Chemistry - Reaction mechanism of combustion of hydrogen
Solution 1:
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:
How explosive is hydrogen gas?
Eugene P. Dougherty1 and Herschel Rabitz1, J. Chem. Phys. 72, 6571 (1980) most comprehensive (theoretical) study on the $\ce{H2-O2}$ system.
Wiberg, Nils: Lehrbuch der Anorganischen Chemie Kap. 3.2 (german) The mechanism is in part taken from this source.
http://arrow.utias.utoronto.ca/~ogulder/ClassNotes3.pdf Thanks to Silvio Levy
Solution 2:
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.