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 : The energy removed must be, -34.67 kJ

Solution :

The process involved in this problem are :

The expression used will be:

where,

= heat available for the reaction =

m = mass of benzene = 125 g

= specific heat of solid benzene =

= specific heat of liquid benzene =

= enthalpy change for fusion =

Molar mass of benzene = 78 g/mole

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

Therefore, the energy removed must be, -34.67 kJ

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

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