Question

How much energy must be removed from a 125 g sample of benzene (C6H6) vapor at 425.0 K to make it a solid at 200.0 K? Use the following steps to get to your answer. ΔHvap = 33.9 kJ/mol ΔHfus = 9.8 kJ/mol Cliq = 1.73 J/g°C Cgas = 1.06 J/g°C Csol = 1.51 J/g°C Tmelting = 279.0 K Tboiling = 353.0 K

1) Determine number of moles the sample contains. ___________________moles

2) How much energy must be removed to cool the gas to its boiling point? ___________J

3) How much energy must be removed to convert the gas to a liquid? ___________J

4) How much energy must be removed to bring the liquid from its ___________J boiling point to its melting point?

5) How much energy must be removed to convert the liquid to a solid? ___________J

6) How much energy must be removed to cool the solid to its final ___________J temperature?

7) What is the total amount of energy, in kJ, needed? ___________kJ

8) If only 20 kJ of heat was removed from the sample, what would its temperature be?

What state or states would it be in? State:__________________________________________________ __________K

Answer 1

1) number of moles the sample = w/mwt = 125/78 = 1.6 mol

2) energy must be removed to cool the gas to its boiling point

   = mgas*Cgas*DT

   = 125*1.06*(425-353)

   = 9.54 kj

3) energy must be removed to convert the gas to a liquid

= mgas*Cgas*DT + n*DHvap

= 125*1.06*(425-353) +1.6*33.9*10^3

= 63.78 kj

4) energy must be removed to bring the liquid from its boiling point to its melting point.

= mliquid*Cliquid*DT

= 125*1.73*(353-279)

= 16.0 kj

5) energy must be removed to convert the liquid to a solid

    = n*DHfus

    = 1.6*9.8

    = 15.68 kj

6) energy must be removed to cool the solid to its final temperature

   = msolid*Csolid*DT

   = 125*1.51*(279-200)

   = 14.911 kj

7) total amount of energy needed = 9.54 +63.78+16+15.68+14.911

             = 119.911 kj

8) if 20 kj is removed , the sample will be at its boling point, and it is in gaseous state.