Q. 12.5.1

a. Use the data in Table 12.5.1 to approximate the enthalpy of sublimation for water. Why is this just an approximation?
b. At 298 K, the standard heat of formation of H$_{2}$O(g) is -241.8 kJ/mol, whereas the standard heat of formation of H$_{2}$O(ℓ) is -285.8 kJ/mol. Use this to calculate the heat of vaporization of water and compare your result with the value in Table 12.5.1.

 Table 12.5.1 Selected Heats of Fusion and Vaporization at the Temperature of the Normal Phase Transition ΔH$_{fusion}$ (kJ/mol) Melting Point (°C) ΔH$_{vap}$ (kJ/mol) Boiling Point (°C) Methane 0.94 -182.5 8.2 -161.6 Ethane 2.86 -182.8 14.7 -88.6 Propane 3.53 -187.6 19.0 -42.1 Methanol 3.16 -97.0 35.3 64.7 Methanol 5.02 -114.3 38.6 78.4 1-Propanol 5.20 -127 41.4 97.2 Water 6.01 0.0 40.7 100.0 Water 2.60 97.82 97.42 881.4 Water 26.11 755 160.7 1390

Verified Solution

You are asked to approximate the enthalpy of sublimation of water and to calculate the heat of vaporization of water.
You are given enthalpies of fusion and vaporization for water and the standard heat of formation for liquid and gaseous water.
a. ΔH$_{sub}$ = ΔH$_{fus}$ + ΔH$_{vap}$ = 6.01 kJ/mol + 40.7 kJ/mol = 46.7 kJ/mol
For water, the heat of fusion is given at 0 °C and the heat of vaporization is given at 100 °C. Because ΔH does vary somewhat with temperature, the calculated value for ΔH$_{sub}$ is only an approximation.
b. H$_{2}$O(ℓ) → H$_{2}$O(g) ΔH° = -241.8 kJ/mol – (-285.8) kJ/mol = 44.0 kJ/mol
The value at 25 °C is larger than the value at 100 °C (40.7 kJ/mol). At the higher temperature, the molecules in the liquid have a higher average energy and are more gaslike (farther apart and less short range order), so less energy should be needed for the conversion.