JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Equilibrium

Starting from one mole of ammonia with degree of dissociation $\alpha$, the equilibrium amounts are ${\text{NH}}_3:(1-\alpha )$, ${\text{N}}_2:\alpha /2$, ${\text{H}}_2:3\alpha /2$, with total moles $1+\alpha$. Writing each partial pressure as (mole fraction)$\times P_T$ and substituting into $K_p=\frac{p_{{\text{N}}_2}^{1/2}{p}_{{\text{H}}_2}^{3/2}}{p_{{\text{NH}}_3}}$ gives $K_p=\frac{3\sqrt{3}}{4}\frac{{\alpha }^2}{1-{\alpha }^2}{P}_T$. With $P_T=\sqrt{3}$ and $K_p=9$ this reduces to $9=\frac{9}{4}\frac{{\alpha }^2}{1-{\alpha }^2}$, so $1-{\alpha }^2=\frac{{\alpha }^2}{4}$ and ${\alpha }^2=0.8$. Then $\alpha =(0.8{)}^{1/2}={\left(\frac{1}{0.8}\right)}^{-1/2}={\left(125\times 10^{-2}\right)}^{-1/2}$, so the value of $x$ is $125$.