Question

Of what order of magnitude is the effect of spontaneous forward-and-backward reaction H2O <=> OH- + H3O+ on the Gibbs Free Energy of H2O (l)?

Answer 1

What you call a forward an backward reaction is what is known in chemistry as an equilibrium reaction:

H2O⟺OH−+H3O+H2O⟺OH−+H3O+

The influence on the Gibbs Free Energy of the system, e.g. 1L1L of water, however is negligible because the equilibrium leans very much to the left.

The equilibrium constant KwKw is given by:

Kw=[OH−]×[H3O+]≈10−14Kw=[OH−]×[H3O+]≈10−14

(The angular brackets represent concentration in mol.dm−3mol.dm−3)

Yes, you're reading correctly: 10−1410−14! In neutral water, of pH=7pH=7:

[OH−]=[H3O+]≈10−7mol.dm−3[OH−]=[H3O+]≈10−7mol.dm−3

The concentration of water (in water) by the way is about 55.6mol.dm−355.6mol.dm−3, so only the tiniest fraction of water is present as OH−OH− and H3O+H3O+, the overwhelming majority is present as H2OH2O. That fraction is about 0.0000004percent0.0000004percent.

The effect of the auto-dissociation of water on the Gibbs Free Energy is therefore almost incalculably small.

Answer 2

Consider a box of liquid water:

One can calculate the Gibbs free energy by considering the enthalpy and the entropy:

ΔG=ΔH−TΔS

For liquid water, the enthalpy is the standard enthalpy of formation plus the energy required to make way for it:

ΔH=ΔHoH2O(l)+PΔV

The entropy term is where I don't know where to start. If one has a box of water at constant temperature and pressure, how do we define the entropy? And more importantly...

when the forward an backward reaction H2O⟺OH−+H3O+ is considered (as opposed to when it is ignored), of what order is change in the Gibbs Free Energy of the system?