The value of ∆G°  for the given reaction would be:
\( 2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{~g}) \rightarrow 2 \mathrm{D}(\mathrm{~g})\)
(Given: ∆U° = – 10.5 kJ and  ∆S° = – 44.1 J K^–1)

Determine the value of  ∆G°, if the equilibrium constant for a reaction is 10:

($\mathrm{R}$ $=$ $8.314$ $\mathrm{J}$ ${\mathrm{K}}^{-1}$ ${\mathrm{mol}}^{-1}$ $;$ $\mathrm{T}$ $=$ $300$ $\mathrm{K}$)

1. \(-5.74 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
2. \(-5.74 \mathrm{~J} \mathrm{~mol}^{-1}\)
3. \( +4.57 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
4. \(-57.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\)