Question

$N_2{O}_4$ dissociates as,$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ at $55{}^{o}C$ and $1.2$ atmosphere, $\%$ decomposition of $N_2{O}_4$ is $50.0\%$.
At what pressure and same temperature, the equilibrium mixture has the ratio of $N_2{O}_4:N{O}_2$ as $1:8$?
(Report the answer upto two decimal)

Answer 1

Case I: $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
At equilibirum: $(1-x)$ $2x$
$p_{N_2{O}_4}=\frac{1-x}{1+x}\times P;p_{N{O}_2}=\frac{2x}{(1+x)}\times P$
$K_p=\frac{(\frac{2x}{1+x}\cdot P)}{(\frac{1-x}{1+x}\cdot P)}=\frac{4x^{2}P}{(1-x^2)}$
Given, $9x=0.50$ and $P=1.2$
$K_p=1.6$ atm
Case II: $N_2{O}_4\rightleftharpoons 2N{O}_2$
$(1-x)$ $2x$
Given, $\frac{(1-x)}{2x}=\frac{1}{8}$
$x=0.8$
Let the new pressure be $P$ atm.
$K_p=\frac{4x^{2}P}{(1-x^2)}=\frac{4\times 0.8\times 0.8\times P}{(1+0.8)(1-0.8)}=1.6$
$P=0.225$ atm