JEE Main 2026 Jan 23 Shift 1, Chemistry Q22: Law of Chemical Equilibrium And Equilibrium Constant

Let initial moles of NH${}_3$ be 1 and degree of dissociation $\alpha$. At equilibrium: $n_{{\text{NH}}_3}=1-\alpha$; $n_{{\text{N}}_2}=\alpha /2$; $n_{{\text{H}}_2}=3\alpha /2$. Total moles $=1+\alpha$.

Mole fractions × $P_T$ give partial pressures:

$K_p=\frac{{\left(\frac{\alpha /2}{1+\alpha }{P}_T\right)}^{1/2}{\left(\frac{3\alpha /2}{1+\alpha }{P}_T\right)}^{3/2}}{\frac{1-\alpha }{1+\alpha }{P}_T}$

Simplify the numerator with $(3\alpha /2{)}^{3/2}\cdot (\alpha /2{)}^{1/2}=\frac{3\sqrt{3}{\alpha }^2}{4}$ and substitute $P_T=\sqrt{3}$:

$K_p=\frac{3\sqrt{3}{\alpha }^2/4}{(1-\alpha )(1+\alpha )}\cdot \sqrt{3}=\frac{9{\alpha }^2/4}{1-{\alpha }^2}$

Set equal to 9: $\frac{9{\alpha }^2/4}{1-{\alpha }^2}=9\Rightarrow 1-{\alpha }^2=\frac{{\alpha }^2}{4}\Rightarrow \frac{5{\alpha }^2}{4}=1\Rightarrow {\alpha }^2=0.8$.

Rewrite as $\alpha =(1/0.8{)}^{-1/2}=(1.25{)}^{-1/2}=(125\times 10^{-2}{)}^{-1/2}$. Hence x = 125.