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Pseudo Order reaction,order reaction
Typology: Lecture notes
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A reaction in which one of the reactants is present in a large excess shows an order different from the
actual order. The experimental order which is not the actual one is referred to as the pseudo order.
Let us consider a reaction
A + B โฏโฏโ products
in which the reactant B is present in a large excess. Since it is an elementary reaction, its rate law can be
written as,
rate = k [A] [B]
As B is present in large excess, its concentration remains practically constant in the course of reaction.
Thus the rate law can be written as,
rate = k โฒ [A]
Where, the new rate constant k โฒ = k [B].
Thus the actual order of the reaction is second-order but in practice it will be first-order.
Therefore, the reaction is said to have a pseudo-first order.
Examples of Pseudo-order Reactions: Hydrolysis of sucrose:
Sucrose upon hydrolysis in the presence of a dilute mineral acid gives glucose and fructose.
12
22
11
2
6
12
6
6
12
6
Sucrose (excess) glucose fructose
If a large excess of water is present, [H 2
O] is practically constant and the rate law may be written
as
Rate = k [C 12
22
11
2
= k
โ
12
22
11
The reaction though of second-order is experimentally found to be first-order. Thus it is a pseudo
first-order reaction.
Let us consider a first order reaction
A โฏโฏโ products
Initial conc. a 0
Final conc. a โ x x
Suppose that at the beginning of the reaction ( t = 0), the concentration of A is a moles litre
after time t , x moles of A have changed, the concentration of A is a โ x.
We know that for a first order reaction, the rate of reaction, dx / dt , is directly proportional to the
concentration of the reactant.
Thus,
๐๐ฅ
(๐โ๐ฅ)
Integration of the expression (1) gives
Where I is the constant of integration.
The constant k may be evaluated by putting t = 0 and x = 0. Thus, I = โ 1n a
Substituting for I in equation (2)
โln
= ๐๐ก โ ln ๐
๐
๐โ๐ฅ
1
๐ก
๐
๐โ๐ฅ
The value of k can be found by substituting the values of a and ( a โ x ) determined experimentally
at time interval t during the course of the reaction.
For a first order reaction,
๐
โ๐๐ก
Where [A] o
is the concentration of reactants at t = 0, and [A] is the concentration of reactant at
timeโtโ
When the reaction reaches the completion, [A] is zero, putting [A] = 0 in the equation (1),
we get,
o
โ๐๐ก
Or ๐
โ๐๐ก
But ๐
โ๐๐ก
= 0 only when t = โ
This means a first order reaction takes infinite time to completion or never ends.
Initial Conc. a 0
Final Conc. a โ x x
Let the initial concentration of A be a moles litre
and after time t , x , moles have reacted. Therefore, the
concentration of A becomes ( a โ x ). The rate law may be written as:
3
3
3
On integration, it gives
1
2 (๐โ๐ฅ)
2
Where I is integration constant. I can be evaluated by putting x = 0 and t = 0. Thus,
1
2 ๐
2
Substituting for I in equation (2)
2
2
2
2
1
๐ก
๐ฅ( 2 ๐โ๐ฅ)
2 ๐
2
.(๐โ๐ฅ)
2
This is the integrated rate equation for a second order reaction.
The units of rate constant for different orders of reactions are different.
For a zero order reaction, the rate constant k is given by the expression,
๐
๐ฅ
๐ก
๐๐๐๐/๐๐๐ก๐๐
๐ก๐๐๐
Thus the units of k are mol L
time
Time may be given in seconds, minutes, days or years.
The rate constant of a first order reaction is given by
๐ก
๐
๐โ๐ฅ
๐ก
[๐ด]
๐
[๐ด]
๐๐๐/๐ฟ
๐ก๐๐๐ร๐๐๐/๐ฟ
Thus the rate constant for the first order reaction is independent of the concentration. It has the unit
time
- 1
The rate constant for a second order reaction is expressed as
Thus the units of k for a second order reactions are mol
- 1
L time
- 1
The rate constant for a third order reaction is
2
2
2
2
2
2
Thus the units of k for third order reaction are mol
- 2
2
time
- 1