Q14 / 150 · MHT CET 2023 (10 May Shift 1) · Chemistry

For the reaction, $3 \mathrm{I}^{-}+\mathrm{S}_2 \mathrm{O}_8^{2-} \rightarrow \mathrm{I}_3^{-}+2 \mathrm{SO}_4^{2-}$, at a particular time $\mathrm{t}$, $\frac{\mathrm{d}\left[\mathrm{SO}_4^{2-}\right]}{\mathrm{dt}}$ is $2.2 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$. What is the value of $-\frac{\mathrm{d}\left[\mathrm{I}^{-}\right]}{\mathrm{dt}}$ ?

  • A
    $1.1 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  • Correct answer
    B
    $3.3 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  • C
    $4.4 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  • D
    $6.6 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Solution

$$ \begin{aligned} & 3 \mathrm{I}^{-}+\mathrm{S}_2 \mathrm{O}_8^{2-} \rightarrow \mathrm{I}_3^{-}+2 \mathrm{SO}_4^{2-} \\ & \text { Rate of reaction }=-\frac{1}{3} \frac{\mathrm{d}\left[\mathrm{I}^{-}\right]}{\mathrm{dt}}=+\frac{1}{2} \frac{\mathrm{d}\left[\mathrm{SO}_4^{2-}\right]}{\mathrm{dt}} \\ & \begin{aligned} -\frac{\mathrm{d}\left[\mathrm{I}^{-}\right]}{\mathrm{dt}} & =\frac{3}{2} \frac{\mathrm{d}\left[\mathrm{SO}_4^{2-}\right]}{\mathrm{dt}}=\frac{3}{2} \times 2.2 \times 10^{-2} \\ & =3.3 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1} \end{aligned} \end{aligned} $$ [Note: In the question, $\frac{\mathrm{d}\left[\mathrm{I}^{-}\right]}{\mathrm{dt}}$ is changed to $-\frac{\mathrm{d}\left[\mathrm{I}^{-}\right]}{\mathrm{dt}}$ to apply appropriate textual concepts.]

Hence, the correct answer is option: 'B'.

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