Question

The degree of dissociation of $N_2{O}_4\left(\alpha \right)$ obeying the equilibrium,
$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$, is approximately related to the pressure at equilibrium by:

• A. $\alpha \propto P$
• B. $\alpha \propto \frac{1}{\sqrt{P}}$
• C. $\alpha \propto \frac{1}{P^2}$
• D. $\alpha \propto \frac{1}{P^4}$

Answer 1

The correct option is A $\alpha \propto P$

$N_2{O}_4\left(g\right)\rightleftharpoons 2{NO}_2\left(g\right)$

$1$ $-$

$1-\alpha$ $2\alpha$

Total moles of gas $=1+\alpha$

$P_{N_2{O}_4}=\frac{P(1-\alpha )}{(1+\alpha )}{P}_{{NO}_2}=\frac{P\left(2x\right)}{(1+\alpha )}$

$K_P=\frac{{P_{{NO}_2}}^2}{P_{N_2{O}_4}}\Rightarrow {\alpha }^2=\frac{K_P}{4P}\Rightarrow \alpha =\sqrt{\frac{K_P}{4P}}$

$\alpha \propto \frac{1}{\sqrt{P}}$