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

14 Torr. Calculate the value of the molar enthalpy of vaporization. Compare this value to that given in the CRC Handbook.​

User Tre
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1 Answer

7 votes

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(P_(2))/(P_(1))=(\Delta H_(vap))/(R) ((1)/(T_(1))-(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(1862)/(101325)}=(\Delta H_(vap))/(8.3145) ((1)/(559)-(1)/(418) )


ln(0.0184)=(\Delta H_(vap))/(8.3145) ((141)/(233662) )


\Delta H_(vap)=(-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:


(55.1)/(71) = 0.78

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

User William Ardila
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