For the reaction $\mathrm{Cl}_{2}+2 \mathrm{I}^{-} \rightarrow \mathrm{I}_{2}+2 \mathrm{Cl}^{-}$, th

Question

For the reaction $\mathrm{Cl}_{2}+2 \mathrm{I}^{-} ightarrow \mathrm{I}_{2}+2 \mathrm{Cl}^{-}$, the initial concentration of $\mathrm{I}^{-}$was 0.2 M and the concentration after 20 min was 0.18 M . The rate of formation of $\mathrm{I}_{2}$ in $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}$ would be\

Professor PC Thomas Classes · Answer

Rate of disappearance of iodide $=\frac{0.2-0.18}{20}=1 	imes 10^{-3} \mathrm{molL}^{-1} \mathrm{~min}^{-1}$ But rate of formation of iodine is $-\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{I}^{-}ight]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{I}_{2}ight]}{\mathrm{dt}}$\ $	herefore \frac{\mathrm{d}\left[\mathrm{I}_{2}ight]}{\mathrm{dt}}=\frac{1 	imes 10^{-3}}{2}=5 	imes 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$\