Question

The compound 2-hydroxybiphenyl (o-phenylphenol) boils at 286 °C under 101.325 kPa and at 145 °C under a reduced pressure of

Answer 1

Answer: $\Delta H_{vap}$ = 55.1 kJ/mol

Explanation: Molar Enthalpy of Vaporization($\Delta H_{vap}$ ) 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:

$ln\frac{P_{2}}{P_{1}}=\frac{\Delta H_{vap}}{R} (\frac{1}{T_{1}}-\frac{1}{T_{2}} )$

$\Delta H_{vap}$ is in J/mol:

1) Temperature in K

$T_{1}=$ 286  +273 = 559K

$T_{2}$ = 145 + 273 = 418K

2) Both pressure in Pa

$P_{1}$ = 101325Pa

$P_{2}$ = 14*133 = 1862Pa

Since molar enthalpy is in Joules, gas constant R is 8.3145J/mol.K

Replacing into the equation:

$ln\frac{1862}{101325}}=\frac{\Delta H_{vap}}{8.3145} (\frac{1}{559}-\frac{1}{418} )$

$ln(0.0184)=\frac{\Delta H_{vap}}{8.3145} (\frac{141}{233662} )$

$\Delta H_{vap}=\frac{-3.9954*1942782.7}{-141}$

$\Delta H_{vap}=55051.02$

$\Delta H_{vap}=55051.02$

$\Delta H_{vap}=55.1$ kJ/mol

Using those values, molar enthalpy is 55.1 kJ/mol

Comparing to the CRC Handbook, which is $\Delta H_{vap}=71$ kJ/mol:

$\frac{55.1}{71}$ = 0.78

The calculated value is 0.78 times less than the CRC Handbook.