Higher chemistry enthalpy question

(Original post by chem1)
Hi all,

I'm struggling with a past paper question using hess' law -

5C + 6H2O + 1/2O2 ----> C5 H11 OH

The question is 'the enthalpy for formation of pentan-1-ol is deltaH = -354 kJ/mol. Using this and the enthalpies of combustion of carbon and hydrogen, calculate the enthalpy of combustion of pental-1-ol'

I thought the answer would be an addition of:

Pentan-1-ol = -354 for formation, so +354 for combustion to gives the intermediate products (CO2 and H2O)

Then, from the intermediate products to 5C and 6H2 you would add...
carbon = -394 for combustion, so 5x394 = 1970
Hydrogen = -286 for combustion, so 6x286 = 1716

Overall --> 354 + 1970 + 1716 = 4040

BUT The answer booklet says that the figures for carbon and hydrogen should be negative (i.e. just the enthalpies of combustion) and I don't understand why.... wouldn't that mean you were adding all the enthalpy changes that lead to the intermediate products, rather than the final products of combustion (on the left side of the given equation??)

Any help would be appreciated, thank you!