Given −dtdMnO4−=4.56×10−3Ms−1 From the reaction given, $\begin{aligned}
& -\frac{1}{2} \frac{\mathrm{dMnO}_4^{-}}{\mathrm{dt}}=\frac{4.56 \times 10^{-3}}{2} \mathrm{Ms}^{-1} \
& -\frac{1}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{1}{5} \frac{d \mathrm{I}_2}{d t} \
& \therefore-\frac{5}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{d \mathrm{I}_2}{d t}
\end{aligned}Onsubstitutingthegivenvalue\therefore \frac{d \mathrm{I}_2}{d t}=\frac{4.56 \times 10^{-3} \times 5}{2}=1.14 \times 10^{-2} \mathrm{M} / \mathrm{s}$