Question

How much energy must be removed from a 125 g sample of benzene (C₆H₆) at 425.0 K to liquefy the sample and lower the temperature to 335.0 K? The following physical data may be useful.

ΔH_vap = 33.9 kJ/mol

ΔH_fus = 9.8 kJ/mol

C_liq = 1.73 J/g°C

C_gas = 1.06 J/g°C

C_sol = 1.51 J/g°C

T_melting = 279.0 K

T_boiling = 353.0 K

(The answer is 67.7 kJ. Please show how you get the answer)

Answer 1

This involves three steps:

Step 1: Benzene gas at 425 K to Benzene Vapor at 425 K

Step 2: Benzene Vapor at 425 to Benzene Liquid at 353 K

Step 3: Menzene Liquid at 353 K to 335 K

Mol. wt of benzene = 78.11

Moles of Benzene = 125/78.11 = 1.6 mol

Energy removed in Step 1= 33.9 kJ/mol * 1.6 mpl = 54.24 kJ

Energy removed in step 2= m*Cgas * (425-353) = 125g * 1.06J/g C * 72 C= 9.54 kJ

Energy removed in step 3 = m * Cliq * (353-335) = 125g * 1.73 J/g C * 18 C = 3.89

Total energy to be removed = 54.24+9.54+3.89 = 67.67 kJ