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It says in my textbook that $\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$, this seems like a displacement reaction to me but then shouldn't it be $\ce{CH4 + O2 -> CO2 + 2 H2}$ since $\ce{4 H}$ is displaced by $\ce{O2}$ and becomes $\ce{2 H2}$ which makes it balanced on its own.

So how is it that $\ce{CH4 + O2 -> CO2 + H2O}$ which is further balanced as my textbook says.Thanks in advance.

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3 Answers 3

up vote 5 down vote accepted

The reaction as you state it is correct only if there will react only one molecule of oxygen. But the reaction describes burning of methane which is supposed to be in the presence of excess of oxygen. Than not only methane is burnt, but also the arised hydrogen.

So in "first" step:

$CH_4 + O_2 \rightarrow CO_2+ 2H_2$

but then the hydrogen will be also combusted in reaction :

$2H_2 + O_2 \rightarrow 2H_2O$

so the overall reaction will be:

$CH_4 + 2O_2 \rightarrow CO_2+ 2H_2O$

As was pointed out in the others answers the real mechanism of a methane combustion at a low oxygen pressure is more complicated:

49th AIAA Aerospace Sciences Meeting including the New Horizons Forum and Aerospace Exposition, 4-7 January 2011, Orlando, Florida

49th AIAA Aerospace Sciences Meeting including the New Horizons Forum and Aerospace Exposition, 4-7 January 2011, Orlando, Florida, http://enu.kz/repository/2011/AIAA-2011-94.pdf

another: COMBUSTION AND FLAME 120:160–172 (2000) Intermediate Species Profiles in Low-Pressure Methane/Oxygen Flames Inhibited by 2-H Heptafluoropropane: Comparison of Experimental Data with Kinetic Modeling, http://www.nist.gov/el/fire_research/upload/Williams-Intermediate-Species-profiles-in-Low-pressure-Methaneoxygen-Flames-inhibited-by-2-heptafluoropropane.pdf

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Thanks, you are the first guy who actually got it. I had another question: This is a displacement reaction or an oxidation reaction? I think it is displacement but I am not sure. –  Abhishek Mhatre Mar 28 at 9:35
Jaroslav, you might want to have a look at mhchem for easier typesetting of chemistry in LaTeX. I like that package a lot! –  Klaus Warzecha Mar 28 at 9:36
@ Klaus Warzecha - thanks I will @ user2468338 - It is an oxidation reaction: An oxidation/ reduction reaction is one where there is a change in the oxidation states of two or more species –  Jaroslav Kotowski Mar 28 at 9:40
Yeah, but displacement reaction is also a type of oxidation reaction right. –  Abhishek Mhatre Mar 28 at 9:44
for example reaction of natrium hydroxide with hydrochloric acid is displacement reaction and not oxidation/reduction reaction. –  Jaroslav Kotowski Mar 28 at 9:57

Look at the enthalpies of formation of $\ce{CO}$, $\ce{CO2}$, and $\ce{H2O}$, then the net enthalpies of reaction. In a deficiency of oxygen, and trimmed by the temperature and kinetics versus equilibrium conditions, are you more likely to obtain $\ce{CO}$ plus $\ce{H2O}$ than free hydrogen? Given less oxygen, soot and water? Note reaction energetics below,


enthalpies of formation

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There's nothing wrong. It's incomplete combustion. It is one of a number of oxidizing reactions that can and will be occurring if you are burning this hydrocarbon in an oxygen-restricted environment. But the dominant reaction will be the one where the lowest energy state is reached. So you will likely wind up with one CO molecule, one water molecule and one hydrogen molecule instead.

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