Question

4. Estimate the heat of vaporization (kJ/mol) of benzene at 25°C, using each of the following correlations and data: a) Trouton’s rule and the normal boiling point b) The heat of vaporization at the normal boiling point and Watson's correlation. c) The Clausius-Clapeyron equation and boiling points at 50 mm Hg and 150 mm Hg. d) Tables B.1 and B.2. Clearly specify the thermodynamic path you are following. e) Find a tabulated value of the heat of vaporization of benzene at 25°C. (Suggestion: Do the same thing you do when you want to find almost any item of information.) Then calculate the percentage errors that result from the estimations of Parts a, b, c, and d.

Answer 1

(a)

Normal boiling point of benzene is 353.2 K

Heat of Vaporization of benzene at the normal boiling point is 30720 J/mol

Watson's equation

H_v1 = Heat of vaporization at T₁ , J/mol

H_v2 = Heat of vaporization at T2 , J/mol

T₁ = Temperature, K

T2 = Temperature, K

Tc = Critical Temperature, K

Let T₁ = 353.2 K, H_v1 = 30720 J/mol, T_c = 562.1 K, T₂ = 25+273.15 = 298.15 K

Substitute these values in the above equation and Solve for H_v2

H_v2 = 33575.53 J/mol

(b)

Clausius-Clapeyron equation

R - Gas constant

P₁ - Liquid vapor Pressure, mmHg

P2 - Liquid vapor Pressure, mmHg

T₁ = Boiling Temperature at P₁, °C

T2 = Boiling Temperature at P2, °C

Boiling temperature can be calculated using antoine equation. We get T₁ = 11.84 °C, T₂ = 35.25 °C

P₁=50 mmHg, P₂=150 mmHg, R=8.314 J/mol K

By substituting these values, we get

Hv = 162.84 J/mol