Question

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

• A. $\Delta G = 0 ; \Delta G^{\circ} = - 11.48 \, kJ \, mol^{-1}$
• B. $\Delta G = 0 ; \Delta G^{\circ} = 11.48 \, kJ \, mol^{-1}$
• C. $\Delta G^{\circ} = - 11.48 \, kJ \, mol^{-1}; \Delta G = 0$
• D. $\Delta G^{\circ} = 11.48 \, kJ \, mol^{-1}; \Delta G = 0$

Answer 1

Answer: (A)

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

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

Given, $K =10^{2}$

$T =27^{\circ} C$ or $300\, K$

on putting the values in Eq (i)

$\Delta G ^{\circ}=-2303 \times 8.314\, JK ^{-1}\, mol ^{-1} \times 300 \times \log 10^{2}$

$\Delta G ^{\circ}=-1148.28\, J\, mol ^{-1}$ or $-11.48\, kJ\, mol ^{-1}$

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