Is negative activation energy possible?

Question

Is negative activation possible? And, in coupled reactions is there any difference?

Because I saw in the paper - Chemical Engineering Science 1996, 51 (11), 2995–2999 - the following conclusion:

Negative activation energies of reaction have been observed [...]. This unusual phenomenon can be predicted from classical theories, as a competition between two consequences of a temperature increase: greater thermal energies available to overcome the energy barrier represented by the intrinsic activation energy, and less adsorption to form reactive intermediates represented by the heat of absorption.

Does it simply mean that although the energy to overcome the activation energy barrier is there but the rate that products are formed is lower?

Answer 1

I have recently encountered this idea of "negative activation energy" as I was reading Chapter 11 of Elements of Physical Chemistry (5th Edition). On p. 252, the kinetics of the following reaction was studied:

$$\ce {2 NO (g) + O2 (g) -> 2 NO2 (g)}$$

A plausible and simple mechanism which was selected for this illustration, involved two steps: Firstly, the formation of the $\ce {N2O2}$ dimer from a collision of the two $\ce {NO}$ reactant molecules. It is then followed by collision of this dimer with an $\ce {O2}$ molecule. By applying the steady-state approximation (i.e. the rate of change of the dimer with respect to time approximately equals $\ce {0}$), the following overall rate constant is obtained:

$$k_\mathrm{r} = \frac{k_1 k_2}{k_1'}$$

$k_\mathrm{r}$ is the overall rate constant, $k_1$ is the forward rate constant for step $1$, $k_1'$ is the reverse rate constant for step $1$ and $k_2$ is the forward rate constant for step $2$.

When temperature increases, all of these individual rate constants for these elementary reactions increase. However, the extent of increase in their values differs between these individual rate constants. For this particular reaction, Atkins & Paula (2009) highlight that as temperature increases, the overall rate constant actually decreases. This is because the rate constant for the dissociation of the dimer increases much more as temperature increases, compared to the other rate constants in the numerator.

Mathematically, if we were to use the Arrhenius relationship for the rate constant for the overall reaction, we would say that the reaction has "negative activation energy" since that is the only way to have the rate constant decreasing in value as temperature increases (as $R, T > 0$), but it is pertinent to keep in mind that the rate of each elementary step does in fact increase as temperature increases. However, the extent to which they individually increase differs between each elementary step.

This is perhaps one way to see how a reaction can be said to have "negative activation energy" overall.

Reference

Atkins, P. W.; Paula, J. Elements of Physical Chemistry (5th ed.). W. H. Freeman & Company, 2009.

Answer 2

No, it's not possible to have a negative activation energy in a simple reaction such as an isomerisation because there is no possible way to draw to potential energy curves to give a negative activation energy. The best you can do is to be activationless, and is commonly observed in electron transfer reactions (see for example J. Am. Chem. Soc. 2002, 124 (19), 5518–5527).

Experimentally, however, negative activation energies can be observed when the reaction comprises several steps some of which are not identified at the time of doing the experiment. For example if each reaction follows an Arrhenius type rate constant, the first may be $k_a = A_a\exp(-E_a/RT)$ and the second $k_b=A_b\exp(-E_b/RT)$, but because there are, say, equilibria involved in the mechanism, what may be observed without realising it is that the rate is a ratio of the two individual rates:

$$\begin{align} k_\mathrm{obs} &\propto \frac{k_a}{k_b} \\ &\propto \frac{\exp(-E_a/RT)}{\exp(-E_b/RT)} \\ &= \exp\left(\frac{-(E_a-E_b)}{RT}\right) \\ &= \exp\left(-\frac{E_\mathrm{obs}}{RT}\right) \end{align}$$

so it is possible for $E_\mathrm{obs} = E_a - E_b$ to be positive or negative.