H2O(g)⇌H2( g)+21O2( g)t=01 molet=teq1−αα2αnT=1+2α≃1(α≪1)kP=PH2OPH2⋅PO21/2=(1−α)P(α⋅P)(2αP)218×10−3=2α3/2α3/2=82×10−3α3=128×10−6α=3128×10−2=5.03×10−2
JEE Main 2025 — Chemistry Physical Chemistry
The equilibrium constant for decomposition of H2O(g)
H2O( g)⇌H2( g)+21O2( g)(ΔG∘=92.34 kJ mol−1)
is 8.0×10−3 at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation (α) of water is ________ ×10−2 (nearest integer value).
[Assume α is negligible with respect to 1 ]
Held on 8 Apr 2025 · Verified 6 Jul 2026.
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