I am aware that combustion reactions are exothermic. The energy stored in the chemical bonds of the products is less than the energy stored in the chemical bonds of the reactants. The difference in energies manifests itself as energy released into the environment. However, this sounds somewhat mystical. How does breaking bonds result in thermal energy, or greater kinetic energy of surrounding molecules? The most satisfying answer that I have seen so far is: the repulsion between like-charge molecules held in a covalent bond is constrained by the covalent bond; once something breaks that covalent bond, the molecules move away from each other, increasing their kinetic energy. Does anyone have further insights on this?

  • $\begingroup$ See also: chemistry.stackexchange.com/q/119093 $\endgroup$ Dec 6 '21 at 16:24
  • $\begingroup$ The forces between the atomic and sub-atomic particles are not "mystical", just complex. For example, the atom nucleus consists of protons that electrically push each other apart, but at the very small scale, the "strong interaction" will keep these together. If you study this kind of physics, you will be able to compute the energy levels of atoms and molecules. $\endgroup$
    – Roland
    Dec 7 '21 at 11:36
  • $\begingroup$ One way of looking at it: Richard Feynman Fire $\endgroup$
    – uhoh
    Dec 7 '21 at 12:04

Better is to say "Energy released by forming bonds of combustion products is bigger than energy needed to break bonds of combustion reactants.".

Particularly breaking $\ce{C-C}$, $\ce{C-H}$ and $\ce{O=O}$ bonds needs less energy than is released by forming $\ce{O-H}$ and $\ce{C=O}$ bonds. The nature of released energy of chemical bonds is electrostatic potential energy being lower than for separated atoms. By other words, breaking the bond does not release energy, it consumes energy, like if you detach 2 magnets, or 2 macroobjects with the opposite charge. It is the forming of bonds that releases energy, via interaction with other molecules while the bond is forming. (*)

When a bond is being created, the bond gets very high vibration energy(**). It has to pass at least part of this energy somewhere else, otherwise the bond breaks again. This is frequently done by the mechanical way, when a moving atoms in this vibrating bond pushing other atom or molecule, passing a part of its energy.

Such a bystanding molecule transforms this energy to its electron excitation, or some form (translation, vibration, rotation) of molecular kinetic energy.

The increased kinetic energy of molecules after reaction contributes to increasing of average kinetic energy of molecules of the mixture being combusted. And, as consequence, to increasing temperature as the statistical measure of the average kinetic energy (per degree of freedom ) for vary large sample of interacting molecules.

E.g. $$\ce{R-CH2-H(slow) + O(slow) -> R-CH2(fast) + O-H(fast)}$$


$$\ce{H(slow) + O(slow) + N2(slow) -> O-H(fast) + N2(fast)}$$

Note that

$$\ce{H(slow) + O(slow) -> O-H(fast) }$$

is not possible due momentum conservation law.

Some of energy can be release as photons too. Some are emitted by excited molecular bonds ($\ce{C-C}$, $\ce{C-H}$), but most of radiation in context of combustion and flame is caused by black body thermal radiation of condensed carbon atom chains, soot or solid particles.

(*) Explanation of quantum chemistry is the topic for thick textbooks, but illustration based on classical analogies would serve well. The proton and the electron in a hydrogen atoms, even if in motion, have lower energy than if they were in rest but free. You need to provide energy to separate them, as they attract each other. Similarly as for planets in the Solar system. When 2 hydrogen atoms are bound in a hydrogen molecule, each electron is around another electrons to repulse each other, but also around another proton to attract each other. The mean squared distance electron-proton is is significantly smaller than the mean squared distance electron-electron or proton-proton, so the total energy is even smaller than for separate atoms. Reaching states with lower total energy of electrostatically bound systems is the source of chemical energy.

(**) Energy of hydrogen molecule - as an example - depends on distance of hydrogen atoms. It decreases with shortening distance, reaches the minimum and that start to raise again when atoms keep getting closer. As the result, there is pushing together if they are too distant and pushing away if they are too close. As the result, hydrogen molecule is able to vibrate along the bond, like if both atoms were connected by a spring, trying to keep them at the particular distance = the length of $\ce{H-H}$ bond. It is just a coarse approximation as the behavious is non-linear. If too close, repulsion raises really steep. If too far, attraction stops raising and starts decreasing toward zero.

  • $\begingroup$ I was posting a comment about IR when I saw that the answer has been edited. Fine! Could you add a warning at the final part (the reaction as collision between slow and fast particles) ? As it is, it might induce readers in mixing statistical ensemble and collisions in well defined systems. (Of course two molecules can react in an quasi vacuum, but then there is no relation to the T). This confusion is the cause of most questions on this topic (although it does not seem the case here, where OP just ask HOWv exothermic reactions heat the surrounding medium) . $\endgroup$
    – Alchimista
    Dec 6 '21 at 9:18
  • 1
    $\begingroup$ @Alchimista Hmm, OK. I have thought it was clear and still think it is correct, but I will reformulate it. $\endgroup$
    – Poutnik
    Dec 6 '21 at 9:26
  • $\begingroup$ I upvote. I wonder if you get what I mean. Basically that if we burn 1 C and 2 O no need to speak of T. The label slow and fast can induce in treating a chemical reaction as a simple well define collision (which can be). $\endgroup$
    – Alchimista
    Dec 6 '21 at 9:29
  • $\begingroup$ @Sure. I have originally written "contributes to increasing temperature", what definitely does not say these few atoms have temperature, but I do agree it may lhave lead to confusion. I have already said multiple times in many questions atoms/molecules do not have temperature. $\endgroup$
    – Poutnik
    Dec 6 '21 at 9:33
  • 5
    $\begingroup$ You do not want me to explain quantum chemistry in a single answer, do you? $\endgroup$
    – Poutnik
    Dec 6 '21 at 18:48

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