At $1 \ bar$ pressure and $310 \ K$ temperature,$25\%$ of $N_2O_4$ decomposes. The reaction is: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.
$(i)$ Find $K_p$.
$(ii)$ At $0.1 \ bar$ pressure and $310 \ K$,what is the percentage of $N_2O_4$ decomposed?

(N/A) For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,let the initial moles be $1$.
At equilibrium,moles are $(1-\alpha)$ of $N_2O_4$ and $2\alpha$ of $NO_2$. Total moles $= 1+\alpha$.
Given $\alpha = 0.25$ at $P = 1 \ bar$.
Partial pressures: $P_{N_2O_4} = \frac{1-0.25}{1+0.25} \times 1 = 0.6 \ bar$ and $P_{NO_2} = \frac{0.5}{1.25} \times 1 = 0.4 \ bar$.
$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(0.4)^2}{0.6} = \frac{0.16}{0.6} = 0.267 \ bar$.
For $(ii)$,at $P = 0.1 \ bar$,$K_p = \frac{(2\alpha)^2 P}{(1-\alpha)(1+\alpha)} = \frac{4\alpha^2 P}{1-\alpha^2}$.
$0.267 = \frac{4\alpha^2 (0.1)}{1-\alpha^2} \implies 0.267 - 0.267\alpha^2 = 0.4\alpha^2$.
$0.667\alpha^2 = 0.267 \implies \alpha^2 = 0.4 \implies \alpha = 0.632$.
Percentage decomposition $= 63.2\%$.