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For given reaction $N_{2} O_{4} ⇌ 2 {NO}_{2}$ reaction starts with both $N_{2} O_{4}$ and ${NO}_{2}$ it was found that in order to attain equilibrium reaction proceed in forward direction. Molecular wt. of mixture at $t = 0$ and at equilibrium are 69 and $50 g m / m o l e$ respectively. Calculate percentage dissociation of $N_{2} O_{4}$ .

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$50 %$

$56 %$

$73 %$

$76 %$

answer is D.

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Detailed Solution

$N_{2} O_{4} ⇌ 2 {NO}_{2}$

At t = 0 x y

At Eq. $x (1 - α)$ $y + 2 x α$

$69 = \frac{22 \times x + 46 y}{x + y}$

$\Rightarrow 69 x + 69 y = 92 x + 46 y$

$\Rightarrow 23 y = 23 x$

$\Rightarrow x = y$

At Eq.

$50 = \frac{x (1 - α) \times 92 + (y + 2 x α) \times 46}{x (1 - α) + y + 2 x α}$

$50 = \frac{(1 - α) \times 92 + (1 + 2 α) \times 46}{1 - α + 1 + 2 α}$

$50 = \frac{92 + 46}{2 + α} = \frac{138}{2 + α}$

$\Rightarrow 100 + 50 α = 138$

$\Rightarrow α = \frac{38}{50} = 0.76 or 76 %$

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