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, $\ce{2 H2O}$.
Answer
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:
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