I would like to discuss this in context of ellingham diagrams. When $\ce{C -> CO}$ line is below metal oxide line, CO is produced. Similarly, in other case $\ce{CO2}$ is produced. But when both lines are below metal oxide, is carbon monoxide produced or carbon dioxide?


If both carbon monoxide and carbon dioxide are thermodynamically favored, then carbon monoxide comes "first" in the sense that the metal oxide first oxidizes carbon to $\ce{CO}$. This is because the reaction is carried out at relatively high temperature where any potential carbon dioxide would itself react with carbon to form the monoxide. If more oxide is available after all the carbon is converted to $\ce{CO}$ and further oxidation to $\ce{CO2}$ is favorable, the latter reaction then occurs.

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  • $\begingroup$ So is it like carbon dioxide forms first but due to high temperature it becomes monoxide on reacting with carbon? Or does monoxide form directly? $\endgroup$ – ModCon Sep 4 '17 at 3:49
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    $\begingroup$ No, I am saying that if any carbon dioxide were to form it would not last in the presence of carbon. $\endgroup$ – Oscar Lanzi Sep 4 '17 at 9:00

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