Question

Estimate the heat of vaporization (kJ/mol) of benzene at a pressure of 120 mm Hg, using each of the following correlations and data: (a) The heat of vaporization at the normal boiling point given in Table B.1, the boiling point at 120 mm Hg as determined from the Antoine equation, and Watson’s correction. Provide your answer with three significant figures.

Answer 1

for Antoine equation

log10(P) = A − (B / (T + C))

constants --> 4.72583 1660.652 -1.461

P = bar, T = K

log(P) = 4.72583 − 1660.652  / (T + -1.461)

for

P = 120 mm Hg = 0.159987 bar

log(0.159987 ) = 4.72583  − 1660.652  / (T + -1.461)

T = -1660.652 / ( log(0.159987 ) -4.72583 ) +1.461

T = 302.2 K

use..

Watson correction

Hv2/Hv1 = ( (Tc-T2)/(Tc-T1))^0.38

Tc = 562.0 K

Hv1 = 33.9 kJ/mol at T1 = 25°C or 298K

Hv2 = ? kJ/mol at T1 = 80.1 °C or 353.1 K

Hv2/33.9 = ( (562.0 -302.2 )/(562.0 -298))^0.38

Hv2 = 33.9*( (562.0 -302.2 )/(562.0 -298))^0.38

Hv2 = 33.6940 kJ/mol --> 3 sig fig --> 33.7 kJ/mol