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

Answer :

(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:

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

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:

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

Therefore, the value of $\Delta H_{vap}$ is 48.6 kJ/mol

(b) The clausius claypron equation is:

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

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 = $8.314\times 10^{-3}kJ/mol.K$

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

$T_2$ = final temperature = ?

Putting values in above equation, we get:

$\ln(\frac{101.3kPa}{1.3kPa})=\frac{48.6kJ/mol}{8.314\times 10^{-3}kJ/mol.K}[\frac{1}{358.5}-\frac{1}{T_2}]\\\\T_2=489.2K$

Therefore, the normal boiling point is 489.2 K

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

$\Delta S_{vap}=\frac{\Delta H_{vap}}{T_b}$

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:

$\Delta S_{vap}=\frac{48.6kJ/mol}{489.2K}=99.3J/K$

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