Naphthalene, c 10 h 8 , melts at 80.2°c. if the vapour pressure of the liquid is 1.3 kpa at 85.8°c and 5.3 kpa at 119.3°c, use the clausius–clapeyron equation to calculate (a) the enthalpy of vaporization, (b) the normal boiling point, and (c) the enthalpy of vaporization at the boiling point.

Answer :

(a) The value of is 48.6 kJ/mol

(b) The the normal boiling point is 489.2 K

(c) The entropy of vaporization at the boiling point is 99.3 J/K

Explanation :

(a) To calculate of the reaction, we use clausius claypron equation, which is:

where,

= vapor pressure at temperature = 1.3 kPa

= vapor pressure at temperature = 5.3 kPa

= Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

= initial temperature =

= final temperature =

Putting values in above equation, we get:

Therefore, the value of is 48.6 kJ/mol

(b) The clausius claypron equation is:

where,

= vapor pressure at temperature = 1.3 kPa

= vapor pressure at temperature normal boiling point = 101.3 kPa

= Enthalpy of vaporization = 48.6 kJ/mol

R = Gas constant =

= initial temperature =

= final temperature = ?

Putting values in above equation, we get:

Therefore, the normal boiling point is 489.2 K

(c) Now we have to determine the entropy of vaporization at the boiling point.

where,

= entropy of vaporization = ?

= enthalpy of vaporization = 48.6 kJ/mol

= boiling point = 489.2 K

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

Therefore, the entropy of vaporization at the boiling point is 99.3 J/K

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