The molar enthalpy of fusion of water is ΔH_{fus} = 6.01\ kJ·mol^{–1} at 0°C, and the molar enthalpy of vaporization of water is ΔH_{vap} = 40.65\ kJ·mol^{–1} at 100°C. Calculate the values of ΔS_{fus} and ΔS_{vap} for water at these two phase transitions at a constant pressure of one atm.
The value of ΔS_{fus} is calculated using Equation 23.8. The melting point of water is 0°C, or 273.15 K; so we have
ΔS_{fus}=\frac{∆H_{fus}}{T_{m}} =\frac{6.01 × 10^3\ J·mol^{–1}}{273.15\ K} = 22.0\ J·K^{–1}·mol^{–1}The value of ΔS_{vap} is calculated using Equation 23.9, with T_{b} = 373.15\ K:
ΔS_{vap}=\frac{∆H_{vap}}{T_{b}} =\frac{40.65× 10^3\ J·mol^{–1}}{373.15\ K} = 108.9\ J·K^{–1}·mol^{–1}Notice that both ΔS_{fus} and ΔS_{vap} are positive, and \Delta S_{vap} > \Delta S_{fus}.