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Question

The variation of equilibrium constant (K) of a certain reaction with temperature (T) is lnK=3.0+2.0×104T given R=8.3 JK−1mol−1, the values of ΔHo and ΔSo are:

A
ΔHo=166 kJ/mol
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B
ΔHo=+166 kJ/mol
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C
ΔSo=+24.9 kJ/mol
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D
ΔSo=24.9 kJ/mol
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Solution

The correct option is C ΔSo=+24.9 kJ/molGiven, lnK=3.0+2.0×104T On differentiating w.r.t. T δln KδT=−2.0×104T2... (1) Also, δln KδT=ΔHoRT2... (2) Equating (1) and (2) ΔHo=−2.0×104×R ΔHo=−2.0×104×8.3 ΔHo=−166 kJ/mol Further, ΔGo=−RT ln K ΔGo=−RT (3.0+2.0×104T) ΔGo=−3RT−2.0×104R Also, (δΔGoδT)p=−ΔSo ΔSo=3R ΔSo=3×8.3 ΔSo=+24.9 Jmol−1K−1

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