# How come I get different Gibbs free energy for the same reduced equation?

I want to calculate the Gibbs free energy of this reaction which can be:

$$\ce{H2(g) + 1/2O2(g) -> H2O(g)}$$

or

$$\ce{2H2(g) + O2(g) -> 2H2O(g)}$$

I am using the entropy values (J/molK): H2(g) - 130.684, O2(g) - 205.138, H2O(g) - 188.7

And enthalpy values (kJ/mol): H2(g) - 0, O2(g) - 0, H2O(g) - 241.8

With the equation ΔG=ΔH−TΔS and temperature 298.15K. How come I get ΔG = -457kJ/mol for the full equation and -228kJ/mol for the reduced equation, shouldn't I get the same Gibbs free energy since they're the same reaction (except one is reduced)?

Shouldn't both be considered "correct" or is one "more correct" than the other?

• Welcome on the ChemSE! We have quite a good Latex support here. Type in $\ce{H2O}$ and you will get $\ce{H2O}$. Look how beautiful became your reaction equation now. :-) – peterh Apr 5 '20 at 0:21
• The equilibrium constant changes, too. The only thing that does not change is the standard cell potential because that is per electron, so to speak, and not per mol of reaction. It is kind of weird that an intensive quantity (i.e. molar Gibbs energy) depends on how you choose the coefficients. At least it does not depend on the amount of substance that reacts. – Karsten Theis Apr 5 '20 at 3:35