One mole of ice is converted into water at 273 K and 1 atm. The entropies of and are 38.0 and 58.0 J — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

At the normal melting point (273 K, 1 atm), the process is in equilibrium, so $\Delta G = 0$. $\Delta G = \Delta H - T\Delta S \implies \Delta H = T\Delta S$. $\Delta S = 58.0 - 38.0 = 20.0\text{ J/K-mol}$. $\Delta H = 273\text{ K} \times 20.0\text{ J/K-mol} = 5460\text{ J/mol}$.