$\mathrm{N}_{2} \mathrm{O}_{4}$ dissociates according to the equation \ $\mathrm{N}_{2} \mathrm{O}_

Question

$\mathrm{N}_{2} \mathrm{O}_{4}$ dissociates according to the equation \ $\mathrm{N}_{2} \mathrm{O}_{4} ightleftharpoons 2 \mathrm{NO}_{2}$ \ If at equilibrium, the degree of dissociation is $33 \%$ at a pressure $\mathrm{P}_{1}$ atm and $40 \%$ at a pressure $\mathrm{P}_{2} \mathrm{~atm}$, the ratio , $\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}$, is equal to\

Professor PC Thomas Classes · Answer

$\mathrm{N}_{2} \mathrm{O}_{4} ightleftarrows 2 \mathrm{NO}_{2}$\ $\mathrm{k}_{\mathrm{p}}=\frac{4 \alpha^{2} \mathrm{P}}{1-\alpha^{2}} \approx 4 \alpha^{2} \mathrm{P}$, since $\alpha \ll 1$\ $\mathrm{k}_{\mathrm{p}}$ is a constant at a given temperature\ As $P$ changes $\alpha$ also changes\ $\alpha_{1}^{2} p_{1}=\alpha_{2}^{2} p_{2} \cdot \alpha_{1}=0.33$ and $\alpha_{2}=0.4$ $$ 	herefore \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\left(\frac{\alpha_{2}}{\alpha_{1}}ight)^{2}=\left(\frac{0.4}{0.33}ight)^{2}=1.5 $$