If a reaction is exothermic it means that the energy state of the products is lower than that of the reactants so this will be the state that 'nature' will naturally strive for. Often you see that stability of molecules is argued in this way (e.g. in one of the answers here), saying that the higher the exothermicity of the decomposition reaction, the more unstable the molecule and thus the more unlikely that it will exist in nature.

My question is whether this reasoning is correct?

The reason I am doubting this, is that the reaction enthalpy $\Delta H_r$ is not the only parameter of importance in the energetics of chemical reactions: the activation energy $\Delta E_a$ can also play a major role. To clarify my point, take a look at the diagram for 2 hypothetical reactions I have made. Reaction 1 has a low activation energy and a low exothermicity, whereas 2 has a high activation energy and a high exothermicity.

Judging solely by exothermicity you would conclude that the reactants of reaction 2 are more unstable, but if you look at the activation energies for the reactions than I would argue that the reactants of reaction 1 are more unstable because they more easily react to form their products. Of course having a high activation energy would at best make the reactants meta-stable, but metastability could last for decades as well given the right conditions.

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So my general question is: can reaction exothermicity be used to say something about the stability of molecules and what is the role of activation energy in this story?


Good question. There exists two types of stability - thermodynamic stability (determined by $\Delta G_{reaction}$, or $\Delta H_{reaction}$ approximately) and kinetic stability (determined by $\Delta G ^{\ddagger}$, or $E_{activation}$ approximately). The reactant compound in your first reaction is thermodynamically more stable than the second, since it sits at a lower potential energy, but it is kinetically less stable, since there is a smaller barrier for decomposition.

Usually context is sufficient to determine what kind of stability is being talked about, though in general I think stability is more often mentioned in the thermodynamic sense than the kinetic, since thermodynamic stability is related to the reaction free energy (a sum of its enthalpy change and a factor containing its entropy change) and is more commonly tabulated and more readily measured than activation free energies.

For example, all explosives are quite thermodynamically unstable (otherwise they wouldn't be very good at exploding!), but we still speak of "stable explosives", such as TNT or RDX and "unstable explosives", such as $\ce{NI_3}$ or $\ce{Hg(CNO)_2}$. Clearly kinetic stability is what is taken into account.

In constrast, when studying acidity and basicity of substances, it's common to justify reactions based on the "stability" of the acids/bases/conjugate species, but in this case, we're talking about thermodynamic stability, as it defines the acidity/basicity constants.

  • $\begingroup$ Cool! So that would mean that the thermodynamic stability will determine whether it is likely for a compound to exist, but that the kinetic stability will determine whether it will 'live on' right?! $\endgroup$
    – Michiel
    May 15 '13 at 12:22
  • 1
    $\begingroup$ @michielm That's correct in general terms. However, thermodynamic stability by itself technically suggests that only a handful of compounds could ever exist (there's only one global energy minimum for each type of atom). Practically everything we know to exist is behind a kinetic barrier to decomposition, so it's a bit tricky to match the purely thermodynamic and the intuitive "likelihood of existence". It's best to take both stabilities into account at once when considering if a compound exists or not. Separating their contributions can be done, but it may not be too useful. $\endgroup$ May 16 '13 at 3:00
  • $\begingroup$ @michielm Also, I just noticed that you defined the reaction enthalpy incorrectly. The correct value is the difference between the enthalpy of the products and the reagents (end state minus initial state), and doesn't take into account the transition state. $\endgroup$ May 16 '13 at 13:05
  • $\begingroup$ mmm, you're right, stupid mistake. I will correct it this evening. Anyway, it doesn't change the story or the relative relations of the 2 reactions $\endgroup$
    – Michiel
    May 16 '13 at 15:10

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