Question

The vapor pressure of liquid chloroform, CHCl3, is 400.0 torr at 24.1 °C and 100.0 torr at –6.3 °C. What is DeltaH of vaporization for chloroform?

Answer 1

Answer:

ΔH vaporization of chloroform is 30.1kJ/mol

Explanation:

It is possible to find ΔH of vaporization of certain compound knowing vapor pressure under 2 different absolute temperatures (In Kelvin) by using Clausius-Clapeyron equation:

$ln\frac{P_2}{P_1}=\frac{DeltaHvap}{R} (\frac{1}{T_1} - \frac{1}{T_2} )$

Where P is vapor pressure. R is gas constant (8.314J/molK) and T absolute temperature of 1, first state and 2, final state.

Absolute temperatures in the problem are:

T₁ = 24.1°C + 273.15 = 297.25K

T₂ = -6.3°C + 273.15 = 266.85K

Replacing:

$ln\frac{P_2}{P_1}=\frac{DeltaHvap}{R} (\frac{1}{T_1} - \frac{1}{T_2} )$

$ln\frac{100torr}{400.0torr}=\frac{DeltaHvap}{8.314J/molK} (\frac{1}{297.25K} - \frac{1}{266.85K} )$

$ln\frac{100torr}{400.0torr}={DeltaHvap}* -4.6x10^{-5}mol/J$

30073J/mol = 30.1kJ/mol = ΔHVap

ΔH vaporization of chloroform is 30.1kJ/mol