Question

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

• A. 7 / 4
• B. 7 / 3
• C. 8 / 3
• D. 8 / 5

Answer 1

Answer: (D)

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

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

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

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