# Exercise 5 | Chemical Thermodynamics

Calculate the values of the molar entropy of fusion and vaporization for water at these two phase transitions at a constant pressure = 1 atm.

Data: ΔHfus [H2O] = 6.01 kJ.mol-1

ΔHvap [H2O] = 40.65 kJ.mol-1

Entropy change on fusion:

ΔSfus = $\frac{∆{\mathrm{H}}_{\mathrm{fus}}}{{\mathrm{T}}_{\mathrm{m}}}$

ΔSfus = molar entropy of fusion (in J.K-1.mol-1)

ΔHfus = molar enthalpy of fusion (in J.mol-1)

Tm = melting point (in K) = 273 K for water

ΔSfus =  = 22.0 J.mol-1.K-1

Entropy change on vaporization:

ΔSvap = $\frac{∆{\mathrm{H}}_{\mathrm{vap}}}{{\mathrm{T}}_{\mathrm{b}}}$

ΔSvap = molar entropy of vaporization (in J.K-1.mol-1)

ΔHvap = molar enthalpy of vaporization (in J.mol-1)

Tb = boiling point (in K) = 373 K for water

ΔSvap =  = 109 J.mol-1.K-1