# Can water synthesis and decomposition be in a dynamic equilibrium?

When heated at extreme temperatures, water can spontaneously decompose.

According to Wikipedia (https://en.wikipedia.org/wiki/Water_splitting):

In thermolysis, water molecules split into their atomic components hydrogen and oxygen. For example, at 2200 °C about three percent of all H2O are dissociated into various combinations of hydrogen and oxygen atoms, mostly H, H2, O, O2, and OH. Other reaction products like H2O2 or HO2 remain minor. At the very high temperature of 3000 °C more than half of the water molecules are decomposed, but at ambient temperatures only one molecule in 100 trillion dissociates by the effect of heat.[15] The high temperatures and material constraints have limited the applications of this approach.

At those temperatures, will the hydrogen and oxygen (and combination of those) still react to form water? If so, could this be considered a dynamic equilibrium?

• I'm intrigued by the statement "at ambient temperatures only one molecule in 100 trillion dissociates by the effect of heat.". It seems the author might be conflating the self-ionization of water and dissociation into hydrogen and oxygen molecules, even though these are quite different processes. At room temperature, the latter process is vastly more suppressed, but in principle still happens. Mar 11, 2022 at 12:41
• At room temperature, $\pu{\Delta G = -228.6 kJ/mol}$. then $\pu{\frac{\Delta G}{RT} = -92.267}$. And $\pu{K = e^{-\frac{\Delta G}{RT}} = 6.48 · 10^{-41}}$ As consequence there is not even $1$ molecule $\ce{H2O}$ out of $1$ mole water which is decomposed into $\ce{H2}$ (or $\ce{H}$) and $\ce{O2}$ (or $\ce{O}$) at room temperature. Mar 11, 2022 at 17:19

The reaction you are looking at is

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

The equilibrium constant can be characterised by the reaction's standard Gibbs free energy, which can be calculated from the enthalpy and entropy.

We have, at standard conditions, $$\Delta H° =\pu{-241.83 kJ mol-1}$$ and $$\Delta S°= \pu{-44.16 J K-1 mol-1}$$. At 25 °C, this gives a large negative value of the Gibbs free energy, so at equilibrium, we almost exclusively have product (we say that the reaction is irreversible).

However, we can calculate the equilibrium constant from $$K=e^{-\frac {\Delta G°} {RT}}$$ and as the entropy is negative, we will get a positive $$\Delta G$$ at some (very high) temperature.

The exact calculation is complicated by the fact that we can't just plug the numbers above into the formula for the Gibbs energy, because the enthalpy and entropy changes significantly over the given temperature range. But if you found the heat capacity data for the three gases, you could calculate $$\Delta G°$$ at 2200°C.

And to answer your question, the reaction can be considered as a dynamical equilibrium, just like any other chemical reaction. And at a sufficiently high temperature, there will actually be a 50-50 equilibrium between the species.