Question

Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 55.8°C, the vapor pressure of CCl4 is 52.7 kPa, and its enthalpy of vaporization is 29.82 kJ/mol. Use this information to estimate the normal boiling point (in °C) for CCl4.

Answer 1

Use the Clausius-Clapeyron equation... 

Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.

$ln(P_1/P_2) = \frac{-\delta H_{vap}}{R} \times [\frac{1}{T_1} - \frac{1}{T_2}]$

ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K) 

------ some algebra goes here ----- 

T = 349.99K ...... or ...... 76.8C