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Question

The equilibrium constant Kc for a reaction is 10. Calculate the value of change in standard Gibbs free energy ΔGo in kJ mol1 at 300 K.

A
5.744
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B
-5.744
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C
-2.744
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D
2.744
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Solution

The correct option is B -5.744
We know that, the relation of Gibbs free energy and equilibrium constant at equilibrium conditions is given by ΔGo=2.303 RT log Kc
Given,
T=300 K
Kc=10
R=8.314 J mol1 K1
Substituting the values we get,
ΔGo=2.303×8.314×300×log 10
ΔGo=5744.14×log 10
ΔGo=5744.14 J/mol (log 10=1)
ΔGo=5.744 kJ/mol


Theory:

Consider for the given reaction:
aA(aq)+bB(aq)cC(aq)+dD(aq)
Gibbs free energy change is given by: ΔG=ΔG+RT ln Q
Where, Q= ReactionQuotient

ΔG=ΔG+2.303.RT log10Q
Where, ΔG is standard Gibbs free energy change

At equilibrium, ΔG=0 and Q=Kc
ΔG=2.303 RT log10Kc
ΔG=RT ln Kc

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