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The vapour density of undecomposed $N_{2} O_{4}$ is 46 . When heated, the vapour density decreases to 24.5 due to its dissociation to $NO_{2}$ . If equilibrium pressure is 1 atm , its $K_{p}$ will be

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Answer:

Solution:

$x = \frac{D - d}{d ( n - 1 )} = \frac{46 - 24.5}{24.5 ( 2 - 1 )} = 0.8775$

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

At equilibrium, $(1 - x)$ $2 x$

$P_{N_{2} O_{4}} = \frac{( 1 - x )}{( 1 + x )} \times 1$

$P_{N O_{2}} = \frac{2 x}{( 1 + x )} \times 1$ .

$K_{p} = \frac{4 x ^{2}}{( 1 - x ^{2} )}$ ; $x = 0.8775$

$K_{p} \approx 13.4$

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Problems involving Kp

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