Question

The enthalpy of fusion for benzene is 127.34 kJ/kg, and its melting point is 5.49 Celcius. Estimate deltaG when 1.00 mole of benzene melts at 25 Celsius.

The optionns are:

A. 8.312kJ

B. -0.696kJ

C. 35.661 kJ

D. -11.301 kJ

Answer 1

Provided an ideal solution is obtained the mole fraction (x2) of solute at saturation is a function of the heat of fusion, the melting point of the solid (Tfus) and the temperature (T) of the solution:

delta H fusion = 127.34 KJ/kg
               = 127.34 J/g
               = 127.34 * 78.11 J/mol since molar mass of benzene = 78.11 g/mol
               = 9946.5 J/mol

ln x2 = -delta H/R * [1/T - 1/Tfus]
Tfus = 5.49 oC= (273+5.49)K = 278.49    K
T = 25 oC = 298 K
ln x2 = -delta H/R * [1/T - 1/Tfus]
      = - 9946.5/8.314 * [1/298 - 1/278.49]
      = 0.281

delta G = - R*T*ln x2
         = -8.314*298*0.281
        = - 696 J
       = 0.696 KJ
Answer: B