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MatroZZZ [7]
3 years ago
9

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 lo

wer 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
Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
4 0

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\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (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=\frac{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\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

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