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Chemical Equilibrium question

2025 · 8 Apr · Shift 2 · Q23
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Chemical Equilibrium question

2025 · 8 Apr · Shift 2 · Q23

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
The equilibrium constant for decomposition of H2O(g)H2O(g)⇌H2(g)+12O2(g)(ΔG∘=92.34 kJ mol−1)\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})H2​O(g)H2​O(g)⇌H2​(g)+21​O2​(g)(ΔG∘=92.34kJ mol−1) is 8.0×10−38.0 \times 10^{-3}8.0×10−3 at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation (α\alphaα) of water is ‾×10−2\underline{\hspace{2cm}}\times 10^{-2}​×10−2(nearest integer value). [Assume α\alphaα is negligible with respect to 1]
Numerical answer
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