At `300K` , the standard enthalpies of formation of `C_(6)H_(5)COOH_((s),CO_(2(g))` and `H_(2)O_((l))` are `-408,-393` and `-286kJ mol^(-1)` respectively. Calculate the heat of combustion of benzoic acid at
`(i)` constant pressure,
`(ii)` constant volume. `(R=8.31J mol^(-1)K^(-1))`

Answer

`-3199.75 kj`

Explanation

Given, `7C (s) + 3H_(2) (g) + O_(2) (g) rarr C_(6) H_(5) COOH (s), Delta H^(0) = -408 kJ`
(i) (i) `C(s) + O_(2) (g) rarr CO_(2) (g)`
`Delta H^(0) = -393 kJ`...(ii)
`H_(2) (g) + (1)/(2) O_(2) (g) rarr H_(2)O (l)`
`Delta H^(0) = -286 kJ`...(iii)
`C_(6)H_(5)COOH (s) + (15)/(2) O_(2) (g) rarr 7 CO_(2) (g) + 3H_(2) O (l) Delta H = ?`.....(iv)
By (ii) `xx 7 + (iii) xx 3 - (i)`
`C_(6)H_(5)COOH (s) + (15)/(2) O_(2) (g) rarr 7CO_(2) (g) + 3H_(2) O (l)`
`Delta H^(0) = [-393 xx 7 - 286 xx 3 + 408]`
`:. Delta H^(0) = - 3201 kJ` Also `Delta H = Delta U + Delta n_(g) RT`
`:gt = 3201 = Delta U + (-0.5) xx 8.314 xx 10^(-3) xx 300`
`:. Delta U = - 3201 + 1.2471 = -3199.75 kJ`