(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

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

(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: $\text{Rate} = -\frac{1}{3} \frac{d[I^{-}]}{dt} = -\frac{d[S_2 O_8^{2-}]}{dt} = \frac{1}{2} \frac{d[SO_4^{2-}]}{dt} = \frac{d[I_3^{-}]}{dt}$.From the expression,the rate of consumption of $I^{-}$ is given by $-\frac{d[I^{-}]}{dt} = \frac{3}{2} \times \frac{d[SO_4^{2-}]}{dt}$.Given $\frac{d[SO_4^{2-}]}{dt} = 0.044 \ mol \ dm^{-3} \ s^{-1}$,we have:Rate of consumption of $I^{-} = \frac{3}{2} \times 0.044 = 0.066 \ mol \ dm^{-3} \ s^{-1}$.

Class 12 Chemistry Chemical Kinetics Rate of a reaction

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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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A
$0.022 \ mol \ dm^{-3} \ s^{-1}$
B
$0.044 \ mol \ dm^{-3} \ s^{-1}$
C
$0.066 \ mol \ dm^{-3} \ s^{-1}$
D
$0.088 \ mol \ dm^{-3} \ s^{-1}$

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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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If the rate of disappearance of $N_2O_5$ in the following reaction is $1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$,the rate of production of $NO_2$ in $mol \ L^{-1} \ s^{-1}$ is:$2N_2O_{5(g)} \longrightarrow 4NO_{2(g)} + O_{2(g)}$

Observe the following reaction: $2 A + B \longrightarrow C$. The rate of formation of $C$ is $2.2 \times 10^{-3} \ mol \ L^{-1} \ min^{-1}$. What is the value of $-\frac{d[A]}{d t}$ (in $mol \ L^{-1} \ min^{-1}$)?

For the reaction $3 A + 2 B \rightarrow C$,which statement is correct $:-$

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If the rate of disappearance of $N_2O_5$ in the following reaction is $1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$,the rate of production of $NO_2$ in $mol \ L^{-1} \ s^{-1}$ is:
$2N_2O_{5(g)} \longrightarrow 4NO_{2(g)} + O_{2(g)}$