Chemistry - Chemical Kinetics Question with Solution | TestHub

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