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.

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

Answer 1

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

`(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)`

The expression used will be:

`\Delta H=[m* c_(p,g)* (T_(final)-T_(initial))]+m* \Delta H_(vap)+[m* c_(p,l)* (T_(final)-T_(initial))]`

where,

`\Delta H` = heat released by the reaction = ?

m = mass of benzene = 125 g

`c_(p,g)` = specific heat of gaseous benzene =
`1.06J/g^oC`

`c_(p,l)` = specific heat of liquid benzene =
`1.73J/g^oC`

`\Delta H_(vap)` = enthalpy change for vaporization =
`33.9kJ/mole=33900J/mole=(33900J/mole)/(78.11g/mole)J/g=434.0J/g`

Molar mass of benzene = 78.11 g/mole

Now put all the given values in the above expression, we get:

`\Delta H=[125g* 1.06J/g.K* (353.0-(425.0))K]+125g* -434.0J/g+[125g* 1.73J/g.K* (335.0-353.0)K]`

`\Delta H=-67682.5J=-67.7kJ`

Therefore, the energy removed must be, -67.7 kJ