Question

The melting point of water at the pressure of interest is 0.00 ℃, and the enthalpy of fusion is 6.010 kJ⋅mol-1. The boiling point is 100. ℃, and the enthalpy of vaporization is 40.65 kJ⋅mol−1. Calculate ΔS for the transformation of the same amount of water.

Homework Answers

Answer #1

Given, the melting point of water, Tf = 0o C = 273 K ; enthalpy of fusion of water , H fus = 6.010 kJ mol-1

Therefore, entropy of fusion of ice, Sfus = Hfus/ Tf = 6.010 kJmol-1/ 273 K = 0.0220 kJmol-1K-1

Considering one mole of water; Sfus = 0.0220 kJ K-1

Again. given that boiling point, Tv = 100oC = 373 K ; enthalpy of vapourisation of water, Hvap = 40.65 kJmol-1

Therefore, entropy of vapourisation of ice, Svap = Hvap / Tv = 40.65 kJmol-1 / 373 K = 0.1089 kJmol-1K-1

Considering one mole of water; Svap = 0.1089 kJ K-1

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