At 1atm benzene freezes at 5.5 o C and its density changes from 0.879 g/cm 3 to

Question

At 1atm benzene freezes at 5.5oC and its density changes from 0.879 g/cm3 to 0.891 g/cm3. Its enthalpy of fusion is 10.59 kJ/mol. Estimate the freezing point of benzene (oC) when pressure increase by 1000 atm.

Hint: Find phase transition molar volume change (unit: m3/mol) using density and molar mass, together with the entropy change (H/T), and then use Clapeyron equation to calculate T when P is given. Refer to homework 4 question 4. Please pay attention to the unit: kJ need to convert to J.

THE ANSWER IS NOT 281.4K (8..25)

Explanation / Answer

Using Clapeyron equation,

deltaP/deltaT = deltaH(fus)/Tm.delltaV(fus)

deltaV(fus) = 78.11 g/mol/0.879 g/cm^3 - 78.11 g/mol/0.891 g/cm^3 = 1.2 cm^3/mol = 1.197 x 10^-6 m^3.mol-1

deltaT = 273 + 5.5 = 278.5 K

delatP/deltaT = 10.59 x 10^3 J/mol/1.197 x 10^-6 m^3/mol x 278.5 K = 3.17 x 10^7 Pa/K

deltaP = 1000 - 1 = 999 atm = 1.012 x 10^8 Pa

delatT = 1.012 x 10^8/3.17 x 10^7 = 3.186 K

So the freezing point of benzene at 1000 atm becomes = 278.5 + 3.186 = 281.7 K

= 8.7 oC

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