Enthalpy Change of Formation? Watch

The change in state for the H2O from liquid to gas in the reaction doesn't matter, it'll just be the enthalpy of formation of gaseous H2O instead of liquid H2O that is factored into the equation you are trying to work out. Hess's law states that the enthalpy change of a reaction is independent of the route taken. Therefore we can use the enthalpy of formation to construct a Hess's law triangle to work out the enthalpy change for this reaction to form the product water from hydrogen gas and oxygen.

Say we create an overall reaction for the formation of water:
2H2(g) + O2(g) -----> 2H2O(g OR l)

Underneath that reaction we use enthalpy of formation to construct a hess's law triangle(imagine the central bold part connecting to the above reaction underlined).
So 2H2(g) + O2(g) <---------2H2(g) + O2(g) ------> 2H2O(g or l or s)
The part in bold is the enthalpy of formation reaction forming the products and reactants (or compounds in reaction). It's this bold part which is the enthalpy of formation which has to be in standard states or conditions, not the reactants and products underlined in the overall reaction. That's why the H2O can be in gas, liquid or even solid state.

By definition, the enthalpy of formation of H2 + O2 will be zero.
The enthalpy of formation of H2O will be different depending on which state it is in.

The enthalpy change of the reaction (ΔH°) = ΣΔH_f° (products) - ΣΔH_f° (reactants)