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:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
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 = ![85.8^oC=[85.8+273]K=358.8K](https://tex.z-dn.net/?f=85.8%5EoC%3D%5B85.8%2B273%5DK%3D358.8K)
= final temperature = ![119.3^oC=[119.3+273]K=392.3K](https://tex.z-dn.net/?f=119.3%5EoC%3D%5B119.3%2B273%5DK%3D392.3K)
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](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B5.3kPa%7D%7B1.3kPa%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B358.5%7D-%5Cfrac%7B1%7D%7B392.3%7D%5D%5C%5C%5C%5C%5CDelta%20H_%7Bvap%7D%3D48616.4J%2Fmol%3D48.6kJ%2Fmol)
Therefore, the value of
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}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
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 = ![85.8^oC=[85.8+273]K=358.8K](https://tex.z-dn.net/?f=85.8%5EoC%3D%5B85.8%2B273%5DK%3D358.8K)
= 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](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B101.3kPa%7D%7B1.3kPa%7D%29%3D%5Cfrac%7B48.6kJ%2Fmol%7D%7B8.314%5Ctimes%2010%5E%7B-3%7DkJ%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B358.5%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D%5C%5C%5C%5CT_2%3D489.2K)
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