A certain liquid has a vapor pressure of 92.0 Torr at 23.0 ∘C and 283.0 Torr at 45.0 ∘C. Calculate the value of Δ𝐻∘vap for this liquid.

Question

Calculate the normal boiling point of this liquid.

Answer 1

Clausius Clapeyron Equation:

ln(P2/P1) = -∆Hvap/R (1/T2 - 1/T1)

P1 = 92.0 torr

T1 = 23 + 273 = 296K

P2 = 283.0 torr

T2 = 45 + 273 = 318K

R = 8.314 J/Kmol

∆Hvap = ?

ln(283/92) = -∆Hvap/8.314 (1/318 - 1/296)

1.12 = -∆Hvap/8.314 (-0.00024)

0.00024∆H = 9.32

∆H = 38,799 J/mol = 38.8 kJ/mol

Normal boiling point suggests a pressure of 1 atm = 760 torr.

Repeat the calculation using any of the above data of P1 and T1 and then use P2 = 760 torr and solve for T2

ln(760/92) = -38,800/8.314 (1/T2 - 1/296)

2.11 = -4667 (1/T2 - 0.0034)

2.11 = -4667/T2 + 15.77

13.66 = 4667/T2

T2 = 341.7K = 68.5ºC = normal boiling point