2
$\begingroup$

I read that a martensitic transformation can travel at the speed of sound in that material. Is that confirmed as being the fastest speed of a chemical reaction (transformation)?

$\endgroup$
  • 7
    $\begingroup$ If the reaction front travels faster, you have an explosive. $\endgroup$ – Karl Feb 13 at 23:22
  • $\begingroup$ @Karl Posted links in his comments here: chemistry.stackexchange.com/q/54282/79678 . Great link, by the way, and I will delete my other comment, since Karl has this! $\endgroup$ – Ed V Feb 13 at 23:27
  • 1
    $\begingroup$ @EdV On second thought, i think it is not clearly impossible that a solid solid transition spreads at a higher velocity. But of course ot cannot be much faster. $\endgroup$ – Karl Feb 13 at 23:40
  • $\begingroup$ @Karl Based on wiki, stishovite is a polymorph of quartz that is produced in hypervelocity meteorite impacts. There is also a wiki article on “shocked quartz”. I have always assumed that melting was not involved, just shock. $\endgroup$ – Ed V Feb 14 at 0:10
  • $\begingroup$ Does a reaction or a phase transition A - - - > B need to propagate in general and in principle? Besides of seeds, this is obscure to me. $\endgroup$ – Alchimista Feb 15 at 7:34
3
$\begingroup$

Certainly, chemical reactions can propagate far faster than the speed of sound. If the reaction emits electromagnetic radiation, e.g. light, and is itself initiated by light, then the speed of propagation is the speed of light in that medium. See this article from Sandia National Labs, or from the Royal Society.

$\endgroup$
  • 1
    $\begingroup$ "If the reaction emits electromagnetic radiation, e.g. light, and is itself initiated by light, then the speed of propagation is the speed of light in that medium" In such a case the propagation speed wouldn't be the speed of light, because you need to account for the time for the reaction itself. I.e., a particle emits a photon, it travels to another particle, a reaction (which takes times) takes place in which that particle emits a photon, that photon then travels to the next particle, and so on. $\endgroup$ – theorist Feb 14 at 6:52
  • $\begingroup$ @theorist, no, that is only the case where the medium is so closely packed that light cannot propagate within it, and then the maximum speed would be far less. Consider a large lump of nitrogen triodide on one counter, and another a meter away. Theoretically (pardon the expression), light from one lump detonating should reach the other in 1/3*10^8 seconds. BTW, the Sandia lab is based on the idea that the detonation is propagated by light. $\endgroup$ – DrMoishe Pippik Feb 14 at 18:13
  • $\begingroup$ The propagation speed of a reaction is generally taken to mean the speed with which a reaction propagates through the reacting medium. That doesn't describe your system, in which you have two lumps of reactant separated by a meter in a darkened room and shine a light on the first lump, leading to the second lump reacting 1/3 *10^8 seconds later. That's simply a trivial consequence of any reaction that both produces, and is initated by, light. In addition, it makes the term "propagation speed" not chemically meaningful, since you now have a propagation speed that is independent... $\endgroup$ – theorist Feb 15 at 5:24
  • $\begingroup$ .... of the specific physical chemistry of the reaction itself (except that that it produces, and is initiated by, light). It is telling that the Sandia article you reference doesn't contain a single use of the terms "propagate" or "propagation"; it only uses "initiate" and "initiation". The Royal Society article is in German, but the English abstract, again, only uses "initiation", not "propagation". Perhaps a better phrasing to describe your set-up is that is allows separated reactants to communicate at the speed of light. $\endgroup$ – theorist Feb 15 at 5:26
  • $\begingroup$ A better example is UV initiated epoxy resin curing. I just don't know if they are that transparent to UV. $\endgroup$ – user2617804 Feb 15 at 14:25

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.