I had some followup questions regarding a previous post I made here regarding the auto-ignition temperature and ASTM E659

  1. For fuel temperature below AIT, we should still have finite reactants above activation energy, reacting, and heating the remaining mixture so additional reactants are above the activation energy. Theoretically as $t \rightarrow \infty$, could the fuel burn itself out like this, regardless of what temperature it is at (obviously the rate will be very different)? If so, is there an implicit rate requirement in defining the AIT that the above chain reaction process has to happen within a short duration?

  2. Given that ignition is defined as when a flash and temperature rise is seen, does this mean that the actual quantity of fuel burnt is unimportant and it is assumed that a significant enough amount is used?

  3. Is the reason for using an open flask in ASTM E659 likely just for simplicity, as opposed to a piston-cylinder arrangement that offers more control over air-fuel mixture?

  • 1
    $\begingroup$ I don't have time for a proper answer, but in 1, it sounds like you've got the right idea. Think of paper browning; over centuries it will oxidize into brittle, ruined material, essentially burning to completion at room temperature air. Here the time limit they establish for a flame to appear is explicit, 10 minutes. They do state that some compounds have a longer delay at a given temperature before they ignite. The reaction rate you refer to would be different for each substance, since each is releasing a different amount of energy. $\endgroup$ Commented Oct 15, 2014 at 12:06

1 Answer 1


To understand the answer to (1), you have to think about what combustion actually is. It isn't a single reaction, with a single set of reactants, and a single set of products. It's actually a whole bunch of tiny steps (thousands) that all occur together. Those reactions happen among a whole bunch of unstable radical species (hundreds). When you put a match to a pool of gasoline, all that energy from the match starts tearing apart bonds in the fuel, leading to the formation of radicals. That's what triggers the initial release of energy, and what creates the cascade of ignition. So theoretically, no, you could never reach combustion products, through the pathway of combustion, because you'd never have that activation energy, the big kick of energy needed to create radicals and keep them alive long enough for them to create a chain reaction.

The answer to (2) is, no, it doesn't matter how much fuel. Remember you're measuring a mass-independent property. The autoginition temperature isn't "5 degrees per gram". It's a single temperature. What matters is the ratio of oxygen to fuel.

As to (3), it's a lot easier to operate an open flask than it is to operate a piston-cylinder, and it doesn't have a substantial impact on the air-fuel ratio if you do it correctly.

  • $\begingroup$ 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. $\endgroup$ Commented Jun 6, 2015 at 3:23
  • $\begingroup$ 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... $\endgroup$ Commented Jun 6, 2015 at 17:14
  • $\begingroup$ @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). $\endgroup$
    – Yandle
    Commented Jun 8, 2015 at 4:39
  • $\begingroup$ 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! $\endgroup$ Commented Jun 8, 2015 at 5:01
  • $\begingroup$ @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? $\endgroup$
    – Yandle
    Commented Jun 12, 2015 at 5:40

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