7
$\begingroup$

I am currently reading a textbook on combustion. They assume that all reactions go to completion, such that post combustion only unburned fuel/air and actual products remain.

Assuming that combustion is lean or perfect (no fuel left unburned) and not neglecting dissociation effects, can reverse reaction actually occur to reform fuel at equilibrium? In other words, is the equilibrium constant infinity in real life?

Improve this question
$\endgroup$
4
  • $\begingroup$ I am also wondering, does the inclusion/exclusion of dissociation effects matter? For instance, if I assume that only CO2 and H2O is formed as products, does this change the answer to the question I have above? $\endgroup$ – Yandle Oct 4 '14 at 18:19
  • $\begingroup$ Complete combustion only forms CO$_2$ and H$_2$O. What other products do you mean by dissociation? $\endgroup$ – Jeffrey Weimer Oct 6 '14 at 14:53
  • 2
    $\begingroup$ Infinite in the sense that you cannot measure any finite backward reaction. Not infinite in theoretical sense, as there is some finit activation energy for the backward reactions. Combustion is a combination of thousands of steps with radical mechanism, which mostly have huge Ks. Also, you should not forget the entropic/information factor: if you burn wood, there will no be reaction producing lignin and complex natural products. Even if energy would allow, the information and selectivity to form lignin is not there. So a backword reaction can lead different products than the initial was! $\endgroup$ – Greg Oct 6 '14 at 15:44
  • $\begingroup$ @JeffreyWeimer Dissociation products I was referring to species are like H2, OH, CO etc. I'm not sure if N or NOx is considered a species of dissociation of combustion products since (I think) it is a primarily result of excess O2 and existing N2 which is technically independent of the fuel-air combustion. I think that dissociation effects would make the reaction even more irreversible? $\endgroup$ – Yandle Oct 10 '14 at 15:39
1
$\begingroup$

An equilibrium constant is never $\infty$ in theory. In practice, large values of $K_{eq}$ (and slow reverse reaction rates) lead to the equivalent of complete combustion.

Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.