NO₂required for a reaction is produced by the decomposition of N₂O₅in CCl₄as per the $2 N_{2} O_{5 (g)} ⟶ 4 {NO}_{2 (g)} + O_{2 (g)}$ equation, The initial concentration of N₂O₅is 3.00 mol L^-1and it is 2.75 mol L^-1 after 30 minutes. The rate of formation of NO₂ is

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$2 .083 \times 10^{- 3} mol L^{-} \min^{- 1}$

$1 .667 \times 10^{- 2} mol L^{-} \min^{- 1}$

$8 .33 \times 10^{- 3} mol L^{-} \min^{- 1}$

$4 .167 \times 10^{- 3} mol L^{-} \min^{- 1}$

answer is A.

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Detailed Solution

$\begin{array}{l} 2 N_{2} O_{5 (g)} ⟶ 4 {NO}_{2 (g)} + O_{2 (g)} t = 0 3 .0 M \\ t = 302 .75 M - \frac{d [N_{2} O_{5}]}{dt} = \frac{- (2 .75 - 3)}{30} = \frac{0 .25}{30} M / \min - \frac{1}{2} \frac{d [N_{2} O_{5}]}{dt} = \frac{1}{4} \frac{d [{NO}_{2}]}{dt} \\ - \frac{d [N_{2} O_{5}]}{dt} = \frac{2}{4} \cdot \frac{d [{NO}_{2}]}{dt} \frac{0 .25}{30} = \frac{1}{2} \cdot \frac{d [{NO}_{2}]}{dt} ∴ \frac{d [{NO}_{2}]}{dt} = 2 \times \frac{0 .25}{30} \\ = 1 .66 \times 10^{- 2} M / \min \end{array}$