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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2025 · 8 April · Evening Shift · Chemistry · Ionic Equilibrium

The equilibrium constant for decomposition of H_2 O(g) H_2 O(~g)...

The equilibrium constant for decomposition of H2O(g)\mathrm{H}_2 \mathrm{O}(\mathrm{g})
H2O( g)⇌H2( g)+12O2( g)(ΔG∘=92.34 kJ mol−1)\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 8.0×10−38.0 \times 10^{-3} at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation (α)(\alpha) of water is ________ ×10−2\times 10^{-2} (nearest integer value).
[Assume α\alpha is negligible with respect to 1 ]

Correct answer

0

Step-by-step solution

H2O(g)⇌H2( g)+12O2( g)t=01 molet=teq1−ααα2nT=1+α2≃1(α≪1)kP=PH2⋅PO21/2PH2O=(α⋅P)(α2P)12(1−α)P8×10−3=α3/22α3/2=82×10−3α3=128×10−6α=1283×10−2=5.03×10−2\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \\ & \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} \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}} \\ & 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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