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 ((P_2)/(P_1))=(\Delta H_(vap))/(R)* ((1)/(T_1)-(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 ((760mmHg)/(98.5mmHg))=(39300J/mole)/(8.314J/K.mole)* ((1)/(303K)-(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`