Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 272.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid. Calculate the normal boiling point of this liquid.

Answer 1

Vapor pressure = 92.0 torr , Temperature = 23.0 ⁰C = 23 + 273. 15 = 293.15 K

And at T = 45.0 ⁰C = 45.0 + 273.15 = 318.15 K , vapor pressure = 272.0 torr

Solution:

Here we use classius clayperon equation

ln P2/ P1 = - Delta H / R ( 1 / T2 – 1/ T1)

Here P2 is the vapor pressure of liquid at T2 , P1 is the vapor pressure at T1.

Delta H is the heat of vaporization. R is gas constant = 8.314 J / ( K mol)

Lets use given value to find out Delta Hvap

Ln ( 272.0 /92.0) = - Delta Hvap / 8.314 ( 1/318.15 – 1/293.15)

1.084 = - Delta Hvap * ( -0.00026805)

Delta Hvap = 4044.05 J / mol