Theoretically, Ea  cannot be negative. so, please go through your calculation again. I think there may be a mistake.

I must disagree!

The negative activation energy can be the consequence of the potential energy surface. In such cases the reaction is barrierless or we can find a submerged potential barrier with a van der Waals complex under the energy level of the reactants.

There are many example:

HBr+CH3

HI+CH3

NO+OCl

HO2+HO2

SO+OH

There are other interesting pecularities of these reactions: in some case the sign of the activation energy can be changed, from negative to positve.

Gourisankar Roymahapatra, I must disagree again!

The activation energy is not equal with the height of the barrier.

The barrier is related to the topology of the potential energy surface,

but the activation energy is a thermodynamical quantity (in a canonical ensemble). Ea includes all dynamical effects, like angular momentum (a centrifugal barrier).   According to the definition:

Ea= -∂ln(k)/ ∂(1/RT)

The Arrhenius plot is not always a straight line.

The transition state is not located always at the saddle point. The TS also exists in these cases, when there are no barrier.

Theoretically, Ea cannot be negative. so, please go through your calculation again.

hello guys,

maybe Ea is actually very close to 0 and the negative value is the consequence of some side reactions almost negligible

for me, the answer would be: if Ea is very close to 0, it correspond to an ultra fast process (e.g. collision beween two radical molecules), on which catalysor is expected to have only a minor influence.

the contrary case would be: a ctalyst influence a reaction whcih is naturally slow with a high activation energy (e.g. the decompositon of an hydroperoxyde) by lowering its activation energy and thus increasing the rate constat value at a given temperature.

regards

Before to discuss something we must clearly define all terms. From this point of view the question is not well formulated. If we mean the apparent activation energy (measured), it could be negative for a complex reaction mechanism. A catalyst could make it more or less negative depending on the range of temperatures, at which E(app)  was determined.

Dear Peter Szabo. You gave examples of elementary bimolecular reactions between two radicals. Do you have any example of catalysis of such reactions? I absolutely disagree with your statement: "the activation energy is a thermodynamical quantity (in a canonical ensemble)." For only this reason I would definitely down-vote your answer , but I do not thanks to comments of Gert Van der Zwan (I up-voted it). The term activation energy belongs to chemical kinetics. Almost all things around us including ourselves are themodynamically unstable. They exists because their decomposition processes have a very high activation energy.  There are two terms in chemistry, thermodynamical and kinetical stabilities.

As a summary, before to discuss something you must clearly define what you are talking about.

Péter Szabó wanted to point out that Ea is not merely the height of a potential energy barrier [a “quantum chemical quantity”] [ or “zero-point energy corrected barrier height” also considering the masses], but it can also have temperature dependence. Furthermore, within the framework of transition-state theory the activation energy has a precise statistical thermodynamic definition (e.g. Tolman’s theorem). Based on this, Ea is a thermodynamic quantity, despite it does not characterize thermodynamic equilibrium. We all agree its absolute value is related to the rate coefficient of a process and thus it is a kinetic parameter ! [In an equilibrium constant the differences of activation energies can be identified.] I recommend using "thermodynamic parameter" and "kinetic parameter" terms for the distinction between the two classes of parameters as the term "thermodynamic quantity" has a very general meaning ! 

Answer to the original question: by definition a catalyst speeds up the reaction, therefore the rate will increase. However activation energy is just one parameter defining the rate (other: entropic effects can also be important), thus I think it can change either ways. 

Very helpful Gert Van der Zwan. I have never considered a negative activation energy. I am a kineticist and tend to break down reactions to their elementary steps. In that light, I always think of activation energies as positive. A negative energy there would result in a transition state that is more stable than the reactants (which is completely unreasonable). I myself have studied several complex reaction systems, and have never seen a negative activation energy. This was a thought provoking post to read first thing in the morning and I look forward to reading through the paper that you attached. Thank you all for your thoughts.

I have given this some more thought. If you have an overall negative activation energy (e.g. it gets slower with increasing temperature) the effect that a catalyst will have on the reaction will depend upon which elementary reaction/reactions it has an effect.

For instance: If you have a system in which your reagent has two forms A and B with equilibrium constant K. Say that B is the more reactive isomer (rate constant kB = 100 x kA) and the equilibrium is shifted to A as temperature increases. In such a system your overall rate would have the form Rate = (kA + kB*K)/(K+1). As you increase temperature your A-B equilibrium will shift to A and your rate will become slower because the is a 2 order of magnitude difference in the reactivity of form A relative to form B. You will then have an apparent negative activation energy from plots relating rate and temperature.

So the question about what the catalyst will do depends upon which reaction it catalyzes. You need to know your system intimately to get to the bottom of this.

Jack,

according to IUPAC definition the Ea is an experimental parameter obtained from Ln(rate) versus 1/RT plot. For the reaction with a complex mechanism this plot is NOT linear and depends on concentrations. That's what you demonstrate in your example. Ea could be negative. For instance, the solubility of dioxygen in water drops with an increase of temperature.If the reaction has the first order with respect to [O2] but is carried out under 1 atm of oxygen, then the overall rate could be lower at higher T.

This is a good question in relation to an often encountered class of reaction in gas-phase or interfacial chemical kinetics: many chemical activation reactions going through a stable adduct and subsequently over a high reaction barrier towards stable final products belong to this class and may be schematically represented by A + B C ==> P (see simplified hand-written potential energy scheme). From the foregoing we all know that negative Ea cannot occur with elementary reactions, but are always a sign of the exisence of a complex reaction mechanism, that is comprising more than a single elementary reaction. The negative T-dependence of the rate CONSTANT or coefficient (NOT the rate!) may be explained by the increase of the rate constant for the back reaction (k-1) as a function of internal (= thermal) energy when the forward reaction is practically T-Independent (A+B bimolecular reaction, small entrance barrier). Now the addition of the catalyst enters into consideration (two cases, "a" and" b" in scheme). For situation "a" k2 becomes competitive with k-1 at a lower energy which makes the reaction LESS negatively T-dependent, for situation "b" the rate constant becomes basically equal to k1, that is T-Independent. Admitted, this is an oversimplification, but the essential elements are there (see also the sparse density of states for the exit channel , the continuum of states for the intermediate C and the density of states intermediate between the entrance and the exit channel). It is impossible to operate exclusively on an enthalpic level, entropies (ratios of densities of states) are an important part of the kinetics!

Result: A lesser negative T-dependence of the rate constant is expected at a concomitantly highly accelerated rate constant upon addition of the catalyst, assumed to only act on the high exit barrier. This is what the catalyst is designed for, namely increase of the reaction rate through action on the rate constant! In the gas phase the total pressure affects the overall kinetics depending on the rate coefficients compared to the collision frequency.

Somebody down-voted my answer. I'm just wondering what is wrong in my answer.

Hi Yurii,

I did not down-vote your answer (take my word for it!), but I know where you went wrong with your answer: In an Arrhenius plot you take the ln or log of the rate CONSTANT, and NOT the RATE vs. 1/T in order to obtain the "apparent" or measured activation energy. (Correspondingly you take the ln or log of the equilibrium CONSTANT in a van't Hoff plot!) This implies that you already know the rate law of your kinetic process, either (pseudo-) first order, second order or complex. It therefore does not make sense to plot the ln of the RATE of a completely unknown process because you won't be able to interprete or evaluate the Ea or A in a reasonable way. In addition, the resulting parameters are not concentration or density invariant if you use rates! For a complex reaction mechanism you will always obtain a result from an Arrhenius plot, either positive or negative values of Ea that correspond to sums and/or differences of Ea of the involved elementary processes. However, you must know the rate law in order to evaluate what the complicated sum/difference means in terms of elementary reactions.

Michel, I absolutely agree with you that for the complex reaction mechanism the Ea(app) measured from Ln(rate) versus 1/RT (or 1/T) does not make any sense! That's my point! I attempted to explain that Ea(app) measured from Ln(rate) versus 1/RT (or 1/T) is NOT interpretable! However, in numerous publications Ea(app) is measured from Ln(rate) versus 1/RT (or 1/T) plot without worrying about the reaction  rate law, which might be complex. Thus, we are in full agreement. If you allow, I would consider your post as in depth explanation of what I attempted to claim. Thanks for your post

Dear Dr Zwan, I did not downvote ur post and I am sorry for being your answer downvoted in my question. I welcome each answer from all the people and value them equally. So, I request everyone not to downvote anyone's answer. Thanks.

Clarification to my example. Surely, the oxygen dissolving is likely to to proceed with an activation energy close to zero. My example was about wrong experimental measurements of activation energy. I saw in numerous publications that the rate of oxidation by dioxygen in solutions were measured under constant PRESSURE of O2. While the solubility of gases decreases with T. Therefore, in this case the reaction rate is measured at different concentrations of O2 in solution. This could result in the negative value of the slope "Ln(Rate) versus 1/RT". This is in full agreement with Gert's interpretation of my example;" if you model it as an activated process, the parameter you measure could well be negative."

Bold face typing. I never do it but it happened for unknown reason and I did not pay an attention.

The first conclusion that can be drawn from a negative activation energy (if the numbers are correct, no uncertainties are given) it is that the reaction mechanism for this process is complex and several elementary reactions are involved in the mechanism. For a mechanism with several steps the observed activation energy is the combination of the activation energies of the different steps and may be a negative value. 

A more negative value obtained in the presence of a catalyst indicates that it opens a new reaction path with a step that is even lower in energy.