ΔG∘=−RTℓnKeq
Given ΔG∘=−9.478KJ/mole
T=495KR=8.314Jmol−1
So −9.478×103=−495×8.314×ℓnKeq
ℓnKeq=2.303
=ℓn10
So Keq=10
Now t=0t=0A(g)2222−x⇌B(g)0x
Keq=[C][B]=22−xx=10
or x=20
So millimoles of B=20
JEE Main 2021 — Chemistry Physical Chemistry
For the reaction A(g)⇌B(g) at 495K, ΔrG∘=−9.478kJmol−1
If we start the reaction in a closed container at 495K with 22 millimoles of A, the amount of B is the equilibrium mixture is ________ millimoles. (Round off to the Nearest Integer).
[R=8.314Jmol−1K−1;ℓn10=2.303]
Held on 16 Mar 2021 · Verified 6 Jul 2026.
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