Question

How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquefy the sample and lower the temperature to 335.0 K? The following physical data may be useful. ΔHvap = 33.9 kJ/mol Cliq = 1.73 J/g°C Tmelting = 279.0 K ΔHfus = 9.8 kJ/mol Cgas = 1.06 J/g°C Tboiling = 353.0 K Csol = 1.51 J/g°C

Answer 1

425 -------------------> 353 ------------------> 335

         1                        2                3

moles of benzene = 125 /78.11 = 1.60

Q1 = m Cp dT = 125 x 1.06 x (425-353) = 9540 J

Q2 = n x ΔHvap = 1.60 x 33.9 = 54.24 kJ = 54240 J

Q3 = m Cp dT = 125 x 1.73 x (353-335) = 3892.5 J

Q = Q1 + Q2 + Q3

Q = 67672.5 J

Q = 67.67 kJ

energy must be removed =67.67 kJ