9
$\begingroup$

Are there any reactions with no activation energies? Our professor just told us there are no chemical cliffs. Is this true?

I read something about nuclear decay as being a reaction with no activation energies. Why is this so, and how does this work?

| improve this question | |
$\endgroup$
  • 5
    $\begingroup$ Bond formation between atoms has no activation energy, at least for simple cases. Consider the reaction between two hydrogen atoms to form hydrogen gas. The potential energy curve only slopes downwards from infinity to some critical value, approximately equal to the bond length. Electron affinities (when exothermic) also likely have no activation energy. $\endgroup$ – Nicolau Saker Neto Oct 28 '14 at 20:34
  • $\begingroup$ @NicolauSakerNeto Very good points. If you find the time it would make a good answer. $\endgroup$ – Philipp Oct 28 '14 at 21:32
  • 1
    $\begingroup$ @Philipp I'm rather busy lately, so I should probably avoid getting into more stuff. If anyone wants to build on my comment, feel free to do so! $\endgroup$ – Nicolau Saker Neto Oct 29 '14 at 0:33
  • 1
    $\begingroup$ As an aside, if you had a "cliff" could you even purify the high energy compound? Seems like it would just "roll off" and spontaneously react. Having activation energies creates a valley that traps molecules in a stable state. $\endgroup$ – user137 Oct 29 '14 at 14:29
  • $\begingroup$ @NicolauSakerNeto I understand that this is indeed case. But wouldn't there be some repulsion between the hydrogen atoms when they approach each other? $\endgroup$ – Tan Yong Boon Nov 2 '17 at 22:48
10
$\begingroup$

Marcus-Hush theory describes electron/charge transfer rates using conventional transition state theory, e.g.:

$\ce{A+ + B -> A + B+}$

Now one might naively assume that there's no activation barrier in a simple electron transfer, but that's not true. Since the geometry of $\ce{A+}$ and $\ce{A}$ as well as $\ce{B}$ and $\ce{B+}$ are not identical, there is a reorganization energy of both the molecular species and the solvent or environment. This reorganization energy ($\lambda$) serves as an activation barrier for the charge transfer.

Now the convenient thing about charge transfer reactions is that you can tune the $\Delta G$ of the reactants.

Marcus predicted that the maximum rate would occur when the $\Delta G^0 = -\lambda$ and thus the $\Delta G^\ddagger = 0$. If the reaction becomes more thermodynamically favorable, you enter the inverted region and the activation barrier goes up and the reaction rate goes down.

This is a long-winded answer to say there are some chemical reactions with no activation barrier, in addition to the other cases here. Usually the key phrase describing these in papers is "barrier-less reactions"

| improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ I'd give a +10 if possible for bringing up electron transfer processes! Btw, Rudolpg Marcus's Nobel lecture can be found here. $\endgroup$ – Klaus-Dieter Warzecha Oct 29 '14 at 9:19
  • 1
    $\begingroup$ have a look at the data in this reply to another question on electron transfer where you can see where the reaction has no barrier. chemistry.stackexchange.com/questions/63545/… $\endgroup$ – porphyrin Dec 10 '16 at 21:35
3
$\begingroup$

So of course it is hard to tell the difference between the case where there is no activation energy and the case where there is a very small activation energy. Given that...

Many reactions, such as simple acid-base reactions are diffusion controlled. This means that reaction occurs at every encounter, and that this is not an activated process. Another example involves benzene. Say we generated the molecule 1,3,5-cyclohexatriene, it would immediately isomerize to benzene (unequal bond lengths, localized pi electrons -> equal bond lengths, delocalized pi electrons). If we put this on a reaction coordinate it might look something like this,

enter image description here

I've always thought of this as a process with a negative activation energy, but maybe zero would be a more accurate description (can you have a negative barrier?). I suspect it applies to many reactions where there is a large thermodynamic difference in stability between product and reactant favoring product and where there is minimal structural reorganization.

| improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ There isn't such a thing as a truly negative barrier, since if the path looked something like your diagram above, it would classify as either a product or a stable intermediate. That said, this diagram gives an example of a barrier-less reaction. $\endgroup$ – Geoff Hutchison Oct 29 '14 at 3:41
  • $\begingroup$ ron, could you please edit your answer to account for the concerns Geoff raised? Thank you! $\endgroup$ – Gaurang Tandon Apr 21 '18 at 5:45
1
$\begingroup$

The issue with some of the proposed reactions of the above posts is activation energy, if its measured in Gibbs free energy, should also subject to entropy, since ΔG=ΔH-TΔS. However the wikipedia article on activation energy (as of this writing) suggests when reaction rates decrease with activation energy, because the molecules were prepared to react at room temperature. The only way for the activation energy equation to work would be through would be for negative activation energy. Reactions like this are simple: i.e. two things smash together and will combine if the potential energy is larger than their momentum. But say for instance you heat this reaction up. You've accelerated these particles (that is, heated them, since heat can be considered kinetic energy) the reactants are so fast and have so much momentum they simply bounce off each other, and so the becomes meaningless. As an aside, you can have 'cheating' occur in normal reactions with hydrogen which exhibits quantum tunneling (because its so tiny). So you can have product form from a reaction while the reactants are way below the activation energy expected. So a smartass rejoinder to your professor's comment would be "sure, cliffs may not exist, but slippery slopes sure do!"

| improve this answer | |
$\endgroup$
  • $\begingroup$ Welcome to Chemistry.SE! As an activation energy is a barrier to reaction progress (e.g., ron's answer) you cannot define a negative activation energy. That's a meaningless statement. If there's no barrier (i.e., it's completely downhill from reactants to products) it's just barrier-less. See my comments about Marcus theory for an example of decreasing reaction rates - with no need for negative activation barriers. $\endgroup$ – Geoff Hutchison Dec 12 '14 at 15:00
1
$\begingroup$

The reaction between two ions to form a molecule has no activation energy as they are attracted to each other and therefore no force, and no energy is required to bring them together. e.g $\ce{H+ + H- -> H2}$

| improve this answer | |
$\endgroup$
0
$\begingroup$

I seem to remember (haven't got the reference on hand) that singlet methylene (|CH2) inserts into the C-H bond with zero activation enthalpy. It's certainly an enormously reactive and high-energy species.

| improve this answer | |
$\endgroup$
0
$\begingroup$

I read something about nuclear decay as being a reaction with no activation energies. Why is this so, and how does this work?

You're asking too much. I can answer you in short:

Nuclear Decay Reactions occur when the nuclear force$^1$ between nucleons is not able to hold the nucleons together.

Now as it can't hold large nucleons, the nucleus starts emitting particles or following special processes to reduce mass or nucleons and reach a stable zone, the choice of process depends on where the nucleon is as per a graph $^3$.

$^1$.one of the four fundamental forces of nature. If there was no nuclear force, everything would repel apart. The reason being, if there was only electrostatic force between protons, the whole nucleus$^{1a}$ would repel away. So another "nuclear force", acts into play, which is blind to the charges, and acts according to another intrinsic property$^{1b}$ called color$^{1c}$ .
So now when nucleons are at the order of 1 fermi$^{1d}$ , the force is attractive and stops acting very (very) quickly (dying).
$^{1a}$ actually, only protons, to be specific
$^{1b}$ like mass, charge, etc.
$^{1c}$ and involves variety of particles called gluons, mesons, hyper-mesons, quarks, etc.
$^{1d}$ named after great scientist, $1$ fm$=10^{-15}$ m
 $^{2}$$\beta^-,\beta^+,\ce{_2^4He}(\alpha)$, K-capture etc.
$^3$ which predicts that is no. of protons = no. of neutrons, then nucleus is stable ,generally, only for lighter nucleus(es) and for heavier ones it bends a little upward like a parabola ($y=x^2$)

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.