# How is the existence of carbon monoxide justified, in spite of the absence of mono-atomic oxygen in nature?

I have been taught that oxygen is found as $\ce{O2}$ in nature.

Then, how does carbon monoxide form if oxygen cannot be found as $\ce{O1}$ in nature?

• Well, you just break the $\ce{O2}$ apart. Ordinarily that's not very favourable, but if carbon is next door to it and says "I want to bind to oxygen", then it can happen. – orthocresol Sep 8 '16 at 11:52
• High temperature may also help to break it apart. – DHMO Sep 8 '16 at 12:31
• Which allotrope of carbon would you like to discuss? – DHMO Sep 8 '16 at 12:33
• Very similar question - chemistry.stackexchange.com/questions/34022/… – Agriculturist Sep 8 '16 at 15:24
• Limiting reagengt -- here oxygen -- may lead to the formation of CO instead. And $\ce{CO2}$ once formed, may react with further carbon to CO, as well. – Buttonwood Feb 25 '17 at 1:40

(Almost) all reactions are not simple additions of compounds to form bigger compounds, but involve breaking of bonds and forming of bonds.

As is with (almost) other reactions, the reaction $\ce{2C + O2 -> 2CO}$ proceeds by breaking the bonds between $\ce{C}$ (it is heavily bonded in nature) and the bonds within $\ce{O2}$, and then re-combining as $\ce{CO}$ among other byproducts. (Note: this is extremely simplified but should be enough for your level.)

In conclusion, here is a very simplified mechanism:

1. $\ce{C -> C*}$
2. $\ce{O2 -> 2O*}$
3. $\ce{C* + O* -> CO}$

In order to form $\ce{CO}$, first you must break the $\ce{O2}$ bond.

Even if $\ce{O2}$ bond energy is as large as nearly 500 kJ/mol , this bond can be broken via lightening in the atmosphere, elevated temperatures or catalytic effect.

When the proper catalysts interact with the $\ce{O2}$ molecule, bond can be broken with less activation barrier than in gas phase via interaction between molecule and potentially energetic surface/nanoparticle sites. This interaction makes easier decomposition via heterolytic or homolytic bond cleavage depending upon a magnitude of the force applied by the catalyst. If the carbon atom exist in the environment and attraction between oxygen atom is very strong together with the thermal excitations or pressure difference or any other effect which enables oxygen to desorb from/diffuse on the catalyst, $\ce{CO}$ molecule can form.

Learn to distinguish between the elements — which always form some sort of elemental structure — and atoms of those elements that can also form compounds with other atoms of other elements.

When your teacher tells you that ‘oxygen typically occurs as $\ce{O2}$’, then they are making a statement about elemental oxygen i.e. a compound formed solely with oxygen atoms. Carbon monoxide is a compound that is formed of one oxygen and one carbon atoms — a statement that is (generally) true for elemental oxygen no longer holds true for compounds that include atoms of oxygen.

Likewise, elemental sulfur is typically $\ce{S8}$ but a number of sulfur compounds contain 1, 2, 3, 4, … sulfur atoms. And a similar case can be made for phosphorus, one modification of which is $\ce{P4}$.

From the article "Combustion" on Wikipedia (after googling "combustion"): "Thermodynamically, the chemical equilibrium of combustion in air is overwhelmingly on the side of the products. However, complete combustion is almost impossible to achieve, since the chemical equilibrium is not necessarily reached, or may contain unburnt products such as carbon monoxide, hydrogen and even carbon (soot or ash). Thus, the produced smoke is usually toxic and contains unburned or partially oxidized products."

So, if you have educts like pure carbon C and pure oxygen O_2, they will always react, if you provide the energy to overcome the reaction barrier, because the product is far more stable than pure C or pure O_2. EDIT: In the reaction process the O2 binds to pure carbon and gets split up process, because it is thermodynamically better to form two CO molecules.

Also, although you may not find single oxygen atoms in nature, O_2 is an aggressive gas, because it reacts spontaneously with most elements and "oxidzes" many materials over time. It may not be spectacular as fluorine, but it is still a very reactive gas.

• I don't think you should answer when you are drunk... jokes aside, you didn't actually answer the question. Your first two paragraphs are not related and your last paragraph is a regurgitation of the question. How does $\ce{CO}$ which have only one oxygen atom form from $\ce{O2}$ which has two oxygen atoms? – DHMO Sep 8 '16 at 13:27
• @user34888 well, I guess I thought of smth, but forgot to write it down – drunk user Sep 8 '16 at 13:46