# Would atomic oxygen and atomic hydrogen result in O2 and H2 only, or some H2O too? [closed]

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:

• $$\ce{O2}$$ and $$\ce{H2}$$ only?
• $$\ce{O2}$$ and $$\ce{H2}$$ with some $$\ce{H2O}$$?
• and how would the process vary with temperature and pressure?
• And I assume creating mostly $$\ce{H2O}$$ 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 $$\ce O$$ and $$\ce H$$)

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). I think I'm more interested in it from a physical chemistry point of view (though without quantum mechanics if possible).

## closed as unclear what you're asking by Mithoron, Todd Minehardt, Martin - マーチン♦Nov 25 '18 at 1:03

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• Thing is, for a guy interested in MO, you're missing crucial thing - energy of your atoms. You could very well have such mixture, but especially in temperature so high that there would be no reaction at all. – Mithoron Nov 25 '18 at 0:34
• @Mithoron I guess energy of atoms can be decided based on temperature, resulting in a distribution of kinetic energies with a mean. Hence, why I'm also interested in knowing what might happen at various temperatures and pressures. – Fi Zixer Nov 25 '18 at 1:08