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AnnZ [28]
3 years ago
5

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

Chemistry
1 answer:
lina2011 [118]3 years ago
8 0

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

Explanation: <u>Molar Enthalpy of Vaporization</u>(\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 <u>Clausius-Clapeyron equation</u>:

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.

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