Respuesta :
Answer: [tex]\Delta H_{vap}[/tex] = 55.1 kJ/mol
Explanation: Molar Enthalpy of Vaporization([tex]\Delta H_{vap}[/tex] ) is the energy needed to change 1 mol of a substance from liquid to gas at constant temperature and pressure.
For the 2-hydroxybiphenyl, there two temperatures and 2 pressures. In this case, use Clausius-Clapeyron equation:
[tex]ln\frac{P_{2}}{P_{1}}=\frac{\Delta H_{vap}}{R} (\frac{1}{T_{1}}-\frac{1}{T_{2}} )[/tex]
[tex]\Delta H_{vap}[/tex] is in J/mol:
1) Temperature in K
[tex]T_{1}=[/tex] 286 +273 = 559K
[tex]T_{2}[/tex] = 145 + 273 = 418K
2) Both pressure in Pa
[tex]P_{1}[/tex] = 101325Pa
[tex]P_{2}[/tex] = 14*133 = 1862Pa
Since molar enthalpy is in Joules, gas constant R is 8.3145J/mol.K
Replacing into the equation:
[tex]ln\frac{1862}{101325}}=\frac{\Delta H_{vap}}{8.3145} (\frac{1}{559}-\frac{1}{418} )[/tex]
[tex]ln(0.0184)=\frac{\Delta H_{vap}}{8.3145} (\frac{141}{233662} )[/tex]
[tex]\Delta H_{vap}=\frac{-3.9954*1942782.7}{-141}[/tex]
[tex]\Delta H_{vap}=55051.02[/tex]
[tex]\Delta H_{vap}=55051.02[/tex]
[tex]\Delta H_{vap}=55.1[/tex] kJ/mol
Using those values, molar enthalpy is 55.1 kJ/mol
Comparing to the CRC Handbook, which is [tex]\Delta H_{vap}=71[/tex] kJ/mol:
[tex]\frac{55.1}{71}[/tex] = 0.78
The calculated value is 0.78 times less than the CRC Handbook.
