Calculate the enthaply of formation of benzene. The enthaply of combustion of benzene is - 3266.0 kj. The enthaples of formation of `CO_(2) and H_(2)O` are -393.1 and -286.0 kj respectively.

Explanation

We are given that
(i) `C_(6)H_(6) + (15)/(2) O_(2) (g) to 6CO_(2)(g) + 3H_(2)O(l) Delta H= - 3266.0 kj`
(ii) `C(s) + O_(2) (g) to CO_(2) (l), Delta H = 393.1 kj`
(iii) `H_(2)(g) + (1)/(2) O_(2) (g) to H_(2)O(l), Delta H = - 286.0kj`
we are to find out the enthalpy of formation of bensence.
6C(s) + `3H_(2)(g) to C_(6)H_(6) (l), Delta H = ?`
Multiply equaiton (ii) by 6 end equation (iii) by 3 and add them.
`6 C(s) + 6O_(2) (g) to 6 CO_(2) (g),Delta H = - 393.1 xx 6 = - 2358.6 kj`
`3H_(2)(g)+ (3)/(2) O_(2) (g) to 3H_(2) O(l),Delta H = - 286 xx 3 = - 858.0` kj
(iv) `6C(s) + 3H_(2) (g) + (15)/(2) O_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l), Delta H = - 3216.6 kj mol^(-1)`
Subtract equaiton (i) from equation (iv)
`6C(s) + 3H_(2) (g) + (15)/(2) O_(2) to 6CO_(2) + 3H_(2)O (l), Delta H = - 3216.6 kj`
`C_(6)H_(6) (l) + (15)/(2) O_(2)(g) to 6CO_(2) (g) + 3H_(2)O (l) , Delta H = - 3226.0 kj`
`3H_(2) (g) + (3)/(2) O_(2) (g) to 3H_(2)O(l) , Delta H = - 286 xx 3 = - 858.0 kj`
`6C(s) + 3H_(2)(g) to C_(6)H_(6) , Delta H = + 49.4kj mol^(-1)`
Thus, enthaply of formation of benzene `(Delta H) + 49.4 kj mol^(-1)`