# Second-step activation energy in SN1 reactions

In SN1 reactions, it's known that there are generally two steps, with the first being the R.D.S step involving carbocation formation and the next having a lower activation energy.

Why does the second step require activation energy at all? It's going from being a carbocation to a stable molecule, which will be more stable than the former, no matter how stabilized a cation it is.

• Barrierless reactions, i.e. reactions without activation energy, are a thing. I don't really want to comment on whether that step is barrierless. But I do want to point out that your general argument seems to be confusing energies of reactants and products (thermodynamics) with activation energies (kinetics). Just because a product is more stable than a reactant, doesn't mean that there is not going to be some kind of activation energy. Apr 6 at 16:26
• Oh, yes, 'rate'-determining step. Kinetics. I get it, thanks. Apr 6 at 16:28
• But @orthocresol: activation energy is a thermodynamic thing, right? That measures stability, not rate of reaction. Which is what the answer seems to be going with, too. Apr 7 at 4:45
• Well, yes, activation energy is an energy, which has to do with stability. But it is the energy of the transition state that is important. You were talking about the energy of the product, which is not relevant. Apr 7 at 8:17
• Let’s also make sure to separate the concept of energy from the concept of thermodynamics. Thermodynamics is concerned with the equilibrium properties of the system, kinetics with the rates of the reaction. That’s all it is; there isn’t any classification where thermodynamics is concerned with energies, and kinetics with other things. Just because something is an energy doesn’t necessarily mean it’s a thermodynamic property. Apr 7 at 8:20

The $$\mathrm{S_N1}$$ reaction has two steps i) the formation of carbocation and 2) the formation of product. (image taken from libretexts)
In the second step an unstable carbocation reacts with the nucleophile. You are asking why this step has to have an activation energy. You can rationalize that by considering the principle of microscopic reversibility. Think about what would happen if the $$\mathrm{S_N1}$$ reaction went in reverse—first the nucleophile leaves the product, forming a carbocation, and then the carbocation joins with the leaving group to give the reactant. Obviously when the nucleophile leaves the prdouct, and carbocation forms, there is an activation barrier for that.