For the reaction; A ( l ) → 2 B ( g ); Δ U =2.1 kcal, Δ S =20 cal K^-1 at 300 K. Hence Δ G in kcal is \ \ \ \.

For the reaction ; $$\begin{aligned} & \mathrm{A}(\mathrm{l}) \rightarrow 2 \mathrm{~B}(\mathrm{~g}) \\ & \Delta \mathrm{U}=2.1 \mathrm{kcal}, \Delta \mathrm{~S}=20 \mathrm{cal} \mathrm{~K}^{-1} \text { at } 300 \mathrm{~K} . \end{aligned} $$ Hence $\displaystyle \Delta \mathrm{G}$ in kcal is \_\_\_\_.