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Chemistry Question on Equilibrium Constant

What will be the value of ΔG\Delta G and ΔG∘\Delta G ^{\circ} for the reaction, A+B⇌C+DA + B \rightleftharpoons C + D at 27∘C27^{\circ} C for which K=102K=10^{2} ?

A

ΔG=0;ΔG∘=−11.48 kJ mol−1\Delta G = 0 ; \Delta G^{\circ} = - 11.48 \, kJ \, mol^{-1}

B

ΔG=0;ΔG∘=11.48 kJ mol−1\Delta G = 0 ; \Delta G^{\circ} = 11.48 \, kJ \, mol^{-1}

C

ΔG∘=−11.48 kJ mol−1;ΔG=0\Delta G^{\circ} = - 11.48 \, kJ \, mol^{-1}; \Delta G = 0

D

ΔG∘=11.48 kJ mol−1;ΔG=0\Delta G^{\circ} = 11.48 \, kJ \, mol^{-1}; \Delta G = 0

Answer

ΔG=0;ΔG∘=−11.48 kJ mol−1\Delta G = 0 ; \Delta G^{\circ} = - 11.48 \, kJ \, mol^{-1}

Explanation

Solution

ΔG∘\Delta G ^{\circ} of a reaction is given by the expression.

ΔG∘=−2.303 RT log K\Delta G ^{\circ} =-2.303\, RT\, log\, K ...(i)

Given, K=102K =10^{2}
T=27∘CT =27^{\circ} C or 300 K300\, K

on putting the values in Eq (i)

ΔG∘=−2303×8.314 JK−1 mol−1×300×log⁡102\Delta G ^{\circ}=-2303 \times 8.314\, JK ^{-1}\, mol ^{-1} \times 300 \times \log 10^{2}
ΔG∘=−1148.28 J mol−1\Delta G ^{\circ}=-1148.28\, J\, mol ^{-1} or −11.48 kJ mol−1-11.48\, kJ\, mol ^{-1}

Now, as the reaction is at equilibrium, hence ΔG=0\Delta G=0.