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Equilibrium Constant and Standard Free Energy Change

The density o...

Question

The density of an equilibrium mixture of $N_{2} O_{4}$ and $N O_{2}$ at 1 atm and 346 K is 1.8 g/L. Calculate $K_{C}$ for the reaction:

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

A

12.8

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B

24

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C

0.36

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D

7.2

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Solution

The correct option is D $7.2$
The average molar mass of the gaseous mixture is $M = \frac{ρ R T}{P}$
$M = \frac{1.8 \times 0.0821 \times 346}{1}$
$M = 51.13$
Let x be the mole fraction of $N_{2} O_{4}$
The mole fraction of $N O_{2}$ is $1 - x$
Th average molar mass of the mixture is $\frac{92 x + 46 (1 - x)}{1} = 51.13$
$46 x = 5.13$
$x = 0.11$
$K_{C} = \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]} = \frac{(0.89)^{2}}{0.11} = 7.2$

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