Consider the reaction 3 I^{-} + S_2 O_8^{2-} | vedclass

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Question

(A) The given reaction is $3 I^{-} + S_2 O_8^{2-} \longrightarrow I_3^{-} + 2 SO_4^{2-}$.According to the rate expression:$Rate = -\frac{1}{3} \frac{d

Vedclass · Accepted Answer

$1.1 	imes 10^{-2} \ mol \ dm^{-3} \ s^{-1}$

Vedclass · Answer

(A) The given reaction is $3 I^{-} + S_2 O_8^{2-} \longrightarrow I_3^{-} + 2 SO_4^{2-}$.According to the rate expression:$Rate = -\frac{1}{3} \frac{d[I^{-}]}{dt} = -\frac{d[S_2 O_8^{2-}]}{dt} = \frac{d[I_3^{-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$.We are given $\frac{d[SO_4^{2-}]}{dt} = 2.2 	imes 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.Equating the terms for $S_2 O_8^{2-}$ and $SO_4^{2-}$:$-\frac{d[S_2 O_8^{2-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$.Therefore,$\frac{d[S_2 O_8^{2-}]}{dt} = -\frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$.Substituting the value:$\frac{d[S_2 O_8^{2-}]}{dt} = -\frac{1}{2} 	imes (2.2 	imes 10^{-2}) = -1.1 	imes 10^{-2} \ mol \ dm^{-3} \ s^{-1}$.

Consider the reaction $3 I^{-} + S_2 O_8^{2-} \longrightarrow I_3^{-} + 2 SO_4^{2-}$. 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 $\frac{d[S_2 O_8^{2-}]}{dt}$?

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