The combustion of one mole 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 liberated. Calculate the standard enthalpy of formation, `Delta_(r) H^( Theta )` of benzene. Standard enthalpies of formation of `CO_(2(g)) and H_(2) O_((l))` are `-393.5` kJ "mol"^(-1) and -285.83 "kJ mol"^(-1)` respectively.

Explanation

The formation reaction of benzene is given by
`6C_("(graphite)") + 3H_(2(g)) to C_(6) H_(6(l)), Delta_(f) H^( Theta ) = (?)….(i)`
Then enthalpy of combustion of 1 mol of benzene is :
`C_(6) H_(6(l)) + (15)/(2) O_(2) to 6CO_(2(g)) + 3H_(2) O_((l)),`
`Delta_(C) H^( Theta ) = - 3267 "kJ mol"^(-1) ...(ii)`
The enthalpy of formation of 1 mol of `CO_(2(g))`,
`C_("(graphite)") + O_(2(g)) to CO_(2(g)) ,`
`Delta_(f) H^( Theta ) = -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(g)) to H_(2) O_((l)),`
`Delta_(f) H^( Theta ) = -285.83 "kJ mol"^(-1) ....(iv)`
Multiplying eq. (iii) by 6 and eq. (iv) by 3 we get:
`6C_("(graphite)") + 6O_(2(g)) to 6CO_(2(g)),`
`Delta_(f) H^( Theta ) =-2361 "kJ mol"^(-1)`
`3H_(2(g)) + (3)/(2) O_(2(g)) to 2H_(2) O_((l)),`
`Delta_(f) H^( Theta )= -857.49 "kJ mol"^(-1)`
Summing up the above two equations :
`6C_("(graphite)") + 3H_(2(g)) + (15)/(2) O_(2(g)) to 6CO_(2(g)) + 3H_(2)O_((l)),`
`Delta_(f) H^( Theta ) = -3218.49 "kJ mol"^(-1) .....(v)`
Reversing equation (ii),
`6 CO_(2(g)) + 3H_(2) O_((l)) to C_(6) H_(6(l)) + (15)/(2) O_(2(g)),`
`Delta_(f) H^( Theta ) = 3267.0 "kJ mol"^(-1) ....(vi)`
Adding equations (v) and (vi), we get,
`6C_("(graphite)") + 3H_(2(g)) to C_(6)H_(6(l)),`
`Delta_(f) H^( Theta ) = 48.51 "kJ mol"^(-1)`