The molar enthalpy of vaporization of benzene at its normal boiling point (80.09

Question

The molar enthalpy of vaporization of benzene at its normal boiling point (80.09°C) is 30.72 kJ/mol. The molar heat capacity of liquid benzene is 136.3 J/molK and does not vary with temperature. The molar heat capacity of gaseous benzene is 82.5 J/molK and also does not vary with temperature. However, both the enthalpy of vaporization and the entropy of vaporization of benzene do vary with temperature. With the above information, determine the molar Gibbs free energy of vaporization of benzene at 75.0°C, at 80.09°C and at 85°C.

Explanation / Answer

Molar enthalpy of vaporization (Hvap,m) = 30.72 kJ/mol = 30720 J/mol

Molar heat capacity of liquid benzene (Cp,l) = 136.3 J/molK

Molar heat capacity of gaseous benzene (Cp,g) = 82.5 J/molK

Molar entropy of vaporization (Svap,m) = Cp,l - Cp,g = 136.3 J/molK - 82.5 J/molK = 53.8 J/molK

T1 = 75.0°C = 75.0 + 273.15 = 348.15 K

T2 = 80.09°C = 80.09 + 273.15 = 353.24 K

T3 = 85°C = 85 + 273.15 = 358.15 K

Molar Gibbs free energy of vaporization, Gvap,m = Hvap,m T Svap,m

1. Molar Gibbs free energy of vaporization at T1, Gvap,m = Hvap,m T Svap,m

= 30720 J/mol - (348.15 K)(53.8 J/molK) = 11989.53 J/mol = 11.99 kJ/mol

2. Molar Gibbs free energy of vaporization at T2, Gvap,m = Hvap,m T Svap,m

= 30720 J/mol - (353.24 K)(53.8 J/molK) = 11715.69 J/mol = 11.72 kJ/mol

3. Molar Gibbs free energy of vaporization at T3, Gvap,m = Hvap,m T Svap,m

= 30720 J/mol - (358.15 K)(53.8 J/molK) = 11451.53 J/mol = 11.45 kJ/mol

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