When benzene, below Figure, is combusted it has the following stoichiometric reaction

**C _{6} H_{6} + 15/2 O_{2} → 6CO_{2} + 3 H_{2}O**

The enthalpy change for this combustion reaction is equal to the sum of the standard enthalpies of formation ΔH^{0} of the products minus the sum of the standard enthalpies of formation of the reactants. Each enthalpy of the formation being multiplied by the stoichiometric coefficient for that molecule. Enthalpy of formation is covered during the first year of your degree.

Rearrange the enthalpy equation to get the enthalpy of combustion of benzene as the subject of the equation.

**Solution**

In order to not get lost in the algebra: (1) expand out the bracket on the right-hand side;

(2) take ΔH^{0} (C_{6} H_{6}) over to the left-hand side by adding ΔH^{0} (C_{6} H_{6}) to both sides of the equation and cancelling out where possible;

(3) take the enthalpy of combustion ΔH^{0}_{comb} to the right-hand side by subtracting ΔH^{0}_{comb} from both sides and then cancelling out where possible.

(4) which then gives the heat of formation of combustion of benzene as the subject of the equation.