For the reaction,3 I_{(aq)}^{-} + S_2 O_{8_{( | vedclass

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Question

(C) The given reaction is $3 I_{(aq)}^{-} + S_2 O_{8_{(aq)}}^{2-} \longrightarrow 2 SO_{4_{(aq)}}^{2-} + I_{3_{(aq)}}^{-}$.The rate of reaction is exp

Vedclass · Accepted Answer

$0.066 \ mol \ dm^{-3} \ s^{-1}$

Vedclass · Answer

(C) The given reaction is $3 I_{(aq)}^{-} + S_2 O_{8_{(aq)}}^{2-} \longrightarrow 2 SO_{4_{(aq)}}^{2-} + I_{3_{(aq)}}^{-}$.The rate of reaction is expressed as: $	ext{Rate} = -\frac{1}{3} \frac{d[I^{-}]}{dt} = -\frac{d[S_2 O_8^{2-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt} = \frac{d[I_3^{-}]}{dt}$.From the expression,the rate of consumption of $I^{-}$ is given by $-\frac{d[I^{-}]}{dt} = \frac{3}{2} 	imes \frac{d[SO_4^{2-}]}{dt}$.Given $\frac{d[SO_4^{2-}]}{dt} = 0.044 \ mol \ dm^{-3} \ s^{-1}$,we have:Rate of consumption of $I^{-} = \frac{3}{2} 	imes 0.044 = 0.066 \ mol \ dm^{-3} \ s^{-1}$.

For the reaction,$3 I_{(aq)}^{-} + S_2 O_{8_{(aq)}}^{2-} \longrightarrow 2 SO_{4_{(aq)}}^{2-} + I_{3_{(aq)}}^{-}$,the rate of formation of $SO_{4_{(aq)}}^{2-}$ is $0.044 \ mol \ dm^{-3} \ s^{-1}$. Calculate the rate of consumption of $I_{(aq)}^{-}$.

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