# 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

#### 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

*universal gas constant*, given in this contex as

Now, it's important to realize that the **normal boiling point** of a substance is measured at an atmoshperic pressure of *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" = - 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))#