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/molHfus = 9.8 kJ/mol
Cliq = 1.73 J/g

Answer 1

Answer : The energy removed must be, -34.67 kJ

Solution :

The process involved in this problem are :

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

The expression used will be:

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

where,

`\Delta H` = heat available for the reaction =
`4.50* 10^3kJ=4.50* 10^6J`

m = mass of benzene = 125 g

`c_(p,s)` = specific heat of solid benzene =
`1.51J/g.K`

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

`\Delta H_(fusion)` = enthalpy change for fusion =
`9.8kJ/mole=9800J/mole=(9800J/mole)/(78g/mole)J/g=125.64J/g`

Molar mass of benzene = 78 g/mole

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

`\Delta H=[125g* 1.73J/g.K* (353-425)K]+125g* -125.64J/g+[125g* 1.51J/g.K* (335-353)K]`

`\Delta H=-34672.5J=-34.67kJ`

Therefore, the energy removed must be, -34.67 kJ