The combustion of one mole of benzene takes place at 298 and 1 atm. After combustion, `CO_(2)(g) and H_(2)O(l)` are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation. `triangle_f H^@` benzene. Standard enthalpies of formation of `CO_2(g) and H_20(l)` are -393,5 kJ mol`""^(-1) and "-285.83 kJ mol"^(-1)` respectively.

Explanation

The formation reaction of benezene is given by :
6C (graphite) `+3H_(2) (g) to C_(6)H_(6) (l)`
`triangle_(f)H^@=?.........(i)`
The enthalpy of combustion of 1 mol of benzene is
`C_(6)H_(6)(l)+(1.5)/2 O_(2) (g) to 6CO_(2) (g)+3H_(2)O (l)`
`triangle_(c) H^@=-"3267kJ mol"^(-1)……..(ii)`
The enthalpy of formation of 1 mol of `CO_2 :`
`"C(graphite) +"O_(2) (g) to CO_(2) (g)`
`triangle_(f) H^@"=-393.5kJ mol"^(-1) ……(iii)`
The enthalpy of formation of 1 mol of `H_2O(l)` is :
`H_(2)(g)+1/2 O_(2) (g) to H_(2)O (l)`
`triangle_(f) H^@=-"285.83kJ mol"^(-1) …….(iv)`
Multiplying egn. (ii) by 6 and eqn. (iv) by 3 we get :
`6C ("graphite") +6O_(2) (g) to 6CO_(2) (g)`
`triangle_(f)H^@=-"2361kJ mol"^(-1)`
`triangle_(f)H^@=-"857.49kJ mol"^(-1)`
Summing up the above two equations :
`"6C ("graphite") +3H_(2) (g)+(15)/(2) O_(2) (g) to 6CO_(2) (g)+3H_(2)O (l)`
`triangle_(f) H^@=-"3218.49kJ mol"^(-1) …..(v)`
Reversing equation (ii),
`6CO_(2) (g)+3H_(2)O(l) to C_(6)H_(6) (l) +(1.5)/2 O_(2)`
`triangle_(f)H^@=-"3267.0kJ mol"^(-1)…….(vi)`
adding equations (v) and (vi), we get
`6C ("graphite")+3H_(2) (g) to C_(6)H_(6) (l),`
`triangle_(f)H^@="48.51kJ mol"^(-1)`