Help please with chemistry

help pleaseee

http://img580.imageshack.us/img580/525/img0840ln.jpg

the set up for your problem suggests that they want it solved using average bond enthalpies
which uses data like from http://www.science.uwaterloo.ca/~cchieh/…
references rarely agree, if your reference has different values for bond energies, use them
(& with B.E.'s tables rarely if ever agree)

bond energies are not the energy of the material,
it actually reflects the stability of the material,
for it relates how much energy is needed to break the bond

much like work done vs. energy left....
the change in bond stability v.s dH reaction are opposites
so we calculate dH using bond energies oppositely from what we do with dHf's
where we would have done:
dH reaction = (dHf products) - (dHf reactants)
instead we get close to the same answer by doing...
dH reaction = (B.E. reactants) - B.E. products)

C5H12 & 8 O2 --> 5CO2 & 6H2O

dH reaction = (B.E. reactants) - B.E. products)
dH reaction = [(B.E 4 C-C) & (B.E. 9 C-H) & (B.E. 8 O=O) ] - [(B.E.10 C=O) & (BE 12 O-H) ]

dH reaction = [(4)(348) & (9)(413) & (8)(495) ] - [(10)(799) & (12)(463) ]
dH reaction = [(1392) & (3717) & (3960) ] - [(7990) & (5556) ]
dH reaction = [9069] - [13,563 ]
dH reaction = - 4494 kJ

your answer, rounded to 3 sig figs is
dH = - 4490 kJ