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Introduction to Le Chatelier's Principle

Density of eq...

Question

Density of equilibrium mixture of $N_{2} O_{4}$ and ${N O}_{2}$ at $1 a t m$ and $384 K$ is $1.84 g {d m}^{- 3}$ . Calculate the equilibrium constant of the reaction, $N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$ .

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Solution

We know that,

$P M_{m i x} = d R T$
$1 \times M_{m i x} = 1.84 \times 0.0821 \times 384$
$M = 29 \times 2$
Vapour density $(d)$ at equilibrium $= 29$

Initial vapour density $= M / 2 = 92 / 2 = 46$

$x = \frac{D - d}{(n - 1) d} = \frac{46 - 29}{29} = 0.586$

$N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)$
$t = 0$ $1$ $0$
$t_{e q .}$ $1 - x$ $2 x$
Total moles $= 1 + x$
$p_{N_{2} O_{4}} = \frac{(1 - x)}{(1 + x)} \times P; p_{N O_{2}} = \frac{2 x}{(1 + x)} \times P$

$K_{p} = \frac{4 x^{2} P}{1 - x^{2}} = \frac{4 \times {(0.586)}^{2} \times 1}{1 - {(0.586)}^{2}} = 2.09 a t m$

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