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Question
The equilibrium constant for the reaction given below is 2.0×10−7 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 mol−1
The equilibrium constant for the reaction given below is 2.0×10−7 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 mol−1
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Solution
The correct option is A 38.48 kJ,−33.6 J mol−1K−1
K=2.0×10−7T=300 KΔH∘=28.4 kJ mol−1ΔG∘=−2.303RTlogK=−2.303×8314×10−3×300×log(2×10−7)=−5.744(log2−7)=−5.744(0.3010−7)=38.48 kJ
From ΔG∘=ΔH∘−TΔS∘⇒TΔS∘=ΔH∘−ΔG∘ΔS∘=ΔH∘−ΔG∘T=28.4−38.48300=−0.0336 kJ mol−1or ΔS=−33.6 J mol−1K−1
K=2.0×10−7T=300 KΔH∘=28.4 kJ mol−1ΔG∘=−2.303RTlogK=−2.303×8314×10−3×300×log(2×10−7)=−5.744(log2−7)=−5.744(0.3010−7)=38.48 kJ
From ΔG∘=ΔH∘−TΔS∘⇒TΔS∘=ΔH∘−ΔG∘ΔS∘=ΔH∘−ΔG∘T=28.4−38.48300=−0.0336 kJ mol−1or ΔS=−33.6 J mol−1K−1
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