Why don't diesel and regular car engines burn all the way to $\ce{CO2}$?

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    $\begingroup$ Most of the carbon-oxygen stuff coming out is CO2. Some isn't, because of detailed balance of the chemistry, inhomogeneous conditions in the piston, etc. $\endgroup$
    – Jon Custer
    Jan 29, 2015 at 20:42
  • $\begingroup$ Seems like another question easily answered with very minor googling. $\endgroup$ Jan 30, 2015 at 2:00

1 Answer 1


I think this is sort of covered by What really happens atomically in an explosion? because the burning of the fuel air mixture is basically an explosion. The point is that there isn't a nice simple answer because the reaction is fearsomely complex.

Actually most of the fuel does burn to carbon dioxide. The proportion of carbon monoxide is quite small. In a normally tuned engine there is enough oxygen to burn all the fuel to carbon dioxide but the fuel/air mixture isn't in the cylinder long enough for the reaction to go to completion. You probably know that all modern cars are fitted with catalytic convertors, and what they do is to complete the reaction as the exhaust gases pass through them.

  • $\begingroup$ But why not run with excess air to make sure it is all burned? This is especially true with Diesel engines which pump out pure carbon $\endgroup$ Jan 30, 2015 at 12:10
  • $\begingroup$ @DirkBruere: Getting the maximum engine efficiency, i.e. the combination of good mpg with high power, is something of a black art as lots and lots of parameters are involved. An obvious parameter is the fuel/air mixture, and it turns out that the optimal fuel/air ratio results in some carbon monoxide being left in the exhaust gases. $\endgroup$ Jan 30, 2015 at 12:18
  • $\begingroup$ Well, yes. That was my question, sort of. If we get more energy by burning to CO2 why do we deliberately stop short of that? What are we optimizing? $\endgroup$ Jan 30, 2015 at 12:22
  • $\begingroup$ @DirkBruere: Generally, if you run lean (high air/fuel) you decrease power and the engine runs hotter so it generates more oxides of nitrogen. I think the mpg are slightly better with a leaner mixture, but the power falls. Exactly why the currently used ratio is considered optimal I don't know. I'm sure a Google would find articles on the subject. $\endgroup$ Jan 30, 2015 at 12:27
  • $\begingroup$ This is why I posted it in physics - it's not really a chemistry question. Those who voted to move it did not seem to understand. $\endgroup$ Jan 30, 2015 at 13:33

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