Confusing definition of mol when considering second order rate equations I am confused by &quot;$\text{mol}$&quot;.
I denote the unit of a physical quantity with square brackets (e.g. $[V]=l$), and the value of a physical quantitiy using curly brackets (e.g. $\left\{ V\right\}$ = 0.01).
A simple second order rate equation for number concentrations $n=N/V$ is given by,
$$\dot{n} = k^*_{on}n^2,\;\;[1]$$
where the $[k^*_{on}]=l/s$.
The same second order rate equation but for molar concentration $c=\frac{N}{N_AV}$, is given by
$$\dot{c} = k_{on}c^2,\;\;[2]$$
where $[k_{on}]=\frac{l}{\text{mol}~s}$.
When transforming equation $[1]$ into equation $[2]$, we get the relation
$$k_{on} = \left\{k_{on}\right\}\frac{l}{\text{mol}~s} = k^*_{on}N_A = \left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{\frac{1}{\left[N_A\right]}s}$$
Hence $\frac{1}{\left[N_A\right]} = \text{mol}$?
But this would be in contraction to the definition I find on wikipedia where it is written that:

In short, for particles 1 mol = $6.02214076×10^{23}$.

Hence, $1~\text{mol} = \left\{N_A\right\}$.
I think the sentence cited above in the wikipedia article is very confusing.
But the wiki article about the Avogadro constant helped me resolve my confusion, which defines $mol$ via the Avogadro constant:
$$N_A = \left\{N_A\right\}[N_A] = 6.022\times10^{23} \text{mol}^{-1}.$$
Hence
$$k_{on} = k^*_{on}N_A = \left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{\frac{1}{[N_A]}s}=\left\{k^*_{on}\right\}\left\{N_A\right\}\frac{l}{mol~s}$$
This definition is also given on the hompege of the National Institute of Standards and Technology (NIST), see here.