I've been reading about how energy is released when new chemical bonds are formed but it's hard to find info on how that typically manifests itself (besides that it's via heat). I read somewhere that the valence electrons will move into lower energy orbits during a bond, but I thought that always created a photon. How and where is the heat actually "created?". When two H atoms form H2, does the new molecule start moving faster (discounted for increased mass)? What happens if there are no surrounding molecules?
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5$\begingroup$ Electronically excited states transition into vibrationally excited states, which desactivate nonradiative by transfering energy to colliding neighbouring atoms, creating heat. $\endgroup$– KarlCommented Aug 11, 2019 at 22:05
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$\begingroup$ chemistry.stackexchange.com/questions/111936/… $\endgroup$– MithoronCommented Aug 11, 2019 at 23:06
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$\begingroup$ chemistry.stackexchange.com/questions/55914/… $\endgroup$– MithoronCommented Aug 11, 2019 at 23:13
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$\begingroup$ chemistry.stackexchange.com/questions/63597/… $\endgroup$– MithoronCommented Aug 11, 2019 at 23:13
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1$\begingroup$ Second link is for duplicate imo. $\endgroup$– MithoronCommented Aug 11, 2019 at 23:14
1 Answer
As pointed out by @Paul in case of the reaction between 2 hydrogens atoms you need a third body to take away the excess energy. This third atom(M) can be another hydrogen molecule or other inert gases(eg. Ar, He, N2).
Or you can have the presence of a wall(eg. gas-phase hydrogens atoms in a container):
Another possibility is adding a reactive gas(eg.oxygen) and in this case, the mechanism is much more complicated with many steps.
All these reactions are discussed in ref.1.
Comment: when two hydrogens atoms combine 4.5eV is released. In the case of an inert third body, I haven't found any quantitative information about how much energy is transferred to the third inert gas. If all of it is transferred to the third body(eg.H2 molecule) as kinetic energy you end up with an H2 molecule moving at 20'800m/s that corresponds to a kinetic temperature of 34'800K!
Another example is the CsBr formation in the presence of a third inert body(R):
From ref.2:
Analysis of the detailed dynamics of recombination at small impact parameters showed that the stabilization of an ion pair occurs through the collision of the third body with the two ions and that the dynamics of energy transfer in this region is almost completely determined by the repulsive interactions of the Xe atom with both the ions.
In the range of impact, the main mechanism of stabilization is energy transfer energy through repulsion between the third body and both the ions or through consecutive collisions of the third body with the ions approaching each other.
Regarding how much energy is transferred to the third inert body:
The amount of energy transferred to the third body is apparently determined by the collision configuration and by the phase of the collision of the ions at the moment of energy transfer.
References
Bennett, J. E., and D. R. Blackmore. 1971. “Rates of Gas-Phase Hydrogen-Atom Recombination at Room Temperature in the Presence of Added Gases.” Symposium on Combustion 13 (1): 51–59.
Kabanov, D. B., and L. Yu Rusin. 2012. “Mechanism of the Direct Three-Body Recombination of Atomic Ions in a Central Collision.” Russian Journal of Physical Chemistry B 6 (4): 475–85.
Stepukhovich, Aleksandr Davidovich, and Viktor Markovich Umanskii. 1969. “Kinetics and Mechanism of Three-Body Recombination of Atoms and Radicals.” Russian Chemical Reviews 38 (8): 590.
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$\begingroup$ What if there isn't a third atom? How does the energy increase manifest itself in the H2 atom? increased kinetic energy? Thanks! $\endgroup$– M DCommented Aug 12, 2019 at 15:29