The Sabatier reaction is as follows:

$$\mathrm{CO_2 + 4\: H_2 \rightarrow CH_4 + 2\: H_2O}.$$

If water is electrolysed

$$\mathrm{2\:H_2O \rightarrow 2\: H_2 + O_2}$$

the global reaction becomes

$$ \mathrm{CO_2 + 2\:H_2 \rightarrow CH_4 + O_2}.$$

Now, this gives an oxygen to methane ratio of 1:1 but both this site and this article state that the ratio is 4:1. Can someone explain what's missing?


It's a 4:1 ratio by mass. One molecule of methane masses 16 daltons, whereas one of dioxygen ($O_2$) masses 32, so two molecules of oxygen massed four times as as much as one of methane.

  • $\begingroup$ Thank you both @leftaroundabout and Steve Linton, I got it now. That being said, I'd need to "accept" both answers the way the information is split. Would you mind "joining" your answers together so that I can accept it please? $\endgroup$ – Jak May 26 '20 at 21:30
  • $\begingroup$ @leftaroundabout fine by me. Happy for you to update your answer and I'll delete mine. $\endgroup$ – Steve Linton May 26 '20 at 22:39
  • $\begingroup$ @SteveLinton just accept this answer, I think it is probably the real reason. Atom-counts aren't really relevant for rocket launches, mass ratio is. I'll leave my answer in place anyway as the explanation for the 2:1 molecule ratio. $\endgroup$ – leftaroundabout May 27 '20 at 9:19

In practice, what you do is actually you start out with water to generate the $\mathrm{H_2}$, rather than just electrolysing the water that comes out in the end. (Well, you'd probably do that to, you'd feed it back into the start.) I.e. $$ \mathrm{CO_2 + 2\:H_2O \to CH_4 + 2\:O_2} $$

That still gives only a 2:1 ratio of $\mathrm{O_2}$ to $\mathrm{CH_4}$. What the article means by “ratio of 4:1 of oxygen to methane” must therefore be the ratio of oxygen atoms to carbon atoms.


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