JEE MAIN - Chemistry (2025 - 8th April Evening Shift - No. 23)
The equilibrium constant for decomposition of $\text{H}_2\text{O(g)}$
$ \text{H}_2\text{O(g)} \rightleftharpoons \text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $
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]
Explanation
$\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}$
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$$\begin{aligned} & \mathrm{n}_{\mathrm{T}}=1+\frac{\alpha}{2} \simeq 1(\alpha \ll 1) \\ & \mathrm{k}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{H}_2} \cdot \mathrm{P}_{\mathrm{O}_2}^{1 / 2}}{\mathrm{P}_{\mathrm{H}_2 \mathrm{O}}}=\frac{(\alpha \cdot \mathrm{P})\left(\frac{\alpha}{2} \mathrm{P}\right)^{\frac{1}{2}}}{(1-\alpha) \mathrm{P}} \\ & \mathrm{P}=1 \\ & 8 \times 10^{-3}=\frac{\alpha^{3 / 2}}{\sqrt{2}} \\ & \alpha^{3 / 2}=8 \sqrt{2} \times 10^{-3} \\ & \alpha^3=128 \times 10^{-6} \\ & \alpha=\sqrt[3]{128} \times 10^{-2} \\ & =5.03 \times 10^{-2} \end{aligned}$$
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