Question

The molar enthalpy of vaporization of hexane (C6H14) is 28.9 kJ/mol, and its normal boiling point is 68.7 degrees Celsius. What is the vapor pressure of hexane in mmHg at 25 degrees Celsius

Answer 1

Using Clausius-Clapeyron Equation:

ln (P2/P1) = (?Hvap / R)[(1/T1) - (1/T2)]

where:
P1 = 1 atm = 760 torr
T1 = 68.73°C = 341.88 K
T2 = 25°C = 298.15 K

P2 = (760 torr)e^(28.9 x 10^3 J/mol / 8.314 J/molK)[(1/341.88 K) - (1/298.15 K)]
P2 = 171 torr