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JEE MAIN 2025
08-04-2025 S2
Question
The equilibrium constant for decomposition of $\mathrm{H}_2 \mathrm{O}$ (g)
$\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\left(\Delta \mathrm{G}^{\circ}=92.34 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
is $8.0 \times 10^{-3}$ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation $(\alpha)$ of water is ________ $\times 10^{-2}$ (nearest integer value).
[Assume $\alpha$ is negligible with respect to 1 ]
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Solution
$\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{g}) f \quad \mathrm{H}_{2(\mathrm{x})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{a})} \\ & \mathrm{t}=0 \quad 1 \mathrm{~mole} \\ & \mathrm{t}=\mathrm{t}_{\mathrm{eq}} \quad 1-\alpha \quad \alpha \quad \frac{\alpha}{2} \\ & \mathrm{n}_{\mathrm{T}}=1+\frac{\alpha}{2} ; \mathrm{l}(\alpha-1) \\ & \mathrm{k}_p=\frac{\mathrm{P}_{\mathrm{H}_2} \cdot \mathrm{P}_{\mathrm{O}_3}^{1 / 2}}{\mathrm{P}_{11 / 0} \quad(\alpha \cdot \mathrm{P})\left(\frac{\alpha}{2} \mathrm{P}\right)^{\frac{1}{2}}} \frac{(1-\alpha) \mathrm{P}}{\mathrm{P}=1} \\ & 8 \times 10^{-3}=\frac{\alpha^{1 / 2}}{\sqrt{2}} \\ & \alpha^{3 / 2}=8 \sqrt{2} \times 10^{-3} \\ & \alpha^3=128 \times 10^{-6} \\ & \alpha=\sqrt[3]{128} \times 10^{-2} \\ & =5.03 \times 10^{-2}\end{aligned}$
Question Tags
JEE Main
Chemistry
Hard
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