Question

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 1

(a) The value of
`\Delta H_(vap)` 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
`\Delta H_(vap)` of the reaction, we use clausius claypron equation, which is:

`\ln((P_2)/(P_1))=(\Delta H_(vap))/(R)[(1)/(T_1)-(1)/(T_2)]`

where,

`P_1` = vapor pressure at temperature
`85.8^oC` = 1.3 kPa

`P_2` = vapor pressure at temperature
`119.3^oC` = 5.3 kPa

`\Delta H_(vap)` = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

`T_1` = initial temperature =
`85.8^oC=[85.8+273]K=358.8K`

`T_2` = final temperature =
`119.3^oC=[119.3+273]K=392.3K`

Putting values in above equation, we get:

`\ln((5.3kPa)/(1.3kPa))=(\Delta H_(vap))/(8.314J/mol.K)[(1)/(358.5)-(1)/(392.3)]\\\\\Delta H_(vap)=48616.4J/mol=48.6kJ/mol`

Therefore, the value of
`\Delta H_(vap)` is 48.6 kJ/mol

(b) The clausius claypron equation is:

`\ln((P_2)/(P_1))=(\Delta H_(vap))/(R)[(1)/(T_1)-(1)/(T_2)]`

where,

`P_1` = vapor pressure at temperature
`85.8^oC` = 1.3 kPa

`P_2` = vapor pressure at temperature normal boiling point = 101.3 kPa

`\Delta H_(vap)` = Enthalpy of vaporization = 48.6 kJ/mol

R = Gas constant =
`8.314* 10^(-3)kJ/mol.K`

`T_1` = initial temperature =
`85.8^oC=[85.8+273]K=358.8K`

`T_2` = final temperature = ?

Putting values in above equation, we get:

`\ln((101.3kPa)/(1.3kPa))=(48.6kJ/mol)/(8.314* 10^(-3)kJ/mol.K)[(1)/(358.5)-(1)/(T_2)]\\\\T_2=489.2K`

Therefore, the normal boiling point is 489.2 K

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

`\Delta S_(vap)=(\Delta H_(vap))/(T_b)`

where,

`\Delta S_(vap)` = entropy of vaporization = ?

`\Delta H_(vap)` = enthalpy of vaporization = 48.6 kJ/mol

`T_b` = boiling point = 489.2 K

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

`\Delta S_(vap)=(48.6kJ/mol)/(489.2K)=99.3J/K`

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