Question

Density of equilibrium mixture of $N_2{O}_4$ and $N{O}_2$ at $1atm$ and $384K$ is $1.84g/d{m}^3$. Equilibrium constant of the following reaction is
$N_2{O}_4\rightleftharpoons 2N{O}_2$

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

Answer 1

Answer: (B)

We know that,
$pm=dRT$
$1\times m=1.84\times 0.0821\times 384$
$m=29\times 2$
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,

$p_{N_2{O}_4}=\frac{1-\alpha }{1+\alpha }\times p;p_{N{O}_2}=\frac{2\alpha }{1+\alpha }\times p$
Thus, $K_p=\frac{4{\alpha }^{2}p}{1-{\alpha }^2}$
$=\frac{4\times (0.586{)}^2\times 1}{1-(0.586{)}^2}$
$=2.09atm$