The vapor pressure of dichloromethane, #"CH"_2"Cl"_2# at #0^@"C"# is 134 mmHg. The normal boiling point of dichloromethane is #40^@"C"#. How would you calculate its molar heat of vaporization?
1 Answer
You would use the Clausius - Clapeyron equation.
Explanation:
The idea here is that you can use the Clausius - Clapeyron equation to estimate the vapor pressure of a liquid at a given temperature provided that you know the vapor pressue of the liquid at another temperature and its enthalpy of vaporization,
Consequently, if you know the vapor pressure of a liquid at two different temperatures, you can use the Clausius - Clapeyron equation to find is enthalpy of vaporization.
The Clausius - Clapeyron equation looks like this
#color(blue)(ln(P_1/P_2) = -(DeltaH_"vap")/R * (1/T_1 - 1/T_2)" "# , where
Now, it's important to realize that the normal boiling point of a substance is measured at an atmoshperic pressure of
#"1 atm " = " 760 mmHg"#
Also, keep in mind that you must use absolute temperature, which is temperature expressed in Kelvin.
So, rearrange the equation to solve for
#DeltaH_"vap" = - ln(P_1/P_2) * R/((1/T_1 - 1/T_2))#
Plug in your values to get
#DeltaH_"vap" = -ln((134 color(red)(cancel(color(black)("mmHg"))))/(760color(red)(cancel(color(black)("mmHg"))))) * (8.314"J mol"^(-1)color(red)(cancel(color(black)("K"^(-1)))))/((1/((273.15 + 0)) - 1/((273.15 + 40)))color(red)(cancel(color(black)("K"^(-1)))))#
#DeltaH_"vap" = "30854.8 J mol"^(-1)#
I'll express the answer in kilojoules per mole and leave it rounded to one sig fig
#DeltaH_"vap" = color(green)("30 kJ mol"^(-1))#