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In fuel-rich combustion, my understanding is that enough fuel reacts and consumes all of the oxygen present. The combustion products $\ce{CO2}$ and $\ce{H2O}$ undergo dissociation until equilibrium is established and one of these is $$\ce{CO2 -> CO + 1/2O2}.$$

From the graph below, the equilibrium mole fraction of combustion products are plotted vs. equivalence ratio. Why is there a sudden rise in $\ce{[CO]}$ as we move into fuel rich combustion? Is this due to the drop in $\ce{[O2]}$ such that $\ce{[CO]}$ must increase to compensate for equilibrium to be established?

enter image description here

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I think it is quite simple: the higher the proportion of the compound that will burn in the mixture, the more oxygen will be required to a complete combustion. If you have a constant amount of oxygen (in the reactor), and more is requires while increasing the amount of fuel, the complete combustion will gradually decrease, and the ratio $CO/CO_2$ produced will increase as a result of that.

As it can be seen on the graphic, from the point that the equivalent ratio of fuel and oxygen is larger than one, the $CO$ starts being formed in a considerable proportion, and over a ratio of 1:3, starts being the major product.

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  • $\begingroup$ In fuel rich combustion, fuel and existing O2 reacts, and CO2 produced dissociates into CO and O2, and the O2 is consumed again since there's fuel left (and produces more CO2 -> more CO etc.), which results in CO accumulation and not very much O2. Equilibrium is established since the large [CO] compensates for the low [O2] in the reverse reaction to form CO2. Is this logic correct? $\endgroup$
    – Yandle
    Nov 20, 2014 at 19:51
  • $\begingroup$ Yeah, I believe that is correct. $\endgroup$
    – Altered State
    Nov 20, 2014 at 23:26
  • $\begingroup$ A followup question, is equilibrium essentially the point where O2 is depleted enough such that the rate in which it reacts with remaining fuel (and forms CO2) is less than the rate in which O2 is reformed from CO2 dissociation? This would imply that as t -> infinity the excess fuel will eventually all be used up? $\endgroup$
    – Yandle
    Nov 22, 2014 at 5:25

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