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 liquify the sample and lower the temperature to 335.0 K? The following physical data may be useful.

?Hvap = 33.9 kJ/mol
?Hfus = 9.8 kJ/mol
Cliq = 1.73 J/g

Answer 1

The sample of benzene contains 125g / 78.11g.mol^-1 = 1.60moles

You can assume that the heat energy is removed from the benzene in three separate steps:

1) In cooling the gaseous sample from 425K to 353K (its boiling point)

2) In liquefying the sample at 353K

3) In cooling the liquid from 353K to 335K

First calculate the energy removed from the benzene at each step,then add them together to find the total amount of energy removed from the benzene.

1) Q = m.c.deltaT = 125 x 1.06 x (425 - 353) = 9540J = 9.540kJ

2)Q = Delta Hvap x 1.6 mol = 33.9 x 1.6 = 54.24 kJ

3) Q = m.c.deltaT =125 x 1.73 x (353 - 335) = 3893J = 3.893kJ

Total = 9.540 + 54.240 + 3.893 = 67.673kJ