[Jens__]

Good evening

So I have a question about standard enthalpy. My teacher gave this example:

Calculate ΔH° of this reaction:
       2H₂ +O₂ --> 2H₂O

And I thought it would be equal to -483.6, since that is the H°_f of water times two.
But then my teacher said that I needed to devide it by two to gain the ΔH° of the reaction.

I don't understand because if there is a reaction that has two products (on the right side of the arrow) and one of them has a coefficient of 1 and the other one a coefficient of 2. Then what is the ΔH° of the reaction? Example: combusion of methane.

Thanks in advance!

Sincerely
Jens

[Enthalpy]

Welcome, Jens__!

I too would take twice the formation ΔH of water. Maybe the teacher meant something else?

By the way, always keep in mind that combustions water can be gaseous or liquid, even if given at room temperature. Not the question here, but eternal error source. Same for the combustion heat.

[mjc123]

If you are asked for the enthalpy of a reaction, that means according to the equation as written. So that means you are right, ΔH_rxn = 2ΔH_f(H₂O). However, you could represent exactly the same chemical reaction by the equation H₂ + 0.5O₂  H₂O, for which ΔH = ΔH_f(H₂O). (Or indeed 436H₂ + 218O₂  436H₂O, for which...). That is why, when you talk about the enthalpy of a reaction, you must specify the equation to which ΔH applies, or the reagent for which ΔH is the enthalpy per mole.