Question

A substance has a normal boiling point of 125.5 oC and a vapor pressure of 35.5 torr at 20.0 oC. Based on this data, calculate the heat of vaporization (in kJ/mol) for this substance.

this the second time I am posting this question, the first answer was wrong. Please help me with the correct answer.

Answer 1

clasius-clepyron equation

ln(P2/P1) = DH0rxn/R[1/T1 - 1/T2]

P2 = 760 torr(atmospheric pressure) , T2 = actual boiling point = 125.5+273.15 = 398.65 k

P1 = 35.5 torr, T1 = 20+273.15 = 293.15 k

DHvap = x kj/mol

ln(760/35.5) = (x/(8.314*10^-3))((1/293.15)-(1/398.65))

x = 28.22

DHvap = heat of vaporization = x = 28.22 kj/mol