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According to the Arrhenius equation the rate of a reaction increases as at higher temperature a greater percentage of molecules have translational kinetic energy greater than or equal to activation energy Wikipedia.

I was recently introduced in molecular potential surfaces and I read that some reactions can proceed when excess kinetic energy is stored into vibrationals degrees of freedom than translational. But according to Wiki it seems that only translational kinetic energy matters. First, I want to understand if the factor:

$$e^{\frac{-E_a}{RT}}$$

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? Also for two reactants $\ce{A}$ and $\ce{B}$ does it calculate the percentage of "pairs" $\ce{A}$, $\ce{B}$ that have this amount of energy or just the percentage of the population with at least this energy?

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 (like collision of two atoms $\ce{A}$ and $\ce{B}$ to form $\ce{A-B})?$

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  • $\begingroup$ In my opinion, the Arrhenius activation energy is an experimental value determined by comparing different reaction rates versus temperature. It is useful for determining reaction rates at any desired temperature. But the nature of this activation energy is not specified. It can be vibrational, translational. Who cares ? $\endgroup$ – Maurice Jan 13 at 15:07
  • $\begingroup$ You may be interested in the answer here chemistry.stackexchange.com/questions/139196/…? $\endgroup$ – porphyrin Jan 15 at 10:09
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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:

"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).

"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?"

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.

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  • $\begingroup$ Thank you for your answer. For A and B does it take into account the percentage of pairs that have this energy or the individuals A and B? Also, I still can't get why a reaction speed up more when for example we use IR radiation to excite the reactants vibrationally. Does it speed it up more because for a brief time the vibrational DOF are more excited than Maxwell-Boltzmann distribution (it takes a little time before the system equilibrates again)? So if the reaction is more sensitive to vibrational DOF it will speed up more (than if we just heated the reactants)? I will check out the book. $\endgroup$ – Anton Jan 13 at 20:55
  • $\begingroup$ They are not mutually exclusive quantities when referring to the rate. When you say "pairs", what you are referring to are transition states (i.e AB contacting each other) Per the TST theory derivation, the rate constant depends on microstate populations among AB, A and B. Remember: the Gibb's free energy barrier, is (del_G of AB - del_G of A - del_G of B). We can use IR to populate A and B into higher energy levels, thereby lowering the activation barrier. On the contrary, if IR were to "excite" AB, we would increase the barrier. $\endgroup$ – chr218 Jan 13 at 21:50

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