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In reading about chemical oscillations such as those that occur in the Belousov–Zhabotinsky reaction (BZ), it's often reported that these reactions were initially not taken seriously, because of a result that a homogeneous system cannot oscillate around its equilibrium point. It is usually claimed that the assumption of detailed balance guarantees this. I want to understand this result, but I can't find any specific references to it.

I understand very well why this argument doesn't apply to the BZ reaction, since the latter doesn't oscillate around its equilibrium point. However, I'm interested in understanding what this argument is, and whether it does really apply to all oscillations around the equilibrium point.

Kondepudi and Prigogine (2002) imply that the argument is as follows: for a single reaction, one cannot move from one side of the equilibrium to the other without passing through the equilibrium. Once the system reaches equilibrium it can't then keep on reacting, as this would contradict the second law, and therefore an oscillatory approach to equilibrium is impossible in such a system.

This of course makes sense for a system that only involves one reaction. However, for a system in which multiple reactions can take place it doesn't seem to be enough. For example, let's imagine that I have reactions $\ce{A <=> B}$, $\ce{B <=> C}$ and $\ce{A <=> C}$, whose kinetics are determined by some complicated set of catalytic effects. Let's also suppose that the Gibbs energies of A, B and C are the same, so that the equilibrium contains all three reagents in a 1:1:1 ratio.

Now it doesn't seem impossible that I could start off with a high concentration of A and watch the following sequence of events:

  1. most of the A is converted into B, leaving a little bit of A unconverted;

  2. most of the B is converted into C, leaving a little bit of B unconverted;

  3. most of the C is converted into A, leaving a little bit of C unconverted;

  4. this cycle repeats, with the unconverted material building up over time until the system converges to its equilibrium concentrations.

If I were to plot the concentration of the three reagents against one another as a 3D plot, it would look something like this (with apologies for my shaky drawing)

enter image description here

The black dot is the equilibrium point to which the concentrations eventually converge. The path is confined to a plane, because the total concentration of A+B+C cannot change. The free energy decreases as the path approaches the equilibrium point, so the second law by itself doesn't seem to rule out this behaviour.

If a system behaved this way, it would be an example of oscillations around equilibrium in a multi-reaction system. My question is, is there a result that says this cannot happen, if we assume the kinetics obey detailed balance at the equilibrium point?

If so, what is it called, how is it derived, and where can I read about it? If not, is there a known example of a chemical system that does approach equilibrium through oscillations in this way? (A 'toy' system will do, as long as the kinetics obey detailed balance.) Note that the BZ reaction is again not an example of this, since its oscillations do not take place around the equilibrium point.

Perhaps part of the argument is that the reasoning described by Kondepudi and Prigogine must hold for each reaction, independently. For the three-reagent system I described, this seems to be enough to rule out oscillations around equilibrium: the $\ce{A <=> B}$ reaction cannot pass through its equilibrium point, regardless of what happens to the concentration of C. But for a four-species system this no longer seems true: then the $\ce{A <=> B}$ reaction can move away from its equilibrium, as long as it's coupled to the $\ce{C <=> D}$ reaction. If oscillations around equilibrium really are impossible in every conceivable system this seems quite non-trivial, and it would be nice to see a proof.

Reference: Dilip Kondepudi and Ilya Prigogine (2002) Modern Thermodynamics: From Heat Engines to Dissipative Structures. Wiley. 2nd edition.

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  • $\begingroup$ Though I do not know of an answer to this, one thing I can say is that the reactions in the example will not proceed the way you state them. Since all have the same or a similar $\Delta{}G$, the reaction does not complete towards B leaving some A. So it might be a ln 'allowed' path, in any system with enough molecules is simply very unlikely to follow it, i.e. it does not happen. $\endgroup$ – Eljee Dec 11 '14 at 13:42
  • $\begingroup$ "Every conceivable system" is for this question "every conceivable kinetics"? $\endgroup$ – John H. K. Jan 2 '15 at 21:37
  • $\begingroup$ @JohnH.K. yes - or to be more precise, every conceivable kinetics that obeys detailed balance, i.e. $k_f/k_r = e^{\Delta G^0/RT}$. (If you ignore detailed balance then it's easy enough to come up with "toy" systems that do oscillate around an equilibrium value.) $\endgroup$ – Nathaniel Feb 6 '15 at 4:14
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The comments have already mentioned one problem with the question: the trajectory of the example system includes both exergonic (OK) and endergonic (not OK for closed systems) stretches. Think about the total Gibbs free energy of the example with time. The regions where the trajectory is moving away from the equilibrium point are endergonic (i.e. free energy is rising) and would not be allowed unless work is being done to the system somehow.

The other problem is ambiguity in what is meant by "around" equilibrium. Passing through the equilibrium point isn't allowed -- almost by the definition of equilibrium. Gibbs free energy would have to rise (i.e. an endergonic process would be required) after reaching equilibrium.

Orbits "around" equilibrium that always spiral closer to it would potentially be OK. But it depends on exactly what variables are in the orbit. Obviously, Gibbs energy must always decrease in an equilibrating system. Concentrations don't necessarily have to. A good graph is this one: case A is disallowed by the second law, but case B isn't.

a is disallowed, b isn't
(source: pojman.com)

The page where I found that graph does a good job of explaining the topic in much more detail: see here. They present chemically feasible reaction mechanisms that lead to oscillations.

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    $\begingroup$ I'm afraid you've rather misunderstood the question. The trajectory I sketched has no endergonic sections and is always moving toward the equilibrium. Your plot at the end is the point of the question. I'm asking for a proof of it for a general multispecies system. $\endgroup$ – Nathaniel Feb 5 '15 at 23:21
  • $\begingroup$ Apologies for misconstruing your question Nathaniel. I still don't really understand what you are asking for. Are the Brusselator, the BZ reaction, etc., not examples of oscillating "around" equilibrium? Can you give a formula or mathematical definition of what you mean by "around equilibrium"? $\endgroup$ – Curt F. Feb 5 '15 at 23:26
  • $\begingroup$ Also, apologies if I misread your graph. Perhaps it's just a limitation of 3D plotting on a 2D screen, but to my eye there are stretches were the red line first gets closer to the equilibrium point, then moves away from it. $\endgroup$ – Curt F. Feb 5 '15 at 23:28
  • $\begingroup$ Yeah, the Brusselator etc. don't oscillate around equilibrium. In my plot the red line is supposed to be constrained to a plane defined by the grey lines, with the equilibrium in the centre. It might get closer according to Euclidean distance, but (my inexact drawing aside) it should always be going downhill according to the free energy, which is a nonlinear function of position. $\endgroup$ – Nathaniel Feb 6 '15 at 4:00
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    $\begingroup$ In terms of a mathematical definition of oscillating around equilibrium, this would basically be that if I linearise the dynamics around the equilibrium point, the leading eigenvalues of the resulting Jacobian matrix are complex. But basically it's the right-hand side of the figure you posted - the point is that in a multispecies system the concentrations could look like the top-right plot, while the free energy still looks like the bottom-left one. $\endgroup$ – Nathaniel Feb 6 '15 at 4:03

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