# Hill function and stoichiometric matrix: confusion about reaction rates

When deriving the Hill function (protein $X$ and $n$ molecules of the signal $S$), my book introduces the following rates:

$$\text{collision rate} = k_{ON} \cdot X_n \cdot S^n\\ \text{dissociation rate} = k_{OFF} \cdot [nSX],$$

where $[nSX]$ is the complex. This translates into the following ODE:

$$\frac{d}{dt} [nSX] = k_{ON} \cdot X_n \cdot S^n - k_{OFF} \cdot [nSX].$$

This seems fine if I follow the $\textit{rules}$ of, for instance, this table.

On the other hand, during the class, my professor introduced the stoichiometric matrix. If we have the reaction $2A + B \leftrightarrow C$, then the corresponding system of ODEs would be:

$$A' = -2r_1A^2B + 2r_2C\\ B' = -r_1A^2B + r_2C\\ C' = r_1A^2B - r_2C.$$

So, my question is: why in this case we put the $2$ in front of the reaction rates while for the Hill function we do not (put the $n$ in front of it)? Is it just because the $n$ molecules of signal $S$ are not consumed nor produced?

If the reaction is $\ce{aA + bB = pP + qQ}$ then we define the rates as
$$-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{p}\frac{d[P]}{dt}=+\frac{1}{q}\frac{d[Q]}{dt}$$