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:
[tex]ln\frac{P_2}{P_1}=\frac{DeltaHvap}{R} (\frac{1}{T_1} - \frac{1}{T_2} )[/tex]
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:
[tex]ln\frac{P_2}{P_1}=\frac{DeltaHvap}{R} (\frac{1}{T_1} - \frac{1}{T_2} )[/tex]
[tex]ln\frac{100torr}{400.0torr}=\frac{DeltaHvap}{8.314J/molK} (\frac{1}{297.25K} - \frac{1}{266.85K} )[/tex]
[tex]ln\frac{100torr}{400.0torr}={DeltaHvap}* -4.6x10^{-5}mol/J[/tex]
30073J/mol = 30.1kJ/mol = ΔHVap
