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Question

For the reaction A(g)B(g) at 495 K, ΔG= 9.478 kJ mol1. If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is millimoles. (Round off to the nearest Integer).
[R=8.314 J mol1K1; ln10=2.303]

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Solution

By using the relation,
ΔG=RTlnKeqWhere,ΔG= 9.478 kJ mol1T=Temperature=495 KR=Universal gas constant=8.314 J mol1K1Substituting the values, we get,9.478×103=495×8.314 lnKeqlnKeq=2.303=ln 10
Keq=10
Now,
A(g)B(g)t=0220t=t22xx
Keq=[B][A]=x(22x)=10x=20

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