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

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


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

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?

  • $\begingroup$ 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. :-) $\endgroup$ – peterh Apr 5 '20 at 0:21
  • $\begingroup$ 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. $\endgroup$ – Karsten Theis Apr 5 '20 at 3:35

The change in Gibbs free energy is written per mole. For reactions, consider it "per mole of reaction." Since the stoichiometry of the second reaction is exactly double the first, "one mole of the first reaction" is half "one mole of the second reaction." Therefore, the value of the change in free energy is doubled as expected.

For an arbitrary reaction, there is no standard.

For the standard free energy of formation of a substance, the canonical reaction should generate one mole of the substance from its constituent elements (the first reaction here).


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