Question

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled
(ii) reduced to half?

Answer 1

Letthe concentration of the reactant be [A] = a
Rate of reaction, R = k[A]²
= ka²
(i) If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be
R = k(2a)²
= 4ka²
= 4 R
Therefore, the rate of the reaction would increase by 4 times

(ii) If the concentration of the reactant is reduced to half, i.e.
\([A] = \frac{1}{2} a\)
then the rate of the reaction would be
\(R = k(\frac{1}{2}a)^2\)

=\(\frac{1}{4}ka^2\)
\(= \frac{1}{4} R\)

Therefore, the rate of the reaction would be reduced to \( \frac{1}{4}^{ th}\)