**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?

**Solution:**

Letthe concentration of the reactant be [A] = *a*

Rate of reaction, R = *k *[A]2

= *ka*2

(i)If the concentration of the reactant is doubled, i.e. [A] = 2*a*, then the rate of the reaction would be

$\mathrm{R}^{\prime}=k(2 a)^{2}$

= 4*ka*2

= 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. $[\mathrm{A}]=\frac{1}{2} a$, then the rate of the reaction would be

$R^{*}=k\left(\frac{1}{2} a\right)^{2}$

$=\frac{1}{4} k a^{2}$

$=\frac{1}{4} R$

Therefore, the rate of the reaction would be reduced to $\frac{1^{\text {th }}}{4}$.