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Arrhenius Acid

Vapor density...

Question

Vapor density of the equilibrium mixture of ${N O}_{2}$ and $N_{2} O_{4}$ is found to be 40 for the equilibrium:
$N_{2} O_{4} ⇌ {2 N O}_{2}$
Calculate
(a) abnormal molecular weight
(b) degree of dissociation
(c) percentage of ${N O}_{2}$ in the mixture

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Solution

Given that $N_{2} O_{4} ⇌ 2 N O_{2}$ density= $40$
At equilibrium $N_{2} O_{4}$ is $1 - X$ ( $92$ = mass)
$N O_{2} = 2 X$ ( $60$ = mass)
$\Rightarrow \frac{n_{i n i t i a l}}{n_{F i n a l}} = \frac{M_{F i n a l}}{M_{I n i t i a l}}$
$\Rightarrow \frac{1}{1 - X + 2 X} = \frac{60}{92}$
$\Rightarrow 1 + X = \frac{92}{60} \Rightarrow X = 0.53$
% of dissociation $0.53 \times 100 = 53$
% of $N O_{2} = 100 - 53 = 47$ %

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