How to distribute the Boltzmann energy to internal and external motion?
The translational energy of a mole of molecules is $\tfrac{3}{2}RT$ which corresponds to an average kinetic energy of $\tfrac{1}{2}M\langle v^2\rangle$.  The average reflects a Boltzmann distribution of speeds
$$f(v)=4\pi\left(\frac{M}{2\pi RT} \right)^{3/2}v^2\operatorname{e}^{-\tfrac{1}{2}Mv^2/RT} $$
i.e. some molecules have very low velocity and some have very high velocities. 

Kinetic energy is constantly being transferred, so if you could monitor its velocity over time its average kinetic energy would be $\tfrac{1}{2}M\langle v^2\rangle=\tfrac{3}{2}RT$

In addition one mole of gas has $\tfrac{3}{2}RT$ of rotational energy ($RT$ for linear molecules), which also reflects an average rotational kinetic energy.  For individual molecules at any given time, the amount of translational and rotational energy is independent.  For example, a molecule can move very slowly while rotating very rapidly – until it collides with another molecule and things change.  However, on average there is an equal amount of translational and rotational energy (for non-linear molecules).