Consider the following reaction: $2\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})\to 2\mathrm{D}(\mathrm{g})$ $\Delta U^{\circ }=-10kJmo{l}^{-1}$ and $\Delta S^{\circ }=-44J{K}^{-1}$ at $298\text{K}$. Identify the correct option with $\Delta G^{\circ }$ for the reaction and the spontaneity of the reaction at $298\text{K}$. (Given: $R=8.31Jmo{l}^{-1}{K}^{-1}$)

Solution & Explanation

\textbf{Answer:} (C) $+0.63568kJmo{l}^{-1}$, non-spontaneous \textbf{Solution:} Change in moles of gas: $\Delta n_g=2-(2+1)=-1$. First convert $\Delta U^{\circ }$ to $\Delta H^{\circ }$: $\Delta H^{\circ }=\Delta U^{\circ }+\Delta n_{g}RT=-10+(-1)(8.31)(298)/1000=-10-2.476=-12.476kJmo{l}^{-1}$. Then $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }=-12.476-298\times (-44)/1000=-12.476+13.112=+0.636kJmo{l}^{-1}$. Since $\Delta G^{\circ }>0$, the reaction is non-spontaneous at $298\text{K}$.