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Say I have two kinds of molecules $A$ and $B$.

The molecules of type $B$ are small and bind onto molecules of type $A$.

(Any number of them may do so, but in practice this number is $\leq 8$.)

We say a molecule of type $A$ is of type $A'$ if it has at least one copy of $B$ bound to it.

More specifically, we say a molecule of type $A$ is of type $A^k$ if it has exactly $k$ copies of $B$ bound to it.

We can easily detect experimentally when a molecule is of type $A'$, but not so easily when a molecule is specifically type $A^3$, for example.

However, it was done at one point, and the researchers gave us a way (given a concentration of $A'$) to determine the individual concentrations of $A^k$'s. (Poisson distribution)

I would like to consider the dissociation reaction $$A^k \rightarrow A^{k-1} + B$$ where $k \geq 1$ and $A^0$ is considered to be $A$.

Because the molecules of $B$ are small, I can assume that the rate of this reaction is independent of $k$ and that the molecules of $B$ dissociate independently of any other molecules of $B$ attached to the same copy of $A$.

My question is: how can I find a constant $c$ so that the rate of the reaction is $r = c[A']$?

Open Season
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  • Is $A'$ the concentration $A$ molecules with at least one $B$ bound, or is it the concentration of $A$ molecules with precisely one $B$ bound? – Curt F. Dec 13 '15 at 04:58
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    In my notation $A^1$ would be the molecules with exactly one $B$ bound. Essentially $A'$ is the collection of all the molecules of type $A^1, A^2, A^3$ etc. – Open Season Dec 13 '15 at 05:33
  • You've written the dissociation reaction as irreversible. Is that what you intend? In that case every reaction is unimolecular so the decomposition kinetics will be a series of exponentials – Curt F. Dec 13 '15 at 18:55

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