Question

What is the heat of vaporization of ethanol, given that ethanol has a normal boiling point of 63.5°C and the vapor pressure of ethanol is 1.75 atm at 780 g.

Answer 1

Explanation:

According to Clausius-Claperyon equation,

       $ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

The given data is as follows.

         $T_{1} = 63.5^{o}C$ = (63.5 + 273) K

                         = 336.6 K

        $T_{2} = 78^{o}C$ = (78 + 273) K

                         = 351 K

         $P_{1}$ = 1 atm,             $P_{2}$ = ?

Putting the given values into the above equation as follows.    

        $ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

       $ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}]$

                      $\Delta H$ = $\frac{0.559}{1.466 \times 10^{-4}} J/mol$

                                  = $0.38131 \times 10^{4} J/mol$

                                  = 3813.1 J/mol

Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.