Question

The degree of dissociation of $N_2{O}_4$ obeying the following equilibrium is approximately related to the pressure at equilibrium by:

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

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

Answer 1

The correct option is B $\alpha \propto \frac{1}{\sqrt{P}}$

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$,

$1-\alpha$ $2\alpha$

$K_P=\frac{PN{O}_2}{P{N}_2{O}_4}$

$K_p=\frac{2{\alpha }^2}{(1+\alpha {)}^2}\times P^2\times \frac{1+\alpha }{1-\alpha }\times P$

$K_p=\frac{4{\alpha }^2\times P}{1+{\alpha }^2}$

${\alpha }^2=\frac{1}{P\times K_P},(1+{\alpha }^{2}isnegligible)$

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