After a discussion about the usefulness of hydrogen fuel for cars with a friend of mine, I wondered what the reaction mechanism for combustion of $\ce{H2}$ was. I only study (well almost) organic chemistry, so I have no idea if the mechanisms from organic chemistry can be applied at all to non-organic molecules.

So we have the following reaction:

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

In most textbooks this would be described as a redox-reaction with the two half reactions:

$$2\ce{H+ + 2 e- -> H2}$$ $$\ce{O2 + 4 H+ + 4 e- -> 2 H2O}$$

But I wonder what the real mechanism is. Based on my organic thinking I thought the following (it's probably completely wrong, but I think it is important to state because someone can correct me much better if I express my thinking than if he has no idea where the misconception started):

If the temperature is high enough a long pair of oxygen could break the $\ce{H-H}$ (the lone pair will be donated to the $\sigma^{*}$ bond) bond and form $\ce{2 H-}$ (each H gets one electron because of their equal electronegativity). $\ce{H-}$ is very unstable and one will immediately react with the empty oxygen orbital, the other will break the $\ce{O=O}$ $\pi$ bond. The electrons in the $\pi$ bond will go to the positively charged oxygen atom (the one that donated its electrons in the first step). Now one oxygen atom is neutral and the other has a positive charge (because the electrons of the $\pi$ bond went for the other oxygen), where the $\ce{H-}$ will attack forming hydrogen peroxide ($\ce{HO-OH}$). The previous process will now repeat, but instead, the $\sigma$ bond between the oxygen atoms will be broken, resulting in two water molecules, $\ce{2 H2O}$.


3 Answers 3


This post deals with the mechanism that is observed in the gas phase. It is of course not as simple as the equation might suggest and you did suspect that already. $$\ce{2H2 + O2 -> 2H2O}$$

This will be divided into many different elementary sub reactions. Any mixture of oxygen and hydrogen is metastable (stable as long as you do not change the conditions). If you provide sufficient energy to overcome the activation barrier, the reaction will take its cause: $$\ce{H2 <=>[\Delta T] 2 H.}$$

The resulting hydrogen atoms are very unstable and will react with anything in reach, but most importantly with $\ce{O2}$ (which is a triplet biradical), forming hydroxyl radicals. This reaction is endothermic (it requires energy). $$\ce{H. + O2 -> HO. + O}$$

The resulting oxygen radicals can again react with $\ce{H2}$ to form more hydrogen radicals. This reaction is also endothermic. $$\ce{O + H2 -> HO. + H.}$$

The hydroxyl radicals can also react with $\ce{H2}$ to form more hydrogen radicals, which is a slightly exothermic (releases energy) reaction.

The net result of those equations leads to a slightly exothermic sum, with high potential: $$\ce{3H2 + O2 -> 2H2O + 2H.}$$

In principle, the hydrogen radicals react as a catalyst. However, this reaction is a highly branched chain reaction, including a lot of radical reactions. Due to this, more and more hydrogen radicals will be produced.

This scheme will eventually end when the concentrations of $\ce{O2,H2}$ will become lower forcing the excess radicals to react with each other. $$\ce{HO. + H. -> H2O}$$

Another possibility is forming as a byproduct hydrogen peroxide, which is a very exothermic reaction: $$\ce{H. + O2 -> HO2.}\\ \ce{2HO2. -> H2O2 +O2}$$

(Also happening but not as important: $\ce{HO. + HO. <=> H2O2}$)

As long as there are hydroxyl and hydrogen radicals present, the peroxide will take part in the chain reaction. However, this reaction also provides the necessary energy to cleave more $\ce{H2}$. peroxide is also easily cleaved again, or reacts with each other: $$\ce{2H2O2 -> 2H2O + O2}$$

This all results in the main product water.

Please keep in mind, that there are many factors, that influence these reactions. It is strongly dependent on pressure, temperature, and surroundings. Surfaces and/ or catalysts involved in this reaction may change it completely.

Further reading:

  • $\begingroup$ I am sure that the combustion of hydorgen gas in the gas phase is by a radical mechanism, I know that alkyl halides such as halon will modify the combustion $\endgroup$ Commented Apr 15, 2018 at 9:11
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    $\begingroup$ @NuclearChemist I am not sure where you are getting at with your comment as the mechanism I describe is a radical one. Obviously if other species are present, the mechanism will change, but that was hardly the point of the question. $\endgroup$ Commented Apr 15, 2018 at 9:17
  • $\begingroup$ The fact that halon tends to modify hydrogen / oxygen flames conforms that the reaction is free radical. Imagine for a momnet if the H2 / O2 reaction occured by a 2+2 pericylic reaction. Then I would expect that halon would have almost no effect on the rate of reaction. Halon works by intercepting a radical in a combustion mechanism to form a lower energy radical which is unable to continue the reaction. I am sure that a hydrogen atom could abstract a Br from CF3Br to form the CF3 radical which I think will have a lower energy than the H atom. $\endgroup$ Commented Apr 15, 2018 at 9:32
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    $\begingroup$ @NuclearChemist I still don't understand where you are going with your comment. You obviously cannot have a pericyclic reaction, the spins don't match. The whole idea of not having a radical mechanism is problematic, as on species is a already a radical. If you really need to consider halogenated compounds, then it is well known that they are relatively easy to split; I wouldn't be too sure that they intercept existing radicals, or are the source of such. $\endgroup$ Commented Apr 15, 2018 at 9:46
  • $\begingroup$ I gave the pericylic reaction idea as an example of a mechanism which is not operating for hydrogen combustion. CH. Kim, OC Kwon, GM. Faeth, JOURNAL OF PROPULSION AND POWER, 2002, volume 18, page 1059 is all about the effect of halon on hydrogen / air flames. $\endgroup$ Commented Apr 15, 2018 at 13:42

The scheme starts with the production of OH radicals since Hydrogen dissociation is very endothermic (432 kJ/mole) compared to $\ce{H2 + O_2 \to 2OH\cdot}$ which is the initiation reaction with $\Delta H^{\mathrm{o}}_{298} = 72$ kJ/mole. There is a propagation reaction and two chain branching reactions and gas phase termination as well as wall termination reactions.

$$\begin{align}\ce{H2 + O_2 &\to 2OH\cdot} \qquad &\text{initiation 72 kJ/mole }\\ \ce{OH\cdot + H2 &\to H2O + H\cdot} \quad &\text{propagation -62 kJ/mole }\\ \ce{H\cdot + O2 & \to OH\cdot + O\cdot} \quad &\text{branching 70 kJ/mole} \\ \ce{O\cdot + H2 &\to OH\cdot + H\cdot} \quad &\text{branching 8 kJ/mole }\\ \ce{H\cdot + O2 + M &\to HO2\cdot + M\cdot} &\quad \text{gas termination -196 kJ/mole }\\ \ce{H\cdot / OH\cdot / HO2\cdot} &\to \text{ wall termination}\\ \end{align} $$

There is strong experimental evidence for $\ce{H\cdot,\, O\cdot,\,OH\cdot}$ and $\ce{\,HO2\cdot}$.

The scheme allows for two explosion limits. The first limit, at low pressure is controlled by surface effects and diffusion to the vessel walls. The second limit, where explosion stops, is determined by the gas phase termination reaction of $\ce{HO2\cdot}$ and is independent of the vessel's size.

Some small amount of peroxide may be formed such as by $\ce{HO2\cdot + H2 \to H2O2 + H\cdot}$ but this acts as an inhibitor $\ce{OH\cdot + H2O2 \to H2O + HO2\cdot}$ because the $\ce{HO2\cdot}$ radical is less reactive that either $\ce{H\cdot}$ or $\ce{OH\cdot}$. Peroxide is perhaps involved in the third explosion limit.

  • $\begingroup$ In your branching steps, $\ce{O^.}$ means $\ce{^.O^.}$ ? $\endgroup$ Commented May 31, 2019 at 15:50
  • $\begingroup$ By convention it is just half of an oxygen molecule. $\endgroup$
    – porphyrin
    Commented May 31, 2019 at 16:01

The combustion of hydrogen can occur both in the gas phase and on the surface of a metal, it can also occur on the surface of platinum. This is used in the hydrogen recombiners which are present in the containment of many nuclear reactors.

Another example of a flameless combustion on a metal surface is the oxidation of solvents such as acetone on the surface of copper metal. Here is one of the many films on youtube showing the reaction. The oxidation of acetone with air using a copper metal surface is much easier to do than the combustion of hydrogen with a platinum surface.

The hydrogen will adsorb onto the metal surface first by physical adsorption and then by chemisorption. This will result in the platinum surface having hydroegn atoms on the surface.

Oxygen molecules from the air will also adsorb onto the platinum, then on the surface a reaction between hydrogen atoms and the oxygen will occur which forms water.

TPD, XPS and LEED were used by Yuichi Ohno and Tatsuo Matsushima to show that oxygen on a platinum surface can be present both as molecules and as atoms, Surface Science, 1991, volume 241, page 47.

If you want to know more then a good refernece would be Study on chemisorption of H-2, O-2, CO and C2H4 on Pt-Ag/SiO2 catalysts by microcalorimetry and FTIR , X. Wang et. al., JOURNAL OF THERMAL ANALYSIS AND CALORIMETRY, 2005, volume 82, page 103.


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