$N_2O_4$ gas is $25\%$ dissociated into $NO_2$ at $37^\circ C$ and $1.0 ext{ atm}$ total pressure.

Question

$N_2O_4$ gas is $25\%$ dissociated into $NO_2$ at $37^\circ C$ and $1.0 	ext{ atm}$ total pressure. Calculate the percentage dissociation of $N_2O_4$ if the total pressure is decreased to $0.1 	ext{ atm}$ at the same temperature. (Give your answer to one decimal place).

Professor PC Thomas Classes · Answer

For the reaction $N_2O_4 ightleftharpoons 2NO_2$, the equilibrium constant $K_p = \frac{4\alpha^2 P}{1-\alpha^2}$. Given $\alpha_1 = 0.25$ at $P_1 = 1.0 	ext{ atm}$, $K_p = \frac{4(0.25)^2(1)}{1 - (0.25)^2} = \frac{0.25}{0.9375} = \frac{4}{15} 	ext{ atm}$. Let the new degree of dissociation be $\alpha_2$ at $P_2 = 0.1 	ext{ atm}$. $\frac{4\alpha_2^2 (0.1)}{1-\alpha_2^2} = \frac{4}{15} \implies \frac{\alpha_2^2}{1-\alpha_2^2} = \frac{1}{1.5} = \frac{2}{3}$. This gives $3\alpha_2^2 = 2 - 2\alpha_2^2 \implies 5\alpha_2^2 = 2 \implies \alpha_2 = \sqrt{0.4} \approx 0.6324$. The percentage dissociation is $63.2\%$.