Timeline for Ozone formation
Current License: CC BY-SA 4.0
22 events
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| Dec 6, 2023 at 9:34 | comment | added | porphyrin | @Martin - マーチン, You are right about OH+O and this was my mistake, I have removed it to avoid any more confusion. These are experiments done in molecular beams at v low pressures where two beams collide and products measured, so not under thermalised conditions. These would be the average of many molec. beam experiments made under many different conditions, such as collision energy, collision angle , vibrational energy and orientation. | |
| Dec 6, 2023 at 9:20 | history | edited | porphyrin | CC BY-SA 4.0 |
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| Dec 5, 2023 at 23:24 | comment | added | Martin - マーチン♦ | I am quite confused. The $\ce{O + O2 ->[M] O3}$ is one step in the ozone cycle, but I cannot remember whether this actually is considered to be an elementary step (I think so). With the examples in the first paragraph though, I am quite positive these deal with gas-phase chemistry and are overall reaction equations and studies. For example $\ce{OH + O -> O2 + H}$ will always go through $\ce{OOH}$, which is a stable, reactive intermediate. | |
| Nov 12, 2023 at 15:40 | vote | accept | user21820 | ||
| Nov 11, 2023 at 15:04 | comment | added | user21820 | I have just read your updated answer. I am reasonably happy with it, thanks! =) | |
| Nov 10, 2023 at 16:53 | history | edited | porphyrin | CC BY-SA 4.0 |
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| Nov 10, 2023 at 12:12 | comment | added | porphyrin | have a look at this article nobelprize.org/uploads/2018/06/porter-lecture-1.pdf which describes a little of what is in the paper. | |
| Nov 10, 2023 at 9:34 | comment | added | user21820 | @porphyrin: I'm not saying I dispute your statements. And yes O3 is a molecule, but why does it not have sufficient freedom in its vibrational modes to store the excess energy? I know for a fact that O3 can have excess energy in its vibration, but nobody has explained why OO2 (or another complex) is necessary for the amount of excess energy involved here, nor how it can even work; why would a weakly bonded complex be stable enough? By the way, regarding ( I + M → IM ), I'm unable to access the cited article, but did they account for adsorption on the container wall? | |
| Nov 10, 2023 at 7:33 | comment | added | porphyrin | species OM if M is an atom or a molecule will have vibrational and rotational energy, i.e. both a diatomic and a polyatomic species has vibrational and rotational energy, there is no basic difference. I don't understand your last posted comment . O3 is a molecule not a complex, OO2 would be the complex with vibrational and rotational energy. I'm not asking you to believe, but its an experimental fact that I+M=IM, leads to IM+I=I2 as a scheme (particularly when M=I), so its not unreasonable to expect the same with O+O2. You will have to search literature for this. | |
| Nov 10, 2023 at 3:46 | comment | added | user21820 | Perhaps you would say that O3 is not a complex, and admits much less variation in vibrational energy. Fine, but why? Even if we pick the reference frame with zero linear momentum and zero angular momentum, the O3 molecule still has other modes of vibration, such as bond angle fluctuation, bond twisting and bond stretching. In a complex, you won't have complete freedom either, so you can't just say "it works". | |
| Nov 10, 2023 at 3:45 | comment | added | user21820 | @porphyrin: Your first sentence does not explain anything, especially since you say "even if M is an atom OM has rotational and vibrational energy". I could by your own reasoning equally claim that ( O + O2 ) has rotational and vibrational energy that can account for the extra energy. Please clearly explain why ( O + O2 → O3 ) is significantly different from ( O + M → OM ). (I'm not satisfied with being told "It is like this so just believe it.".) | |
| Nov 8, 2023 at 19:08 | comment | added | porphyrin | @user21820, the OM complex can form quite easily, even if M is an atom OM has rotational and vibrational energy as well as translational kinetic energy. Momentum (a vector) will also be conserved. Experiment shows that molecules are more effective than atoms in forming a complex, more vibrational modes and more polarisability for electronic interaction . Such a (short lived) complex is the only possible explanation for the $\ce{I + I}$ reaction as observed by experiment, i.e. experiment shows that $\ce{I + M =IM, IM + I\to I2 + M}$ is the mechanism. (Porter Proc. Roy. Soc. v261, p28, 1961) | |
| Nov 8, 2023 at 16:24 | comment | added | Poutnik | Let us continue this discussion in chat. | |
| Nov 8, 2023 at 15:46 | comment | added | user21820 | @Poutnik: What... how?? If you can explain, please post an answer, thanks! I need to go now, but I'll look at it when I get back. But I think you're wrong.. I know quite enough (classical) mechanics.. | |
| Nov 8, 2023 at 15:45 | comment | added | Poutnik | It does go to M :-) . 2 objects can have any momentum (and therefore kinetic energy) each if they cancel each other, | |
| Nov 8, 2023 at 15:44 | comment | added | user21820 | You're not getting the point! In the COM reference frame, ( O + M ) has zero momentum, and hence also after forming a complex. But it initially had positive energy, and where does that go? | |
| Nov 8, 2023 at 15:43 | comment | added | Poutnik | @user21820 M is supposed to be inert, being needed for mechanical reasons to honor conservation laws. It is not some catalyst in chemical sense. It serves just as a ball to be pushed. | |
| Nov 8, 2023 at 15:41 | comment | added | user21820 | @Poutnik: Then explain the problem with ( O + M → OM ), which again has only one product. | |
| Nov 8, 2023 at 15:41 | comment | added | user21820 | And by the way none of the reactions you mentioned in your first paragraph has the conservation problem, since there are enough products to balance the equations. | |
| Nov 8, 2023 at 15:41 | comment | added | Poutnik | The essential is one versus two product entities. The former needs M, the latter does not. All other examples have two products. 2 H -> H2 needs some M too. | |
| Nov 8, 2023 at 15:40 | comment | added | user21820 | So you are claiming that with ( O + M → OM ), the extra energy can somehow be absorbed by the complex formation? I don't get it at all. We still have too much extra energy. (Take the reference frame centred at the centre of mass of the ( O + M ), and after complex formation it has 0 velocity...) | |
| Nov 8, 2023 at 15:15 | history | answered | porphyrin | CC BY-SA 4.0 |