Answer to Question #62998 in General Chemistry for Tharshika thanabalasingham

Question

How much energy must be removed from a 125 g sample of benzene (molar mass = 78.11 g
mol-1) at 425.0 K to liquify the sample and lower the temperature to 335.0 K? The following
physical data may be useful:

Answer 1

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

Aassuming that the heat energy is removed from benzene in three separate steps:
1) Cooling gaseous sample from 425 K to 353 K (till its boiling point)
2) Liquefying the sample at 353 K
3) Cooling the liquid from 353 K to 335 K
Sum of energies removed from benzene at each step will indicate the amount of energy removed from benzene.

1) Q = m·c·ΔT = 125 x 1.06 x (425 - 353) = 9540J = 9.540 kJ
2) Q = Δ Hvap x 1.6 mol = 33.9 x 1.6 = 54.24 kJ
3) Q = m·c·ΔT =125 x 1.73 x (353 - 335) = 3893J = 3.893 kJ
Total = 9.540 + 54.240 + 3.893 = 67.673 kJ