Use bond energies to calculate the enthalpy of combustion of methanol in kj/mol

CH3OH + 2-3 O2 rightarrow CO2 + 2H2O C – H = 414 C – O = 360 O – H = 464 O – O = 142 C – O = 799 H – O = 434 H-C-H-H – – O – H 3(414) + 360 + 464 3/2 (142) (C-H) (C-O) (O-H) 0-0) -2(799) – 4(464) (C – O) (H – O) = – 1175KJ

Answer

Well bond breaking requires energy while bond making releases
energy. Look up the bond energy and mean bond energy for the bonds
in the reactants and products.
Consider the reaction
CH4 + 2O2 —> CO2 + 2H2O
Bond breaking (endothermic)
Each mole of CH4 has 4 moles of C-H bond. So look up the mean bond
energy for C-H bonds and multiply the value by 4.
Each mole of O2 has 1 mole of O=O bonds. 2 moles O2 has 2 moles of
O=O bonds. Look up the bond enthalpy value for O=O and double
it.
Add the values for these two bond breaking steps together to give
the total energy needed.

Bond making (exothermic)
Each mole CO2 has 2 moles of C=O bonds in it (O=C=O). Use the C=O
energy and double it.
Each mole H2O has 2 moles of O-H bonds. 2 moles H2O has 4 moles of
O-H. Lookup the O-H energy and multiply it by 4.
Add these values together.

Now the energy for the overall reaction is the difference
between what you put in and what you get back. As this reaction is
exothermic, you must get more back than you put it. So subtract the
values and you have the value.

You need to apply this to your reaction.
Hint
CH3OH has 3 moles of C-H bonds, 1 mole of C-O bonds and 1 mole of
O-H bonds.
1.5 moles O2 has 1.5 moles of O=O bonds