Question

Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 58.2°C, the vapor pressure of CCl4 is 57.0 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

$$ln\frac{P_1}{P_2}=\frac{\Delta H_{vap}}{R}\times(\frac{1}{T_2}-\frac{1}{T_1})$$

$T_1(K) = 273.15+ 58.2 = 331.35 K$

$P_1 = 57.0 kPa$

$\Delta H_{vap} = 29.82 kJ/mol = 29820 J/mol$

$P_2 = 101.325 kPa$

$R= 8.314 \frac{J}{mol\times K}$

$$ln\frac{57.0}{101.325} =\frac{29820}{8.314} (\frac{1}{T_2}-\frac{1}{331.35})$$

$$T_2= 349.9 K$$

$$T_2(^\circ C) = 349.9-273.15 = 76.75 ^\circ C$$