<p>It depends all on the reaction rates between all the <span class="math-container">$n$</span> different atoms and molecules. So you need to know all the <span class="math-container">$n\times n$</span> reaction cross sections in order to be able to solve the <span class="math-container">$n$</span> equations for the <span class="math-container">$n$</span> unknowns. In addition you may have to include the production and loss rates due to the radiation which could result from the covalent bonding and which in turn could dissociate other molecules. Without knowing each of the reaction coefficients it is impossible to say what the eventual outcome will be. Just one particularly strong or weak reaction channel could affect the whole system as everything is coupled to some degree.</p>

CC BY-SA 4.0
2023-10-20T18:39:47.563Z
2023-10-20T18:39:47.563Z
Thomas
<p>I have read about the actual formation of interstellar molecules; although interstellar chemistry is very different from solvent-based chemistry on earth, I understand it to be driven by the presence of dust grains, and by the constant ionization by cosmic rays, which facilitates exchange reactions.</p>
<p>Now I'm wondering about this hypothetical case:</p>
<ul>
<li>we start with a hot atomic gas with a density similar to the interstellar medium, with the same type of composition we can observe now (74% H, 24% He, O, C, N, Fe, Mg, Si, S,...)</li>
<li>we let that gas cool down, there isn't any radiation nor any dust.</li>
</ul>
<p>Without the effects that I mentioned above, covalent bonds could only form when collisions happen between two atoms.</p>
<p>Of course, the more exothermic a formation reaction is, the more likely a collision between two atoms to lead to a covalent bond.</p>
<p>By the time the gas is completely cold, how much of it should I expect to be molecular?</p>
<ul>
<li>Would I only see O2, N2 and the likes?</li>
<li>Would molecules like CH4 ever get to form? I understand its enthalpy of formation is negative, but I don't know if that's the case starting with just atomic C and atomic H?</li>
<li>Would metals ever bond with anything, or is the probability of formation of Ferric/Magnesium oxides too low given the low density of metals and oxygen?</li>
</ul>

CC BY-SA 4.0
2023-10-22T02:37:46.707Z
Have a read at this: https://academic.oup.com/mnras/article/437/1/9/992280
Nilay Ghosh
<interstellar-medium><astrochemistry><molecules>
Molecule formation when an atomic gas cools down
SgtJohn74
# Molecule formation when an atomic gas cools down

(Asked by: username_0 on 2023-10-18)

I have read about the actual formation of interstellar molecules; although interstellar chemistry is very different from solvent-based chemistry on earth, I understand it to be driven by the presence of dust grains, and by the constant ionization by cosmic rays, which facilitates exchange reactions.

Now I'm wondering about this hypothetical case:

* we start with a hot atomic gas with a density similar to the interstellar medium, with the same type of composition we can observe now (74% H, 24% He, O, C, N, Fe, Mg, Si, S,...)
* we let that gas cool down, there isn't any radiation nor any dust.

Without the effects that I mentioned above, covalent bonds could only form when collisions happen between two atoms.

Of course, the more exothermic a formation reaction is, the more likely a collision between two atoms to lead to a covalent bond.

By the time the gas is completely cold, how much of it should I expect to be molecular?

* Would I only see O2, N2 and the likes?
* Would molecules like CH4 ever get to form? I understand its enthalpy of formation is negative, but I don't know if that's the case starting with just atomic C and atomic H?
* Would metals ever bond with anything, or is the probability of formation of Ferric/Magnesium oxides too low given the low density of metals and oxygen?

**Tags:** interstellar-medium, astrochemistry, molecules
> **Comments:**
> **username_1**: Have a read at this: https://academic.oup.com/mnras/article/437/1/9/992280  

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## 1 answers

### Answer by username_2 on 2023-10-20. Score: 2

It depends all on the reaction rates between all the $n$ different atoms and molecules. So you need to know all the $n\times n$ reaction cross sections in order to be able to solve the $n$ equations for the $n$ unknowns. In addition you may have to include the production and loss rates due to the radiation which could result from the covalent bonding and which in turn could dissociate other molecules. Without knowing each of the reaction coefficients it is impossible to say what the eventual outcome will be. Just one particularly strong or weak reaction channel could affect the whole system as everything is coupled to some degree.


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