As mentioned in this question, the idea of chemical potential energy explains how the thermal/kinetic energy of particles can change in a reaction; energy is taken in / released by the bonds broken / formed.

However, consider the reaction A -> B + C occurring in a closed system of just these single particles.
A reaction such as this is often endothermic. Therefore, the kinetic energy of the products must decrease. However, it is impossible to remove kinetic energy from the system without changing momentum (unless B and C are travelling in opposing directions to begin with, which is not the case. If they split in opposite directions, that will have already resulted in a KE gain).

How can this happen without violating the principle of conservation of momentum?

EDIT: Here is an explanation showing that it does not work if B and C travel in opposite directions:

  • $\begingroup$ Another point is that the average KE relates to T for a statistical ensemble. Your standing still A molecule would be in your picture at 0 K. It is not clear what you mean, at least to me. The point I want to make is that KE will decrease unless you postulate the starting KE is zero. $\endgroup$
    – Alchimista
    Feb 10, 2018 at 10:05
  • $\begingroup$ Or put it like this: B + C --> A is not an elastic collision at all. It is again related to the previous Q you have mentioned. $\endgroup$
    – Alchimista
    Feb 10, 2018 at 10:20
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    $\begingroup$ I'm voting to close this question as off-topic because it's not about chemistry. $\endgroup$
    – Mithoron
    Feb 10, 2018 at 13:52
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    $\begingroup$ @Mithoron It's about the crossover between chemistry and physics. I could post it on the physics Stack Exchange, but what's to stop someone over there saying the same thing? I'm asking what part of the behavior of molecules allows the laws of chemistry to agree with the laws of physics, because at the moment I can't see that they do. Sounds like a chemistry question to me. $\endgroup$ Feb 10, 2018 at 16:54
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    $\begingroup$ Is air hotter when the wind is blowing? $\endgroup$
    – Zhe
    Feb 12, 2018 at 22:58

4 Answers 4


How do you measure temperature? The thermometer has to be in thermal equilibrium, so you need to send the thermometer with the object. But that means that temperature, for our purposes, is going to be invariant between different reference frames. No matter how you're moving, the reading on the thermometer is the same.

This is clearly not the case with translational kinetic energy of the bulk material. In the rest frame of the object moving with constant velocity, the kinetic energy is zero.

Therefore, you should not mix up the thermal kinetic energy of the system with the translational kinetic energy of the system. They can't be the same thing. In other words, when the exothermic system loses energy, you don't get to subtract that from the translational kinetic energy.

See also these discussions: here and here.

I would also like to point out that you've just described a rocket. And rocket science, while complicated, is still obeying conservation of energy and momentum.

  • $\begingroup$ Translational kinetic energy of the system is irrelevant. Look at one molecule (in an inertial reference frame, if you'd like); can it decompose? $\endgroup$ Feb 13, 2018 at 8:32
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    $\begingroup$ @IvanNeretin I think we're saying the same thing. Translational kinetic energy is irrelevant when you consider the thermal energy, even if that's derived from random translational kinetic energy. $\endgroup$
    – Zhe
    Feb 13, 2018 at 13:59

That's an excellent question, and your reasoning proves exactly what it seems to be proving: that such a scenario is not possible. A molecule can't just up and transform its own kinetic energy into some endothermic reaction within itself, since the conservation of momentum forbids that. Whether or not the molecule snaps in two (or maybe more) parts, the total kinetic energy must become lower, and the momentum must remain the same. There is no way to reconcile these contradicting demands.

How are endothermic decomposition reactions possible, then?

By looking at the bigger picture, that's how. Say, a molecule bumps into a wall or another molecule. Its momentum is lost; the constraint no longer applies. (The total momentum of the system is of course still conserved, but the system is now bigger than just one molecule). Now the molecule may bounce off like a tennis ball and fly away at the same speed (that's what we call an elastic collision). Or... it may react.

So it goes.

  • $\begingroup$ Actually you should correct the final paragraph as for momentum is always conserved. $\endgroup$
    – Alchimista
    Feb 13, 2018 at 9:10
  • $\begingroup$ And no, I want to make clear that I amnot he downvoter. Normally I would have just suggested to you a correction. Even more so after our opinion exchange. $\endgroup$
    – Alchimista
    Feb 13, 2018 at 9:13
  • $\begingroup$ OK, got it. Maybe I should have made it more clear. $\endgroup$ Feb 13, 2018 at 9:16
  • $\begingroup$ Ivan your answer is clear. Just look at the final paragraph $\endgroup$
    – Alchimista
    Feb 13, 2018 at 9:17

Temperature is not a vector quantity though the KE of an individual particle is

The apparent problem here arises because of a confusion between the bulk view of a reaction and the view from the perspective of individual molecules.

Individual molecules have kinetic energy and, in reactions, the kinetic energy may change if chemical energy is released or absorbed during a specific reaction (this is a conversion between potential energy in chemical bonds and kinetic energy). But this doesn't translate into the bulk perspective. Temperature is, in one sense, the average kinetic energy of the molecules in the system. Different molecules have different kinetic energies (Boltzmann distribution of them, in fact) but the net vector of those kinetic energies is zero as they are moving in random directions. Moreover, they are constantly exchanging their kinetic energy in random collisions with other molecules.

If the reaction vessel is not closed (so the products can escape) this point of view is incomplete. The net kinetic energy of the products will no longer be zero (the molecules can escape through the open part of the vessel taking their kinetic energy with them). The molecules can strike the closed end of the vessel and bounce off transferring momentum to the vessel and this is not compensated by by molecules striking the other end of the vessel as there isn't one. This is how rockets work, in effect translating the chemical energy released by a reaction into kinetic energy which is ultimately transferred into the kinetic energy of the rocket.


This answer is meant to put some order to my comments that, lets fragmented, appeared cryptic to the OP.

Moreover I try to answer without editing. The involved calculations just require second degree equations thus every reader can easily figure them.

The OP is forced to confusion because he assumed that the kinetic energy KE of one particle must be reflected in temperature T. This is a wrong assumption as for in a statistical ensemble of particles, a number of them have KE lower or higher than the average value, which is the one related to the macroscopic value T of the system.

See, for instance, https://physics.stackexchange.com/questions/65690/can-a-single-molecule-have-a-temperature

where all the answers deserve attention but the most voted one suffices to the current discussion.

The wrong assumption, once made, drives OP to think that the dissociation A --> B + C, being endothermic, results in a lower T of the products, or in other words the process takes energy at cost of the KE of the particles.

This is true for an ensemble in which collisions occur but obviously not in the system under consideration.

The only way for A dissociating is to input energy into the balance.

The system is easily treated supposing molecule A at rest, either isolated or in the ideal box as in the question.

Nothing happens to it unless the enthalpy of reaction DeltaH is inputted to the system in a way or another, let us assume is already in the box if we want to keep it closed and isolating.

In this case the conservation of momentum requires

mBvB = mCvC

and the conservation of energy

DeltaH = KEB + KEC.

There is not a disequality (forced by the vicious assumption endothermic process => T decreases => KE decreases) but just energy conservation.

In the case of A moving with an initial speed the treatment is the same. It suffices to treat the system in a new reference frame applying galilean relativity.

The thermochemistry and the physics of the process remains exactly the same.

In short, the term (a - 5)^2 sketched as example by the OP equals DeltaH.

A more realistic scenario is as follows. Let us open a small hole to the box and shoot A through it with a certain speed. As soon as A enter the box the aperture is closed and the box is isolating.

Now we can wait until A collides to the wall. Assuming KE of A is enough, the collision with the wall*** (not the following dissociation!) can be treated as an elastic one which solely triggers the dissociation reaction itself. In this case the DeltaH is taken at expense of kinetic energy, but indeed there is a collision transferring the required amount of energy. In a way the scenario is more similar to that of a statistical system of particles (equipartition and microscopic interpretation of T do not hold, tough and still).

The equations are now***

mAvA = mBvB + mCvC


KEA - DeltaH = KEB + KEC

Ideally B and C will recombine along their trajectory giving back A and so on. However no work no heat can be extracted.

Although the real molecule(s) have other channels to store/transfer energy, the above arguments shall answer the question.

***for the treatment of the collision to the wall, and specifically for the reversal of velocities, the trick is as for the elastic bouncing of a small ball against a rigid and much more massive body.

  • $\begingroup$ Temperature is irrelevant. Ensembles are irrelevant. Look at one molecule. It has some kinetic energy and some momentum. Can it decompose? $\endgroup$ Feb 13, 2018 at 8:33
  • $\begingroup$ @Ivan Neretin don get your comment. It is basically my answer. I had to explain why OP was wrong in assumption. Of course the molecule can't decompose by virtue of just kinetic energy , wasn't clear from my text? $\endgroup$
    – Alchimista
    Feb 13, 2018 at 8:44
  • $\begingroup$ @Ivan Neretin. I copy and paste from Answer above. " Nothing happens to it unless the enthalpy of reaction DeltaH is inputted to the system in a way or another,". $\endgroup$
    – Alchimista
    Feb 13, 2018 at 8:51
  • $\begingroup$ Then you answer contains a lot of unnecessary statements, and, while technically right, may still appear cryptic to a less knowledgeable reader. $\endgroup$ Feb 13, 2018 at 8:57
  • $\begingroup$ @Ivan Neretin. My answer is just your answer worded as such to answer OP. Going through the wrong assumption he made and correcting his p and KE balances. You could have just not commented under it. $\endgroup$
    – Alchimista
    Feb 13, 2018 at 9:02

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