Question

Density of equilibrium mixture of $N _{2} O _{4}$ and $NO _{2}$ at $1 \,atm$ and $384 \,K$ is $1.84 \,g / dm ^{3}$. Equilibrium constant of the following reaction is
$N _{2} O _{4} \rightleftharpoons 2 NO _{2}$

• A. 1.98 atm
• B. 2.09 atm
• C. 2.36 atm
• D. 1.48 atm

Answer 1

Answer: (B)

We know that,
$p m =d R T $
$1 \times m =1.84 \times 0.0821 \times 384$
$m =29 \times 2$
Vapour density $(d)$ at equilibrium $=29$
Initial vapour density $=\frac{m}{2}=\frac{92}{2}=46$
Therefore, degree of dissociation is
$\alpha=\frac{D-d}{(n-1) d}=\frac{46-29}{29}=0.586$
From reaction,

$p_{ N _{2} O _{4}}=\frac{1-\alpha}{1+\alpha} \times p ; p_{ NO _{2}}=\frac{2 \alpha}{1+\alpha} \times p$
Thus, $ K_{p}=\frac{4 \alpha^{2} p}{1-\alpha^{2}}$
$=\frac{4 \times(0.586)^{2} \times 1}{1-(0.586)^{2}}$
$=2.09 \,atm$