ΔG∘=ΔH∘−TΔS∘
ΔG∘=−29.8kJmol−1+0.1×298kJmol−1=0
\Delta G= \Delta G^{\circ}+RTlnQ at Equilibrium \Delta G=0 & Q=k
ΔG∘=−RTln(k)
ifln(k)=0,k=e0
∴K=1
JEE Main 2016 — Chemistry Physical Chemistry
For the reaction,
A(g)+B(g)→C(g)+D(g),ΔHo and ΔSo are, respectively, −29.8kJmol−1 and −0.100kJK−1mol−1 at 298 K. The equilibrium constant for the reaction at 298 k is:
Held on 9 Apr 2016 · Verified 6 Jul 2026.
1.0×10−10
10
1
1.0×1010
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