For the reaction,3 I^{-} + S_2 O_8^{2-} right | vedclass

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(B) The given reaction is $3 I^{-} + S_2 O_8^{2-} \rightarrow I_3^{-} + 2 SO_4^{2-}$.According to the rate law expression,the rate of reaction is give

Vedclass · Accepted Answer

$3.3 	imes 10^{-2} \ mol \ dm^{-3} \ s^{-1}$

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(B) The given reaction is $3 I^{-} + S_2 O_8^{2-} ightarrow I_3^{-} + 2 SO_4^{2-}$.According to the rate law expression,the rate of reaction is given by: $	ext{Rate} = -\frac{1}{3} \frac{d[I^{-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$.We are given $\frac{d[SO_4^{2-}]}{dt} = 2.2 	imes 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.Rearranging the expression to find $-\frac{d[I^{-}]}{dt}$: $-\frac{d[I^{-}]}{dt} = \frac{3}{2} 	imes \frac{d[SO_4^{2-}]}{dt}$.Substituting the given value: $-\frac{d[I^{-}]}{dt} = \frac{3}{2} 	imes 2.2 	imes 10^{-2} = 3.3 	imes 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.

For the reaction,$3 I^{-} + S_2 O_8^{2-} \rightarrow I_3^{-} + 2 SO_4^{2-}$,at a particular time $t$,$\frac{d[SO_4^{2-}]}{dt}$ is $2.2 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$. What is the value of $-\frac{d[I^{-}]}{dt}$?

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