[benzophile]

Hi everyone,
I'm not really sure where to post this, but these questions arose when I studied organic chemistry 2 years ago, so here goes. My first question is about the equilibrium of reactions. As far as I know (after taking g chem), every reaction has an equilibrium point, even if it is highly small or highly large. But say you have a mol of O₂ and a mol of H₂ and let them react to form water. At 25 C I believe K is so high that it says there is less than 1 molecule of oxygen and hydrogen at equilibrium. In this case, since you cannot have a fraction of a molecule, does the reaction go 100% to completion and stay this way, or does one molecule of hydrogen and oxygen pop up now and then and then react to form 100% water shortly after? Hope that makes sense.... I was also wondering if every reaction can, at least in theory, go backward (if every equilibrium is a dynamic equilibrium), or if some mechanisms can only go one way. Some funkier mechanisms like ozonolysis and Bayer-Villager oxidation seem to only be able to go in one direction, but maybe I'm wrong.

[Hunter2]

If you mix oxygen with hydrogen nothing happens. You need some activation energy. Only you have a spark or some heat source the reactions starts....boomm.  The reaction goes 100% to water. 

[curiouscat]

I'd say it is always dynamic. In theory I don't think there can be any process that is strictly one directional.

[benzophile]

hunter I'm aware of the kinetics of the reaction, but thanks for trying. I also thought that, curiouscat, but I have no proof or backup for it.

[curiouscat]

hunter I'm aware of the kinetics of the reaction, but thanks for trying. I also thought that, curiouscat, but I have no proof or backup for it.

I don't know what you mean by "proof"; but I'll try: Take this Fig. ; In your case the reverse barrier is just very high. That makes ΔG very negative (and hence Keq. very large). What physical reason would there be for a molecule to not jump in the reverse direction?

[orgopete]

With the graph, I thought Keq would be determined by the difference in initial and final, that activation energy determines how fast or slow it is. For example, in the presence of a spark, the reaction does not go backwards, or simply reach equilibrium. If a catalyst was added that reduced the activation energy, the equilibrium would not be altered, though the speed of the reaction would be increased.

I presume that in a nuclear reactor, water is homolyzed and hydrogen and oxygen form. Then a spark and boom. If anyone know this, please comment.

[curiouscat]

With the graph, I thought Keq would be determined by the difference in initial and final, that activation energy determines how fast or slow it is. [...] If a catalyst was added that reduced the activation energy, the equilibrium would not be altered, though the speed of the reaction would be increased.

Correct.