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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 Ho=28.40 kJ mol1.

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 kJmol1 So=33.6 JK1mol1
The 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[2.0×107]
=+38479.8 J mol1
=+38.48 kJ mol1
Also, Go=HoTSo
So=HoGoT
So=28.4038.48300
So=0.0336 kJ K1mol1=33.6 J K1mol1

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