First, let me stress that the activation energy is not a well defined microscopic quantity, but a convenient parameter in which we can hide our ignorance on the details of the many different possibilities for the individual reactions. 

This having been said, the importance of quantum effects at a given temperature can be assessed by comparing the corresponding thermal energy with the characteristic energies for the different statistics, that is, the zero point energy for enclosing the particle in a given volume, which, for Fermions is known as the Fermi energy and for Bosons as the condensation energy.

As an example, let us take the Fermion case. For electrons in sodium metal, with a particle density of 2.65 x 10^28 /m^3, the Fermi energy is 3.24 eV and the Fermi temperature is 37'700 K, proportional to (N/V)^(2/3) and inversely proportional to the mass of the particle. I am not an expert in solution chemistry (if the concept makes sense  at all at such low temperatures), but I think it is fair to assume that, just on steric grounds, the number density of the reacting molecules is at least two orders of magnitude smaller, which divides the sodium result by a factor of the order of 20. Then, we know that the mass of the nucleon is 1836 times larger than that of the electron, and already for a small molecule like methane, this means division by further factor, this time of the order of 30'000. So you see that the degeneracy temperature at which quantum effects become important is below 0.1 K. Exactly the same reasoning holds for the boson case.

In summary, there is no need to change anything in your treatment.

Best regards, René Monnier

Dear Peter,

there is another and "simple" way to look at your question. All three statistics merge to a single one (Boltzmann) , when condition (Ei-chem.pot.)/k.T >>1 (~more than 3), where Ei is the energy of a particle(quantum or otherwise) and chem.pot, is the chemical (or electrochemical potential - I think your case) of your particle at energy Ei. This is so even in the case of one-particle, quasi-classical approximation, like in semiconductors.

I said "simple", because, depending on the system under investigation, it might be that the calculation of the el.chem.potential is not as trivial as is the case for monocrystalline Silicon (electrical transport and deep levels/defects).

I have recently noticed (others have probably noticed this long before me !) that the description of the electrical response from deep levels/defects in a semiconductor is formally described by the same type equations as electrochemical reactions.

In summary then, the Fermi Dirac and/or Bose-Einstein statistics should be used when the particle energies in question are close to the corresponding chemical(electrochemical) potential. 

I hope this might help

With best regard

Petr

The form you give is only obvious (and generally correct)  in the classical limit, where only the initial distribution of collision energies (a simple-to-calculate Maxwell distribution involving the relative mass) need to be considered. In the quantum case one must not only take into account the full initial distributions of the two particles -- of energies E1 and E2, moving in such a way that their collision energy is Ecoll -- but also of their energies  E'1 and E'2 after the collision. This is so because Pauli blocking may prevent or reduce collisions to some final states, and Bose enhancement may encourage some collisions. If the collision cross-section depends on the scattering angle it is not even in principle possible to reduce the rate formula to the form you have given.

You should look up some textbooks on quantum transport processes for more information. Sorry I cannot suggest any textbooks; I have only studied the case of relativistic collision processes (from research papers).

Note that whether a temperature should be considered high or low does not really depend on its value compared to room temperature. It depends on the density of the particles, and their mass also: More precisely of whether the parameters xi = di * λi3, (i=1, 2) are small or large. Where di is the density of particle of type i, and λi is its thermal de Broglie wavelength. Hence, you should first estimate the value of these parameters, and only consider more complicated calculations if they are close to one or larger.

Dear Peter,

Let me focus your attention on case of the statistic for the gas.

The atoms of the gas are supposed in-discernable. For this reason generally people apply them the Bose-Einstein statistic. But as soon as we consider the volume of the atoms the hypothesis of in-discernable particles is wrong.

The statistical problem can be solve like this: How much important the absolute temperature T may be, it is the mean thermal energy U per particle which play a fundamental role in statistics. This leads to determine D(E,U) the density of probability per unit of energy for the energy E of the particles. E is the energy above the energy of condensation.

The problem is to show that the total energy of a given volume of gas shall be calculated by summing, on small segments of energy, the number of atoms with energy on these different segments. For a number N of atoms, this sum should be equal to N times the average U value per atom of thermal energy. To respect the average value of the energy; It should also be that the total probability on the set of the possible energy values must be equal to unity.

To solve the problem it should be noted that the energy exchanges that produce high or low energy particles are less probable than those who leave particles in the vicinity of the average value. There is therefore a maximum of statistical weight determining the probability value around average value.  On the other hand it is their own volume which forbids to two particles to occupy the same place but not the exclusion principle of Pauli.

The study of the variations of UD(E,U) as a function of E/U shows that the highest values are in the vicinity of energy zero. This result is normal because if one atom has a very high energy, it should be that a large number of other atoms have ceded a large part of their energy in order to respect the average U. Taking the parallel with money this is to say that it takes a lot of poor to make one rich.  See “Perturbations and Statistical Distribution of the Thermal Energy” on my site.

I have given different experimental proof in paramagnetism, superconductivity and so on.

A comment to the question:

What kind of quantum statistic should be used in the calculation of thermal rate constant at ultralow temperatures?

The thermal rate coefficient can be obtained from the reactive cross section (σ(Ecoll)):

k(T) = c(T)×∫ P(T,Ecoll)Ecollσ(Ecoll)dEcoll

where Ecoll is the relative collision energy and c(T) is a constants at a given temperature and P(T,Ecoll) is the statistical weight.

In normal case Boltzmann statistic is used for the calculation of statistical weights. But Boltzmann statistic is valid when the temperature is high and the particles are distinguishable. At ultralow temperatures (T< 10K) we should use the appropriate quantum statistic (Fermi or Bose).

What kind of quantum statistic should be used in the collision of a

radical[spin = 1/2] + closed shell molecule (spin=0)

at ultralow temperatures?

What is the form of P(T,Ecoll) in this case?

End-of-question

Dear Péter Szabó , you have gotten a couple of important answers. My comments are as follows:

1.     What do you mean by reactive cross section ?

2.     What do you mean by ultralow temperature ?

You keep talking of particles that, from the context, seem to be classical systems with spin, one S=1/2 the other S=0. But a chemical reaction is better seen as a change of quantum state.

But then, there is need to define the spectrum of this systems as well as the composite.

I hope you understand that a ill-posed problem has no solution. Using the answers given to you, try to put the ideas of yours in shape.

The question implies quantum scattering. And take this as a possible answer.