Question

The instantaneous rate of disappearance of the $Mn{O}_4^{-}$ 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 of $I_2$ is:

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

Answer 1

The correct option is D $1.14\times {10}^{-2}M{s}^{-1}$

$\text{Rate of reaction}=\frac{\text{Rate of disappearance of}Mn{O}_4^{-}}{2}=\frac{\text{Rate of appearance of}{I}_2}{5}$

Rate of appearance of $I_2=\frac{5}{2}$ rate of disappearance of $Mn{O}_4^{-}$

$=\frac{5}{2}\times 4.56\times {10}^{-3}Mse{c}^{-1}=1.14\times {10}^{-2}Mse{c}^{-1}$.

Hence, option D is correct.