Is it possible to chemically convert methane into propane?

If so, what is involved and is it economical?

  • $\begingroup$ Also, can you please describe what you mean by involved? The catalyst (if any)? The process? What? (The answer could cover all of them, though) $\endgroup$ – M.A.R. ಠ_ಠ Mar 24 '15 at 21:04

Yes, it can be done by mimicking the conditions in the earth's mantle.

It was shown by Alexander Goncharov and coworkers [1] by placing methane in a diamond anvil cell, applying a pressure of more than 2 GPa and heating everything to a temperature around 1000-1500 K with a laser.

Is it economical? Hardly, but very interesting anyway!

  1. Kolesnikov, A.; Kutcherov, V. G.; Goncharov, A. F. Nature Geosci 2009, 2 (8), 566–570. DOI 10.1038/ngeo591.
  • $\begingroup$ Can we use WURTZ reaction to convert methane to propane? $\endgroup$ – NeilRoy Mar 25 '15 at 3:22
  • 1
    $\begingroup$ @NeilRoy You might want to have a look at this older question for possible problems with the Wurtz reaction. $\endgroup$ – Klaus-Dieter Warzecha Mar 25 '15 at 5:16

Generally, this reaction is run in reverse.

Hydrogen reformers are used to generate hydrogen by the following reaction scheme:

$$\ce{CH4 + H2O -> CO + 3H2}~~~\mathrm{(Endothermic, 1000C)}$$

This is frequently followed by the shift reaction

$$\ce{CO + H2O -> CO2 + H2}~~~\mathrm{(Exothermic, 300C)}$$

Hydrogen reformers that run on heavier hydrocarbons than methane are frequently fitted with a pre-reformer in which the reaction starts at a lower temperature. In the pre-reformer the following reaction occurs, amongst others, including some that generate hydrogen:

$$\ce{C3H8 + 2H2 -> 3CH4}$$

Whether this reaction is useful is debatable, but it does indeed occur because it is mildly exothermic and has no change in the number of moles of gas.

Running the reaction in the way that you want would require the removal of some of the hydrogen, to push the equilibrium in favour of propane. This might be useful in the case of stranded gas: i.e natural gas which is economic to extract (or which is extracted as a byproduct of oil extraction) but is not economic to ship.

Frequently stranded gas is converted to methanol for shipping. Synthesis gas (a mixture of $\ce{H2}$ and $\ce{CO}$ is an intermediate step.) Methanol can be converted to heavier hydrocarbons such as gasoline if desired.

However, as mentioned in the same reference we can also perform a more direct conversion of natural gas to heavier hydrocarbons via the Fischer-Tropsch process (using the synthesis gas step but not the methanol step). Main article

Partial oxidation of the natural gas eliminates the excess hydrogen and ensures the correct balance of of $\ce{H2}$ and $\ce{CO}$ in the synthesis gas to ensure heavier hydrocarbons are formed. We have two reactions that can vary the ratio of $\ce{H2/CO}$ in the feed to the Fischer-Tropsch process:

$$\ce{2H2 + O2 -> H2O}$$

Consumption of $\ce{H2}$ in preference to consumption of $\ce{CO}$, decreases $\ce{H2/CO}$ ratio.

$$\ce{CO + H2O -> CO2 + H2}$$

Shift reaction, increases $\ce{H2/CO}$ ratio.

This composition control and several other parameters enable the desired mixture of hydrocarbons to be obtained from the Fischer-Tropsch process, though a single compound such as propane is rarely if ever obtained.

In summary, $\ce{3CH4 -> C3H8 + 2H2}$ is not possible under normal conditions, but $\ce{3CH4 + O2 -> 2H2O + \mathrm{a~mixture~of~hydrocarbons~of~average~formula~C3H8}}$ is.

Adding air to the mixture for oxidation is problematic because it adds nitrogen which must be purged away later, but I am guessing it is much cheaper than running a pressure swing absorption plant to remove excess hydrogen.

Here's an example of a gas to liquid hydrocarbon plant. I'm guessing it doesn't make propane, because heavier hydrocarbons are more valuable and easier to ship, but the principle remains the same.


protected by Community Jan 3 '16 at 17:17

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