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
Feb 15, 2016

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, #DeltaH_"vap"#.

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

#P_1# - the vapor pressure of the liquid at a temperature #T_1#
#P_2# - the vapor pressure of the liquid at a temperature #T_2#
#R# - the universal gas constant, given in this contex as #"8.314 J mol"^(-1)"K"^(-1)#

Now, it's important to realize that the normal boiling point of a substance is measured at an atmoshperic pressure of #"1 atm"#. You can express this in mmHg by using the conversion factor

#"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"#

#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))#