodine in potassium iodide solution is also pipetted with a standard solution of
sodium thiosulphate to obtain volume of sodium thiosulphate that reacts with it. The
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volume of sodium thiosulphate required is plotted against the time elapsed since the start
of the reaction.

- The graph obtained is a straight line with a negative gradient. This shows that the reaction
is zero order with respect to iodine.
𝐶𝐻3 𝐶𝑂𝐶𝐻3(𝑎𝑞) + 𝐼2(𝑎𝑞) ⇌ 𝐶𝐻3 𝐶𝑂𝐶𝐻2 𝐼2(𝑎𝑞) + 𝐻𝐼(𝑎𝑞)

Experiment to show that decomposition of hydrogen peroxide is first order.

Procedure:
- A known volume of hydrogen peroxide is placed in a beaker followed by a known volume
of sodium hydroxide solution.
- A known volume of a catalyst of iron (III) chloride is added and a stop clock is
simultaneously started.
- Samples of the reaction are pipetted and added to a conical flask containing excess dilute
sulphuric acid (to prevent further decomposition of hydrogen peroxide) at given time
intervals.
- These samples are then titrated with a standard solution of potassium permanganate from
the burette.
- The concentration of hydrogen peroxide at a given time interval is calculated by backward
calculation and a graph of concentration against time is plotted or a graph of rate against
time is plotted.

Conclusion

- The reaction is found to be first order with respect to hydrogen peroxide.

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For esterification;
Procedure
- A given amount of the ester is added to a given amount of water in a beaker. The mixture
is heated to a given temperature i.e. above 40°C.
- A few drops of concentrated sulphuric acid are added and the stop clock is started
simultaneously.
- The beaker is shaken for some minutes and at measure time intervals, known volumes of
the reaction mixtures are pipetted into conical flask and cooled in ice cold water to prevent
the reaction.
- Then amount of carboxylic acid formed is then titrated with a standard solution of sodium
hydroxide using phenolphthalein indicator.
- A graph of concentration of sodium hydroxide is then plotted against time.

Graphical determination of orders of reaction

If a graph of concentration is plotted against time,
i. For first order, the half-life is constant each time the “initial” concentration is reduced
to half its previous value.

ii. For second order’ the half-life doubles each time the “initial” concentration is reduced
to half its previous value.

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Worked Examples
1. The table below shows the kinetic data that were obtained for the hydrolysis of the ester,
methyl ethanoate in acid solution
Time (min) 0 3600 7200 10800 14400
Concentration of ester 0.2405 0.1565 0.1045 0.0685 0.455
−3
(𝑚𝑜𝑙𝑑𝑚 )
a. Plot a graph of concentration of the ester, methyl ethanoate against time.

b. Determine the half-life of the reaction.

Solution

2. For the gas phase (Dimerization) reaction at 3000°𝐶 , 2𝐶2 𝐹4 → 𝐶4 𝐹8 the following
concentration of 𝐶2 𝐹4 were obtained.
t (s) 0 250 750 1750 3750
−3
[𝐶2 𝐹4 ] (𝑚𝑜𝑙𝑑𝑚 ) 0.0500 0.0250 0.0125 0.00625 0.00312

Plot a graph of concentration against time, determine the half-life and order of reaction
giving a reason for your answer.

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Solution

From the graph, the half-life doubles each time the “initial” concentration is reduced to
half its previous value. This behavior indicates a second-order reaction.

3. The experimental data for decomposition of N2 O5 ; [N2 O5 ⟶4𝑁O2 + O2 ] in gas phase at

318𝐾 are given below;
t (s) 0 400 800 1200 1600 2000 2400 2800 3200
[𝑁2 𝑂5 ]𝑚𝑜𝑙𝑑𝑚 −3
163 136 114 93 78 64 53 43 35
a. Plot [N2 O5 ]against t.

b. Find the half-life period for the reaction.

Solution

From the graph, Time corresponding

163
to the concentration, =
2
81.5 𝑚𝑜𝑙 𝑑𝑚−3 is the half-life. From
the graph, the half-life is 1450s also
the successive half-lifes are constant
implying that the reaction is first
order.

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FACTORS AFFECTING RATES OF REACTIONS

The way factors affect rates of chemical reactions is explained basing on the theories of chemical
reactions (collision theory and transition state theory).
a) Collision theory
It’s also known as the Arrhenius theory and proposed by Swedish scientist, Svante
Arrhenius.
In a reaction between two gaseous substances, a molecule of one gas must collide with a
molecule of the other gas for a reaction to occur. Collision theory is that if every collision
results into reaction, then rates of reaction are equal to the frequency of collision between
activated molecules.
The total number of collisions (𝑍) between molecules in a gas per unit volume per unit
time can be calculated.
−𝐸
Rate of reaction(𝑐𝑚−3 𝑠 −1 ) = frequency of collision between molecules = 𝑍𝑒 ⁄𝑅𝑇
Where 𝑍 − total number of collisions
𝐸 − activation energy
𝑇 − temperature
The theory therefore states that in order for molecules to react, they must collide in a
favourable orientation and with enough energy to react.

b) Transition state theory

It is also known as the activated complex theory. It considers the nature of molecular
collisions.
For example in a reaction: 𝐴𝐵 + 𝑋 𝐴 + 𝐵𝑋
When 𝐴𝐵 and 𝑋 are far away from each other, they have some energy. When 𝑋
approaches 𝐴𝐵 with sufficient energy, its electrons begin to overlap with those of 𝐴𝐵 so
that some bonding between 𝐵 and 𝑋 begins to form. At the same time the 𝐴𝐵 bond
becomes longer and weaker.

As the 𝐵𝑋 bond becomes stronger and the 𝐴𝐵 bond becomes weaker, an activated
complex or transition state; [𝐴 𝐵 𝑋] is formed. The activated complex then decomposes
into 𝐴 and 𝐵𝑋.

In general; 𝐴𝐵 + 𝑋 [𝐴 𝐵 𝑋] 𝐴 + 𝐵𝑋

This changes in energy is represented on an energy profile/ enthalpy profile/ reaction

profile or potential energy diagram ( a sketch of potential energy against reaction
coordinate/ reaction pathway).

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i) Reaction profile for a nucleophilic substitution bimolecular reaction.

Consider a reaction between 1 −bromopropane and aqueous sodium hydroxide
which is exothermic.
𝐶𝐻3 𝐶𝐻2 𝐶𝐻2 𝐵𝑟 + 𝑁𝑎𝑂𝐻 → 𝐶𝐻3 𝐶𝐻2 𝐶𝐻2 𝑂𝐻 + 𝑁𝑎𝐵𝑟

Mechanism
𝑁𝑎𝑂𝐻 → 𝑁𝑎+ + 𝑂𝐻 −

Activated complex
𝐶𝐻3 𝐶𝐻2 𝐶𝐻2 ⋯ 𝐵𝑟
𝐶𝐻3 𝐶𝐻2 𝐶𝐻2 − 𝐵𝑟 slow step [ ⋮ ] fast 𝐶𝐻3 𝐶𝐻2 𝐶𝐻2 𝑂𝐻 + 𝐵𝑟 −
𝑂𝐻

𝑂𝐻 −
Rate = 𝑘[𝐶𝐻3 𝐶𝐻2 𝐶𝐻2 𝐵𝑟][𝑂𝐻 −]
- The reaction is a first order with respect to 𝑂𝐻 −and 𝐶𝐻3 𝐶𝐻2 𝐶𝐻2 𝐵𝑟
- The molecularity of the reaction is 2 hence bimolecular (nucleophillic substitution
bimolecular reaction, 𝑆𝑁2 ).
- The reaction coordinate is as shown below.

𝐸𝑎 is the activation energy for the reaction (can also be called the energy barrier). 𝐸𝑏 is
activation energy for backward reaction and ∆𝐻 𝜃 is the enthalpy of reaction

ii) Reaction profile for a nucleophilic substitution unimolecular reaction.

Consider a reaction between 2 −bromo−2 −methylpropane and aqueous sodium
hydroxide.
This reaction takes place via a reactive intermediate. The reactive intermediate is
preceded and also followed by a transition state, the reaction is exothermic.

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Mechanism
slow
(𝐶𝐻3 )3 𝐶 − 𝐵𝑟 (𝐶𝐻3 )3 𝐶 + + 𝐵𝑟 −

fast
(𝐶𝐻3 )3 𝐶 + (𝐶𝐻3 )3 𝐶 − 𝑂𝐻

𝐵𝑟 −

In the above mechanism the slowest step involves one reactant therefore the
molecularity of the reaction is one hence (nucleophilic substitution uni-molecular
reaction, 𝑆𝑁1. The energy level diagram for the reaction is shown below

𝐸𝑎1 is the activation energy for the first step.

𝐸𝑎2 is the activation energy for the second step.

Activation energy is the minimum energy which is required to be possessed by reactants

in order to react and form products.
Activated complex is the intermediate unstable compound formed immediately once the
reactants have absorbed the activation energy available.
This unstable compound undergoes bond re-organization by releasing some of the energy
to form the final stable product. The activated complex can also be called a transition
state.
A rate determining step is the slowest step in the reaction mechanism of a chemical
reaction that proceeds through more than one step.

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Using the theories of reaction to explain factors affecting rates of reaction

The following factors affect the rate of a chemical reaction:
- Concentration of the reactants in solution
- Temperature
- Pressure for gaseous reactants
- Particle size/ surface area for solid reactants
- Catalyst

A. The effect of pressure for gaseous reactants on rate of reaction.

Increase in pressure of gaseous reactants increases the rate of chemical reaction
because the gas molecules are pushed closer together, hence collide more frequently
and react more rapidly.

B. The effect of particle size/ surface area of solid reactants.

The larger the surface area (the smaller the particle size), the higher the rate of reaction
because smaller reactant particles have a higher chance of colliding with each other.

C. The effect of a catalyst.

A catalyst is a substance which alters the rate of a chemical reaction but remains
unchanged in quantity at the end of the reaction or at equilibrium.
The presence of a catalyst increases the rate of a chemical reaction. This is because a
catalyst provides an alternative pathway/ mechanism with lower activation energy.

Below is an energy profile for both a catalyzed and an uncatalyzed exothermic reaction.

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Below is an energy profile for both a catalyzed and an uncatalyzed endothermic

reaction.

Types of catalysts
A catalyst can be homogeneous or heterogeneous.
A homogeneous catalyst is one which is in the same phase as the reactants while A
heterogeneous catalyst is one whose phase is different from that of the reactants.

How catalysts work

A homogeneous catalyst works by lowering the activation energy of the rate
determining step by forming an intermediate. It is generally used for reactions in
aqueous solution. The catalyst reacts with one of the reactants to form an intermediate
which the reacts with the other product to form the final product, regenerating the
catalyst.
Consider a reaction between 𝐴 and 𝐵 in presence of catalyst 𝐶 to form product 𝑋.
𝐴 + 𝐵 𝑋
The reaction takes place in via faster steps as shown below
𝐴 + 𝐶 𝐴𝐶
𝐴𝐶 + 𝐵 𝑋+𝐶
Examples of homogeneous catalyst:
 Dilute sulphuric acid as a catalyst in iodination of propanone or in the reaction
between hydrogen peroxide and iodide ions.
 Oxidation of vanadium (III) ions to vanadium (IV) ions by Iron (III) ions using
copper (II) ions as catalyst.
 Esterification of ethanol by ethanoic acid in the presence of sulphuric acid

The heterogeneous catalysts used are commonly transition metals in processes like
hydrogenation (nickel), cracking and polymerization. It’s the surface of the catalysts
that helps in catalysis. They work by using their empty d–orbitals to form weak
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temporary bonds with the reactant molecules. This weakens the internal bonding of
the reactant molecules and also increases the concentration of the reactant molecules
at the surface of the catalyst thereby reducing the activation energy hence increasing
the rate of reaction.
Examples of heterogeneous catalysts:
- Finely divided iron in the Haber process.
- Nickel in hydrogenation of alkenes.
- Combination of hydrogen and iodine vapour using platinum catalyst

D. The effect of concentration of the reactants in solution on rate of reaction.

Increase in concentration of the reactants increases the rate of reaction. This is because as
concentration increases, the reacting molecules/ ions become closer to each other thus
colliding more frequently. The greater the number of collisions, the greater the rate of
reaction.

Worked examples
a) The rate equation for the reaction between substances 𝑋, 𝑌 and 𝑍 is given
Rate = 𝑘[𝑋]2 [𝑌][𝑍]
State how the rate of reaction would change if:
i) Concentration of 𝑋 is doubled while concentrations of 𝑌 and 𝑍 are kept
constant
Solution
Using Rate = 𝑘 [𝑋]2 [𝑌][𝑍]
New rate = 𝑘[2𝑋]2 [𝑌][𝑍] = 4𝑘[𝑋]2 [𝑌][𝑍]
The rate of reaction quadruples (rate of reaction increases four times)
ii) Concentrations of all the reactants are halved
Solution
From Rate = 𝑘 [𝑋]2 [𝑌][𝑍]
1 2 1 1
New rate = 𝑘 [2 𝑋] [2 𝑌] [2 𝑍]
1 1 1
= 4 × 2 × 2 𝑘 [𝑋]2 [𝑌][𝑍]
1
= 16 𝑘 [𝑋]2 [𝑌][𝑍]
The rate of reaction decreases by 16 times
iii) Concentration of 𝑋 is kept constant while both concentrations of 𝑌 and 𝑍
are trebled.
Solution
Using Rate = 𝑘 [𝑋]2 [𝑌][𝑍]
New rate = 𝑘[𝑋]2 [3𝑌][3𝑍] = 9𝑘[𝑋]2 [𝑌][𝑍]
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The rate of reaction increases by nine times

iv) Concentration of 𝑋 is halved, that of 𝑌 doubled and that of 𝑍 kept
constant.
Solution
From Rate = 𝑘 [𝑋]2 [𝑌][𝑍]
1 2
New rate = 𝑘 [2 𝑋] [2𝑌][𝑍]
1
= 4 × 2𝑘 [𝑋]2 [𝑌][𝑍]
1
= 2 𝑘 [𝑋]2 [𝑌][𝑍]
The rate of reaction is halved (decreases by 2 times)

v) the temperature is increased to double the rate constant.

New rate = 2𝑘 [𝑋]2 [𝑌][𝑍]
The rate of reaction doubles
b) The rate equation for 2𝐴 + 𝐵 → 𝐶 is given by Rate= 𝑘[𝐴] [𝐵]2 . How will the
rate of reaction alter if concentration of;
i. 𝐴 is doubled but concentration of 𝐵 is kept constant
ii. Concentration of 𝐵 is halved but concentration of 𝐴 is kept constant
iii. Both Concentration of 𝐴 and concentration of 𝐵 are doubled
Solution:
i) 𝑅𝑎𝑡𝑒 = 𝑘[𝐴] [𝐵]2
= 𝑘[2𝐴] [𝐵]2
= 2𝑘[𝐴] [𝐵]2 New rate is twice the initial rate
𝐵 2
ii) 𝑅𝑎𝑡𝑒 = 𝑘[𝐴] [ 2 ]
1
= 4 𝑘[𝐴] [𝐵]2 New rate is ¼ times the initial rate
iii) 𝑅𝑎𝑡𝑒 = 𝑘[𝐴] [𝐵]2
= 𝑘[2𝐴] [2𝐵]2
= 𝑘[2𝐴] [𝐵]2 New rate is 8 times the initial rate

E. The effect of temperature on rate of reaction.

The higher the temperature, the higher the kinetic energy of the ions or molecules of
reactants. The ions or molecules hence move at a higher speed and collide more frequently
and with more energy. The higher the collision frequency, the higher the rate of reaction.
The rate constant is also only constant at a fixed temperature. Therefore, increase in
temperature increases the value of the rate constant. The rate of reaction doubles for
every 10℃ or 10K increase in temperature.
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Dufile Seed SS S6 Chemistry term I, 2024

The Arrhenius theory suggests that a molecule would only react if it had higher than the
average energy (if it is activated, has gained the activation energy) Arrhenius found the
experimental relationship between the rate constant and temperature.
The Arrhenius equation is therefore given as;

The Arrhenius Equation

𝐸𝑎
𝑘 = 𝐴𝑒 −𝑅𝑇 ………………………….(a)
Where A is the frequency factor/ Arrhenius constant, 𝑅 is the molar gas constant, 𝐸𝑎 is the
activation energy, 𝑅 is molar gas constant and 𝑇 is absolute temperature (K)
From equation (a)
𝐸𝑎
ln 𝑘 = ln (𝐴𝑒 −𝑅𝑇 )
𝐸𝑎
ln 𝑘 = ln 𝐴 + 𝐼𝑛(𝑒 −𝑅𝑇 )
𝐸𝑎 1
ln 𝑘 = ln 𝐴 − …………………….(*)
𝑅 𝑇

Also
𝐸𝑎
2.303 log10 𝑘 = 2.303 log10 𝐴 − 𝑅𝑇
𝐸𝑎 1
log10 𝑘 = log10 𝐴 − 2.303𝑅 (𝑇) ………………………….(**)

1
Using equation (*), a graph of 𝐼𝑛𝑘 against can be plotted.
𝑇
1
A graph of 𝐼𝑛 𝑘 verses 𝑇

𝐼𝑛 𝐴

𝐸𝑎
In 𝑘 Slope−
𝑅

1
0 (𝐾 −1 )
𝑇
1
Using equation (**), a graph of 𝑙𝑜𝑔10 𝑘 against 𝑇 can be plotted.

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Dufile Seed SS S6 Chemistry term I, 2024

1
A graph of 𝑙𝑜𝑔10 𝑘 verses 𝑇

log10 𝐴

𝐸𝑎
log10 𝑘 Slope− 2.303𝑅

1
0 (𝐾 −1 )
𝑇

Note
From the Arrhenius equation:
- The rate constant of a reaction increases with increase in the temperature of the reaction.
- The rate of reaction increases with increase in temperature
- At the same temperature, the higher the activation energy, the slower the reaction
- The higher the activation energy, the lower the rate constant hence the lower the rate of
reaction.
From the graph, the activation energy 𝐸𝑎 for the reaction can be obtained. Let 𝑘1 and 𝑘2 bethe
rate constant of a reaction at temperatures 𝑇1 and 𝑇2 for which 𝑘1 > 𝑘2 and 𝑇1 > 𝑇2 , then;-
𝐸𝑎 1
log10 𝑘1 = − 2.303𝑅 (𝑇 ) + log10 𝐴 ……………………………. (b)
1

𝐸𝑎 1
log10 𝑘2 = − 2.303𝑅 (𝑇 ) + log10 𝐴 ……………………………. (c)
2

Equation (c) – (b), we get;-

𝐸𝑎 1 1
log10 𝑘2 − log10 𝑘1 = − 2.303𝑅 (𝑇 − 𝑇 )
2 1

𝑘 𝐸𝑎 1 1
log10 (𝑘2 ) − log10 𝑘1 = 2.303𝑅 (𝑇 − 𝑇 )
1 1 2

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Dufile Seed SS S6 Chemistry term I, 2024

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Dufile Seed SS S6 Chemistry term I, 2024

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Prepared by Azudi George E-mail: georgeazudi@gmail.com
Dufile Seed SS S6 Chemistry term I, 2024

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Prepared by Azudi George E-mail: georgeazudi@gmail.com
Dufile Seed SS S6 Chemistry term I, 2024

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Dufile Seed SS S6 Chemistry term I, 2024

END

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