JEE Main 2013 — Chemical Kinetics Question with Solution

Given $-\frac{d{\mathrm{MnO}}_4^{-}}{dt}=4.56\times 10^{-3}{\mathrm{Ms}}^{-1}$ From the reaction given, $\begin{aligned} & -\frac{1}{2} \frac{\mathrm{dMnO}_4^{-}}{\mathrm{dt}}=\frac{4.56 \times 10^{-3}}{2} \mathrm{Ms}^{-1} \\ & -\frac{1}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{1}{5} \frac{d \mathrm{I}_2}{d t} \\ & \therefore-\frac{5}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{d \mathrm{I}_2}{d t} \end{aligned}$ On substituting the given value $\therefore \frac{d{\mathrm{I}}_2}{dt}=\frac{4.56\times 10^{-3}\times 5}{2}=1.14\times 10^{-2}\mathrm{M}/\mathrm{s}$