Question

The enthalpy of vapourization of benzene(l) to benzene(g) is +33.0 kJ/mol.

a)From the BE values in Table 3.1, and again the enthalpy of atomization of C(s), +717 kJ/mol, estimate the enthalpy of formation of benzene(l). (Hint: now the formation of benzene(g) has to be followed by the reaction benzene(g)®benzene(l), the opposite of vapourization).

Answer 1

benzene(l) benzene(g) ,Hvap=33.0KJ/mol

or, benzene(g) benzene(l) ,H1=-33.0KJ/mol...............(1)

C(s)C(g) ,Hatomization=717 KJ/mol...............(2)

Also, H2(g)2H(g) ,Hatomization=436.4 KJ/mol.....................(3)

Benzene formation :

6C(g)+6H(g) ---> C6H6(g) ,H=?

Using Hess's law, equation (2)*6 +eqn (3)*3 gives

6C(g)+6H(g) ---> C6H6(g) ,H=(6*717+3*436.4)=5611.2 KJ/mol

now ,6C(g)+6H(g) --->C6H6(g) ,H=5611.2 KJ/mol..............(4)

So,equation (1)+eqn (4) gives

6C(g)+6H(g) --->C6H6(l) ,Hf=5611.2 KJ/mol-33.0 KJ/mol=5578.2 KJ/mol= enthalpy of formation of benzene(l)