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

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Answer :

Alleei Alleei · Answer 1 · 2023-11-08T02:24:56-05:00

Answer : The normal boiling point of ethanol will be, [tex]348.67K[/tex] or [tex]75.67^oC[/tex]

Explanation :

The Clausius- Clapeyron equation is :

[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]

where,

[tex]P_1[/tex] = vapor pressure of ethanol at [tex]30^oC[/tex] = 98.5 mmHg

[tex]P_2[/tex] = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

[tex]T_1[/tex] = temperature of ethanol = [tex]30^oC=273+30=303K[/tex]

[tex]T_2[/tex] = normal boiling point of ethanol = ?

[tex]\Delta H_{vap}[/tex] = 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:

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

[tex]T_2=348.67K=348.67-273=75.67^oC[/tex]

Hence, the normal boiling point of ethanol will be, [tex]348.67K[/tex] or [tex]75.67^oC[/tex]