1. Chemical kinetics deals with the rates of chemical reactions and factors that affect reaction rates. It examines how fast reactions occur and the mechanisms of reactions. 2. Reaction rates can vary significantly, from fractions of seconds to years. Third-order reactions involve three molecules and are also called termolecular reactions. 3. The rate of a reaction is affected by temperature - higher temperatures provide more energy and increase reaction rates by producing more effective collisions between reactant molecules.

1. Chemical Kinetics. Prof. Jadhav Swapnil S.
Introduction:
Chemical kinetics deals with the rates of chemical reactions i.e., how fast a
reaction occurs? Such studies help to understand the mechanism through which
the reactants are converted to product.
It is observed that some reactions occur within a fraction of second, whereas
some reactions take years together for completion.
Consider the following chemical changes, which occur at different speeds.
i) Rusting of iron:- It is a very slow reaction.
It may take days to months or years together to undergo complete change.
ii) Digestion of food:- It is a reaction with medium speed.
Usually the food is digested in 3 to 4 hours time.
iii)The formation of a white precipitate of silver chloride, AgCl from a
solution containing chloride ions, Cl-
by adding aqueous solution of AgNO3
occurs in a fraction of second. It is a very fast reaction.
The study of chemical kinetics deals with the qualitative and quantitative
study of: a) The rates of reaction.
b) The factors affecting rate of reaction.
c) The mechanisms of reactions.
It also explains why some of the thermodynamically feasible reactions
occur slowly; or do not occur unless initiated by applying suitable conditions.
For example, burning of wood is a spontaneous or feasible process
according to thermodynamic laws. But wood cannot burn itself. It starts
burning only after igniting it.
* Third Order Reactions:-
The reaction in which three molecules are take part in reaction is called
Third Order Reaction.
Hence, Third Order Reactions are also called termolecular reactions.
Eg.
1) 3A ⟶ product . 2) 2A + B ⟶ product. 3) A + B + C ⟶ product.
Derivation of rate constant :-
Let us consider a simple third order reaction having equal concentration
of all reactants.
3A ⟶ product.
Let, A = initial concentration of reactant ‘A’.
x = concentration of reactant ‘A’ at time ‘t’.
(a–x) = concentration of reactant ‘A’ after time ‘t’.
The rate law is written as
= K. (a–x)3
……… (1) K = 3rd
order constant.
Rearranging eq.(1) we get,

2. ( – )³
= K. dt ……… (2)
Integrating eq.(2)
∫ ( – )³
= K ∫
( – )²
= Kt + C ……… (3) C = Integration constant
At t = 0 , x= 0 & we get, C =
²
Put value of C in eq.(3)
( – )²
= Kt +
²
Or Kt =
( – )²
–
²
Ie. K = [
( – )²
–
²
]
Thus,
K = (
( – )
²( – )²
) ……… (4)
Eq(4) represents the expression for rate constant of third order reaction.
* Characteristics of 3rd
order reaction:-
1) The velocity constant (k) depends on the unit of concentration terms.
2) Half–Life of Reaction:
(Que:- Show that for 3rd
order reaction, the required to complete any definite
fraction of the reaction is inversely proportional to the square of the initial
concentration of reactant.)
The time (t1/2) for completion of half the reaction can be calculated as,
We have,
K = (
( – )
²( – )²
)  t = (
( – )
²( – )²
) ……… (A)
When
t = t1/2 , x = a/2 eq.(A) becomes,
t1/2 = (
( – )
² ( – )²
)  t1/2 = (
( )
² ( )²
)
t1/2 = (
(
²
)
² ( )
)  t1/2 = ²
ie. t1/2 α ²
Thus, Half–Life time of Reaction is inversely proportional to the square
of the initial concentration of reactant.
Note: “Time required for the concentration of a reactant to decrease to half
its initial value is called half life of reaction.”

3. 3) Unit of k:-
We have, K = (
( – )
²( – )²
)
K = ×
( ) × ( )
( )² × ( )²
K = ×
( )²
ie K =
( )²
×
K =( )⁻² ( )⁻¹
K = (mol/litre)-2
. (min)-1
ie. K = mol-2
.litre2
.min-1
For nth
order ( )¹⁻ⁿ ( )⁻¹
In C.G.S. unit k expressed in mol-2
.litre2
.min-1
In SI. unit k expressed in mol-2
.(dm3
)2
.s-1
or mol-2
.(dm6
).s-1
* Examples of third order reaction:-
1) Reaction of nitric oxide with oxygen or hydrogen or chlorine / bromine.
a) 2NO (g) + O2 (g) ⟶ 2NO2(g)
b) 2NO (g) + Cl2 (g) ⟶ 2NOCl(g)
c) 2NO (g) + H2 (g) ⟶ N2O (g)+ H2O (g)
2) Oxidation of ferrous sulphate in water.
3) 2FeCl3 (aq) + SnCl2 (aq) ⟶ 2FeCl2 + SnCl4
4) The Reaction between benzoyl chloride and alcohol in ether solution.
5) The Reaction between iodite and ferric ions in aq. Solution.
Fe+3
(aq) + 2I–
(aq) ⟶ product
6) The decomposition of hypobromous acid in the pH
range 6.4 to 7.8
* Method to determine the order of reaction:
“The number of molecules or atoms whose concentration changes during
the reaction is called Order of reaction.”
Or “The number of molecules or atoms whose concentration determines the
rate of reaction is called Order of reaction.”
(1) Van’t Hoff’s differential method (1884):-
This method involves determination of rates (dc/dt) by measuring slopes
of concentration (c) vs. time (t) curves
For nth
order reaction, we have, Rate =
–
= K.cn
…. (1). C = concentration.

4. For concentrations, c1 & c2 we have,
– ₁
= k. cn
1 ……. (2)
– ₂
= k. cn
2 …... (3)
Taking ratio of the eqns
(2) & (3) …... (4)
Taking logarithm of the
Eq. (4) …... (5)
Thus order of reaction (n)
is calculated as
(2) Integrated rate expression method:-
(1) In this method, the values of ‘x’ at various time interval ‘t’ determined
experimentally. (x = amount of reactant decomposed).
(2) These values substituted in rate constant equations of first, second and
third order reactions.
(3) (i) For 1st
order reaction, k =
.
log10
( – )
(ii) For 2nd
order reaction,
k =
( – )
(For equal conn
) k =
.
( – )
log10
( – )
( – )
(For unequal conn
)
(iii) For 3rd
order reaction k =
( – )
² ( – )²
(For equal conn
)
(4) The order of reaction (n) is determined by the equation which gives
satisfactory constant value of velocity constant (K).
(5) This is method of trial and error but it is extensively used.
(3) Half-life method (Fractional change method) or (method of
equifractional parts).
In this method time (t) taken to complete a definite fraction of the reaction
is calculated.
If t = time required for completion of definite fraction of reaction.
a = initial concentrations of a reactant.

5. Then,
a) For 1st
order reaction, t α
°
ie t is independent on initial concentration.
b) For 2nd
order reaction, t α
c) For 3rd
order reaction, t α
²
d) In general for nth
order reaction, t α
ⁿ⁻¹
If, t1 & t2 times required for completion of same fraction of reaction with
different initial concn
a1 & a2 respectively.
And ‘n’ is order of reaction.
Then,
t1 α
₁ⁿ⁻¹
and t2 α
₂ⁿ⁻¹
[
₁
₂
] = [
₂
₁
]n–1
Taking log
log10 [
₁
₂
] = (n–1) log10[
₂
₁
]
₁₀[
₁
₂
]
₁₀[
₂
₁
]
= n–1 ie n =1 +
₁₀[
₁
₂
]
₁₀[
₂
₁
]
……… (A)
To get values of a1 & a2 plot graph of ‘x’ against ‘t’
And from eq.(A) ‘n’ can be calculated.
Effect of temperature on the rate of reaction:-
Rise in temperature  (1) Initiate the reaction.
(2) Increase the rate of reaction.
(A) Temperature coefficient:-
It has been found that generally rate of reaction and rate (velocity)
constant increases with increase in temperature.
For homogeneous reaction, rate and rate (velocity) constant of reaction get
approximately doubled or tripled for every 100
C rise in temperature. This is
generally expressed in the form of temperature coefficient.
“The ratio of rate constants of a reaction at two different temperatures which
differ by 100
c is called temperature coefficient.”
Temperature coefficient =
( )
≈ 2 or 3
Where,
Kt = velocity constant at t0
c. Kt+10 = velocity constant at t0
c

6. (B) Arrhenius Equation:-
Arrhenius suggested as simple relationship between the rate constant (k)
and the temperature (T)
k=A.e–Ea/RT
…….... (1)
This is called the Arrhenius equation.
A = constant called frequency factor. K = velocity constant. R = gas constant.
Ea= activation energy. T = Absolute temperature. E = logarithmic base = 2.718
Alternately we can write, =
²
……. (2)
Integrating eqn
(2), we get, Assume Ea as constant.
loge k =
–
+ C or log10 k =
–
.
+ C’ …. (3) C & C’ are constants
Now, log10 k =
–
.
× + C’  Is a eqn
of straight line y = mx + c
Hence, graph of loge K vs 1/T, is a straight line.
Slop = m =
.
 Ea = – 2.303 × R × slop
On integrating eqn
(2) between the limit
K = K1 at T = T1 and K = K2 at T = T2 We get,
log10
₂
₁
=
.
[
₁
–
₂
]
 Ie. log10
₂
₁
=
.
[
₂– ₁
₁. ₂
] From this equation, Ea can be calculated.
(C) Energy of activation (Ea):-
According to concept of activation, reactant does not pass directly to
product.
Arrhenius suggested that, before react, colliding molecule must be
activated by absorbing minimum amount of energy called Activation energy.
As temperature increases the number of such active molecules also increases.
Thus, “the minimum amount of energy required for the collision between
the molecules to be effective is called Energy of activation.”
Energy of activation (Ea) can be calculated by Arrhenius equation as
=
²
Or log10
₂
₁
=
.
[
₂– ₁
₁. ₂
]
Energy of activation (Ea) depends on the nature of reactants.
Slow reactions have high Ea. Fast reactions have low Ea.

7. Collision Theory Of Chemical Kinetics (Or) Kinetic Molecular Theory Of
Rates Of Reactions:- (Max Trautzb and William Lewis)
1) This theory is based on kinetic theory of gases.
2) According to this theory, to occur chemical reactions there should be
collision between reacting molecules.
3) However all the collisions are not effective to produce chemical change.
4) Hence Arrhenius suggested that, before formation of product colliding
molecule must be activated by absorbing minimum amount of energy called
Activation energy (Ea), So that they can pass over energy barrier existing
between reactants and product. (fig A)
5) If reacting molecule colloids with insufficient energy can’t pass over the energy
barrier. However if the reacting molecule colloids with sufficient energy can pass
over the energy barrier and get activated. This activated molecule then reacts to
form product. (Fig B)
6) This Threshold Energy or minimum energy necessary to allow a
reaction to occur is called energy of activation. Thus energy of activation is the
minimum energy required for the collision between the molecules to be effective.
7) The magnitude of Ea depends on the nature of reactants.
8) For slow reaction energy of activation is high while for fast reaction
energy of activation is low.

8. 9) As temperature increases, effective collision increase and hence
activated molecule increases. Thus rate of chemical reaction increases with
increase in temperature.
Transition state theory or Activated complex theory or Theory of absolute
reaction rate:- (Henry Erying 1935)
1) This theory is applicable to gas and liquid phase reaction. It is very
complicated theory.
2) According to this theory, “Before reacting molecules changes into
products, they form transition state or activated complex which is unstable
and decompose to form product.”
3) In this theory it is supposed that, as two reacting approach each other,
their potential energy increases and reaches to maximum. (fig).
4) This lead to the formation of activated complex. This activated complex
is unstable and decomposes to form product or collapse back into reactants. (Fig)
5) Example: Consider the reaction X + YZ  XY + Z
Initially P.E. (E1) of the system is unaffected because X and YZ are far away
from each other. When X approaches to YZ, the P.E. of system increases and
reaches to maximum (fig), which corresponds to activated complex X-Y-Z. This
activated complex is unstable and decomposes to form product XY and P.E. drop
to E2
X + YZ X-Y-Z XY + Z
Reactants Activated complex Product
Energy of activation Ea = P.E. of activated complex - P.E. of reactants.
Note: The minimum amount of energy required by the colliding molecules to
yield the products is called Threshold Energy.