# What is the enthalpy of vaporization for a compound that has a vapor pressure of "135 torr" at 52^@ "C" and "24.3 torr" at 0^@ "C"?

May 22, 2015

Your compound's enthalpy of vaporization will be close to 24.3 kJ/mol.

To determine the enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, you can use the Clausius-Clapeyron equation. There are several equivalent forms in which you can write this equation, so I'll just use the one I remember

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

${P}_{1}$, ${P}_{2}$ - the vapor pressure at 273 K and 325 K, respectively;
$\Delta {H}_{\text{vap}}$ - the enthalpy of vaporization;
${T}_{1}$, ${T}_{2}$ - the two temperatures given to you.

So, plug in your values into the above equation and solve for $\Delta {H}_{\text{vap}}$

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

DeltaH_"vap" = (ln((135cancel("torr"))/(24.3cancel("torr"))) * 8.314"J"/("mol" * cancel("K")))/((1/273 - 1/325)cancel("K"^(-1))

$\Delta {H}_{\text{vap" = (1.7148 * 8.314"J"/"mol")/(0.0005861) = "24324.9 J/mol}}$

Rounded to three sig figs and expressed in kJ, the answer will be

DeltaH_"vap" ~~ color(green)("24.3 kJ/mol")

Note that this is only an estimate, under the assumption that in the pressure and temperature range examined, the enthalpy of vaporization is essentially linear, but that is not always the case.

(3rd paragraph here)
http://en.wikipedia.org/wiki/Enthalpy_of_vaporization