Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 377.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid.

Calculate the normal boiling point of this liquid.

Answer 1

A)
P1 = 92 torr
T1 = 23.0 oC = (23 + 273.1) K = 296.1 K

P2 = 377.0 torr
T2 = 45.0 oC = (45 + 273.1) K = 318.1 K

use:
ln (P2/P1) = (ΔH°vap /R) * (1/T1 - 1/T2)
ln (377/92) = (ΔH°vap/8.314) *(1/296.1 - 1/318.1)
ΔH°vap = 50205 J/mol
= 50.205 KJ/mol

B)
P1 = 92 torr
T1 = 23.0 oC = (23 + 273.1) K = 296.1 K

At boiling point, P = 760 torr
P2 = 760 torr
T2 = ?

use:
ln (P2/P1) = (ΔH°vap /R) * (1/T1 - 1/T2)
ln (760/92) = (50205/8.314) *(1/296.1 - 1/T2)
(1/296.1 - 1/T2) = 3.5*10^-4
T2 = 330.3 K
= (330.3 - 273.1) oC
= 57.2 oC

Answer: 57.2 oC