Question

The instantaneous rate of disappearance of $Mn{O}_4$ ion in the following reaction is $4.56\times 10^{-3}M{s}^{-1}$
$2Mn{O}_4^{-}+10I^{-}+16H^{+}\to 2M{n}^{2+}+5I_2+8H_{2}O$
The rate of appearance $I_2$ is :

• A. $4.56\times 10^{-4}M{s}^{-1}$
• B. $1.14\times 10^{-2}M{s}^{-1}$
• C. $1.14\times 10^{-3}M{s}^{-1}$
• D. $5.7\times 10^{-3}M{s}^{-1}$

Answer 1

Answer: (B)

Given $-\frac{dMn{O}_4^{-}}{dt}=4.56\times 10^{-3}M{s}^{-1}$
From the reaction given,
$-\frac{1}{2}\frac{dMn{O}_4^{-}}{dt}=\frac{4.56\times 10^{-3}}{2}M{s}^{-1}$
$-\frac{1}{2}\frac{dMn{O}_4^{-}}{dt}=\frac{1}{5}\frac{d{I}_2}{dt}$
$\therefore -\frac{5}{2}\frac{dMn{O}_4^{-}}{dt}=\frac{d{I}_2}{dt}$
On substituting the given value
$\therefore \frac{d{I}_2}{dt}=\frac{4.56\times 10^{-3}\times 5}{2}=1.14\times 10^{-2}M/s$