Dear Zakir,

An elementary reaction can not have a negative activation energy: it must be zero or positive. However, a reaction mechanism that is composed of several steps may have a negative activation energy. Here is a link that may be useful.

http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Activation_energy.html

With my best regards

Prof. Bachir ACHOUR

 Dear Bachir Achour Sir,

Thank you for your quick response. Sir, when i was carrying out the kinetic studies of esterification between non-edible oil and alcohol the activation energy backward reaction is found to be negative. Can you please tell me whether it is possible or not

Dear Bachir Achour,

I must disagree! This topic is my favourite research area.  Negative activation energy is possible even for elementary reactions. In these cases there are attractive forces between reactants, there are no barrier, or the barrier is submerged. This means you must find complexes between the reactants and submerged barrier.

It is common demagogic misconceptions that can not have negative activation energy of elementary reactions. In most of the cases these reactions are barrierless or there are complexes on the potential energy surfaces that complicates the description of the kinetic characterization of the elementary step. Generally we use transition state theory to calculate the rate coefficients  of elementary processes. In such cases it is not  unequivocal how to use TST, because there are no equilibrium (the assumptions of the TST do not fulfil)

See the following reactions:

OH+HI

OH+HBr

O+H2

OH+O

alkyl radicals+ HX(X=Cl,Br,I)

alkyl radical + OH

alkyl radical + HO2

etc....

Dear Peter, what about

"an absolute rate constant of the OH + C2H6 reaction using k(ethane) = 1.0 × 10-11 exp(−1094/T) cm3/molecule s"

DOI: 10.1021/jp983273d

Dear Yuri V Geletii,

OH + radical reactions were mentioned but there are cases when OH + alkane/alkene reactions also show negative activation energy. Moreover discrepancies can be found between different experiments. One of the measurements provides positive others provide negative activation energies for the same reactions (one of the most prominent example is the HX(X=Br,I) + CH3 reaction). It is very hard to ensure the proper thermal conditions in experiments: in most of the cases the radicals are prepared with flash-photolysis which can results vibrationally and rotationally hot products. The reactivity of these non-thermalized molecules can be different, and it can modify the obtained experimental result. Furthermore it is a very important which temperature range is considered. There are a plenty of elementary reaction where the sign of the activation energy is changing with the temperature: at high temperatures Ea is positive but at low temperatures it is negative (eg. the HI/Br+alkyl radical reactions ). According to the latest (in the last 30 years) experimental and theoretical studies  negative activation energy can be found for many radical+molecule and radical+radical reactions. If you want I can attach a bunch of references about this topic.

Dear Peter, 

I'm not an expert in this area. As far as I understand the negative activation energy is explained by formation of a pre-reactive complex or by two transition states. In this case the overall reaction can be written as two separate reactions, each of them can be called an elementary reaction. Therefore, is the overall reaction HO. + RH elementary, or includes two elementary reactions? 

The IUPAC definition of an elementary reaction:

A reaction for which no reaction intermediates have been detected or need to be postulated in order to describe the chemical reaction on a molecular scale. An elementary reaction is assumed to occur in a single step and to pass through a single transition state.

Dear Yurii V Geletii,

In some case (eg. the reaction of ionic species) there are no barrier what results a negative activation energy: there are purely attractive forces that guide the reactants to each other. If you give more and more translational energy (or higher temperature) they can not feel each other. This dynamical picture can be refined if we consider the rising centrifugal barrier as the energy is growing (what is repulsive):

                                                                  Vcf = Ecoll×b2/r2

where b is the impact parameter, r is the distance between the centre of masses, Ecoll is the relative translational energy of reactants. As the centrifugal barrier rises, the effective potential energy surface (V(r)+Vcf(r)) will be less attractive. The reactive channel is open if

                                                                    Ecoll  > V(r) + Vcf(r)

In this case there are no potential-barrier, the transition state (the dividing surface which separates the reactants and products) can be put on the centrifugal barrier, that's position is moving with Ecoll (or temperature). You need a variational TST if you want assume this dynamical process with statistical rate theory

In the neural (radical+molecule or radical+radical) reactions you can find pre-reactive complex (sometimes more than one) with a submerged barrier (barrier with "negative" energy), as you wrote. The description of such kind of system is more tricky. You need more assumptions,  if you want to use TST. In this case you must assume a thermally stable complex. It means that the energy flows fast between the modes of the complex: we assume effective energy redistribution in complex. A long lifetime is need for this effective energy redistribution. So, the first main assumption of TST is a complex with a long lifetime with statistical nature. In this case the process can be described with the RRKM theory (the unimolecular version of TST). If we use two TST (one before the complex, another one after the complex put on the vicinity of the submerged barrier), it will be almost equivalent with the RRKM theory or with the third theory (SACM, statistical adiabatic channel model). The two-TST and SACM model implicitly assumes the existence of the stable-complex with statistical nature. The second main assumption is the "point of no return" feature of transition state. If the trajectories crossing the TS, then there will be products from reactants. TST assumes that there are no recrossing.

In most of the cases you can not find good transition state, where the two mentioned conditions are fulfilled. TST is only an assumptthon. The IUPAC definition is not good, because it uses the TST.

I perform dynamical calculations for such kind of systems on global ab initio potential energy surfaces. Dynamical results can give definite answer for this question not the statistical rate theories.  As I see, there are no stable complex during the collisions. The dynamical picture is similar to the barrierless reactions, where there are no complex.  I think these reactions with pre-reactive complexes are real elementary steps, but their dynamical picture is pretty complex.

Peter, 

thanks for clarification. I've found numerous wrong IUPAC definitions

Dear All,

Thank you for your valuable inputs and guidance. I found the fault in my calculations and now I am getting the positive value of activation energy for the esterification case. 

Thank you all once again.

Dear Zakir Hussain 

Activation energy of the reaction between iodide atoms in toluene is about -14 kJmol-1. Because the following mechanism has proposed for this reaction:

   I + M = I….M                                   (1)

   I + I…M =  I2 + M                            (2)

The, the rate equation is as follows: 

  Rate = k2[I…M][I]                             (3)

 By substituting for equilibrium constant of the reaction (1) in equation (3) we will have:

 Rate = k2K1[I]2[M]=kobs[I]2[M]          (4)

 Then, we have:

 kobs= k2K                                             (5)

by using Arrhenius equation, we finally have:

Ea(obs) = Ea2 + ΔH1

It is clear that the equation (1) is exothermic. If |ΔH1| >Ea2, then Ea(obs) will be negative. As can be seen, activation energy can be negative for a reaction with at least two steps with an equilibrium step before the rate determining step.

You are studying the esterification of an non-edible oil with some alcohol, and do you really study the process of simultaneous esterification of several higher aliphatic acids with some alcohol (probably with methanol or ethanol). You must therefore realize that as a result of the mathematical description of kinetic data you will receive no activation energy value but the value of some lumped parameter which may only be treated as the average activation energy. Activation energy refers to a single chemical reaction, and in the case of a chemical reaction system, to each separately - which is sometime forgotten.

That's to start with.

You received a negative activation energy value and you are wondering why. There are several possible reasons for both the way the experiment is conducted and the mathematical description of its outcome. One can mention here: not achieving equilibrium, too little mixing intensity, change in reaction mixture volume, errors in chemical analysis, computational errors - not including the fact that exponential factor and activation energy in Arrhenius equation are correlated, etc, etc. ……. Although in your case the explanation may be different. In my opinion you should look for it in the sentence:

“Assumption: B is in large excess so we assume forward reaction is first order”

You can obviously conduct a chemical reaction with a large excess of one substrate (of course, you have to be able to cheaply separate this substrate from the reaction mixture after the process). The other way, however, is to obtain the product as a result of a given chemical reaction, and another to examine the kinetics of this reaction.

It is true that, if the A + B < -- > C + D reaction is carried out with a large excess of B, say 20:1 or more, then its rate under these conditions does not depend on concentration of B. It may, but it does not have to be true that the rate of above reaction depends linearly on the concentration of substrate A. Whether it is a first order reaction to A or a different order, it must be experimentally verified. The reaction order is influenced not only by the type of reaction but also by the catalyst used.

Activation energy is not determined on the basis of unverified assumptions, but on the basis of a properly determined kinetic equation. However, in order to obtain an equation with the rank of kinetic equation, in addition to determining the effect of temperature on the reaction rate (for one molar ratio of substrates), the experiments should also be performed for several molar substrate ratios, say 3:1, 4:1, 5:1 etc. Only on the basis of such equation it is possible to determine the correct values of kinetic parameters. If you did not perform the experiment in such way, the values of parameters you obtained have no physical meaning.

Negative activation energy for a given chemical reaction means that the rate of this reaction decreases with increasing temperature. Is that possible in case of hydrolysis? - I doubt it.

Regards

Generally.

There are known processes whose overall rate may decrease with increasing temperature - however, this should not be equated with negative activation energy.

During the freezing of biological material, within a certain temperature range below 273 K, the rate of enzymatic reactions in cells increases as the temperature decreases. This is due to an increase in the concentration of substrates limiting the rate of these reactions along with the phase transition of water to ice.

The rate of individual polymerization stages: initiation, propagation and termination increases with temperature. Chemical reactions may, however, be accompanied by a sharp increase in the apparent viscosity of the reaction mixture limiting the mobility of the molecules.  etc.

When we take into account DFT calculations, initial and final states should be checked and if possible, they or your consideration about your problem should be updated in accordance with the previous results. Today, I had the same problem.

Zakir Hussain

No, it makes no physical sense to have negative activation energy. Rather in most cases, such a situation can be taken to be zero activation energy.

What is the role of charge carrier type ? in the sign of activation energy .

As it is known n=np+ne so the number of charge carriers is the sum of both hole and electrons. With temperature may occur change in the type of charge carriers from hole into electrons and vice versa. The question is this change of type of charge carrier has any contribution in the sign of activation or not?

If for the forward reaction the activation energy is positive, for the backward reaction it can be positive or zero. It is determined by the potential energy along the reaction path.

When talking about the activation energies in rate coefficients, first of all, it should be mentioned that those coefficients are calculated by averaging corresponding cross-sections with the equilibrium distribution function. Reaction cross-sections are zero below some threshold energy E*. Averaging means integration over energy E from 0 to infinity. Since the cross-section is zero for E < E*, we can integrate from E* to infinity. By changing the variable E’= E-E* the integration is performed over E’ from 0 to infinity. The equilibrium distribution f=Q^{-1}\exp(-E/kT) will be transformed to Q^{-1}\exp(-E’/kT) \exp(-E*/kT). Thus, the expression for rate coefficient contains the factor \exp(-E*/kT), where E* is the reaction threshold and, therefore, is definitely positive. From this expression it is seen that E* can be interpreted as an activation energy. The same conclusion can be reached when analyzing the potential energy along the reaction path. The activation energy is connected with the height of the potential barrier, which can be positive or equal to zero. In some cases (usually for condensation-like reactions), rate coefficients demonstrate decrease when temperature increases. This is convenient to approximate by the factor \exp(-E**/kT) with negative E**, but here E** is just a parameter for fitting the experimental data. Such representation can be used in the narrow temperature range, since at low temperatures it gives infinitively large values for the rate coefficients.

As mentioned by Péter Szabó, negative activation energy is connected with transition complex (transition state) formation. At the same time, Peter mentioned that TST cannot be applied, because of non-equilibrium. Existence of the transition complex needs more detailed consideration. Non-equilibrium contradicts the definition of the rate coefficients, since under non-equilibrium no rate coefficients can be introduced. Moreover, when we are talking about the rate coefficient, we should consider not an elementary act of the reaction (which is characterized by the cross-section), but the kinetics of the process. If a lifetime of the TS is shorter than a mean time between collisions, the distribution function can not be introduced for its characterization (and, therefore, its behavior cannot be characterized by the rate coefficient). This is because the kinetic equation (equation for the one-particle distribution function) can be derived from the hierarchy of the N-particle distribution functions by the Bogoliubov’s method (by averaging over the fast processes) only for the metastable bounded complexes whose lifetime is not shorter than the time between collisions.

Concerning non-equilibrium impact on the reaction rates please see [Yu.Gorbachev, O.Kunova, G.Shoev, A non-equilibrium dissociation and vibrational relaxation model for computational fluid dynamics simulations of flows with shock waves. Phys. Fluids 33, 126105 (2021); https://doi.org/10.1063/5.0062628], references therein and my answer to the question https://www.researchgate.net/post/Is-the-reaction-rate-constant-kT-of-a-chemical-reaction-affected-by-the-kind-of-a-third-inert-species-under-the-constant-temperature-condition#view=61b70993ca6cfc191962069b).

Therefore, we meet an inconsistency while considering the negative activation energy.

Yes, if you approximate the rate constant of a chemical reaction with Arrhenius equation. No, if you speak about Theory.