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

the formation reaction of benezene is given by :
6C (graphite)`+eH_(2)(g)rarrC_(6)H_(6)(I),Delta_(f)H^(Ө)=?.....(i)`
The enthalpy of combustion of 1 mol of benzene is :
`C_(6)H_(6)+(15)/(2)O_(2)rarr CO_(2)(g):`
`" " Delta_(f)H^(Ө)=-393.5 kJ "mol"^(-1)....(ii)`
The enthalpy of formation of 1 mol of `CO_(2)(g)` :
C(graphite)`+O_(2)(g()rarrCO_(2)(g):`
`" " Delta_(f)H^(Ө)=-393.5 kJ "mol"^(-1).....(iii)`
The enthalpy of formation of 1 mol of `H_(2)O(l)` is :
`H_(2)(g)+(1)/(2)O_(2)rarrH_(2)O(1),`
`" " Delta_(c)H^(Ө)=-285.83kJ "mol"^(-1)....(iv)`
multiplying eqn. (iii) by 6 and eqn. (iv) by 3 we get:
6C (graphite)`+6O_(2)(g)rarr6CO_(2)(g):`
`" " Detla_(f)H^(Ө)=-2361 kJ "mol"^(-1)`
`3H_(2)(g)+(3)/(2)O_(2)(g)rarr3H_(2)O(1),`
`" " Detla_(f)H^(Ө)=-857.49 kJ "mol"^(-1)`
Summing up the above two equations :
6C (graphite)`+3H_(2)(g)+(15)/(2)O_(2)(g)rarr6CO_(2)(g)+3H_(2)O(1),`
`" " Delta_(f)H^(Ө)=-3218.49kJ "mol"^(-1)...(v)`
Reversing equation (ii),
`6CO_(2)(g)+3H_(2)O(1)rarrC_(6)(1)+(15)/(2)O_(2),`
`" " Delta_(f)H^(Ө)=-3267.0kJ "mol"^(-1)...(vi)`
Adding equations (v) and (vi), we get
6C (graphite) `+eH_(2)(g)rarrC_(6)H_(6)(1),`
`" " Delta_(f)H^(Ө)=-48.51 kJ "mol"^(-1)...(iv)`