Question

138 gm of $N_2{O}_4\left(g\right)$ is placed in a $8.2L$ container at $300K$. If the equilibrium vapour density of mixture was found to be $30.67$, then: $[R=0.082L.atm.mo{l}^{-1}.K^{-1}]$

• A. Degree of dissociation of $N_2{O}_4$ is $0.25$
• B. $K_p$ of $N_2{O}_4\rightleftharpoons 2N{O}_2\left(g\right)$ is $9$ atm
• C. Total pressure at equilibrium is $6.75$ atm
• D. The density of equilibrium mixture is $16.83gm/litre$

Answer 1

Answer: (B), (C), (D)

The density of equilibrium mixture is $16.83gm/litre$
$K_p$ of $N_2{O}_4\rightleftharpoons 2N{O}_2\left(g\right)$ is $9$ atm
Total pressure at equilibrium is $6.75$ atm

$N_2{O}_4\rightleftharpoons 2N{O}_2$
$a$ 0 -------------at $t=0$
$a(1-\alpha )2a\alpha$ ...............at $t=t$

Vapour density$=\frac{46}{1+\alpha }=30.67$
So, $1+\alpha =1.5\Rightarrow \alpha =0.5\Rightarrow 50$%
Number of moles initially $=\frac{138}{92}=1.5mol$
Total pressure $=\frac{1.5\times 1.5\times 0.082\times 300}{8.2}=6.75atm$
So, $K_p=\frac{4{\alpha }^2}{1-{\alpha }^2}P=9atm$
and for density of mixture $=\frac{138}{8.2}gm/L=16.83gm/L$