From the following heats of combustion ,?

Question

CH3OH(l) + 3/2O2(g)----> CO2(g)+2H20(l)
Delta Hrxn= -726.4kJ/mol

C(graphite)+O2(g) ----> CO2(g)
Delta Hrxn= -393.5kJ/mol

H2(g) + 1/2O2(g)----->H2O(l)
Delta Hrxn= -285.8kJ/mol

calculate the enthalpy of formation of methanol (CH3OH) from its elements:
C(graphite)+2H2(g)+1/2O2(g)
----->CH3OH(l)

If anyone knows how to do this please explain to me step by step how to solve it. Thanks so much, i really need the help.

Answer 1

Write the first equation backwards (remembering to change the sign of Delta Hrxn) to get CH3OH on the right. Double the third equation (remembering to double the sign of Delta Hrxn) to get 2H2 on the left. Then add the three equations together (also adding the Delta Hrxn values). You will end up with the fourth equation once you have cancelled 3/2O2 from each side of the equation. Delta Hrxn for the fourth equation is the sum of the vales for the three equations you added together.

Answer 2

It's enthalpy of combustion of C + (2 x enthalpy of combustion of H2) - enthalpy of combustion of CH3OH.

A straightforward Hess Law Cycle.