${NO}_{2}$ required for a reaction is produced by the decomposition of $N_{2} O_{5}$ in ${CCl}_{4}$ as per the equation $2 N_{2} O_{5} ⇌ 4 {NO}_{2} + O_{2}$ The initial concentration of $N_{2} O_{5}$ is ${lit}^{- 1}$ after $30$ minutes. The rate of formation of ${NO}_{2}$ is

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$1.667 \times 10^{- 2} {molelit}^{- 1} \min^{- 1}$

$4.167 \times 10^{- 3} {molelit}^{- 1} \min^{- 1}$

$2.083 \times 10^{- 3} {mole lit}^{- 1} \min^{- 1}$

$8.33 \times 10^{- 3} {mole lit}^{- 1} \min^{- 1}$

answer is A.

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Detailed Solution

$2 N_{2} O_{5} ⇌ 4 {NO}_{2} + O_{2}$

$Δ [N_{2} O_{5}] = {[N_{2} O_{5}]}_{final} - {[N_{2} O_{5}]}_{initial}$

$= 2.75 - 3.00$

$= - 0.25 \frac{Δ [N_{2} O_{5}]}{dt} = \frac{- 0.25}{30} =$

$\frac{Δ [{NO}_{2}]}{dt} = - \frac{4}{2} \times \frac{d [N_{2} O_{5}]}{dt}$

$= \frac{4}{2} \times \frac{0.25}{30}$

$= 1.667 \times 10^{- 2} {molelit}^{- 1} \min^{- 1}$

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