Question

$N_2{O}_4$ is dissociated to 33% and 40% at total pressure $P_1$ and $P_2$ atm respectively. Then the ratio $\frac{P_1}{P_2}$ is :

• A. $\frac{7}{4}$
• B. $\frac{7}{3}$
• C. $\frac{8}{3}$
• D. $\frac{8}{5}$

Answer 1

The correct option is D $\frac{8}{5}$

Reaction:

$N_2{O}_4\rightleftharpoons 2N{O}_2$

$1$ $0$

$1-\alpha$ $2\alpha$

$\sum n=1-\alpha +2\alpha$

$=1+\alpha$

$K_p=\frac{P_{[N{O}_2{]}^2}}{P_{\left[N_2{O}_4\right]}}$

$K_p=\frac{{\left(n_{N{O}_2}\right)}^2}{n_{N_2{O}_4}}\times {\left[\frac{P}{\sum n}\right]}^1$

For 33% dissociation : $K_p=\frac{{(2\times 0.33)}^2}{0.67}\times \left[\frac{P_1}{1.33}\right]$

For 40% dissociation : $K_p=\frac{{(2\times 0.40)}^2}{0.60}\times \left[\frac{P_2}{1.40}\right]$

$\therefore \frac{P_1}{P_2}=1.56\simeq 1.60=\frac{8}{5}$