Question

$N_2{O}_4$ is $25$% dissociated at ${37}^{o}C$ and $1atm$ pressure. Calculate (i) $K_p$ and (ii) the percentage dissociation at $0.1atm$ and ${37}^{o}C$.

Answer 1

(i) $N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$
Initial: $1$ $0$
At equilibrium: $(1-x)$ $2x$
$\therefore$ Total moles $=(1-x)+2x=(1+x)$
$p_{N_2{O}_4}=\left(\frac{1-x}{1+x}\right)P;p_{{NO}_2}=\frac{2x}{(1+x)}.P$
Given, $x=0.25$ and $P=1atm$
$p_{N_2{O}_4}=\left(\frac{1-0.25}{1+0.25}\right)\times 1=0.6atm$
$p_{{NO}_2}=\frac{2\times 0.25}{(1+0.25)}\times 1=0.4atm$
$K_p=\frac{{\left(p_{{NO}_2}\right)}^2}{p_{N_2{O}_4}}=\frac{0.4\times 0.4}{0.6}=0.267atm$
(ii) Let the degree of dissociation of $N_2{O}_4$ at $0.1atm$ be '$\alpha$', then,
$p_{N_2{O}_4}=\left(\frac{1-\alpha }{1+\alpha }\right)\times 0.1;p_{{NO}_2}=\frac{2\times \alpha }{(1+\alpha )}\times 0.1$
$K_p=\frac{{\left(\frac{2\times \alpha }{(1+\alpha )}\right)}^2\times {\left(0.1\right)}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)\times 0.1}=\frac{0.4{\alpha }^2}{(1-{\alpha }^2)}$
$0.267=\frac{0.4{\alpha }^2}{(1-{\alpha }^2)}$
$\alpha =0.632$
Hence dissociation percent of $N_2{O}_4=63.2$%