[fizixer]

Let's assume we have atomic oxygen and atomic hydrogen in a closed chamber, the temperature and pressure of which we're able to monitor (and control). Number of hydrogen atoms are twice that of oxygen.

Would this result in spontaneous formation of:

- O2 and H2 only?

- O2 and H2 with some H2O?

- and how would the process vary with temperature and pressure?

- And I assume creating mostly H2O is the hardest (though I would be interested in a comment on this as well).


and how could one carry out analysis on paper to figure out what would happen? (I'm from a physics background, and my chemistry knowledge is a bit rusty. Though I do have a good background on thermodynamics, enthalpy, Helmholtz free, Gibbs free energies, etc, etc).

For the sake of brevity and focus of this question, we can ignore the question of how atomic oxygen and hydrogen came about in the first place. (This assumption is even more acceptable in the special case of carrying out molecular dynamics computer simulation, something I'm interested in, because we don't have to worry about how to form atomic O and H).

edit: I guess I should clarify that I'm interested in a first principles analysis of this problem (since I'm mainly interested in MD). Also what if we had two other species of atomic gases (Ar and Ne, Ar and O, Ar and H, etc, etc). So I'm more interested in the physical chemistry point of view (though without using too much quantum mechanics if possible).

[Borek]

To have atomic oxygen/hydrogen you need either extremely low pressure or high temperature (or both) or some high energy radiation (somewhere in the UV range) capable of splitting up the molecules. In such a mixture I would expect both recombination and water synthesis, as what happens depends on which atoms collide (which is completely random). However, all three processes will produce enough energy to initiate further reaction between H₂ and O₂ (and let's not forget that this energy was most likely already there if the atomic gases were present). So perhaps initially some small amounts of diatomics would be produced, but they would not survive for long.

If the mixture is stoichiometric final stable state is just water, unless the temperature is high enough for the thermal decomposition.

[Enthalpy]

Provided the heat can escape, you get water.
If not, you also get significant amounts of H, O, OH and more. This happens already in H₂+O₂ flames, and will worsen as atomic O and H make the flame much hotter.

[snorkack]

Provided the heat can escape, you get water.

Or to the contrary?
If heat is hindered from escaping, then any formed H₂, O₂, O₃ or H₂O₂ would undergo repeat reactions towards the most stable H₂O.
If, however, heat can escape and your temperature control freezes the flame, then you should observe recombination to the less stable compounds alongside water.

[Enthalpy]

The compounds unusual at room temperature are in equilibrium in a hot flame. This is not a matter of reaction speed, which is extremely snappy in an H₂ and O₂ flame.

As an illustration of the reaction speed, H₂ and O₂ combustion at rockets is accurately computed with a "shifting equilibrium" where the flame is supposed at equilibrium all the way down the nozzle as the gas expands and cools. The flow takes <1ms from the chamber to the exit.

Here atomic H and O would make an even hotter flame, which creates even more of the radicals and compounds unusual at room temperature.

So: if heat stays, the temperature is hot, and much of H, O and OH exist there. If heat escapes and the temperature drops, the equilibrium shifts to more usual compounds like H₂O.

Here's the composition of the flame, at 1atm before any expansion, of liquid 2H₂+O₂, computed by Propep:

0.62  H₂O
0.14  H₂
0.10  OH
0.06  H
0.05  O₂
0.03  O
plus other minor species, indeed ppm H₂O₂ and ppb O₃.

which explains why engines run fuel-rich: better some unburnt H₂ and H that add little mass than unused O₂ and O.

I convinced Propep to run on atomic 4H + molecular O₂, the 1atm 3770K flame composition is
0.40  H
0.20  O
0.15  H₂
0.13  OH
0.09  H₂O
0.04  O₂
plus ppm of HO₂ and ppb of H₂O₂ and O₃

[snorkack]

So: if heat stays, the temperature is hot, and much of H, O and OH exist there. If heat escapes and the temperature drops, the equilibrium shifts to more usual compounds like H₂O.

Here's the composition of the flame, at 1atm before any expansion, of liquid 2H₂+O₂, computed by Propep:

0.62  H₂O
0.14  H₂
0.10  OH
0.06  H
0.05  O₂
0.03  O
plus other minor species, indeed ppm H₂O₂ and ppb O₃.

Equilibrium shifts, but it does not mean that there is time to establish equilibrium for any temperature. If you suddenly cool the mixture to a temperature where H₂O₂ and O₃ are stable, what are the 0.10 OH going to do? They have a better chance of encountering one of the 0.10 OH than 0.06 H.

[Enthalpy]

Show me a practical way to cool the flame so quickly to achieve that.

OH encountering OH: that will improbably make H₂O₂. Conservation of energy, of momentum and so on. More generally, reactions where 2 molecules or radicals produce 1 rarely happen. It can be a simplified writing, but more species are usually involved. Here the energy of 2OH (2*39kJ) largely suffices to produce H₂O (-242kJ) and expel O (+249kJ).

Anyway, until you find some fantastic means, you can completely forget the idea of freezing an equilibrium that exists at 3700K. The species like H, OH, O are so reactive that a chemical reaction happens at nearly every encounter, that is, at the ps to ns time scale. Ultra-fast cooling means the ms timescale rather.