My initial thoughts on this are that you are viewing the Maxwell-boltzmann $velocity$ distribution the wrong way. Let me explain,
In nature there are three main distributions that "things" follow:
1) Boltzmann
2) Bose-Einstein
3) Fermi-Dirac
The first one is a classical result while the latter two are the result of quantum mechanical effects (symmetry of the wavefunction with respect to permutation). These three distribution functions describe the "way" energy is distributed among the energy levels of a molecule/atom/generic particle.
Perhaps a better way to phrase it would be that the macrosystem can be separated into lots of tiny microsystems. Each of these tiny microsystems can have their own particular "state". The individual microstates will combine in a collection or $ensemble$ to give the overall macrostate observed on our scale.
At this point we can vary our definition of how to define the ensemble and indeed there are quite a few buzz terms to learn. A $canonical$ ensemble is one in which the number of particles in each microstate ($n_i$), the volumes of the microstates, and their temperatures are all the same. If you choose the $micro-canonical$ ensemble then you would have $n_i$, $V_i$ and the energy of the $i$th state as being kept constant but not the temperature. A $grand-canonical$ ensemble has constant $n_i$, $V_i$ and chemical potential $\mu _i$.
Now depending on what you choose you will get slightly different equations, but they are all variations on a general unifying theme:
"The frequency distribution of a particular state or outcome out of all those states or outcomes possible, is proportional to
\begin{equation}
e^{-\frac{E}{KT}}
\end{equation}
Which is know as the Boltzmann factor. $E$ is the energy of the state, which varies from state to state."
You have already come across this factor in kinetic theory. !!AS A PARTICULAR EXAMPLE OF A NATURAL PHENOMENA THAT OBEYS THE BOLTZMANN DISTRIBUTION!!.
You can clearly see that the velocity distribution function, the frequency of having a particular velocity, is proportional to the Boltzmann factor, in this case the particular energy we are considering is the kinetic energy, so let $E=\frac{1}{2}mv^2$.
\begin{equation}
f(v)=\sqrt{\frac{m}{2\pi KT}}e^{-\frac{mv^2}{2KT}}
\end{equation}
The Maxwell velocity distribution is just a special case of a Boltzmann distribution.
With this in mind I'm sure you can see that many different phenomena will be described by this distribution. If you wanted to put the Gibbs energy in there (indeed as we do in transition state theory (a personal favourite of mine)) then we can.
This brings me to your next question which is a little easier to answer. The Activation energy can readily be described by a Gibbs potential as indeed it usually is. I think this inherently answers your third question.
If you have any questions I will try to answer them when I can! :)
