Question

Consider the phase diagram of NH Triple point: 195.410 K and 0.06921 atm Normal boiling point:-33.342 PC Normal freezing point: -77.728 °C Density of liquid: 0.618g/mL Density of solid: 0.817 g/mL (a) Determine the heat of fusion, the heat of vaporization the heat of sublimation at near the triple point. (b) Determine the entropy change of fusion, vaporization and sublimation at near the triple point. (c) Plot the phase diagram in the temperature range T: [-90 °C, 130 °C]. For pressure (in unit of atm), use In (p) other than p.

Answer 1

The standard entropy of fusion of ammonia ($\Delta$S_fus) = 28.93 J/(mol·K)

The standard entropy of vaporization of ammonia ($\Delta$S_vap) = 97.41 J/(mol·K)

(a) The heat of fusion ($\Delta$H_fus) = $\Delta$S_fus * freezing point = 28.93 J/(mol·K) * (-77.728 + 273.15) K = 5653.56 J/mol or 5.654 kJ/mol

The heat of vaporization ($\Delta$H_vap) = $\Delta$S_vap * boiling point = 97.41 J/(mol·K) * (-33.342 + 273.15) K = 23359.70 J/mol or 23.36 kJ/mol

The heat of sublimation ($\Delta$H_sub) = ($\Delta$H_fus) + ($\Delta$H_vap) = 5.654 + 23.36 = 29.014 kJ/mol