Chemistry: Enthalpies of formation problem?

Question

Given the following heats of combustion.
CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) H°rxn = -726.4 kJ
C(graphite) + O2(g) CO2(g) H°rxn = -393.5 kJ
H2(g) + 1/2 O2(g) H2O(l) H°rxn = -285.8 kJ

Calculate the enthalpy of formation of methanol (CH3OH) from its elements.
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I don't know how to set this problem up and/or solve it. Please explain how you set up the problem and determined the answer.

Answer 1

It goes like this: What you want is the *delta*H of the reaction,

2C + 4H2 + O2 ===> 2CH3OH

So you add up the *delta*H's of the reactions given like this. Keep in mind that the *delta*H of a reverse reaction is -*delta*H of the forward one.

2C + 2O2 ===> 2CO2 *delta*H = 2(393.5kJ)
4H2 + 2O2 ===> 4H2O *delta*H = 4(-285.8kJ)
3O2 + 2CO2 + 4H2O ===> 2CH3OH *delta*H = 2(+726.4kJ)

So add the equations up. Cancel all the CO2's and H2O's on the left and right. Add up all the kJ's. Voila!