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?

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))`