Calculate the entropy change accompanying conversion of 1 mole of ice at 273 K and 1 atm pressure into steam at 373 K and 1 atm pressure, given that at 273 K, the molar heat of fusion of ice is 6 kJ/mol and at 373 K, the molar heat of vapourization of water is 40.60 kJ/mol. Also assume that the molar heat capacity in the temperature range 373-273 K remains constant as 75.2 J/K per mol.
Answers
The entropy change accompanying conversion of 1 mole of ice at 273 K and 1 atm pressure into steam at 373 K and 1 atm pressure, given that at 273 K, the molar heat of fusion of ice is 6 kJ/mol and at 373 K, the molar heat of vapourization of water is 40.60 kJ/mol is given as follow:
- The total entropy is given by the relation.
- ∆S = ∆S1+∆S2+∆S
Where ∆S1= Entropy change due to fusion of ice
∆S2= Entropy change for heating of 1 mole of water from 273K to 373K
∆S3= Entropy change due to conversion of water at 373K itno steam
3. Thus
4. ∆S1 = ∆Hf / ∆Tf = 6 KJ(mol)^-1/ 273K
= 21.9780 J K^-1 mol^-1
5. ∆S2 = ∫(Cp/ T)dT from 273K to 373K = ∫(75.22/T)dT with same limits
= 75.22 J K^-1mol^-1 × ln(373/273)
= 23.4728 J K^-1mol^-1
6. ∆S3 = ∆Hv/Tv = 40.60 KJ mol^-1/ 373K = 108.8471 J K^-1 mol^-1
7. ∆S = ∆S1+∆S2+∆S3
=(21.9780+25.4728+108.8471)JK^-1mol^-1
= 156.2979 J K^-1 mol^-1