Can water molecules be heated enough (2500°C? 3000°C?) to be separated into hydrogen and oxygen by thermolysis via a microwave emitter?

I've searched around but it appears this hasn't been researched? So I'm guessing it's either a very inefficient process that hasn't been given any thought, or it has been tried and not proven useful.

I believe there should be many different approaches to how to achieve this (eg. a very small water amount put inside an EM resonating chamber with a magnetron attached), but I wonder if there's any limiting factor why there would be a top temperature achievable by microwave heating of water (eg: how does the heating effect vary according to the water sample's phase/pressure-temperature).

  • $\begingroup$ You would get a high temperature mixture of oxygen and hydrogen ions, which would convert back to water when the temperature dropped. $\endgroup$ Feb 25, 2022 at 0:16

1 Answer 1


In modern chemical theory we deal with Chemical Equilibria. In the case of the decomposition of water:

$$\text{H}_2\text{O}(g) \leftrightarrows \text{H}_2(g)+\frac12 \text{O}_2(g)$$

We define an Equilibrium Constant $K$: $$K=\frac{[\text{H}_2]\times[\text{O}_2]}{[\text{H}_2\text{O}]}$$

where the bracketed quantities are concentrations.

Nernst teaches us that:

$$\Delta G=-RT\ln K\tag{1}$$

where $\Delta G$ is the change in Gibbs Free Energy (left to right).

But from $(1)$ we glean that:

$$K=\exp{\left(-\frac{\Delta G}{RT}\right)}\tag{2}$$

This reference gives the Standard Giggs Free Energy of the Decomposition of water as:

$$\Delta G=+229\,\mathrm{kJ/mol}$$

This being $> 0$ , the value for $K$ will be $\ll 1$ and the values for $[\text{H}_2]$ and $[\text{O}_2]$ very small.

From $(2)$ can be gleaned that as $T$ increases, $K$ becomes larger (assuming $\Delta G$ in temperature invariant, which is not $100 \,\text{%}$ true) and the concentrations on the Right hand side also increase (as expected).

However, this is not a practical way of splitting water into hydrogen and oxygen because the moment you lower the temperature again, the equilibrium 'shifts back' to the left and the hydrogen and oxygen recombine to water.

The only way to avoid this recombination would be to physically separate the hydrogen and oxygen at the thermolysis high temperature. There are some well-known thermochemistry systems that rely on that method but thermolysis of water just isn't one of them.

  • $\begingroup$ I see. I guess I'm trying to shoehorn a few sci-fi-ish hypothesis I've got into my poor understanding of high-energy particle physics. I'm trying to come up with a way to "instantly" turn a (small) volume of water into an ignitable mix of H and O (regardless of inefficiency of the input energy required), but most methods I try reading/asking about end up with a plasma state that does not appear to be combustable. $\endgroup$ Mar 4, 2022 at 12:57
  • $\begingroup$ A 'plasma state' implies a gas at very high temperature (like $>20000\,\mathrm{K}$ or so) At such temperatures dissociation of water into hydrogen and oxygen is 'complete'. But this plasma isn't combustible. Lowering its temperature leads to reconstitution of the water. Thanks for accepting! $\endgroup$
    – Gert
    Mar 4, 2022 at 13:30
  • $\begingroup$ Can't plasma state be achieved by adding (a large) electrical charge too, though? $\endgroup$ Mar 4, 2022 at 15:50
  • $\begingroup$ Also, when you say "reconstitution of water", do you mean it'd spontaneously combust when temperature lowers enough? Or water would form without any further heat being released? $\endgroup$ Mar 4, 2022 at 15:50
  • $\begingroup$ You could call it combustion, I suppose. The heat to decompose water (say per mol) is EXACTLY the same as the heat of reaction (hy. + ox. = water). You bet back EXACTLY what you put in. Conservation of energy. $\endgroup$
    – Gert
    Mar 4, 2022 at 16:05

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