Consider the following reaction: $\ce{2A(g) + B(g) -> 2D(g)}$ $\Delta U^\circ = -10\ kJ\,mol^{-1}$ and $\Delta S^\circ = -44\ J\,K^{-1}$ 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: $R = 8.31\ J\,mol^{-1}K^{-1}$)

Solution & Explanation

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