The combusition of 1 mol of benzene `(C_(6) H_(6))` takes place at `298 K` and 1 bar pressure. After combustion, `CO_(2) (g)` and `H_(2) O (1)` are produced and `3267 kJ` of heat is liberated. Calculate the standard enthaply of formation, `Delta_(f) H^(@)` of benzene. Standard enthapies of formation of `CO_(2) (g)` and `H_(2) O (1)` are `- 393.5 kJ mol^(-1)` and `- 258.83 kJ mol^(-1)`, respectively.
Strategy : Apply Eq. the mathematical form of Hesis's law, to the combustion reaction of 1 mol of benzene. Remember `Delta_(f) H^(@)` for `O_(2) (g)` is zero by convention. We are give `Delta_(1) H^(@)` and `Delta_(f) H^(@)` values for all substance except `C_(6) H_(6) (1)`. We can solve for this unknown.

Explanation

The thermochemical equation for the combustion of one mole of benzene is:
`C_(6) H_(6) (1) + (15)/(2) O_(2) (g) rarr 6 CO_(2) (g) + 3 H_(2) O (1)`,
`Delta_(C ) H^(@) = - 3267 kJ mol^(-1)`
Applying Eq. we can write
`Delta_(C )H^(@)=[(6 mol CO_(2),g)Delta_(C )H^(@)(CO_(2),g)+(3mol H_(2)O,l)Delta_(f)H^(@)(H_(2)O,l)]-[(1 mol C_(6)H_(6))Delta_(f)H^(@)(C_(6)H_(6),l) + (15//2 mol O_(2), g) Delta_(f) H^(@) (O_(2), g) - 3267`
`= [(6) (-393.5) + (3) (-285.83)] - [Delta_(f) H^(@) (C_(6) H_(6), g) + 0]`
Rearranging to solve for `Delta_(f) H^(@) (C_(6) H_(6), g)`,we have
`Delta_(f) H^(@) (C_(6) H_(6), g) = (3267) - (2361) - (857.49)`
`= 48.51 kJ mol^(-1)`