Question

$138gm$ of $N_2{O}_4$ (g) is placed in $8.2L$ container at $300K.$ The equilibrium vapour density of mixture was found to be $30.67$. Then: $(R=0.082Latmmo{l}^1{K}^{-1})$

• A. $\alpha =$ degree of dissociation of$N_2{O}_4=0.25$
• B. $K_{p}of{N}_2{O}_4\rightleftharpoons 2N{O}_2\left(g\right)$ will be $9atm$.
• C. Total pressure at equilibrium $=6.75atm$
• D. The density of equilibrium mixture will be $16.83gm/litre.$

Answer 1

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

The correct options are
B $K_{p}of{N}_2{O}_4\rightleftharpoons 2N{O}_2\left(g\right)$ will be $9atm$.
C Total pressure at equilibrium $=6.75atm$
D The density of equilibrium mixture will be $16.83gm/litre.$

138 gm of $N_2{O}_4$ (g) is placed in 8.2L container.
$N_2{O}_4\rightleftharpoons 2N{O}_2$.
$t=0a$
$ta(1-\alpha )2a\alpha$
Vapour density $\frac{46}{1+\alpha }=30.67$
So $1+\alpha =1.5,\alpha =0.5=50$%.
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 $=\frac{138}{8.2}gm/L=16.83gm/L$