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?
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
$\mathrm{R}^{\prime}=k(2 a)^{2}$
= 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. $[\mathrm{A}]=\frac{1}{2} a$, 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 $\frac{1^{\text {th }}}{4}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.