At 298 K, the standard enthalpy of combustion of sucrose is and the standard Gibb's energy of the re — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

At 298 K: $\Delta G = \Delta H - T\Delta S \implies -6333 = -5737 - 298\Delta S \implies \Delta S = \frac{5737 - 6333}{-298} = 2\text{ kJ/K-mol}$. The maximum non-expansion work obtainable is $-\Delta G$. At $29^\circ\text{C}$ (302 K): $\Delta G_{302} = \Delta H - T_2\Delta S = -5737 - 302(2) = -6341\text{ kJ/mol}$. Work at 298 K $= 6333\text{ kJ}$. Work at 302 K $= 6341\text{ kJ}$. Additional work $= 6341 - 6333 = 8\text{ kJ}$.