Why is it possible to use free energy functions on Boltzmann factor? When the Boltzmann factor is derived on the canonical ensemble, one gets the exponential function of the energy:
$\large{e^{-\frac{E}{k_b T}}}$
However, some texts (it is very common on physical-chemistry, for exemple) just assume that you may use Gibbs free energy on the factor instead: 
$\large{e^{-\frac{G}{k_b T}}}$
But what is the justification for this?
 A: I am new to this, so if I am missing something obvious, my apologies. Also, I have not dealt with TD from the chemical or biology perspective.
Assume a given energy, $e^{-E/k_BT}$, provides the relative probability of a particular microstate. Now assume that we are concerned with a isobaric system, so E goes to H, (the enthalapy) and use the equation  $e^{-H/k_BT}$.
For the canonical ensemble, we need to take account, at a given energy, of the different microstates that will be present. 
$S = k_B ln \Omega$ therefore $\Omega = e^{S/k_B}$
So multiplying these together , to determine the probability that the system, at a fixed energy, is in a particular state:
$$\Omega \cdot e^{-H/{k_B}T} = e^{S/k_B} \cdot e^{-H/k_BT} = e^{TS/{k_B}T} \cdot e^{-H/k_BT} = e^{-(H-TS)/{k_B}T} = e^{-G/{k_B}T}$$
So the overall probability is actually proportional to $e^{-G/{k_B}T}$
So the  Gibbs free energy takes into account both energy (in the Boltzmann factor) and entropy, the number of possible arrangments. (by virtue of $\Omega$)
