JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Equilibrium

For ${\mathrm{H}}_2\mathrm{O}(\mathrm{g})\rightleftharpoons {\mathrm{H}}_2(\mathrm{g})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})$, let the degree of dissociation be $\alpha$. Starting from one mole of water, the equilibrium amounts are $(1-\alpha )$ of ${\mathrm{H}}_2\mathrm{O}$, $\alpha$ of ${\mathrm{H}}_2$ and $\alpha /2$ of ${\mathrm{O}}_2$, with total moles close to one because $\alpha \ll 1$. At a total pressure of 1 bar the partial pressures equal the mole fractions, so $p_{{\mathrm{H}}_2}=\alpha$, $p_{{\mathrm{O}}_2}=\alpha /2$ and $p_{{\mathrm{H}}_2\mathrm{O}}\approx 1$. Hence $K_p=\frac{p_{{\mathrm{H}}_2}{p}_{{\mathrm{O}}_2}^{1/2}}{p_{{\mathrm{H}}_2\mathrm{O}}}=\frac{\alpha (\alpha /2{)}^{1/2}}{1}=\frac{{\alpha }^{3/2}}{\sqrt{2}}$. Setting this equal to $8.0\times 10^{-3}$ gives ${\alpha }^{3/2}=8\sqrt{2}\times 10^{-3}$, so ${\alpha }^3=128\times 10^{-6}$ and $\alpha =\sqrt[3]{128}\times 10^{-2}=5.04\times 10^{-2}$. The degree of dissociation of water is $5\times 10^{-2}$.