Question

The standard enthalpy of fusion of water is 6.01 kJ/ mol, the molar heat capacity at a constant pressure of liquid water can be assumed to be constant at 75.5 J/mol*K, thoughout the liquid phase, and the standard enthalpy of vapoziration is 40.7 kJ/mol at the boiling point. Calculate the total entropy change (in SI units), for heating one mole of water, starting with ice just below the melting point and going to just above the boiling point at standard pressure.

Answer 1

                  ∆S1                   ∆S2                  ∆S3           ∆S4              ∆S5
1 mol H2O(S) -1^oC ----> H2O(S) 0^oC ----> H2O(l) 0^oC -----> H2O(l) 100 ^oC -----> H2O(g) 100^oC ---> H2O(g) 101^oC

∆S₁ = n.Cp.ln(T_f/T_i) = (1mol x 36.9J/mol.K)ln (273.15/272.15) = 0.135 J/K

∆S₂ = n.∆H_fus/T = (1molx 6.01 x1000 J)/ mol/273.15 K = 21.966 J/K

∆S₃ = n.C_pm.ln(T_f/T_i) = 1mol x 75.5 J/mol*K ln (373.15/273.15) = 23.553 J/K

∆S4 = n.∆H_vap/T = (1 mol x 40.7 x1000 J/mol)/373.15 K = 109.07 J/K

∆S4 = n.Cp.ln(Tf/Ti) = (1mol x 36.5 J/mol.K)ln(374.15/373.15) = 0.977 J/K

Total entropy ∆S = ∆S1 + ∆S2 + ∆S3 + ∆S4 + ∆S4
               = 0.135 J/K+ 21.966 J/K + 23.553 J/K + 109.07 J/K + 0.977 J/K
               = 155.7 J/K
              
Therefore, the total entropy change = 155.7 J/K