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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

User Minerva
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1 Answer

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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

User Adi Sutanto
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