[professordad]

The complete combustion of 1 mol of 2,3,4-trimethylpentane to CO₂ (g) and H₂O (g) leads to ΔH° = -5069 kJ. 

Calculate ΔH_f for C₈H₁₈ (which is 2,3,4-trimethylpentane).


So what I did was use the balanced combustion reaction:

2C₈H₁₈ + 25O₂ 16CO₂ + 18H₂O, ΔH° = -5069 kJ

The reactions for CO₂ have ΔH° = -393.5 kJ/mol and the reactions for H₂O have ΔH° = -285.83 kJ/mol.  Here we have 16 moles of CO₂ and 18 moles of H₂O so [tex]16 \cdot -393.5 + 18 \cdot 285.83 = 6296 + 5144.9 = 11441[/tex] mol. 

How do I account for the 2 moles of 2,3,4-trimethylpentane?  How would I proceed?

Thanks in advance.

[ramboacid]

We can use Hess's Law to calculate ΔH_f for C₈H₁₈.

Hess's Law says the total reaction enthalpy will be the difference of the ΔH_f of the products and the ΔH_f of the reactants. So you can set up an expression involving each of the heats of formation:

16ΔH_f (CO₂) + 18ΔH_f (H₂O) - 2ΔH_f (C₈H₁₈) = -5069 kJ

The heat of formation of elemental oxygen is zero, so I didn't include it explicitly in the above equation.

Now, substituting your values of the heats of formation of water and carbon dioxide, you should be able to solve for the heat of formation of C₈H₁₈.