Ok, so I learned about the equilibrium constant. Now, I've seen that the equilibrium constant of burning is extremely small $(K \ll 1)$. here, I have a question. you see, $K$ is still NOT 0, which means that the forward reactions happen at least a tiny bit. Then, shouldn't we see some parts of anything burning at least a little bit?

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    $\begingroup$ Where did you see that burning was an equilibrium process? One cant 'un-burn' something once its burned, there is no reversibility to the process. $\endgroup$
    – NotEvans.
    Commented Jul 17, 2016 at 11:13
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    $\begingroup$ Actually, if you look at the Wikipedia page for 'combustion' you can see that the combustion process is an equilibrium process (but the K is too small that we can 'consider' it as an irreversible process ) $\endgroup$
    – Danny Han
    Commented Jul 17, 2016 at 11:17
  • $\begingroup$ my question here is, the K of burning(at room temperature)is too tiny that it can be considered irreversible, but if at least a tiny tiny bit of something did the burning, shouldn't we be able to see it? $\endgroup$
    – Danny Han
    Commented Jul 17, 2016 at 11:18
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    $\begingroup$ @NotNicolaou Well, every reaction is technically an equilibrium. The thing he's missing is activation energy. $\endgroup$
    – Superbest
    Commented Jul 17, 2016 at 23:10
  • $\begingroup$ Why do you think that things don't burn? Just because it doesn't start of a chain reaction ("a fire") doesn't mean it's not burning. One of the things that's very slowly killing you is oxidative stress, which can be considered "burning". And you've probably seen rusted iron before - that's also the result of the slow "burning" of iron in our oxidative atmosphere, though it's not quite the same as burning coal in a fire (a hunk of iron doesn't burn that way very well at room temperature - iron or aluminiumdust, on the other hand, is very flammable indeed). $\endgroup$
    – Luaan
    Commented Jul 18, 2016 at 11:31

2 Answers 2


The equilibrium constant for combustion of organic matter in air with oxygen is not small, but extremely large ($K_\mathrm{eq} \gg 1$), as is expected from a reaction that is simultaneously very exothermic and (usually) increases entropy due to the formation of more gaseous molecules than the input oxygen.

The major reason carbon-based life can exist at ambient temperature in an oxygen atmosphere is purely kinetic, not thermodynamic. You, the tree outside and everything else made of carbon is right now undergoing continuous combustion. However, in the absence of catalyst, this process is too slow below a couple hundred degrees Celsius for it to be self-sustaining. More technically, combustion of organic matter is a highly exergonic process, but the activation energy is high. The meagre amount of heat generated by the handful of molecules reacting is too quickly diluted into the surroundings, and the reaction does not accelerate and spiral out of control (a fire, as described by the eternal Feynman).

Very luckily for us, Life figured out this vast untapped source of chemical energy held back by kinetics approximately three billion years ago and developed a whole metabolic process to extract this energy in a stepwise fashion using catalysis, which we call aerobic respiration. Without it, multicellular organisms could well never had evolved.

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    $\begingroup$ Note that simple piles of coal may become warm enough by this process to start a fire! (It takes time, and the right levels of moisture, piece size, etc.) $\endgroup$ Commented Jul 17, 2016 at 17:17
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    $\begingroup$ I believe that paper turning yellow/brown over time as it ages is due to this--is this true? If so, it might make a good example in your answer. $\endgroup$
    – msouth
    Commented Jul 17, 2016 at 18:22
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    $\begingroup$ @msouth I think that the reason paper turns yellowish over time is mostly due to the presence of impurities left over in the paper after manufacturing (e.g. acids) and air pollution, which cause its slow decomposition. This is especially true of older paper. Of course, paper still combusts ever so slowly, but I have no idea how long it would take for this process alone to discolour paper. $\endgroup$ Commented Jul 17, 2016 at 21:48

Yes it the pesky activation energy that keeps us together!

It is quite illuminating to calculate the rate of a reaction with different activation energies using a simple approach, such as the Arrhenius equation at a constant temperature; $\mathrm{k=A*exp(-\Delta E/RT)}$. If we assume that the pre-exponential term A (the rate constant at zero activation energy or infinite, i.e. very high, temperature ) is $\mathrm{10^{13} s^{-1}}$ then the following values are obtained with different activation energies in kJ/mol.

$\Delta E$ =10, $k = 1.810^{11} $ s$^{-1}$
$\Delta E$ =40, $k = 1.110^{6} $ s$^{-1}$
$\Delta E$ =80, $k = 0.12 $ s$^{-1}$
$\Delta E$ =100, $k = 0.000038 $ s$^{-1}$ lifetime = 7.2 days
$\Delta E$ =140, $k = 4*10^{-12} $ s$^{-1}$ lifetime 7555 years
$\Delta E$ =180, $k = 4*10^{-19} $ s$^{-1}$ lifetime $6.9*10^{10}$ years
$\Delta E$ =200, $k = 1.5*10^{-22} $ s$^{-1}$ lifetime $2.1*10^{14}$ years

The age of the universe is approx $1.4*10^9$ years so even moderate activation barriers produce exceptionally slow reactions. Very many types of bonds have dissociation energies above 200 kJ /mol, e.g. C-C 350, C-H 413, but this is not the same as the activation energy which must be smaller than this, and by how much depends on the particular reaction. In unfolding a protein, the activation energy is essentially the dissociation (unfolding) energy, so if proteins are to last any appreciable time the activation energy has to be at least 100 kJ/mol.


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