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Question

The equilibrium constant for the reaction given below is 2.0×107 at 300 K. Calculate the standard free energy change for the reaction.
PCl5(g)PCl3(g)+Cl2(g)
Also, calculate the standard entropy change if,
ΔH=28.40 kJ mol1

A
38.48 kJ,33.6 J mol1K1
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B
48.48 kJ,33.6 J mol1K1
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C
38.48 kJ,43.6 J mol1K1
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D
48.48 kJ,43.6 J mol1K1
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Solution

The correct option is A 38.48 kJ,33.6 J mol1K1
K=2.0×107T=300 KΔH=28.4 kJ mol1ΔG=2.303RTlogK=2.303×8314×103×300×log(2×107)=5.744(log27)=5.744(0.30107)=38.48 kJ
From ΔG=ΔHTΔSTΔS=ΔHΔGΔS=ΔHΔGT=28.438.48300=0.0336 kJ mol1or ΔS=33.6 J mol1K1

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