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Consider a box of liquid water:

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

$$ \Delta G = \Delta H - T\Delta S $$

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

$$ \Delta H = \Delta H^{o}_{H_2O(l)} + P\Delta V $$

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 $H_{2}O \iff OH^{-} + H_{3}O^{+} $ is considered (as opposed to when it is ignored), of what order is change in the Gibbs Free Energy of the system?

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  • $\begingroup$ Would Chemistry be a better home for this question? $\endgroup$ – Qmechanic Aug 18 '16 at 9:15
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What you call a forward an backward reaction is what is known in chemistry as an equilibrium reaction:

$$H_{2}O \iff OH^{-} + H_{3}O^{+}$$

The influence on the Gibbs Free Energy of the system, e.g. $1\:\mathrm{L}$ of water, however is negligible because the equilibrium leans very much to the left.

The equilibrium constant $K_w$ is given by:

$$K_w=[OH^{-}]\times [H_{3}O^{+}] \approx 10^{-14}$$

(The angular brackets represent concentration in $\mathrm{mol.dm^{-3}}$)

Yes, you're reading correctly: $10^{-14}$! In neutral water, of $pH=7$:

$$[OH^{-}]=[H_{3}O^{+}] \approx 10^{-7} \:\mathrm{mol.dm^{-3}}$$

The concentration of water (in water) by the way is about $55.6\:\mathrm{mol.dm^{-3}}$, so only the tiniest fraction of water is present as $OH^{-}$ and $H_{3}O^{+}$, the overwhelming majority is present as $H_{2}O$. That fraction is about $0.0000004\:\mathrm{percent}$.

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

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