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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!

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  • $\begingroup$ 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? $\endgroup$ – theforestecologist Sep 20 '18 at 16:57
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    $\begingroup$ @theforestecologist combustion reaction mechanisms can be quite complex. Here is a paper about a simplified model of propane combustion: web.anl.gov/PCS/acsfuel/preprint%20archive/Files/… $\endgroup$ – Tyberius Sep 20 '18 at 18:10
  • $\begingroup$ 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 en.wikipedia.org/wiki/Combustion#Reaction_mechanism to help you out. ;-) $\endgroup$ – Karl Sep 20 '18 at 20:59
  • $\begingroup$ @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] $\endgroup$ – theforestecologist Sep 20 '18 at 21:27
  • $\begingroup$ 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. $\endgroup$ – Karl Sep 20 '18 at 21:56
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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 another hydrogen off 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. Very 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 radical grows rapidly, the combustion is gaining speed.

  • Step 4

As temperature rises, 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 it's burning.

  • 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 other reactions. Creating 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.

  • 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 (formation of $\ce{CO2}$, ...). Especially 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 a depolymerisation of polyethylene.)

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.

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  • $\begingroup$ 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. $\endgroup$ – Karl Sep 21 '18 at 23:32
  • $\begingroup$ This question chemistry.stackexchange.com/questions/14704/… has a nice answer on hydrogen combustion, which adds a few liteature reads. $\endgroup$ – Karl Dec 22 '18 at 19:33

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