I have designed a molecule using hydrocarbons A and B. I am getting the formation energy slightly positive (~0.04 eV). As per my experience, any positive formation energy means that the designed molecule will be not stable. Is there any way to synthesize a molecule for which the theoretical formation energy was predicted to be positive?
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$\begingroup$ +1 But are you asking: "is there any way" to synthesize such a molecule (like the last sentence in your question body) or "how can we" synthesize such a molecule (as in your question's title)? If you're asking "how to do" a wet-laboratory procedure, maybe Chem.SE is a better choice (we can wait for Tyberius or another Chem.SE expert to chime in about that)? If you're just asking about whether or not it's possible, someone here might be able to help you. $\endgroup$– Nike DattaniApr 23, 2021 at 4:49
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$\begingroup$ Yes, I am asking as described in the question body. $\endgroup$– asthaApr 23, 2021 at 5:18
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1$\begingroup$ As you point out, it won't be stable, but sure, you can synthesize it. Drug companies synthesize things all the time that end up being unstable and going bad on the shelf. There is alot of work done to avoid this. Since it happens plenty accidently, it can also be done on purpose. $\endgroup$– B. KellyApr 23, 2021 at 21:21
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3$\begingroup$ It should be made clear that positive formation energy usually refers to the formation from atoms, which is probably not you are doing. Instead, you will most likely be using different compounds. In this case, 1) use clouds with even higher energies as starting point, 2) if not possible, you can always add some external heat, or even better, 3) continuously remove the product. It is taught in organic chemistry that most reactions are reversible, so if it's exothermic in one direction, it's endothermic in the other, and that reaction proceeds just fine too. $\endgroup$– user430Apr 24, 2021 at 22:51
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1$\begingroup$ This seems to be a blind spot of quite a number of scientists - I've heard of a paper incorrectly rejected for this reason. It is perfectly possible to for a chemical system to have a positive enthalpy of formation. Well know examples include all nitrogen oxides, ethylene, acetylene, benzene, human beings. As noted above just because it is thermodynamically unstable WRT to the elements, that says nothing about the kinetic stability, and you don't have to make it from the elements. $\endgroup$– Ian BushNov 1, 2021 at 8:03
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The theoretical formation energy is a little ambiguous here, depending on the way you calculate it. Is it a 0 K calculation (for example using DFT)? or have you already accounted for the entropic effects?
If it is a 0 K calculation then it is indeed difficult to synthesize this particular compound. However, it is not impossible since you are still taking into account only the thermodynamics of the system rather than the kinetics. So you have to figure out some kinetic parameters to lock in your metastable configurations. This can be done by quick annealing or changing your precursors.
However, if you are already accounting for entropy in your calculations. You can go check through lower temperatures if your formation energy becomes negative.
Furthermore, formation energies are usually calculated with reference to a certain precursor. So do check if the formation energies are properly calculated or consider changing the precursors.
