Question:
For a dimerization reaction, $2 \mathrm{~A}(\mathrm{~g}) \rightarrow \mathrm{A}_{2}(\mathrm{~g})$, at $298 \mathrm{~K}, \Delta \mathrm{U}^{\Theta}=-20 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{S}^{\Theta}=-30 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, then the $\Delta \mathrm{G}^{\Theta}$ will be ________________J.
Solution:
$(-13538)$
From $\Delta \mathrm{H}^{\circ}=\Delta \mathrm{U}^{\circ}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$
$\Delta \mathrm{H}^{\circ}=-20 \times 1000-1 \times 8.314 \mathrm{~J} / \mathrm{mol} . \mathrm{K} \times 298 \mathrm{~K}$
$=-22477.57 \mathrm{~J}$
$\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}=-22477.57-(298 \times-30)$
$=-13538 \mathrm{~J}$