9
$\begingroup$

I watched a video showing an orange solution that goes to clear and then back to orange (and so forth.) The reaction goes through a number of cycles before it will no longer oscillate. This tells me that the overall energy in the system dissipates over time, but how does such a reaction work?

What is happening that causes this oscillation in concentration?

$\endgroup$
  • 2
    $\begingroup$ If you are concerned in conservation of energy in BZ-reactions, it's all fine: you can imagine the reaction as a physical pendulum whose oscillations are damping over time. More or less complete mechanism includes over 80 steps (L. Gyorgyi, T. Turanyi, R.J. Field. Mechanistic Details of the Oscillatory Belousov-Zhabotinskii Reaction. J. Phys. Chem. 1990. 94 (18) 7162-7170 DOI: 10.1021/j100381a039). If you don't have access to the paper, I can try to summarize the core principles behind in an answer. $\endgroup$ – andselisk Jul 20 '17 at 6:11
  • 1
    $\begingroup$ Have a look at the Oxford Chemistry Primer by Scott ' Oscillations, waves and Chaos in Chemical Kinetics'. This is a good starting point to understand these types of reactions. $\endgroup$ – porphyrin Jul 20 '17 at 7:48
  • 2
  • 2
    $\begingroup$ chemistry.stackexchange.com/questions/267/… $\endgroup$ – Mithoron Jul 20 '17 at 13:40
8
+50
$\begingroup$

I worked on a project using this reaction fairly recently. There are a significant number of intermediate species involved in the reaction with a significant number of autocatalytic steps.

In order to obey the rules of free energy, the reaction may not oscillate about its equilibrium as a pendulum does- there is always a monotonic decrease in reactant concentration and an increase in product concentration, but the multitude of intermediates oscillate in concentration as the reaction switches its reliance on certain sections of the mechanism (dependent upon the instantaneous concentration of stuff in the mixture). The indicator used dictates the colour changes- I used ferroin, and so when the concentration of Ce(4+) (the catalyst used) is high the mixture is red, and when the concentration of bromate is high the mixture is blue.

It should be noted that the reaction will always come to a stop unless you use a continuously stirred flow reactor in which products are removed and reactants added as the reaction proceeds.

There are several mechanisms such as the FKN, MBM, GTF and Oregonator. The Oregonator is by far the most simple but lacks chemical accuracy, the former 3 display real chemical conversions yet still fail to perfectly model the reaction. The mechanisms can be modelled using coupled differential kinetic equations, which I did during my project in Mathematica.

The simplest general reaction scheme for a chemical oscillator is given by the Brusselator:

  1. A → X
  2. 2X + Y → 3X
  3. B + X → Y + D
  4. X → E

In this scheme, A represents the reactions and E the products. X is autocatalytic (it catalyses its own production).

Another interesting example comes into play in population dynamics, the Lotka-Volterra equations (also known as predator-prey equations). They can be used to quantify the changes in populations in a predator-prey system such as foxes and rabbits. The reaction itself has also been suggested as a very primitive form of chemical order- interesting stuff.

Disclaimer: Sorry if this seems a little off-topic; the details of the specific BZ mechanisms are far less interesting than the concept of the oscillators themselves. If you want to look into the actual mechanisms though, start with the FKN model, it's fairly easy to digest and even model.

$\endgroup$
-4
$\begingroup$

Perhaps the simplest oscillating ''Gedankenexperiment'':

1:  A -> B
2:  B -> C
3:  C -> A

Now we say that A inhibits reaction 2, B inhibits reaction 3, and C inhibits reaction 1. If you start with any pure substance A,B or C, the next step never starts before the previous is fully completed.

Of course this needs an energy source, otherwise the last step could never run:

1:  A     -> B
2:  B     -> C
3:  C + O -> A + OO
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.