How to use Hess's Law to calculate enthalpy for the reaction?

Use Hess' Law to calculate the enthalpy of formation of methanol (CH3OH)?

• Methanol (wood alcohol) is extremely useful as a fuel, as a solvent, and as an antifreeze. However, it is also extremely poisonous; your body changes it into formaldehyde and formic acid after it is ingested. Drinking as little as 10 mL will cause permanent blindness, and at 30 mL, methanol consumption can be fatal. For this reason, methanol is also used to "denature" ethanol, thereby exempting it from liquor taxes; ironically, "moonshiners" used to add methanol to their spirits to give it "a little extra kick" Given the following: CH3OH (l) + 1.5O2 (g) yields CO2 (g) +2H2O (l) Delta H = -726.4 kJ/mol C(graphite) + O2 (g) yields CO2 (g) Delta H = -393.5 kJ/mol H2 (g) + 0.5O2 (g) yields H2O (l) Delta H = -285.8 kJ/mol Use Hess' Law to calculate the enthalpy of formation of methanol (CH3OH): C(graphite) + 2H2 (g) + 0.5O2 (g) yields CH3OH(l) PLEASE HELP! any help would be wonderful.

• Answer:

  First reverse the first equation and get CO2 + H2O-->CH3OH + 1.5O2 delta H=+726.4kj/mol Next double the 3rd equation and get 2H2 + 1O2--> 2H20 delta H = 571.6kj/mol Now add the 3 reations together to get the final heat of formation of methanol The delta H = +238.7kj/mol