Question

Find the heat of vaporization in kJ mol-1 for compound X. The compound is found to have a vapor pressure of 15.47 kPa at 10.0 °C and 22.25 kPa at 18.0 °C.

Answer 1

The enthalpy of vaporization could be calculated with using of the following expression:

$\ln{\frac{P_{2}}{P_{1}}}=-\frac{\Delta H}{R} (\frac{1}{T_{2}}-\frac{1}{T_{1}});$

Reorganizing the formula we should get the following:

$\Delta H=-\frac{R*ln{\frac{P_{2}}{P_{1}}}}{(\frac{1}{T_{2}}-\frac{1}{T_{1}})};$

$\Delta H=-\frac{8.31 \frac{J}{mol K} ln{\frac{22.25kPa}{15.47kPa}}}{(\frac{1}{291K}-\frac{1}{283K})}=31090 \frac{J}{mol};$

Thus, the heat of vaporization possesses the opposite sign or Q_vap. = -31.09 kJ/mol.