
		Electrolyis apparatus used to generate oxygen on the MIR space station (photo courtesy of NASA)

Synthesizing Fuel:

Methane for Nothing and the Oxygen is Free?*

The first step in the Sabatier process is exothermic.

CO₂ (g) + 4H₂ (g)

CH₄ (g) + 2H₂O (g)

H = -165.0 kJ

(1)

Since the reaction occurs rapidly enough to be useful at about 400 ºC, some source of heat will be needed to get the reactor to its initial temperature. However, since the reaction produces energy, once the initial temperature is achieved, the reaction can provide its own heat to continue the reaction. Does this mean we can actually get energy for nothing from this reaction? No, because there is more to this process. While the methane could be liquefied and stored for use as a fuel for the return trip, we won't be able to burn it without an oxidizer. For that we need to break down the water, and that takes energy.

The electrolysis of water

The water produced in step (1) could be liquified, and used for human consumption; however, most of it would be electrolyzed to produce some more hydrogen and oxygen gas.

2H₂O (g) 2H₂O (l)		H = -88.0 kJ		(2)
2H₂O (l) 2H₂ (g) + O₂ (g)		H = +571.6 kJ		(3)

The oxygen would be liquified and used as fuel, as well as for human consumption, and the hydrogen would be recycled back through reaction (1). Thus, overall, adding equations (1), (2) and (3) gives the net reaction of:

CO₂ (g) + 2H₂ (g)

CH₄ (g) + O₂ (g)

H = +318.6 kJ

(4)

Since overall this reaction is quite endothermic, it would require a large energy source, perhaps a nuclear reactor, or a large solar panel array. So, in fact, we are not getting our energy to leave the Martian surface for nothing. Actually, these reactions are just a way of changing one kind of energy – nuclear or solar – which we currently can't use for space transportation, into chemical energy which we can release rapidly in a combustion reaction in a rocket.

Also notice that only one oxygen molecule is created for every methane molecule created. Since the reaction between CH₄ and O₂ that will be used to propel our rocket home

CH₄ (g) + 2O₂ CO₂ (g) + 2H₂O (g)

requires two oxygen molecules for complete combustion, you have two choices:

using the Sabatier process, make twice as much methane as required in order to get enough oxygen, and throw one half of the methane away. This is wasteful of hydrogen, which had to be brought at great cost from earth, and so is an impractical solution.
find another reaction that will produce more oxygen from the available raw materials.

This can be done by running the water gas shift reaction in reverse.

CO₂ (g) + H₂ (g)

CO (g) + H₂O (g)

H = +41.2 kJ

(5)

then electrolyzing the water to produce more oxygen gas, and hydrogen.

H₂O (g) H₂O (l)		H = -44.0 kJ		(6)
H₂O (l) H₂ (g) + 1/2O₂ (g)		H = +285.8 kJ		(7)

Provided there are no leaks of hydrogen, there is no net use of hydrogen in the overall reaction (5), (6) and (7):

CO₂ (g)

CO (g) + 1/2O₂ (g)

H = +283.0 kJ

(8)

Doubling reaction (8) and adding to reaction (4) gives the overall process of:

2CO₂ (g) 2CO (g) + O₂ (g)	H = +566.0 kJ	(9)
CO₂ (g) + 2H₂ (g) CH₄ (g) + O₂ (g)	H = +318.6 kJ	(4)
3CO₂ (g) + 2H₂ (g) 2CO (g) + CH₄ (g) + 2O₂ (g)	H = +884.6 kJ	(10)

The CO could actually be used as a fuel together with the CH₄ (g); however, it would more likely just be vented away into the Martian atmosphere. So, overall equation (10) tells us that: for every 884.6 kJ of energy we use, we can turn 4 g of hydrogen gas which we brought from earth, into 16 g of methane, and 64 g of oxygen gas.

This is a mass leverage of 4:80. We get 20 times the mass of rocket propellants by utilizing a small quantity of resources brought from earth, and a large amount of resources found in the Martian environment.

Equation (10) is very endothermic. We are only able to produce the methane and oxygen propellants, at a large cost in energy. The energy must either come from earth in the form of a nuclear reactor, or be collected in large solar panel arrays on Mars. These processes only work because we are converting one kind of energy, probably nuclear or perhaps solar, into chemical energy which we can release rapidly to launch our rocket. It's not methane for nothing, and the oxygen's not free!

The Sabatier and reverse water gas shift reaction system

The closed-loop Sabatier and reverse water gas shift reaction system

The problem with all of this is reaction (5).

CO₂ (g) + H₂ (g)

CO (g) + H₂O (g)

H = +41.2 kJ

(5)

As you will see in the next chapter, this reaction has a very small equilibrium constant. This means that there will be very small amounts of useful H₂O in the products, unless we apply our knowledge of le Châtelier's principle. We need to drive the reaction to the right. We can do this by applying a stress. Since the quantity of CO₂ (g) is virtually limitless in the Martian atmosphere, it would make sense to run the reaction with the greatest possible amount of CO₂ present in the reaction container. We can also use the maximum amount of H₂ available (since it it not really consumed, but easily recycled), and remove the water as quickly as it is formed. As well, since the reaction is endothermic, a high temperature favors higher product yields. Applying all of these conditions will allow high yields in a closed loop reaction system.

Repairing a Martian rover (courtesy of NASA) Can this be done? Probably. The Sabatier reaction has been known about since the early 1900's, and the other reactions are used extensively in the petroleum industry. However, the technology is complex, and the failure of recent Martian satellite probes have shown that it is very difficult to build technology that will function 100% reliably in the hostile environment of space.

Apologies to Dire Straits for the bad play on their lyrics.