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# 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 △Ho=28.40 kJ mol−1.

A
Go=40 kJmol1 So=30 JK1mol1
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B
Go=38.48 kJmol1 So=33.6 JK1mol1
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C
Go=45 kJ mol1 So=35 J K1mol1
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D
Go=42 kJ mol1 So=40 JK1mol1
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Solution

## The correct option is B △Go=38.48 kJmol−1 △So=−33.6 JK−1mol−1The formula which relate Gibbs free energy ΔG0 and equilibrium constant Keq is ΔG0=−2.303×R×T×log10[Keq] △Go=−2.303×8.314×300 log10[2.0×10−7] =+38479.8 J mol−1 =+38.48 kJ mol−1 Also, △Go=△Ho−T△So ∴△So=△Ho−△GoT △So=28.40−38.48300 △So=−0.0336 kJ K−1mol−1=−33.6 J K−1mol−1  Suggest Corrections  0      Similar questions  Explore more