A reaction is second order with respect to a reactant. How is the rate of rection affected if the concentration of the reactant is
(i) Doubled
(ii) Reduced to 1/2?

Solution

Let the reaction A → B is a 2nd order reaction w.r.t A and conc. of A is ‘a’ mol/L, then rate of reaction can be written as:

\frac{dx}{dt} = k [A]^{2} = {ka}^{2}

(i) When conc. [A] is doubled
i.e., [A’] = 2a mol/L
Then new rate of reaction

\frac{d' (x)}{dt} = k [2 a]^{2} = 4 {ka}^{2} = 4 \frac{dx}{dt}

Thus rate of reaction will become four times where concentration is doubled.

(ii) Similarly, when conc. of A is reduced to

\frac{1}{2}

i.e., [A] is a/2 then new rate of reaction,

\frac{d' (x)}{dt} = k {[\frac{a}{2}]}^{2} = \frac{1}{4} {ka}^{2} = \frac{1}{4} \frac{dx}{dt}

The rate of reaction will become one-fourth of the initial rate of reaction.

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