Respuesta :
Answer :
(a) The value of [tex]\Delta H_{vap}[/tex] 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 [tex]\Delta H_{vap}[/tex] of the reaction, we use clausius claypron equation, which is:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]P_1[/tex] = vapor pressure at temperature [tex]85.8^oC[/tex] = 1.3 kPa
[tex]P_2[/tex] = vapor pressure at temperature [tex]119.3^oC[/tex] = 5.3 kPa
[tex]\Delta H_{vap}[/tex] = Enthalpy of vaporization = ?
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature = [tex]85.8^oC=[85.8+273]K=358.8K[/tex]
[tex]T_2[/tex] = final temperature = [tex]119.3^oC=[119.3+273]K=392.3K[/tex]
Putting values in above equation, we get:
[tex]\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[/tex]
Therefore, the value of [tex]\Delta H_{vap}[/tex] is 48.6 kJ/mol
(b) The clausius claypron equation is:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]P_1[/tex] = vapor pressure at temperature [tex]85.8^oC[/tex] = 1.3 kPa
[tex]P_2[/tex] = vapor pressure at temperature normal boiling point = 101.3 kPa
[tex]\Delta H_{vap}[/tex] = Enthalpy of vaporization = 48.6 kJ/mol
R = Gas constant = [tex]8.314\times 10^{-3}kJ/mol.K[/tex]
[tex]T_1[/tex] = initial temperature = [tex]85.8^oC=[85.8+273]K=358.8K[/tex]
[tex]T_2[/tex] = final temperature = ?
Putting values in above equation, we get:
[tex]\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[/tex]
Therefore, the normal boiling point is 489.2 K
(c) Now we have to determine the entropy of vaporization at the boiling point.
[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T_b}[/tex]
where,
[tex]\Delta S_{vap}[/tex] = entropy of vaporization = ?
[tex]\Delta H_{vap}[/tex] = enthalpy of vaporization = 48.6 kJ/mol
[tex]T_b[/tex] = boiling point = 489.2 K
Now put all the given values in the above formula, we get:
[tex]\Delta S_{vap}=\frac{48.6kJ/mol}{489.2K}=99.3J/K[/tex]
Therefore, the entropy of vaporization at the boiling point is 99.3 J/K
