For the reaction, $3 \mathrm{I}^{-(\mathrm{aq}, \mathrm{~s})}+\mathrm{S}_2 \mathrm{O}_{8(\mathrm{aq})}^{2-}…

For the reaction, $3 \mathrm{I}^{-(\mathrm{aq}, \mathrm{~s})}+\mathrm{S}_2 \mathrm{O}_{8(\mathrm{aq})}^{2-} \longrightarrow 2 \mathrm{SO}_{4(\mathrm{aq},)}^{2-}+\mathrm{I}_{3(\mathrm{aq},)}^{-}$ rate of formation of $\mathrm{SO}_{4(\mathrm{aq} .)}^{2-}$ is $0.044 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$. Calculate rate of consumptiuon of $\mathrm{I}_{(\mathrm{aq} .)}^{-}$.
  1. $0.022 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  2. $0.044 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  3. $0.066 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$
  4. $0.088 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Solution

$3 \mathrm{I}_{\text {(aq) }}^{-}+\mathrm{S}_2 \mathrm{O}_{8(\mathrm{aq})}^{2-} \longrightarrow 2 \mathrm{SO}_{4(\text { aq })}^{2-}+\mathrm{I}_{3(\mathrm{aq})}^{-}$
Rate of reaction $=-\frac{1}{3} \frac{\mathrm{~d}\left[\mathrm{I}^{-}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{~S}_2 \mathrm{O}_8^{2-}\right]}{\mathrm{dt}}=\frac{1 \mathrm{~d}\left[\mathrm{SO}_4^{2-}\right]}{2}=\frac{\mathrm{d}\left[\mathrm{I}_3^{+}\right]}{\mathrm{dt}}$ $\therefore \quad$ Rate of consumption of $\mathrm{I}^{-}=-\frac{1}{3} \frac{\mathrm{~d}\left[\mathrm{I}^{-}\right]}{\mathrm{dt}}=\frac{1 \mathrm{~d}\left[\mathrm{SO}_4^2\right]}{2}$ $=\frac{3 \times 0.044}{2}=0.066 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$

Asked in: MHT CET 2024 (04 May Shift 1)

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