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The instantaneous rate of disappearance of the ${MnO}_{4}^{-}$ ion in the following reaction is $4.56 \times 10^{- 3}$ ${Ms}^{- 1} .$ Then, the rate of appearance of $I_{2}$ is:
$2 {MnO}_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2 {Mn}^{2 +} + 5 I_{2} + 8 H_{2} O$

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By Expert Faculty of Sri Chaitanya

a

1.14 \times 10^{- 3} {Ms}^{- 1}

b

5.7 \times 10^{- 3} {Ms}^{- 1}

c

4.56 \times 10^{- 4} {Ms}^{- 1}

d

1.14 \times 10^{- 2} {Ms}^{- 1}

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detailed solution

Correct option is D

Given reaction is,

$2 {MnO}_{4}^{-} + 10 I^{-} + 16 H^{+} \to 2 {Mn}^{2 +} + 5 I_{2} + 8 H_{2} O$

Disappearance of the ${MnO}_{4}^{-}$ ion,

$\begin{aligned} - \frac{1}{2} \frac{d [{MnO}_{4}^{-}]}{d t} = + \frac{1}{5} \frac{d [I_{2}]}{d t} \end{aligned}$

So, the rate of appearance of $I_{2}$ is:

$\frac{d (I_{2})}{d t} = \frac{5}{2} \times 4.56 \times 10^{- 3}$

$\Rightarrow \frac{d (I_{2})}{d t} = 1.14 \times 10^{- 2} {Ms}^{- 1}$

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