Question

The value of $K_p$ for the equilibrium of the reaction $N_2{O}_4\left(g\right)\leftrightharpoons 2N{O}_2\left(g\right)$ is $2$. Calculate the percentage dissociation of $N_2{O}_4$ at a pressure of $0.5atm.$

• A. $71$
• B. $50$
• C. $25$
• D. $88$

Answer 1

The correct option is A $71$

$N_2{O}_4\rightleftharpoons 2{NO}_2$

$1$ $-$ $K_P=\frac{{\left(n_{{NO}_2}\right)}^2}{n_{N_2{O}_4}}\times \frac{P}{{\alpha }_n}$

$1-\alpha$ $2\alpha$

${\alpha }_n=1-\alpha +2\alpha$ $2=\frac{{\left(2\alpha \right)}^2}{1-\alpha }\times \frac{0.5}{1+\alpha }$

$=1+\alpha$ ${\alpha }^2=0.5\Rightarrow \alpha =0.71$

% dissociation $=0.71\times \frac{100}{100}=71$%