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))`

Explanation

Given:
`7C(s) +3H_(2)(g) +O_(2)(g) rarr C_(6)H_(5)COOH(s) DeltaH^(Theta) =- 408 kJ …(i)`
`C(s) +O_(2)(g) rarr CO_(2)(g),DeltaH^(Theta) =- 393 kJ …(ii)`
`H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l),DeltaH^(Theta) =- 286 kJ …(iii)`
`C_(6)H_(5)COOH(s) +15//2O_(2)(g) rarr 7CO_(2)+3H_(2)O(l)DeltaH = ? .....(iv)`
by (iii) `xx7 + (iii) xx3 - (i)`, we get
`C_(6)H_(5)COOH (s) +15//2O_(2)(g) rarr 7CO_(2)(g) +3H_(2)O(l)`,
`DeltaH^(Theta) = [-393 xx 7-286 xx3 +408]`
`:. DeltaH^(Theta) =- 3201 kJ`
Also `DeltaH = DeltaU + DeltanRT`
`:. -302 =DeltaU +(-0.5) xx 8.314 xx 10^(-3) xx 300`
`(Deltan =-0.5` and `R = 8.314 xx 10^(-3) kJ)`
`DeltaU =- 3201 +1.2471 =- 3199.7529 kJ mol^(-1)`