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I am trying to figure out how is it possible to reconcile the reality of exothermic reactions, which means that kinetic energy is transferred to (heats) the surrounding matter, with the principle of conservation of momentum, in particular when the product is a single molecule.

I have been exploring the question but found no sufficiently explanatory answer to the question of how the bond energy that is released may create movement (heat). The nearest to an answer that I can imagine is that the reactants enter some transitory combined state of vibration and that further contact with some other molecule results in the separation of each of them in opposite directions with equal but opposite momenta increments thus preserving the overall momentum, while at the same time reducing the vibration and stabilizing the product molecule.

If that were the correct answer (at least in some cases), I would like to know a bit more about the details of the process, which I suppose entail some description about how binding energy is transferred to that vibration.

In any case, it would be interesting to know whether is it actually possible for a single carbon atom and oxygen molecule to react (if they collide with the necessary energy) and produce carbon dioxyde or not, be it for the reason above or another one, since if they are in isolation, the transfer of vibration energy cannot be realized and thus the reaction could not be completed, and then I suppose that would eventually end with the spontaneous separation of the components.

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  • $\begingroup$ First of all, you are asking several questions at the same time. We generally encourage focusing on one, which also helps to get quality answers. Also, what do oyu mean by " I would like to know a bit more about the details of the process, which I suppose entail some description about how binding energy is transferred to that vibration."? Do you you specific questions about the PES of this system? $\endgroup$
    – Greg
    Sep 14, 2020 at 2:13
  • $\begingroup$ I agree that formally there are several questions, but in my mind it's a single one, which may be summarized as: how can it be the proccess (in isolation or not) by which bond energy is converted into movement --macroscopically, heat-- in exothermic reactions while conserving momentum? And, sorry, I don't know what PES refers to... $\endgroup$
    – scifriend
    Sep 14, 2020 at 3:33
  • $\begingroup$ It seems you would benefit from reading about astrochemistry. There are books and presumably websites where similar matters are thoroughly discussed. $\endgroup$
    – Nicolau Saker Neto
    Sep 14, 2020 at 4:51
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    $\begingroup$ There is no single molecule of C. That being said, the concern is legitimate and quite important, so I'd ask everyone to refrain from the close votes. $\endgroup$
    – Ivan Neretin
    Sep 14, 2020 at 5:59
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    $\begingroup$ Aha, I found it - The Physics and Chemistry of the Interstellar Medium. Chapter 4, Chemical Processes, contains a wealth of information. $\endgroup$
    – Nicolau Saker Neto
    Sep 14, 2020 at 12:28

2 Answers 2

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$\ce{C + O2}$ is awfully complicated, so let's just pretend you've asked this:

In a single act of the reaction $\ce{H. + H .-> H2}$, how is momentum conserved?

That's a legitimate concern all right. After all, we are taught that this reaction does happen instantly, once given a chance, and that's in fact true. Also, we know that it releases a lot of heat. Now, heat is nothing but the motion of molecules; how does energy convert to the motion of one molecule as a result of one single reaction act?

It doesn't. The conservation of momentum forbids that, just as you reasoned. Chemical reactions are collective phenomena. Nobody cares about a single molecule.

Now what really happens to a single molecule which has just formed as a result of the mentioned reaction? That's really simple: the molecule is vibrating wildly, ready to break apart. Is has just enough energy to do so. It will do so half of the times, or maybe more often. But that doesn't matter. What matters is that sometimes the vibrating molecule will hit another molecule and sent both of them flying away in opposite directions, thus releasing a part of its energy and becoming more or less stable.

An emission of a photon is also an option, but that's another story.

So it goes.

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  • $\begingroup$ Thank you very much again! I suppose that one can conclude that in isolation exothermic reactions are essentially unstable if not mostly impossible if the product is a single molecule. I'd also tend to agree that photon emission would be if only marginal, taking as reference black body radiation. Nevertheless, I am still intrigued about the details of the transitory, dynamic process by which the binding energy converts to that excess vibrational energy... maybe I could try some readable description? $\endgroup$
    – scifriend
    Sep 14, 2020 at 12:15
  • $\begingroup$ @scifriend When the two atoms are colliding and forming a bond the potential energy of the system can be described as a function of the atomic distance, a curve, that has a minimum around bonding distance: demonstrations.wolfram.com/EnergyLevelsOfAMorseOscillator When the bond is formed the energy is coming from this potential energy difference between the minimum and the r -> infinity values. Imagine a sleigh that goes down to the minimum: it would go up again on the other side. Rolling back and force around the minimum is the vibration you are looking for. $\endgroup$
    – Greg
    Sep 14, 2020 at 13:38
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    $\begingroup$ Not sure exactly about C + O2, but in general the photon emission path is nowhere marginal. This is essentially what happens in upper layers of the atmosphere after the sunset. There are UV-broken molecules that want to recombine. There are few molecules around to "termalize" with. That's why the upper atmosphere slightly glows at night. $\endgroup$
    – fraxinus
    Sep 14, 2020 at 15:20
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When two isolated atoms collide the total energy and momentum must remain with the two atoms so both are conserved overall. In fact in a reaction such as $\ce{H\cdot + H\cdot <=> H2}$ the hydrogen molecule only lasts for a few femtoseconds. This is because even though the bond is formed the atoms will still approach one another (total energy being constant, potential energy becomes more negative and kinetic energy more positive) and rebound as the atoms become very close. The 'molecule' only lasts for a single vibrational period, i.e. a few femtoseconds.

If, however a third body is present, say an inert molecule or atom and this collides with the nascent $\ce{H2}$ molecule then some energy can be taken away from the $\ce{H2}$ and it becomes stabilised. At this point it may radiate away some energy or suffer further collisions and so become thermalised. What happens depends on the relative rate constants for these processes.

In the atom-diatomic collision, e.g. $\ce{F + D2<=> D + DF}$, overall, total energy and momenta is again preserved if there are no other species involved. However, in this case the $\ce{D2}$ has translational, rotational and vibrational energy, this is then partitioned among the products depending on the nature of the potential energy surface describing the approach of the reactants and that of the products. Such 'reactive scattering' has been extensively studied in the gas phase under high vacuum conditions and in molecular beams. See Polanyi & Woodall, J. Chem. Phys. 57, 1574, (1972); Polanyi & Schreiber, Faraday Disc. Chem. Soc. 62, 267, (1977) and textbook by Steinfeld, Francisco & Hase, Chapter 9,' Chemical Dynamics & Dynamics'(Prentice Hall 1999);Levine & Bernstein 'Molecular Reaction Dynamics and Chemical Reactivity' (OUP 1987).

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  • $\begingroup$ So does that mean H + H —> H2 won’t happen in the gas phase? As you’d need some kind of solid body to transfer energy to? Or is an excited H2 molecule colliding into another atom in the mixture enough to dissipate its energy? $\endgroup$
    – Josh Mitchell
    Sep 18, 2020 at 20:46
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    $\begingroup$ Some energy has to be taken away so colliding with another molecule or atom can do this and make a molecule. Collision with a surface will also remove energy. Most likely here though is that reaction occurs by a different mechanism which is atoms diffusing together when on the surface. $\endgroup$
    – porphyrin
    Sep 19, 2020 at 6:53
  • $\begingroup$ @porphyrin Thank you very much for your illustrative answer and references. However, I am still intrigued about how the chemical energy in, say, the O2 bonds can be released into vibrational energy in the CO2 product, even if the reaction is, as Ivan Neretin said, terribly complicated. Is that type of energy conversion a well described process? Could someone give a summary description or a reference? $\endgroup$
    – scifriend
    Sep 20, 2020 at 3:19
  • $\begingroup$ Have a look for articles on LEPS potentials, and atom-diatomic reactions or the text books in my answer. The important point is that energy in translation vibration and rotation are not isolated from one another but are 'coupled' so that energy can flow between them. $\endgroup$
    – porphyrin
    Sep 20, 2020 at 7:03
  • $\begingroup$ @scifriend Carbon and Oxygen are complicated because there are so many effects that make a mess of any single, simple way to explain it. A simple model without explanatory power is not of any use to anyone. Carbon and Oxygen react through intermediary species, with rate determining steps (forbidden transitions), radicals and the product is rarely carbon dioxide, that is the final product of a series of unstable carboxides. You won't find a nice way to track what happens to momentum... I think... $\endgroup$
    – Stian
    Dec 2, 2020 at 12:11

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