# Can carbon dioxide be reduced to carbon monoxide and oxygen to produce energy?

There are lots of questions about reducing or burning CO2 to carbon and oxygen to solve climate change, but of course that wouldn't work because it takes a lot of energy. But carbon monoxide is more stable than the dioxide, so could CO2 be split into CO and oxygen to create more heat? Of course this is a bad idea since it'd produce a toxic gas but is it at least theoretically possible?

Apologies if I'm misusing chemistry terms or concepts; please inform me in the comments section.

• It will take a lot of energy too. – Ivan Neretin Sep 2 '20 at 22:49
• CO could be captured more easily than CO2 – TLDR Sep 2 '20 at 23:08
• It will take energy to break that C=O bond. I can't imagine that there's an easy way to do this on an industrial scale. Cool idea, though, I'm sure the chemists of the future will be working on something like this! – Mitchell Jacky Sep 3 '20 at 2:50
• Related reading: Syngas is used as a fuel, and consists mainly of H2 and CO - this suggests that the thermodynamics are unfavourable (as confirmed by the answers below). The article suggests some ways that related reactions might be useful – Chris H Sep 3 '20 at 9:41
• Forget carbon monoxide. Why not take it all the way back to gasoline? – user253751 Sep 3 '20 at 10:17

Unfortunately, the question as stated is thermodynamically impossible. Let's look at the proposed reaction:

$$\ce{CO2(g) -> CO(g) + O(g)}$$

This reaction is simply a bond dissociation (specifically, a carbon-oxygen covalent double bond is broken). We can look up the enthalpy change associated with it. From a table of values on Wikipedia, we find in the row for carbon dioxide that this reaction has an enthalpy change of $$\mathrm{+532\ kJ\ mol^{-1}}$$ at $$\mathrm{298\ K}$$. The proposed reaction is therefore heavily endothermic. That is to say, it must absorb energy. Interestingly, it's true that the extreme strength of the bond in carbon monoxide has a measurable effect, making this process more favourable than expected. However, it is still overall extremely unfavourable, and therefore requires a large input of energy. I stress that this is unavoidable, no matter how fancy your machine - if the end result is the reaction stated above, then you must pay the energy cost somehow.

Part of the problem, though, is that we have monoatomic oxygen as a product, which is a very reactive, high energy species - it doesn't actually exist except in special conditions. A simple adjustment therefore is to have molecular dioxygen, $$\ce{O2}$$ (the kind in the atmosphere that you breathe). The reaction then becomes:

$$\ce{2 CO2(g) -> 2 CO(g) + O2(g)}$$

So what's the enthalpy change associated with this reaction? Looking up another table, this turns out to be $$\mathrm{+283\ kJ\ mol^{-1}}$$ at $$\mathrm{298\ K}$$. Again, this reaction is endothermic, though much less so than the first one. Regardless, once more this reaction is an energy sink.

If you want more visceral confirmation of this fact, consider the following. It is well known that pure carbon monoxide burns in an oxygen atmosphere. The reaction is self-sustaining and releases considerable heat. If you pay close attention, the reaction in the video is the exact inverse of the second equation. By chemical thermodynamics, if the combustion of $$\ce{CO}$$ to $$\ce{CO2}$$ releases heat, then it is necessarily true that cleaving $$\ce{CO2}$$ to form $$\ce{CO}$$ and $$\ce{O2}$$ will consume energy.

As a last point, there are ways to make the production of $$\ce{CO}$$ from $$\ce{CO2}$$ feasible, but it requires changing the products. For example, if hydrogen gas is used as a reagent, the following becomes possible:

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

The enthalpy change for this reaction is $$\mathrm{+41\ kJ\ mol^{-1}}$$ at $$\mathrm{298\ K}$$ , which is still endothermic, but approaching the break-even point. This is not too surprising, as hydrogen gas can behave as a reductant, and the bonds in water molecules are strong, pushing the reaction forwards. Let us make one last tiny modification:

$$\ce{CO2(g) + H2(g) -> CO(g) + H2O\color{red}{(l)}}$$

By assuming the water produced is in the liquid state as opposed to a gas, the reaction surrenders a little more energy, and the calculated reaction enthalpy becomes $$\mathrm{-3\ kJ\ mol^{-1}}$$ at $$\mathrm{298\ K}$$. This reaction is very mildly exothermic, which is to say it releases heat (admittedly, so little that it's within margin of error, and slightly different conditions could make the reaction overall endothermic).

If you're not dead-set on having carbon monoxide as a product, then there are further options still. For example, here is the complete reduction of $$\ce{CO2}$$ to methane ($$\ce{CH4}$$), a considerably exothermic process with a reaction enthalpy of $$\mathrm{-253\ kJ\ mol^{-1}}$$ at $$\mathrm{298\ K}$$:

$$\ce{CO2(g) + 4H2(g) -> CH4(g) + 2 H2O(l)}$$

Methane is not an ideal product, as it too is a greenhouse gas and is a low value chemical feedstock due to its abundance and relative lack of useful chemistry. There is much more interest in conversion of $$\ce{CO2}$$ to compounds such as methanol $$\ce{CH3OH}$$ and formic acid $$\ce{HCOOH}$$. These two particular reactions are also exothermic.

There are several issues with using hydrogen reduction of $$\ce{CO2}$$ as a carbon capture strategy to combat climate change, but perhaps the main one is a real-world factor: most hydrogen we currently produce is derived from fossil fuels, notably from the partial combustion of fossil methane (natural gas) with water at high temperatures, known as steam reforming. Therefore, while alternative sources of hydrogen gas using renewable low-carbon intensity energy are not available, this is a poor strategy to remove anthropogenic $$\ce{CO2}$$ from the atmosphere.

• All true, good answer. But if we had readily available hydrogen gaz, I don't think we be burning fossil fuels and producing CO2 in the first place. So, we can assume we don't have hydrogen available. – Jeffrey Sep 3 '20 at 17:44
• And let's not forget that capturing the carbon dioxide is one of the biggest hurdles in the first place. Sure, it's accumulating in the atmosphere, but in the bulk, it's still just a couple hundred ppm - you need to process ridiculous amounts of air to extract the carbon dioxide, and that adds massive costs to anything you do with the carbon dioxide afterwards. Of course, there were interesting alternative proposals - e.g. feeding the exhaust from thermal power plants directly through vats of algae, recovering much of the fuel with the help of sunlight; not without its own problems, though. – Luaan Sep 4 '20 at 8:24
• @Luaan: That would be more than a "interesting alternative proposal". The Dutch commonly run greenhouses on the heat and CO2 of small power plants. It accounts for 10% of the total Dutch electricity production. It's a bit less impressive when you account for the fact that these greenhouses are also illuminated at night, which uses that same electricity. In effect, these greenhouses turn natural gas into food - the other byproduct of burning natural gas is water, which like CO2 is also consumed by the plants. – MSalters Sep 4 '20 at 15:20
• @MSalters That's pretty much what I meant, yeah. We're very far from being anywhere near to a closed loop, but there's certainly at least some cases where it's profitable (money and ecology likewise). Even the fact that they're illuminated at night isn't as bad as it sounds, given that night electricity is still a bit of a waste product - the electricity would have been produced in much the same quantity anyway. – Luaan Sep 6 '20 at 6:11

You’re looking at bond dissociation energies. They, however, do not give a good picture. A better place to start looking is the standard enthalpy of formation. The linked Wikipedia article provides an extensive list of compounds but only two matter:

• $$\displaystyle\Delta_\mathrm fH^0 (\ce{CO}) = \pu{-110.525 kJ/mol}$$
• $$\displaystyle\Delta_\mathrm fH^0 (\ce{CO2}) = \pu{-393.509 kJ/mol}$$

Next would be invoking Hess’ law which is based on the fact that enthalpy is a state function and thus the route getting from A to be in the reaction $$\ce{A->B}$$ does not matter, only the enthalpies of A and B. In mathematical terms, this is usually summed up as:

$$\Delta H^0_\text{reaction} = \sum \Delta_\mathrm f H^0_\text{products} - \sum \Delta_\mathrm f H^0_\text{reactants}\tag{1}$$

Plugging the values of a hypothetical reaction that goes only from carbon dioxide to carbon monoxide (and hitherto ignoring other reactants or products) gives us:

$$\ce{CO2 -> CO}\tag{A}$$ \begin{align}\Delta H^0_\text{(A)} &= \pu{-110.525kJ/mol} - (\pu{-393.509kJ/mol})\\ &= \pu{+282.984kJ/mol}\end{align}\tag{2}

This is an endothermic reaction and quite greatly so.

‘But wait!’, I hear you exclaim. ‘I’m producing oxygen too, aren’t I?’
Indeed you are. In simplest terms, you would be producing oxygen gas ($$\ce{O2}$$). And this does not help you, as the standard enthalpy of formation of elements is, by definition, zero. (At this point, we also nicely see that the reverse reaction – combustion of carbon monoxide to give carbon dioxide – is energetically favourable.)

That is not to say this reaction is impossible. You would need something else to put in – the reaction (A) as is isn’t balanced anyway. A real reaction would have to look something like this:

$$\ce{CO2 + X -> CO + Y}\tag{B}$$

Obviously, Y will have to be related to X but contain more oxygen. X and Y would then need to be carefully balanced to give an overall negative reaction enthalpy. One such example would be the creation of (liquid) water whose enthalpy of formation is $$\pu{-285.8 kJ/mol}$$ according to the same Wikipedia article linked above. As water would be produced by adding hydrogen gas to the reaction, the reactant side only gains another zero.

More complicated systems are evidently possible, the reactions and equations will just get more and more complicated. The key takeaway is, however, that the reaction as proposed is endothermic and you need some sort of driving force (something more exothermic) to make it happen.

Note that my calculations all centre around enthalpy. To determine whether a reaction will actually occur under a given set of conditions, the important value to check is Gibbs Free Energy. However, enthalpies usually provide a very good starting point.

• Your values regarding the hydrogen reduction of CO2 made me look around a bit more carefully. According to ACS Energy Lett. 2018, 3, 1938−1966, equation (1) on page 1943 shows a value of $\mathrm{+41.2\ kJ\ mol^{-1}}$, which is corroborated by Top Curr Chem (Z) 2018, 375, 23. It seems I had the right numerical value, but swapped the sign. I'll have to partially revise my answer. – Nicolau Saker Neto Sep 3 '20 at 8:53
• Also after rewriting my answer, I believe all our values are in agreement (to within fractional errors) - water being produced in the liquid state drives the reaction enthalpy to almost zero, which agrees with your calculations. You really can trust Hess's law! – Nicolau Saker Neto Sep 3 '20 at 9:31
• @NicolauSakerNeto Ah, wonderful, I was scared where this minor discrepancy might have come from but glad to see we agree even numerically in the end. – Jan Sep 3 '20 at 10:25

To get heat, you need to go the other way (add $$O_2$$ to $$CO$$). In steel mills, $$CO$$ is a waste byproduct of the blast furnaces that gets used as fuel. It gets burned in boilers and makes high pressure steam. The steam spins turbines and makes megawatts of electricity. It's not a great fuel compared to methane, but you can't beat the price.