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I understand that combustion of hydrocarbons (such as alkanes) produces energy by breaking bonds:

$$\ce{C_nH_{2n} + $\frac{3n}{2}$ O2 -> nCO2 + n H2O}$$

And I can find dozens of explanations of how combustion works in an engine.

What does the chemical reaction (not equation) of hydrocarbon polymer combustion actually look like?

In other words, how is a hydrocarbon polymer actually broken down to form energy?

  • What process actually breaks up the polymer into its respective monomers (methane?) for combustion to take place?

Please explain. I would also love a visual (like this or this ) if someone could find one!

orthocresol
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theforestecologist
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  • Related: Is there a reaction type/name for the splitting of octane into its parts prior to combustion, or does this process all occur at once in an engine? – theforestecologist Sep 20 '18 at 16:57
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    @theforestecologist combustion reaction mechanisms can be quite complex. Here is a paper about a simplified model of propane combustion: https://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/37_2_SAN%20FRANCISCO_04-92_0960.pdf – Tyberius Sep 20 '18 at 18:10
  • All common combustions at not too high temperatures if I remember correctly are radical chain reactions. Those should be known from your high school chemistry class, or you should have caught up in college. But, there is wikipedia https://en.wikipedia.org/wiki/Combustion#Reaction_mechanism to help you out. ;-) – Karl Sep 20 '18 at 20:59
  • @Karl radical chain reactions may very well have been taught in my HS/college chemistry classes, but it's been a few too many years to remember. FYI I'm not a chemist, I'm a biology professor -- so excuse my unspecific language as I'm barking up a tree I [clearly] remember less than optimal about. [btw, I would suspect you would not be treated so rudely as a chemist asking Q's over at Bio.SE] – theforestecologist Sep 20 '18 at 21:27
  • I'm sure I wouldn't, because I still know my high school biology. Mostly. ;-) No, sorry. Its just that the answer here would require a solid chemical base knowledge to understand. But, your question makes it clear you don't have that available. Please forgive if that leads to some exasperation. Look it up, there's a good chance you have learned about this at university and will remember. – Karl Sep 20 '18 at 21:56
  • @Karl, I actually went back to my college chem textbook, and after a reasonably thorough search found no evidence of radical chain reactions or the discussion of combustion of hydrocarbons in any great detail. So it's not that I don't remmeber, it's that It was not included in my course. The only mention of chain reactions (1x) was in reference to atmospheric ozone, which I know well. So now I have an idea of what you mean by chain reactions. Lack of knowledge in this specific topic is not indicative of my base chemical knowledge anyay. I simplified the question because of a previous comment. – theforestecologist Sep 20 '18 at 22:35
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    It is not a polymer, because there are no reasonable constituent "monomers" which you could put together to form these long alkanes, unlike e.g. a protein where you can join amino acids via condensation reactions. They are just standalone alkanes. I edited to reflect this and to put a more descriptive equation. Anyway, there is a reason why the mechanism isn't typically covered (even at undergraduate level): it's *very* messy. There are radical chain reactions which are easy to understand, but this is not one of them. [Now, that doesn't mean you can't get an answer - it is just a warning.] – orthocresol Sep 20 '18 at 23:24
  • @orthocresol thanks for the edit, clarification ,and heads up. However, I do want to point out that some sources (e.g., encyclopedia Britannica) do refer to alkenes as monomers and their chains as polymers. Given the repeatable nature of these chains, you could see where my confusion arose... Still, I'm curious how these chains are broken (and what we call that reaction) prior to combustion of the constituent carbons. – theforestecologist Sep 21 '18 at 02:31
  • Higher alkanes, neglecting their end groups, are not per se distinguishable from polyethylene. @orthocresol The messiness is what makes these reactions so fascinating. At least for people who are into unorderly (=polymer) chemistry. ;-) Btw. thanks for fixing up my answer. – Karl Sep 21 '18 at 23:38

1 Answers1

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The following, very general radical reactions apply to any common combustion reaction of organic material. Details vary largely.

  • Step 1

An energy spark of whatever kind turns triplet oxygen into its high energy singlet state$^*$, which typically strips a hydrogen atom off of some organic molecule

$\ce{R-CH3 + O2^* -> R-CH2. + HOO.}$

Now you have two radicals.

  • Step 2

The perhydroxy radical strips off another hydrogen somewhere,

$\ce{R-CH3 + .OOH -> R-CH2. + HOOH}$

while the alkyl radicals react with more oxygen (self-catalysed)

$\ce{R-CH2. + O2 -> R-CH2-OO.}$

$\ce{R-CH3 + .OO-CH2-R -> R-CH2. + HOO-CH2-R}$

, this incorporates $\ce{-OOH}$ hydroperoxide motives into more and more molecules. This is in itself a linear process, with a constant number of radicals. Similar to free-radical polymerisation or organic radical-substitution reactions, and very much the same as formation of the infamous ether hydroperoxides.

  • Step 3

These peroxides are not stable at elevated temperature, and their O-O bond tends to split homolytically

$\ce{R-OO-H -> R-O. + .OH}$

Now the number of radicals grows rapidly, the combustion is gaining speed.

  • Step 4

As temperatures rise (see step 5!), esp. larger organic molecules start to break up also by themselves

$\ce{R-CH2-CH2-R -> R-CH2. + .H2C-R}$

(That's also a possible initiation step.) You get more radicals, which can react with more oxygen to form peroxides. Now the chain reaction snowballs!

  • Step 5

$\ce{HOOH -> HO. + .OH}$ gives more hydroxy radicals (also from step 3), which turn into water

$\ce{HO. + H3C-R -> H2O + .H2C-R}$

leaving another radical behind. Producing water is highly exothermic, this fuels the reactions creating radicals (in steps 3+4+5). Making radicals costs a lot of energy.

Steps 2-5 (it's really not a sequence) run in parallel and repeat as they produce more radicals and peroxides all the time. The reaction is usually limited by the inflow of fresh oxygen.

  • Step 6

Termination: Two carbon radicals can recombine, but at high temperature this does not really happen, because the energy from the recombination has to go somewhere. When temperatures are too low, this makes coke and smoke.

This picture is by no means complete. $\ce{CO2}$ would likely occur via decarboxylisation of $\ce{R-COO.}$ radicals, etc. Organic radicals at high temperatures have more internal reactions where they split off smaller molecules, e.g. $\ce{R-CH2-CH2. -> R. + H2C=CH2}$. (The latter can be seen as depolymerisation of polyethylene.) A radical in the middle of a chain splits it in half ($\ce{R-CH2-CH2.-R -> R. + H2C=CHR}$, increasing the volatility.

And remember that this picture is terribly simplified. I'm by no means an expert on this subject, and the experts also have not agreed on all details yet, especially since it varies a lot.

Karl
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    My selection of "steps" is not very convincing. Feel free to arrange the reactions in a different manner. The gist is they are all happening in parallel and producing more educts for each other. – Karl Sep 21 '18 at 23:32
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    This question https://chemistry.stackexchange.com/questions/14704/reaction-mechanism-of-combustion-of-hydrogen?answertab=active#tab-top has a nice answer on hydrogen combustion, which adds a few liteature reads. – Karl Dec 22 '18 at 19:33
  • No alkanes have the formula $\ce{C_nH_{2n}}$ as given in the question. Alkanes may be $\ce{C_nH_{2n+2}}$, but not $\ce{C_nH_{2n}}$, which is more probably an alkene, or a cycloalkane. The theory developed by Karl is OK, but it used $\pu{R}$ to avoid this difficulty, where $\pu{R}$ is all the rest of the alkane molecule when $\ce{CH3}$ has been removed – Maurice Jul 04 '23 at 19:32