Question

A certain liquid has a vapor pressure of 92.0Torr at 23.0∘C and 273.0 Torr at 45.0∘C. Calculate the value of ΔH∘vapm for this liquid in kj/mol. Calculate the normal boiling point of this liquid boiling point ∘C

Answer 1

T1 = 23 + 273 = 296K

P1 = 92 torr

T2 = 45 + 273 = 318K

P2 = 273 torr

ln (P2 / P1 ) = ΔH∘vap / R [1 / T1 - 1/ T2 ]

ln (273 / 92) = ΔH∘vap / (8.314 x 10^-3) [1/296 - 1/ 318]

ΔH∘vap   = 38.7 kJ / mol

at normal boiling point : P2 = 760 torr

                                      T2 = ?

ln (P2 / P1 ) = ΔH∘vap / R [1 / T1 - 1/ T2 ]

ln (760 / 273) = 38.7 / (8.314 x 10^-3) [1 / 318 - 1/ T2 ]

2.20 x 10^-4 = [1 / 318 - 1/ T2 ]

T2 = 342 K

normal boiling point temperature = 68.9 ^oC