The combustion of `1 mol` of benzene takes place at `298 K` and `1 atm`. After combustion, `CO_(2)(g)` and `H_(2)O(l)` are produced and `3267.0 kJ` of heat is librated. Calculate the standard entalpy of formation, `Delta_(f)H^(Θ)` of benzene
Given: `Delta_(f)H^(Θ)CO_(2)(g) = -393.5 kJ mol^(-1)`
`Delta_(f)H^(Θ)H_(2)O(l) = -285.83 kJ mol^(-1)`.

Explanation

Combustion of 1 mol of benzene takes place as follows
`C_(6)H_(6)(l)+7 1//2 O_(2) (g)rarr 6CO_(2)(g)+3H_(2)O(l)`
`Delta_(c )H^(Θ)=-3267.0 kJ mol^(-1)`
`Delta_(r)H^(Θ)=Delta_(c )H^(Θ)`
`=Sigma[a Delta_(f)H^(Θ)("Products")]-Sigma[Delta_(f)H^(Θ) ("reactants")]`
or `-3267.0 kJ mol^(-1)=6(Delta_(f)H^(Θ)(CO_(2)(g))+ 3(Delta_(C )H^(Θ)(H_(2)O(l))-1(Delta_(f)H^(Θ)(C_(6)H_(6)(l))-7 1//2(Delta_(f)H^(Θ) (O_(2)(g))`
`-3267 kJ mol^(-1)=(6)(-393.5 kJ mol^(-1))+ (3)(-285.83 kJ mol^(-1))-(1)(Delta_(f)H^(Θ)(C_(6)H_(6)(l))-0`
or `Delta_(f)H^(Θ)C_(6)H_(6)(l)=[6(-393.5)+ 3(-285.83)+ 3267] kJ mol^(-1)`
`=[-2361.0-857.49 + 3267.0] kJ mol^(-1)`
`=48.51 kJ mol^(-1)`