What is the probability that two chemical species of binding energy $E$ will be bound?

Question

This may seem like a very simple question, but I've been agonising over it for days. What is the probability, $p$, that two chemical species with binding energy $E$, will be bound.
My first instinct is that
begin{equation} tag{1}label{1} p = Ae^{-beta E} end{equation}
with A some normalisation factor. This would give the correct behaviour ($pto1$ as $Eto -infty$ and $pto 0$ as $Eto 1$). But if this is the case, what is the form of $A$.
However, for
begin{equation}tag{2}label{2} p = 1 - e^{beta E} end{equation}
we see the same behaviour.
I have had a look in some literature for this, and have become even more confused, seeing both eqref{1} and eqref{2} used.

Bert Barrois · Answer

The answer is not a simple function of binding energy E.  What you need to know is the equilibrium constant  $[AB]/[A][B]=K=exp (-beta {{G}_{0}})$, where ${{G}_{0}}$ denotes the free energy difference at reference concentrations.  Doubling the concentration of unbound A & B will quadruple the concentration of bound AB.
K is proportional  (not equal) to $exp (beta E)$ as you suspected, but there is a further factor that pertains to entropy.  In gas phase, the entropy difference can be calculated via the semi-classical Sackur-Tetrode equation, which drags in Planck’s constant, cubed.  The situation in liquid phase is not conceptually different, but solvation energies complicate things.

What is the probability that two chemical species of binding energy $E$ will be bound?

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