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)

The equilibrium constant for a reaction is 10. The value of $∆\mathrm{G}°$ will be:

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

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