Question

The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling point of ethanol from this data.

Answer 1

Answer : The normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of ethanol at $30^oC$ = 98.5 mmHg

$P_2$ = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

$T_1$ = temperature of ethanol = $30^oC=273+30=303K$

$T_2$ = normal boiling point of ethanol = ?

$\Delta H_{vap}$ = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})$

$T_2=348.67K=348.67-273=75.67^oC$

Hence, the normal boiling point of ethanol will be, $348.67K$ or $75.67^oC$