Timeline for Additional Questions Regarding the Auto-Ignition Temperature
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 13, 2015 at 5:00 | comment | added | Yandle | @charlesreid1 My apologies I worded that poorly. I was referencing to this diagram where enthalpy is shown as a difference between the activation energies of forward and reverse. Just from a rate perspective if the reverse rate >> forward rate, at the same T, from re-arranging the Arrhenius equation I arrive at Ea_rev << Ea_fwd which is the case for an endothermic reaction and the source of my confusion. | |
| Jun 12, 2015 at 16:53 | comment | added | charlesreid1 | no - you're confusing thermodynamics and kinetics. reaction enthalpy changes and rate constants are independent. | |
| Jun 12, 2015 at 15:08 | comment | added | Yandle | @charlesreid1 If the reverse rate constant is greater than the forward, doesn't that imply that the enthalpy change of the forward reaction is positive (as Ea_rev < Ea_fwd), which implies an endothermic reaction (assuming T is constant) while combustion is exothermic? | |
| Jun 12, 2015 at 8:41 | comment | added | charlesreid1 | "forward rate constants... are all greater than zero" - yes, but there are reverse rate constants too, that are much, much larger, so at equilibrium your reactants are always reactants. There's no "spontaneous" combustion, just chemical reactions. | |
| Jun 12, 2015 at 5:40 | comment | added | Yandle | @charlesreid1 What I'm confused about is that, if we have an closed system of fuel-air, T and Ea are finite, and per Arrhenius equation forward rate constant and reactant concentration are all greater than zero. From this I expect a net formation of products (given that initial product concentration is zero) until equilibrium. I get the logic in your explanation but I also don't know the flaw in my logic. Also, isn't AIT technically a point where fuel spontaneous combust? | |
| Jun 8, 2015 at 5:01 | comment | added | charlesreid1 | A fuel-air mixture does require an ignition source to combust. a spark is just a big shot of energy to the system, transferred from one set of (say, metal) molecules to another set of (say, methane) molecules. that energy starts tearing apart molecules to create radicals, which initiates combustion. it isn't a spontaneous process at all. otherwise, all fuel in existence would spontaneously combust! | |
| Jun 8, 2015 at 4:39 | comment | added | Yandle | @charlesreid1 (1) A little confused. The answer implies that fuel-air require an ignition source to or else reaction will not occur. But aren't oxidation reactions spontaneous so we should expect the products to form so long as T (and thus forward rate constant) is greater than zero? My current though is that as t -> infinity, enough fuel will react such that equilibrium will reached (and depending on the heat of reaction there should be very little fuel remaining). | |
| Jun 6, 2015 at 17:14 | comment | added | charlesreid1 | That's getting into the realm of metaphysics. The mass of fuel you'd need would be greater than the mass of carbon in the universe. Sounds like navel-gazing to me... | |
| Jun 6, 2015 at 3:23 | comment | added | Nicolau Saker Neto | Regarding (1), given infinite time, every finite kinetic barrier will eventually be overcome at any finite temperature, no matter how high the barrier is, or how close to $0\ \mathrm{K}$. For example, it would take at most approximately $10^{1500}$ years for nuclei to spontaneously fuse into iron. Any chemical reaction would have a much, much lower kinetic barrier, and would happen in a much shorter timescale. Whether a puddle of fuel inside an atmosphere of oxygen at $300\ \mathrm{K}$ could have had enough time to oxidize completely during the current lifetime of the Universe, I don't know. | |
| Jun 6, 2015 at 1:49 | history | answered | charlesreid1 | CC BY-SA 3.0 |
