# 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?

Feb 15, 2016

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, $\Delta {H}_{\text{vap}}$.

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 ${\text{8.314 J mol"^(-1)"K}}^{- 1}$

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

$\text{1 atm " = " 760 mmHg}$

Also, keep in mind that you must use absolute temperature, which is temperature expressed in Kelvin.

So, rearrange the equation to solve for $\Delta {H}_{\text{vap}}$

$\Delta {H}_{\text{vap}} = - \ln \left({P}_{1} / {P}_{2}\right) \cdot \frac{R}{\left(\frac{1}{T} _ 1 - \frac{1}{T} _ 2\right)}$

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)))))

$\Delta {H}_{\text{vap" = "30854.8 J mol}}^{- 1}$

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))