Question

How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and lower the temperature to 205.0 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

At 322 K, benzene is still a liquid, but it freezes at 279 K. Therefore, the only phase-transition that we are interested in is:

C6H6(l) ---> C6H6(s)

There are two types of energy that need to be considered.
a) The specific heat of liquid that occurs while the temperature is above the freezing point.
b) Hfus at the freezing point until all of the benzene turns into solid.
c) The specific heat of the solid until you reach your desired temperature

The very first thing you need to do before you start mingling with the energies is to calculate the number of moles of benzene that you have:

moles = (94.4 g C6H6) x [(1 mole C6H6)/(78.11 g C6H6)] = 1.21 moles C6H6

Alright, now let's calculate the amount of energy lost at the three regimes:

a) (94.4 g) x (1.73 J/g*K) x (279 K - 322 K) = - 7.02 kJ

b) - (9.8 kJ/mol) x (1.21 moles C6H6) = - 11.9 kJ

c) (94.4 g) x (1.51 J/g*K) x (205 K - 279 K) = - 10.5 kJ

Thus, the total energy lost to lower 94.4 g of benzene to 205 K is - 29.4 kJ