Activation energy and degrees of freedom
This depends upon the theory used to model the reaction; there are two major ones.
Collision Theory
If your molecules may be assumed to be spherical particles with no rotational or vibrational DOF, then collision theory appropriately describes your system. In this case, only translational energy is accounted for. The rate expression may then be derived from the Maxwell-Boltzmann kinetic theory of gases, where the energy of each particle is $\frac{1}{2}mv^{2}$.
I. Filot has a decent explanation of this.
Transition State Theory
Unfortunately, most reaction systems are much more complicated; the reactants experience not only translational, but also rotational and vibrational DOF. This must be accounted for in the rate expression. Again, I. Filot has a good explanation for this.
A general derivation includes the following:
Consider a bimolecular reaction:
$$A+B \rightarrow C$$
We assume there exists a transition state (TST), which defines the highest point along a minimum energy path on the potential energy surface. Moreover, we assume that our reactants are quasi-equilibrated with this TST; the rate of the reaction depends upon the frequency of reactants overcoming this barrier $(k_{f})$. Then:
$$A+B \leftrightarrow C^{\ddagger} \rightarrow C$$
Assuming this reaction occurs in the gas phase under ideal conditions, we can approximate their fugacity using the partial pressure of each reactant. Let us also assume that the reference partial pressures are unity. We derive an equilibrium constant between the TST and the reactants:
$$K^{\ddagger}=\frac{P_{C^{\ddagger}}}{P_{A}P_{B}} = exp(-\Delta G^{\ddagger}/RT) = exp(-\Delta H^{\ddagger}/RT)exp(\Delta S^{\ddagger}/R))$$
Our rate expression becomes:
$$r = K^{\ddagger}k_{f} P_{A}P_{B} = [(k_{f}) exp(\Delta S^{\ddagger}/R))exp(-\Delta H^{\ddagger}/RT)]P_{A}P_{B} = kP_{A}P_{B}$$
Our rate constant is:
$$k = (k_{f}) exp(\Delta S^{\ddagger}/R))exp(-\Delta H^{\ddagger}/RT) = (A)exp(-\Delta H^{\ddagger}/RT) \equiv (A)exp(-E_{a}/RT)$$
Where the partition functions defining the enthalpy and entropy include translational, rotational and vibrational DOF.
Therefore, to answer your questions:
&quot;First, I want to understand if the factor $e^{-E_{a}/RT}$ takes into account only the percentage that have the necessary kinetic energy in translational degrees of freedom or in general the percentage with at least this energy irrespective if this energy is stored in a combination of various degrees of freedom?
No, it takes into account all DOF. In transition state theory, both the activation energy and pre-exponential factor consider the translational, rotational and vibrational DOF of the transition-state irrespective of the DOF providing the energy (This is directly tied to the canonical partition function, which is used in defining the enthalpy and entropy. It only cares about the total microstate energy and not from which DOF it comes from).
&quot;Secondly, if some reactions speed up depending in which type of degrees of freedom (vibrational or translational) are excited, does this mean that the Arrhenius equation even for an elementary bimolecular reaction holds only if the reactants have only translational degrees of freedom?&quot;
Reactions do not speed up depending on which type of DOF are excited. However, certain DOF speed up the reaction much more than other DOF. For example, based on the derivation I have provided, the pre-exponential factor scales exponentially with the transition-state entropy.
$$S_{total} = S_{translations}+S_{rotations}+S_{vibrations}$$
The DOF with the largest contribution to the entropy are translations. In theory, we may increase the rate of the reaction by increasing the vibrational entropy of our transition state. But, we may increase it to a much larger extent by increasing it's translational entropy.