I only have a GCSE in Chemistry but I have discovered a lot of resources myself. One thing I cannot find an answer to is this: What is physically responsible for the different compounds you can get with the same types of atoms.

For example, if you were to mix a bunch of carbon and hydrogen what exact process determines if you get methane, butane, ethyne etc.? Is it the different concentrations of carbon and hydrogen? Then what is the physical explanation? If there is 4 times more hydrogen than carbon might a carbon atom more likely find itself surrounded by four hydrogens by chance and then bond with them to form methane? Similarly if there is equal carbon and hydrogen is it just morel likely that ethyne will form? If this is the case then in general if you mix carbon and hydrogen in different concentrations would you always get different concentrations of methane, butane, ethyne as products just by chance? For example if there is 4 times more hydrogen than carbon then is there still a small chance that at least some ethyne will form?

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    $\begingroup$ Statistics plays a role, but it is very heavily weighted by the "energy landscape" of a mixture of materials under a given set of conditions, which is an extremely complex, very high-dimensional potential energy surface. $\endgroup$ Dec 22, 2016 at 13:11
  • $\begingroup$ It's hard to get my head around the concept of energy landscapes and potential energy surfaces. Is there a way you can sum up what you said in simpler terms? Are there any trends which can roughly be followed even if they aren't accurate? $\endgroup$
    – Jonathan
    Dec 22, 2016 at 16:13
  • $\begingroup$ In general its a huge question but specifically (a) you will need to determine what products are energetically feasible for a given ratio of reactants (b) what energy are you are going to give the reactants to make them react and (c) probably most important, of the products that can be formed, chemical kinetics (reaction rates) will determine these and they may not be the therm0dynamically most favourable $\endgroup$
    – porphyrin
    Dec 22, 2016 at 19:37

2 Answers 2


Okay, I'm sure there's a much more direct, down-to-earth answer that you'd find more helpful, but I'm going for something different. To stop myself from writing a book chapter, there will be some hand-waving and cut out details. With more time and thought it could be unpacked further, but hopefully I'll at least be able to give you a taste.

Right, let's forget chemistry and try to deduce the product for the following simple reaction:

$$\ce{H2 + O2 -> ?}$$

To start off, we establish some basic rules:

  1. Molecules are formed from the combination of integer amounts of atoms
  2. Atoms cannot be formed or destroyed (i.e., the number of atoms is a conserved quantity)
  3. There is a quantity (let's call it potential energy) inherent to each kind of free atom and molecule
  4. The Universe tends towards the lowest possible potential energy

With nothing else to guide us, we are forced to consider every possible product (isomers neglected for simplicity):

$$\ce{H2 + O2 -> H + O + H_2 + HO + O_2 + H_3 + H_2O + HO_2 + O3 + H4 +...}$$

Now, if the point is to minimize the total potential energy of the system, we can develop a heuristic (which will turn out to be wrong, but entertain the idea for now). First, find the specific atom or molecule among the listed products with the lowest possible energy,$^1$ then make as much of it as allowed by stoichiometry. When no further can be made, look for the next lowest energy molecule, then make as much of that as possible, and iterate the process until no lower energy molecules can be made.

That's nice and all, but without knowing which molecules have lower or higher energy, you don't get anywhere. Here, we either resort to some hefty quantum mechanical calculations, or empirical data. By doing some thermochemical experiments (measuring enthalpies of formation), it turns out that $\ce{H_2O}$ is the molecule with the lowest energy of the list by quite a margin, so it's going to be the main product. Indeed that is what is observed empirically when hydrogen is burnt in oxygen.

So we know that water is the lowest energy product for the reaction, but what happens if we mix $\ce{H2}$ and $\ce{O2}$ with a ratio different from that in water? Say you have two moles of $\ce{H2}$ and two moles of $\ce{O2}$, does that mean you now get $\ce{H2O2}$ or potentially some other products? In fact, this is not the case. The heuristic above still applies to minimize the potential energy of the system, so $\ce{H_2O}$ is preferentially produced. This shouldn't come as too much of a surprise, because the low energy of $\ce{H_2O}$ is an inherent feature of the molecule, and is entirely independent of how much $\ce{H2}$ and $\ce{O2}$ you decide to mix in an experiment.

Then what happens to the excess $\ce{O2}$? Again following the heuristic, it will seek out the lowest energy molecule available. Since there's no more hydrogen left to react, it has a more limited number of possibilities. It could conceivably still react and turn into other molecules (for example, $\ce{O3}$ is perfectly acceptable), but it turns out that of the remaining candidates, $\ce{O2}$ is the molecule with the lowest energy, so the reaction just stops there, and we get a mixture of $\ce{H_2O}$ and left over $\ce{O2}$.

Okay, we've made some progress, but we need to talk about some important complications. In the real world, at any temperature above absolute zero, you can actually retain a small amount of molecules with a higher potential energy. All of those lower energy molecules can randomly gain a bit of thermal energy, and every so often that causes them to react and turn into a higher energy molecule. If the lowest energy molecules are much lower in energy than anything else (as is the case for $\ce{H_2O}$ among the combustion products of $\ce{H2}$ and $\ce{O2}$), then this effect is very small and can be neglected. However, in many cases there isn't such a clear-cut winner, and you end up getting a mixture of different products with similar energies. This is especially true as you get into more complicated chemistry.

Unfortunately, even after all this we're still missing a very big piece of the puzzle. Everything I've mentioned so far is a part of thermodynamics, and that alone cannot always predict the product of a reaction. Allow me to prove it. According to everything I've argued, $\ce{H2}$ and $\ce{O2}$ will always react to form primarily $\ce{H2O}$ since it has the lowest energy, and $\ce{H2O2}$ can never be a significant product. But hydrogen peroxide is made in very large amounts industrially, so how do they do it? A quick search on the internet provides the answer, the anthraquinone process, whose net reaction is simply:

$$\ce{H2 + O2 -> H2O2}$$

What gives? What we've been missing is the concept of kinetics. For many transformations in nature, chemical or otherwise, even if the final state has a lower energy than the initial state, they are often separated by an energy barrier. Even if there can be a net release of energy, the transformation is hindered because to start off it needs a net input of energy.

This is the case even for the reaction $\ce{H2 + O2 -> H2O}$. If you just mix the gasses together at room temperature, they will not react even after a century. Either some initial energy needs to be supplied (a flame, a spark, etc.) or the energy barrier needs to be decreased by using a catalyst (Pt), after which the reaction will proceed in a fraction of a second.

When you consider all possible reactions, and all the energy barriers separating the different molecules, you get the very high-dimensional potential energy surfaces I mentioned in my comment. All our work in Chemistry boils down to finding reliable paths in (or making our own by actively modifying) the crazily complex energy landscapes and exploiting them for our purposes.

$^1$ Technically, the lowest amount of potential energy per amount of atoms the molecule is made of. You can think of it as the lowest potential energy per gram of molecules.

Here's something which might also help. Is there a case where all this potential energy and kinetic stuff doesn't skew the results, and you can tease out the purely statistical effect? Sure, let me present a typical case.

Consider the following transesterification reaction (poorly drawn because I don't currently have access to ChemDraw):

enter image description here

Both sides of the equation have an alcohol and an ester. The alcohols are very similar to each other, and the same can be said for the esters. Because of this, there's barely any difference between the potential energy of the products and the reagents.

Now consider taking a mole of the starting ester and alcohol. In the beginning, there is only ethanol and no methanol, so statistically you'd expect the reaction to happen in the direction shown. As the reaction progresses, the amount of ethanol available is reduced, disfavouring further generation of products, and the presence of methanol now actually allows the reaction to also go the other way around. By pure statistics, you'd expect the reaction to stop when there's about the same amount of methanol and ethanol, half a mole each, and this is what is observed. In short, simple transesterification reactions tend to have equilibrium constants very close to 1 (e.g. this article).

  • $\begingroup$ Thanks for the answer! Very comprehensive. So you mentioned that you'd have to do a load of quantum calculations to find out the molecule with lowest energy. Are there any basic rules of thumb which aren't entirely accurate but hold some truth to determine molecules with lower energy? For example, is bond energy anything to do with it? I'd imagine that double bonds are higher energy and atoms might prefer to make single bonds perhaps? $\endgroup$
    – Jonathan
    Dec 22, 2016 at 20:52
  • $\begingroup$ Without quantum calculations, the rules of thumb are pretty much what make general chemistry. The bond energies are very important, yes, but not all-encompassing (e.g., some reactions will happen spontaneously even though they absorb energy and make higher energy products). There is no clear rule whether multiple bonds or single bonds are better, at least for the second row elements, though heavier elements tend to favour single bonds. For example, $\ce{N2}$ is an exceptionally low energy molecule, and it has a triple bond. Meanwhile, $\ce{P2}$ also has a triple bond, but is very reactive. $\endgroup$ Dec 23, 2016 at 10:36
  • $\begingroup$ Ultimately, you can't run. If you want to learn chemistry, you'll have to familiarise yourself with how the elements behave by reading some of their basic properties including typical reactions, and looking at tabulated empirical values of multiple properties (ionization energies, electron affinities, electronegativities, proton affinities, equilibrium constants, etc) to help compare different compounds. $\endgroup$ Dec 23, 2016 at 10:41
  • $\begingroup$ Yeah I wish I had time to learn it all before my stage 0. I'm doing a PhD and my background is more informatics / complex systems. I have to write a detailed research proposal for my stage 0 which is due very soon. They want a super high level of specificity, to the point where they expect you to have the answers to a lot of your research before you've actually done it! I'm doing a computational chemistry course next semester and I'll be teaching myself a lot of chemistry. But the stage 0 is before that! Hopefully I can convince them. Thanks anyway! $\endgroup$
    – Jonathan
    Dec 24, 2016 at 12:57

The products you can have from a reaction depend on many things, not only their concentrations. Also, you can hardly mix only free atoms of different elements. You are thinking on a situation where you mix carbon free atoms with hydrogen free atoms, and that is not the usual situation. The answer is about conditions, for instance, one important factor could be pressure: in situations with high pressures, it is more likely to form compounds that have less volume per mass unit. In the end, the compounds formed depend on a combination of two factors: thermodynamic and kinetics. Atoms will form the compounds in which they have the minimum energy, as much as the energy barriers can be overcome.

  • $\begingroup$ I disagree that atoms will always form the compounds in which they have the minimum energy: that is very much not the case in many examples. For example if you have species, A, B, and C, and compound AB is lowest energy but can (even rarely) dissociate back into A + B, but--if AC is formed even for a second, AC is removed from the system (such as as a solid or as a gas), then over time AC will eventually be the primary product, even though it is not lower energy than AB. Perhaps this is what you meant by "as much as the energy barriers can be overcome", but this is not clear. $\endgroup$
    – NMJD
    Dec 22, 2016 at 17:01
  • $\begingroup$ Could you explain a few terms for me? I'm unsure what volume per mass unit means exactly, is that molar volume? By that did you mean smaller compounds are created at higher pressure? So in terms of thermodynamics and kinetics do you mean that for example at high temperatures molecules/atoms move more quickly so then bonds are broken from higher energy collisions and others can form etc? I have heard about the minimum energy thing. Does this mean that single bonds are prefered over double ones because double ones have more potential energy? $\endgroup$
    – Jonathan
    Dec 22, 2016 at 18:45
  • $\begingroup$ Yeah, by energy barriers I mean that a reaction can be kinetically controlled. If molecules/atoms have no energy enough, then will rather form a less stable compound, as you explained. $\endgroup$
    – AngBon
    Dec 26, 2016 at 17:35
  • $\begingroup$ Also, when I was talking about the pressure in gas reactions, I meant that the compounds formed at high pressures are those occupying less volume. See if you had the reaction: 2 SO2 + O2 <--> 2 SO3, you would get: $\endgroup$
    – AngBon
    Dec 26, 2016 at 17:38
  • $\begingroup$ you would get at high pressures, more SO3, because two molecules of SO3 occupy less volume than two of SO2 and one of $\endgroup$
    – AngBon
    Dec 26, 2016 at 17:41

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