The first step in any problem with a chemical reaction is to make sure it's balanced, which this reaction is. To find the enthalpy of a reaction, we need to know the enthalpies of formation for all species in the reaction. 2CH4(g) + 3O2(g) −> 2CO(g) + 4H2O(l)

Once we know the enthalpies of formation, we can set up our calculation by taking the enthalpies of formation of the product and subtracting the enthalpies of formation of the reaction. Remember that the enthalpy of formation of elements in their standard state is always zero. We also need to include the coefficients by multiplying each enthalpy of formation by the substance's coefficient from the balanced equation.

To keep things less cluttered, I'll use a capital D to represent the delta symbol (triangle).

DH(rxn) = 2 DH(CO(g)) + 4 DH(H2O(l)) - 2 DH(CH4(g)) - 3 DH(O2(g))

Now, we need to find the enthalpies of formation for each substance. Note that the phase does matter. Liquid water and gas phase water will have different enthalpies of formation. Since the enthalpy of formation of water isn't given in the problem, we'll need to check another source for that information (see link below).

DH(rxn) = 2(-283 kJ/mo) + 4(-285 kJ/mol) - 2(-890 kJ/mol) - 3(0)

Watch your signs carefully as you work out the math!

DH(rxn) = -566 - 1140 + 1780 - 0

DH(rxn) = 74 kJ/mol

Now we know that the standard enthalpy for the reaction given was 74 kJ/mol.