Exercise 5 | Chemical Thermodynamics

General Chemistry 3 - Exercise 5

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 = HfusTm

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 = 6.01 × 103273 = 22.0 J.mol-1.K-1

 

Entropy change on vaporization:
 

ΔSvap = HvapTb

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 = 40.65 × 103373 = 109 J.mol-1.K-1