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

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Question

(C) The given balanced chemical equation is: $3 I_{(aq)}^{-} + S_2 O_{8_{(aq)}}^{2-} \longrightarrow I_{3_{(aq)}}^{-} + 2 SO_{4_{(aq)}}^{2-}$ Accordin

Vedclass · Accepted Answer

$0.011 \ mol \ dm^{-3} \ sec^{-1}$

Vedclass · Answer

(C) The given balanced chemical equation is: $3 I_{(aq)}^{-} + S_2 O_{8_{(aq)}}^{2-} \longrightarrow I_{3_{(aq)}}^{-} + 2 SO_{4_{(aq)}}^{2-}$ According to the rate law expression for the reaction,the rate of reaction is given by: $Rate = -\frac{1}{3} \frac{d[I^-]}{dt} = -\frac{d[S_2 O_8^{2-}]}{dt} = \frac{d[I_3^-]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$ Given that the rate of formation of $SO_4^{2-}$ is $\frac{d[SO_4^{2-}]}{dt} = 0.022 \ mol \ dm^{-3} \ sec^{-1}$. Equating the rates: $\frac{d[I_3^-]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt}$ $\frac{d[I_3^-]}{dt} = \frac{1}{2} 	imes 0.022 \ mol \ dm^{-3} \ sec^{-1} = 0.011 \ mol \ dm^{-3} \ sec^{-1}$

$t/s$	$0$	$50$	$100$	$150$	$200$	$300$	$400$	$700$	$800$
$[C_{4}H_{9}Cl]/mol\ L^{-1}$	$0.100$	$0.0905$	$0.0820$	$0.0741$	$0.0671$	$0.0549$	$0.0439$	$0.0210$	$0.017$

For the reaction,$3 I_{(aq)}^{-} + S_2 O_{8_{(aq)}}^{2-} \longrightarrow I_{3_{(aq)}}^{-} + 2 SO_{4_{(aq)}}^{2-}$,the rate of formation of $SO_4^{2-}$ is $0.022 \ mol \ dm^{-3} \ sec^{-1}$. What is the rate of formation of $I_{3_{(aq)}}^{-}$?

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