Pseudo Reaction - Definition, Example, Formula and Pseudo First Order Reaction

The sum of the overall concentration of reactants is called the order of the reaction. Depending upon the concentration of reactants, the order of the reaction can vary. For example, consider a reaction below:

A and B are reactants, and C and D are products. The concentration of A is given by x, that of B by y, C by c and D by d. then, the order of the above reaction will be Order of reaction = x + y

This is in accordance to rate law.

Pseudo reaction

A first-order reaction which is actually a second-order reaction but is considered first-order due to the concentration of either of its reactants is known as a pseudo-first-order reaction. It usually occurs when one of the reactants has a much higher concentration than the other, but the overall change in its concentration is so less that it is considered constant. For example, consider the above reaction. If the concentration of A and B is one each, i.e. concentration of A (x) is equal to 1 and the concentration of B (y) is also 1, then the overall concentration would become 2. This means that the order of reaction would be two, which results in making it a second-order reaction. But if the concentration of B (y) is much higher than x and the change in the concentration is very small, it is treated as constant. Thus, the reaction, which was of second-order but due to assumptions in concentration, becomes a pseudo reaction of the first order.
Generally, the first-order pseudo reaction is a higher-order reaction that is approximated to appear as a first-order reaction. Therefore, a reaction of nth order which can be approximated to look like a reaction of some other order is called a pseudo reaction.

Half-life in a pseudo 1st order reaction

The time duration taken by the concentration of the reactants to reduce to half of its original value is known as the half-life of the pseudo reaction. For example, consider a reaction.

The concentrations of the reactants A and B are shown in [] brackets. So the concentration of A will be [A], the initial concentration of A is [a], and the concentration which is much more than that of A is shown as [B]. Let ‘t1/2’ be the half-life time, then:

[A] = [a] e^-[B]kt 

[A]/ [a] = e^-k’t 

On taking natural log on both sides of the first order pseudo equation, we get

L_n ([A]/ [a]) = - k’t

Since it is for half life, the concentration of A becomes ½ [a]. Further solving the equation, we get

L_n (1/2 [a]/ [a]) = l_n (1/2)

L_n (1/2 [a]/ [a]) = - k’t_1/2

From the previous equations we know that

K^’ = k. [B]

Also we know

[B] = [b] 

where [b] is the initial concentration of B.

And                                                                           - l_n (1/2) = l_n /2

L_n /2 = k [b] t_1/2

t_1/2 = L_n /2 k [b]