The Boltzmann distribution is a probability density function which expresses the probability of finding a particle in an energy state $\epsilon$ while in thermal equilibrium given a specific temperature. The Boltzmann distribution can be written as:
$$ p(\epsilon)=C\cdot\exp{\left(\frac{-\epsilon}{k_\mathrm BT}\right)} $$
I have noticed that it is possible to derive the Arrhenius equation by slightly modifying the Boltzmann distribution. We simply have to express the Boltzmann Distribution in terms of the universal gas constant R by multiplying the Avogadro’s number.
$$ p\left(\epsilon\right)=C\cdot\exp{\left(\frac{-\epsilon}{k_\mathrm BT}\right)}=C\cdot\exp{\left(\frac{-\epsilon\cdot N_\mathrm a}{k_\mathrm BT\cdot N_\mathrm a}\right)}\iff p\left(E\right)=C\cdot\exp{\left(\frac{-E}{RT}\right)} $$
Then normalize the function.
$$ C\int_0^\infty{\exp{\left(\frac{-\epsilon}{k_\mathrm BT}\right)}\,\mathrm dE=1}\Longleftrightarrow C=\frac1{RT} $$
Finally, integrate the function from $E_a$ to $\infty$ to find the ratio of particles with sufficient energy.
$$ \frac1{RT}\int_{E_\mathrm a}^\infty\exp{\left(\frac{-E}{RT}\right)\,\mathrm dE}=\exp\left(\frac{-E_\mathrm a}{RT}\right) $$
This gives the Arrhenius exponential term.
However, I’m rather unsure about the validity of this demonstration, as the Boltzmann distribution fails to model gas particle behaviour conceptually. Since the Boltzmann distribution is decreasing over its entire domain, this suggests that given any gas particle, it has the highest probability to have energy equal to zero. In other words, any gas particle will most likely be immobile.
Given this inconsistency, is the above proof rigorous or is it just mathematical luck? If it is luck, then what is a more rigorous way of deriving the exponential term in the Arrhenius equation?