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Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 8 April, Evening Shift — Question 18

The equilibrium constant for decomposition of H2O(g)\mathrm{H}_{2} \mathrm{O}(\mathrm{g})

H2O(g)⇌H2( g)+12O2( g)(ΔG0=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}^{0}=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−\times 10^{-} 2{ }^{2}

(nearest integer value) [Assume α\alpha is negligible with respect to 1]

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

H2O(g)⇌H2(g)+12O2(g)\text{H}_2\text{O}(g) \rightleftharpoons \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) Initial mole1−−Moles at equil.1−ααα2\begin{array}{lccc} \text{Initial mole} & 1 & - & - \\ \text{Moles at equil.} & 1 - \alpha & \alpha & \frac{\alpha}{2} \end{array} Equilibrium pressure=1 bar\text{Equilibrium pressure} = 1 \, \text{bar} Kp=(α2+α)(α/22+α)1/2(1−α2+α)=8.0×10−3K_p = \frac{\left(\frac{\alpha}{2+\alpha}\right) \left(\frac{\alpha/2}{2+\alpha}\right)^{1/2}}{\left(\frac{1-\alpha}{2+\alpha}\right)} = 8.0 \times 10^{-3} Kp=α⋅(α/2)1/2⋅(2+α)1/221/2⋅(2+α)1/2⋅(1−α)=α3/223/2(1−α)(2+α)1/2−1/2=α3/22(1−α)K_p = \frac{\alpha \cdot (\alpha/2)^{1/2} \cdot (2+\alpha)^{1/2}}{2^{1/2} \cdot (2+\alpha)^{1/2} \cdot (1-\alpha)} = \frac{\alpha^{3/2}}{2^{3/2} (1-\alpha)} (2+\alpha)^{1/2 - 1/2} = \frac{\alpha^{3/2}}{\sqrt{2} (1-\alpha)} (α2+α)(α2(2+α))1/2(1−α2+α)=α⋅α1/221/2(2+α)1/2(1−α)(2+α)=α3/2(2+α)1/22(1−α)\frac{\left(\frac{\alpha}{2+\alpha}\right) \left(\frac{\alpha}{2(2+\alpha)}\right)^{1/2}}{\left(\frac{1-\alpha}{2+\alpha}\right)} = \frac{\alpha \cdot \alpha^{1/2}}{2^{1/2} (2+\alpha)^{1/2} (1-\alpha)} (2+\alpha) = \frac{\alpha^{3/2} (2+\alpha)^{1/2}}{\sqrt{2} (1-\alpha)} (α2+α⋅1)(α2(2+α)⋅1)1/2(1−α2+α⋅1)=α⋅α1/2⋅(2+α)1/22(2+α)1/2(1−α)=α3/22(1−α)\frac{\left(\frac{\alpha}{2+\alpha} \cdot 1\right) \left(\frac{\alpha}{2(2+\alpha)} \cdot 1\right)^{1/2}}{\left(\frac{1-\alpha}{2+\alpha} \cdot 1\right)} = \frac{\alpha \cdot \alpha^{1/2} \cdot (2+\alpha)^{1/2}}{\sqrt{2} (2+\alpha)^{1/2} (1-\alpha)} = \frac{\alpha^{3/2}}{\sqrt{2} (1-\alpha)} Kp=(α2+α)(α/22+α)1/2(1−α2+α)=α(α/2)1/2(2+α)1/2(2+α)(1−α)=α3/22(2+α)1/2(1−α)K_p = \frac{\left(\frac{\alpha}{2+\alpha}\right) \left(\frac{\alpha/2}{2+\alpha}\right)^{1/2}}{\left(\frac{1-\alpha}{2+\alpha}\right)} = \frac{\alpha (\alpha/2)^{1/2} (2+\alpha)^{1/2}}{(2+\alpha) (1-\alpha)} = \frac{\alpha^{3/2}}{\sqrt{2} (2+\alpha)^{1/2} (1-\alpha)} α⋅(α/2)1/21−α(12+α)1−1/2−1=α3/22(1−α)(2+α)1/2\frac{\alpha \cdot (\alpha/2)^{1/2}}{1-\alpha} \left( \frac{1}{2+\alpha} \right)^{1 - 1/2 - 1} = \frac{\alpha^{3/2}}{\sqrt{2} (1-\alpha)} (2+\alpha)^{1/2} α⋅(α2)1/21−α=α3/22(1−α)=8.0×10−3\frac{\alpha \cdot (\frac{\alpha}{2})^{1/2}}{1-\alpha} = \frac{\alpha^{3/2}}{\sqrt{2} (1-\alpha)} = 8.0 \times 10^{-3} α3/22=8.0×10−3(assuming α≪1)\frac{\alpha^{3/2}}{\sqrt{2}} = 8.0 \times 10^{-3} \quad (\text{assuming } \alpha \ll 1) α3/2=82×10−3\alpha^{3/2} = 8\sqrt{2} \times 10^{-3} α=(82×10−3)2/3\alpha = (8\sqrt{2} \times 10^{-3})^{2/3} =(23⋅21/2×10−3)2/3=(27/2×10−3)2/3=27/3×10−2=22⋅21/3×10−2= (2^3 \cdot 2^{1/2} \times 10^{-3})^{2/3} = (2^{7/2} \times 10^{-3})^{2/3} = 2^{7/3} \times 10^{-2} = 2^2 \cdot 2^{1/3} \times 10^{-2} =4×(1.26)×10−2=5.04×10−2≈5×10−2= 4 \times (1.26) \times 10^{-2} = 5.04 \times 10^{-2} \approx 5 \times 10^{-2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
8 April, Evening Shift
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Gibbs Free Energy - Relation with Equilibrium, Metallurgy and Electrochemistry

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