Enthalpy of combustion of benzene is `-3267 kJ mol^(-1)`. Calculate enthalpy of formation of benzene, given enthalpy of formation of `CO_(2)` and water are `- 393.5 kJ mol^(-1)` and `-285.83 kJ mol^(-1)`.

Explanation

`C_(6)H_(6)(l) + (15)/(2)O_(2)(g) rarr 6CO_(2)(g) + 3H_(2)O(l) , Delta H = -3267 kJ mol^(-1)`
`Delta_(f)H^(@)` of `CO_(2) = -393.5 kJ mol^(-1)`
`Delta_(r )H^(@)` of `H_(2)O = -285.83 kJ mol^(-1)`
`Delta_(r )H^(@) = SigmaH_(p)^(@) - Sigma - H_(R )^(@)`
`Delta_(c )H^(@) = (6 xx H_(CO_(2))^(@) + 3 xx H_(2)O)^(@) - (H_(C_(6)H_(6))^(@) + (15)/(2) xx H_(O_(2))^(@))`
`-3267 = (6 xx -393.5) + (3 xx -285.83) - [H_(C_(6)H_(6)) + (15)/(2) xx 0]`
`-3267 = -2361.0 - 857.49 - H_(C_(6)H_(6))`
`H_(C_(6)H_(6)) = -3218.49 + 3267`
`H_(C_(6)H_(6)) = 48.51 kJ mol^(-1)`.