Titan, a moon of Saturn, has readily available water ice ($\ce{H2O}$) and methane ($\ce{CH4}$) on its surface, at a frigid $\pu{93.7 K}$.

It would be significant if these (or possibly, other available hydrocarbons) could be reacted in a self-sustaining way, releasing usable energy. I presume the best way of doing this would be to react by the simple following two step process:

  1. Split water into hydrogen and oxygen: $$\ce{2 H2O → 4 H2 + O2}$$
  2. Burn oxygen and methane: $$\ce{CH4 + 2 O2 → CO2 + 2 H2O}$$

Keeping in mind the starting temperature, does this sequence of reactions operate at a net energy release (that is, is it theoretically possible to use this reaction to power itself, with energy leftover)? If not, could other hydrocarbons* found on Titan fit the bill?

*Methane, ethane, propane, and benzene exist. Others may as well.


1 Answer 1


The thermodynamic problem is not so much with the hydrocarbons reacting with O2 as it is with forming O2 from water. That reaction is energetically uphill. A big hill. In fact water is a waste product of your second reaction. The reason the second reaction is energetically favorable is due to the high stability of CO2 and water relative to the reactants, methane and oxygen.

Would it make sense that you should be able to then convert the water from your second reaction into O2 and H2 in a way that would consume less energy than it took to make the water in the first place? That is almost equivalent to what you are proposing by starting out with the water already present on Titan.

The prohibitive energetics of your reactions apply on Earth as well as Titan. You need a process to split the water to burn with hydrocarbons that will consume less energy than the energy you will get back from burning the hydrocarbons. You could conceive of using solar power as the energy source for converting water into O2 and H2. Or other energy sources. But that wasn't your question, and you definitely won't get the energy back that you would require via your second reaction. And good luck with my solar energy idea on Titan ;)

  • $\begingroup$ So, just to be sure I understand your answer: The energy released by reaction 2 is not enough to offset reaction 1, even if the water from reaction 2 is used. Is the story similar for other hydrocarbons? $\endgroup$
    – geometrian
    Commented Jan 13, 2017 at 8:30
  • $\begingroup$ Exactly. And yes, by all the same arguments it is true for all hydrocarbons. In fact, methane would be the most energetically favorable for that oxidation reaction because it is the most reduced hydrocarbon. $\endgroup$
    – airhuff
    Commented Jan 13, 2017 at 8:32
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Syngas $\endgroup$
    – Mithoron
    Commented Jan 13, 2017 at 13:47

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