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Question

For the reaction
A(l)2B(g)
ΔU=2.1 kcal,ΔS=20 cal/K at 300 K
Hence ΔG in kcal is

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Solution

ΔH=ΔU+ΔngRT
ΔH=2100+(2×2×300) (R=2calK1mol1)
=3300 cal
ΔG=ΔHTΔS
ΔG=3300(300×20)=2700 cal=2.70 kcal

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