Question

At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.

Short answer

Enthalpy of vaporization of ethanol is calculated to be 41.32kJ/mol.

Step-by-step solution

Define the vapour pressure  

The enthalpy of vaporization is basically defined as the amount of energy required by a liquid to transform some quantity of liquid into its gaseous form. This quantity is a function of pressure. To solve the given question, the following expression can be used,

\(\ln \left( {\frac{{{{\rm{P}}_2}}}{{{{\rm{P}}_1}}}} \right) = \frac{{ - \Delta {{\rm{H}}_{{\rm{vap}}}}}}{{\rm{R}}}\left( {\frac{1}{{\;{{\rm{T}}_2}}} - \frac{1}{{\;{{\rm{T}}_1}}}} \right)\)

 Identify the enthalpy of vaporization for ethanol.

P₁ and P₂ represent vapor pressure and T₁ and T₂ represent temperatures. \(R\) is the gas constant that takes the value of \(8.314\,{\rm{J}}/\) mol.K. Using values from the question and taking the temperatures in kelvin scale,

\(\begin{aligned}{}\ln \left( {\frac{{{{\rm{P}}_2}}}{{{{\rm{P}}_1}}}} \right) = \frac{{ - \Delta {{\rm{H}}_{{\rm{vap}}}}}}{{\rm{R}}}\left( {\frac{1}{{\;{{\rm{T}}_2}}} - \frac{1}{{\;{{\rm{T}}_1}}}} \right)\\\ln \left( {\frac{{53.5}}{{5.95}}} \right) = \frac{{ - \Delta {H_{{\rm{vap }}}}}}{R}\left( {\frac{1}{{336.5}} - \frac{1}{{293}}} \right)\end{aligned}\)

\(\begin{aligned}{}\ln (8.957983) = \frac{{ - \Delta {{\rm{H}}_{{\rm{vap }}}}}}{{\rm{R}}}( - 0.0004412)\\\Delta {{\rm{H}}_{{\rm{vap}}}} = \frac{{2.19255{\rm{R}}}}{{0.0004412}}\\ = 41316.547\;{\rm{J}}/{\rm{mol}}\\ = 41.32\;{\rm{kJ}}/{\rm{mol}}\end{aligned}\)

Hence, the enthalpy of vaporization of ethanol is calculated to be \(41.32\;{\rm{kJ}}/{\rm{mol}}\)