For the phase transition of liquid water to steam at 1 atm external pressure, the standard enthalpy — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

At the boiling point, the liquid and vapor phases exist in thermodynamic equilibrium, meaning $\Delta G = 0$. Since $\Delta G = \Delta H - T\Delta S$, setting this to zero yields $T = \Delta H / \Delta S = 40.63 \times 10^3 \text{ J/mol} / 108.8 \text{ J/K-mol} = 373.43 \text{ K}$.