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

`7 C_(s) + 3 H_(2 (g)) + O_(2 (g)) to C_(6) H_(5) COO H_((s)) H^(@) = -408 kJ`
`implies Delta H_(f)^(@) (C_(6) H_(5) COOH) = -408 kJ`
`C_((s)) + O_(2(g)) to CO_(2) (g) " " DeltaH^(@) = -393 kJ`
`implies DeltaH_(f)^(@) (CO_(2)) = - 393 kJ`
`H_(2 (g)) + 1//2 O_(2(g)) to H_(2) (l) " " DeltaH^(@) = -286 kJ`
`implies Delta H_(f)^(@) (H_(2)O) = -286 kJ`
`C_(6)H_(5) COOH_((s)) + 15//2 O_(2 (g)) to 7 CO_(2 (g)) + 3H_(2) O(l)`
`implies Delta H_(c)^(@) (C_(6) H_(5) COOH) = 7 Delta H_(f)^(@) CO_(2) + 3 DeltaH_(f)^(@) (H_(2)O) - 4 Delta H_(f)^(@) (C_(6)H_(5) COOH)`
`= 7 xx (-393) + 3 xx (-286) + 408 Delta H_(c)^(@) (C_(6)H_(5) COOH) = -3609 + 408 `
`= -3201 ` kJ/mol
`implies ` enthalpy of combustion at constant pressure
`= -3201 kJ mol^(-1)`
Also ,
`Delta H = Delta U + Delta ng RT -3201 = Delta U + (0.5) xx 8.31 xx 10^(-3) xx 300`