Heat of vaporization help?

Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 ∘C and 366.0 Torr at 45.0 ∘C.

Calculate the value of Δ𝐻∘vap for this liquid and the boiling point.

Answer 1

Clausius-Clapeyron Equation:

ln(P2/P1) = - ∆Hvap/R (1/T2 - 1/T1)

P1 = 92.0 torr

P2 = 366.0 torr

T1 = 23C + 273 = 296K

T2 = 45C + 273 = 318K

R = 8.314 J/K-mol

Solve for ∆Hvap

ln(366/92) = -∆Hvap/8.314 (1/318 - 1/296)

1.38 = -∆Hvap/8.314 (0.003145 - 0.003378)

1.38 = 0.000233∆Hvap/8.314

∆Hvap = 49,242 J/mol = 49.2 kJ/mol

Normal boiling point would be when atmospheric pressure is 1 atm or 760 torr...

P1 = 92 torr

P2 = 760 torr

T1 = 296K

T2 = ?

∆Hvap = 49.2 kJ/mol

R = 0.008314 kJ/Kmol (note changed units to match those of ∆Hvap)

Solve for T2:

ln(760/92) = -49.2/0.008314 (1/T2 - 1/296)

2.11 = -5918 (1/T2 - 0.003378)

2.11 = -5918/T2 + 19.99

T2 = 331K = 58ºC