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The instantaneous rate of disappearance of $\mathrm{MnO}_4^{-}$ion in the following reaction is $4.56 \times 10^{-3} \mathrm{Ms}^{-1}$
$2 \mathrm{MnO}_4^{-}+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{I}_2+8 \mathrm{H}_2 \mathrm{O}$
The rate of appearance $\mathrm{I}_2$ is :
Options:
$2 \mathrm{MnO}_4^{-}+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{I}_2+8 \mathrm{H}_2 \mathrm{O}$
The rate of appearance $\mathrm{I}_2$ is :
Solution:
2668 Upvotes
Verified Answer
The correct answer is:
$1.14 \times 10^{-2} \mathrm{Ms}^{-1}$
$1.14 \times 10^{-2} \mathrm{Ms}^{-1}$
Given $-\frac{d \mathrm{MnO}_4^{-}}{d t}=4.56 \times 10^{-3} \mathrm{Ms}^{-1}$
From the reaction given,
$\begin{aligned}
& -\frac{1}{2} \frac{\mathrm{dMnO}_4^{-}}{\mathrm{dt}}=\frac{4.56 \times 10^{-3}}{2} \mathrm{Ms}^{-1} \\
& -\frac{1}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{1}{5} \frac{d \mathrm{I}_2}{d t} \\
& \therefore-\frac{5}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{d \mathrm{I}_2}{d t}
\end{aligned}$
On substituting the given value
$\therefore \frac{d \mathrm{I}_2}{d t}=\frac{4.56 \times 10^{-3} \times 5}{2}=1.14 \times 10^{-2} \mathrm{M} / \mathrm{s}$
From the reaction given,
$\begin{aligned}
& -\frac{1}{2} \frac{\mathrm{dMnO}_4^{-}}{\mathrm{dt}}=\frac{4.56 \times 10^{-3}}{2} \mathrm{Ms}^{-1} \\
& -\frac{1}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{1}{5} \frac{d \mathrm{I}_2}{d t} \\
& \therefore-\frac{5}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{d \mathrm{I}_2}{d t}
\end{aligned}$
On substituting the given value
$\therefore \frac{d \mathrm{I}_2}{d t}=\frac{4.56 \times 10^{-3} \times 5}{2}=1.14 \times 10^{-2} \mathrm{M} / \mathrm{s}$
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