Ozone formation

Question

I have read about the formation of natural atmospheric ozone on Wikipedia, where it is claiming that the ozone creation step requires an extra molecule in order to conserve momentum and energy:
$$\ce{O + O2 + A -> O3 + A}$$

I understand that the stated reason implies that it is impossible to have $$\ce{O + O2 -> O3},$$ but I do not understand why it cannot be:
$$\ce{O2 + O2 -> O3 + O}.$$

Specifically, why does the source claim that we have complete cleavage of the $\ce{O2}$ double-bond before the $\ce{O^{..}}$ radical attacks an $\ce{O2}$ molecule? Why is it not possible that after one bond is cleaved we have an $\ce{^.O−O^.}$ radical reacts with an $\ce{O2}$ molecule directly to form $\ce{O3}$ and an $\ce{O^{..}}$ radical? In particular:
$$\ce{O=O + ^.O−O^. -> O=O^+−O^- + O^{..}}$$

This not only solves the momentum+energy conservation problem, but also seems simpler. The other proposal requires 3 molecules to meet at the same time and does not really explain how the momentum/energy gets transferred.

Answer 1

The $\ce{O + O2 + M \to O3 + M }$ does appear written as such in many texts and is unusual in that other atom-diatom reactions that have been studied do not seem to need any $M$, such as $\ce{OH + Cl \to HCL +O}$, $\ce{ H +NO2 \to OH +NO}$, $\;\ce{F + H2 \to HF + H}$, $\;\ce{O +HCl\to OH + Cl}$ and many others. How a reaction proceeds depends both on the shape of the potential surface between the four species involved as the reaction evolves and on how much rotational and vibrational energy the diatomic has and the mutual approach velocity.

There could be two reasons for writing the reaction with $M$ given that a three body collision reaction is very unlikely to happen on statistical grounds and that is either that the nascent ozone has so much energy that it needs a collision with a third body to remove this energy to stabilise it, or that a transient complex is initially formed such as $\ce{O + M \to OM}$ or $\ce{O2 + M\to O2M}$ which then reacts with the remaining species to give products. The $\ce{I + I \to I2}$ reaction proceeds via an $IM$ intermediate, for example.

(Any linear momentum considerations can be accounted for by different velocities after reaction and changes in rotational energy levels for angular momentum conservation. The $\ce{O3 + O \to O2 + O2 }$ reaction has quite a large activation energy $\sim 18$ kJ/mol so the reverse reaction probably has a larger one making it even slower that this one).

Update in response to questions

The energy the two species have for reaction is relative to their difference in velocity, this may be quite small relative to any dissociation or activation energy. At $400$ K the average thermal energy is only $\approx 265\;\mathrm{cm^{-1}}$. The ozone and oxygen vibrational (B=rotational) (all frequencies are in wavenumbers) O2, 2063 (B=1.4) and O3, 1110,705,1042 (B=3.5,0.4,0.39) so the molecules are vibrationally 'cool' and rotationally 'hot'. The O3 dissociates at $\approx$ 11700, i.e. a huge energy relative to thermal energy, but not to UV photon energy.

When then O+O2 species approach (or any A+B species) they experience one another's potential before collision (they are not hard spheres) this potential is, approximately, of a Lennard-Jones type so the species are attracted initially and move towards one another at a distance many times typical bond length. At close range repulsion can occur. A reaction may occur provided the energy is not too great or too small and the approach trajectory correct or the O and O2 will partially orbit and then separate. (see figure from https://chemistry-maths-book.com/chapter-11/). Typically this last event is the most likely.

If reaction can happen the transient species $\ce{OO2}$ forms. The presence of the extra O atom will weaken the O2 bond and energy is transferred into new vibrations (this will take only a few picoseconds) and reaction may be complete, most of the time however the transient species falls apart with a bit of energy having being transferred between the departing species. If an O3 molecule is formed it has all the energy of the collision but this is small compared to dissociation energy and so the molecule will be stable but initially be vibrationally and rotationally 'hot'. This will remain so until a non-reactive collision with an other species removes some energy and thermalises the new O3 molecule. How long this takes depends on the pressure, but at experimental pressures of a few torr, will occur within tens of nanoseconds

The impact parameter is the distance away from the line of centres. This diagram is for spheres, but should give the general idea even for a slightly no spherical molecule such as O2.

Answer 2

Evaluating $\ce{2 O2 -> O3 + O}$ singlet oxygen hypothesis:

• The molecule $\ce{O2}$ – also called triplet oxygen – has in its standard state a double bond, but it already is a biradical CH SE: why-is-o2-a-biradical. It is paramagnetic and has 2 unpaired electrons.
• The next state with higher energy is the singlet oxygen, is paramagnetic and fits the classical idea about double-bonded biatomic molecule. It is formed by specific chemical reactions and is very reactive.
• The opposite reaction $\ce{O3 + O -> 2 O2}$ is fast.
• Singlet oxygen is not $\ce{\cdot O-O\cdot}$, but rather $\ce{O=O}$, while the triple oxygen is like $\ce{O÷O}$.
• Single oxygen in upper atmosphere is not formed from triplet oxygen, but by photolysis of ozone.
• Molecular entities can emit a photon, but it is a process much slower than the resonance break.
• $\ce{O2->[UV]2 O}$ is supported by the spectroscopic evidence of oxygen UV spectrum, when absorption below about $\pu{240 nm}$ confirms the breaking of oxygen molecules to atoms, as the needed energy is equivalent to $\pu{241 nm}$. There are not shown any local maxima at higher wavelengths, suggesting some lower energy photon excited oxygen molecule.

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The third object mechanism:

Reminding the known:
In reference frame of zero total momentum, without $\ce{M}$, resulting $\ce{O3}$ from assumed $\ce{O2}$ + $\ce{O}$ collision has zero momentum and zero translational energy. Due angular momentum conservation, it has comparable rotational energy as original molecule $\ce{O2}$. Plus lower potential energy due bonding. Something does not fit. It is like when a skateboarder runs through an U-ramp and ends up on the other side free. Molecules do one vibrational period, bounce and part away.

If there is any inert entity $\ce{M}$, both $\ce{M}$ and $\ce{O3}$ are parting with the same but opposite sufficient momentum, according to conversion of energy released by bonding to translational energy. Like we when a skateboarder hits something and does not make it to the other side, being "bonded to the U ramp".

When there are two products, one plays simultaneously the role of $\ce{M}$, like in case $\ce{NO + O3 -> NO2 + O2}$.

Involving $\ce{M}$ does not mean that collisions of $\ce{O}$, $\ce{O2}$ and $\ce{M}$ occur at the same time point. It means one collides when the other two are still attached during resonance meeting before parting.

Resonance can be understood by 2 hydrogen atoms, approaching each other. They reach maximum kinetic and minimum potential energy at the distance equal nuclei distance of a hydrogen molecule. But the motion continues, kinetic energy decreases to zero, reaching maximum potential energy. Then the motion continues like if reversed in time. In fact, in is one vibrational period on the maximum energy level with one side amplitude in infinity (and then some).

Without existence of resonance, there would be no carbon in universe and no life based on carbon compounds. In fact there are even 2 resonances in row:
$\ce{2 ^4He <<=> ^8Be^{*}<<=>[+^4He]^{12}C^{*}->^{12}C + \gamma }$. These resonances have many orders shorter life than resonances of molecular entities. See Beryllium-8 and Triple-alpha_process.

As a curiosity, see Atomic hydrogen welding.