For the phase transition of liquid water to steam at external pressure, the standard enthalpy of vap — Thermodynamics and Thermochemistry Chemistry Question
Question
For the phase transition of liquid water to steam at $1 \text{ atm}$ external pressure, the standard enthalpy of vaporization ($\Delta_{vap}H$) is $40.63 \text{ kJ mol}^{-1}$ and the corresponding entropy of vaporization ($\Delta_{vap}S$) is $108.8 \text{ J K}^{-1}\text{mol}^{-1}$. Determine the precise equilibrium temperature in Kelvin at which the Gibbs free energy change ($\Delta G$) for this physical transformation is exactly zero. (Round the answer to one decimal place).
💡 Solution & Explanation
At the boiling point, the liquid and vapor phases exist in thermodynamic equilibrium, meaning the Gibbs free energy change $\Delta G = 0$. By definition, $\Delta G = \Delta H - T\Delta S$. Setting this to zero yields the relation $T = \frac{\Delta H}{\Delta S}$. Converting enthalpy to compatible units (Joules): $T = \frac{40.63 \times 10^3 \text{ J mol}^{-1}}{108.8 \text{ J K}^{-1} \text{mol}^{-1}} = 373.4375 \text{ K}$. Rounded to one decimal place, this is $373.4 \text{ K}$.