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If two particles collide and they are under activation energy, why do they JUST “bounce apart”? Isn’t the activation energy steadily decreasing if those particles continuously collide?

I always viewed the activation energy as a "barrier" that prevents collisions from producing a reaction, if sufficient energy is not met. I thought that if particles collide and they do not have enough energy to cause a reaction, the "barrier" gets weakened. Therefore, I found it surprising that this "barrier" doesn't get weakened, and the activation energy isn't reduced.

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    $\begingroup$ Atoms and molecules obviously don't wear out. Not in a minute, or a year, or a million years. If there's no reaction, they stay exactly the same. $\endgroup$ – Karl Apr 29 at 1:14
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    $\begingroup$ Reactants have to jump over the barrier, not bulldoze through it. Think high jump rather than American football. $\endgroup$ – Karsten Theis Apr 29 at 1:15
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This is the nature of the quantum world. An action, at some level, either takes place or does not... there is no half-way state. Another example is the photoelectric effect. Planck and Einstein explained the requirement for at least a minimum energy of a photon before it can raise an electron to a higher energy level. A million photons just under that energy will not* raise an electron to a higher level, but just a single photon of the required energy (or higher) can bump up the electron.

Caveat: There is also the uncertainty principle, allowing electrons to tunnel to a higher level, or for multi-photon absorption to cause ionization, but they're not as common as single-photon absorption and ionization.

If you're bothered by this non-intuitive behavior, you're not alone. Bohr, Bose, Boltzmann, Einstein, and others were forced to this theory through experimental results and the horrors of the ultraviolet catastrophe. Those who fail to appreciate the ultraviolet catastrophe shall be made, energetically, to walk the Planck.

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  • $\begingroup$ (To reader:) Note, however, that classical mechanics would also predict the same behavior, so this isn't a uniquely quantum phenomenon. $\endgroup$ – a-cyclohexane-molecule Apr 29 at 13:31
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As the two species collide their electrons come closer and as they are similarly charged the energy to 'push them together' increases. Some times they just 'bounce off' one another but occasionally when their mutual energy is large enough the repulsion can be overcome and reaction occurs. This is a rearrangement of electrons and nuclei to make a new molecule. As stated in other answers the barrier does not wear down any more than the electron's charge changes, which is not at all.

Recall that by the Boltzmann distribution, arising from the random nature of collisions, generally speaking, most pairs of molecule will not not sufficient kinetic energy when they collide so that only a fraction of pairs of molecules can react on collision.

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@Karl has the right answer. Molecules do not "wear" like macroscopic objects do, and it seems your intuitive angle toward this is to think they do, since you talk of the activation energy "decreasing" with repeated collisions as though such are "softening" them up. Quantum mechanics is not needed; rather, it's more to do with the fact that they are very elementary objects, just a couple levels above fundamental particles.

What we call "wear" on macroscopic objects is really just the loss of molecules or aggregates of molecules from them, and thus results from the fact that they are made of such smaller constituents. To remove constituents from an already-microscopic molecule would mean losing atoms and/or electrons, as those are the only more elementary constituents, and that is effectively the definition of (one type of) a chemical reaction hence, if there is not enough energy to make a reaction go, nothing "wears", and if there is, then indeed they do get "used up" - that's the whole point of chemistry!

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A variety of models have been concocted to describe the dependence of gas-phase reaction rates on the energy of collisions between two molecules. At its simplest you have the empirical Arrhenius equation, the form of which is predicted by collision theory, without need to invoke quantum mechanics. At the very least there are two important aspects to such a model: (1) translational (kinetic) energy can be transferred to ro-vibrational (internal) degrees of freedom of the molecules, and (2) the total energy is conserved. The conversion of translational into vibrational energy may result in breaking of an existing bond. The minimum energy required to do that is what we call the "activation energy". A simple analogy to illustrate this is that of a roller coaster: only if a wagon has enough kinetic energy to get to the top of a hill (where the potential energy is greater - akin to the vibrational energy in the molecular model) will it be able to get over the hill and continue on its path. Otherwise the wagon will roll back the way it came.

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