Question

N₂ + 3H₂ $\rightleftharpoons$ 2NH₃

Initial moles 1 3 0

Equilibrium moles 1 – x 3 – 3x 2x

From question, $\frac{2x}{4 - 2x} = a \Rightarrow x = \frac{2a}{1 + a}$

Now, $K_{P} = \frac{(2x)^{2}}{(1 - x)(3 - 3x)^{3}}.\left( \frac{P}{4 - 2x} \right)^{- 2} = \frac{4x^{2}(4 - 2x)^{2}}{27(1 - x)^{4}.P^{2}}$

Or, $\sqrt{K_{P}} = \frac{2x.(4 - 2x)}{\sqrt{27}(1 - x)^{2}.P} = \frac{2.\left( \frac{2a}{1 + a} \right).\left( 4 - \frac{4a}{1 + a} \right)}{\sqrt{27}.\left( 1 - \frac{2a}{1 + a} \right)^{2}.P} = \frac{32a}{\sqrt{27}.(1 - a)^{2}.P}$

$\therefore\frac{a}{(1 - a)^{2}}\alpha P$