required for a reaction is produced by decomposition of in as by equation

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Solution

Q.
${NO}_{2}$ required for a reaction is produced by decomposition of $N_{2} O_{5}$ in ${CCl}_{4}$ as by equation

$2 N_{2} O_{5 (g)} \to 4 {NO}_{2 (g)} + O_{2 (g)}$

The initial concentration of $N_{2} O_{5}$ is 3 mol $L^{- 1}$ and it is 2.75 mol $L^{- 1}$ after 30 minutes.

The rate of formation of ${NO}_{2}$ is $x \times 10^{- 3}$ mol $L^{- 1}$ $\min^{- 1}$ , Value of $x$ is ________ . (nearest integer) [2024]

Ans.
(17)

   ${Rate of dissociation of N}_{2} O_{5} :$    $- \frac{Δ [N_{2} O_{5}]}{t} = - \frac{(2.75 - 3) {mol L}^{- 1}}{30 min} = \frac{1}{120} \frac{mol}{L \min}$

   $By stoichiometry of the reaction, 2 N_{2} O_{5} (g) \to 4 {NO}_{2} (g) + O_{2} (g)$

   $Rate of reaction = - \frac{1}{2} \frac{Δ [N_{2} O_{5}]}{t} = \frac{1}{4} \frac{Δ [{NO}_{2}]}{t}$

   $\Rightarrow \frac{Δ [{NO}_{2}]}{t} = 2 \times \frac{Δ [N_{2} O_{5}]}{t}$

   $\frac{Δ [{NO}_{2}]}{t} = 2 \times \frac{1}{120} \frac{mol}{L \min} = 16.67 \times 10^{- 3} \frac{mol}{L \min} \approx 17 \times 10^{- 3} \frac{mol}{L \min}$

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