I wonder about the implications of substances having an energy distribution around a mean, as opposed to them all having the same energy level, on the spontaneity of reactions.

2 somewhat separate questions:

  • Can a reaction with a high activation energy, but which in the end has a negative Gibbs free energy change at certain temperature, occur spontaneously even though at this temperature the energy level of the reactants is lower than the activation energy? If I understand correctly, the answer would be yes because the energy of the reactants is just an average of a broader distribution, and so some small fraction of the reactants would attain the $E_a$ even if on average $E_a > E_{\text{reactants}}$?

  • Wouldn't it also be possible that some non-spontaneous reactions (with a positive change in Gibbs free energy) occur spontaneously? I mean we know that if we invest $\Delta G$ energy, we will force the non-spontaneous reaction to occur. But what if some small fraction of the reactants get to the energy level they would have had if $\Delta G$ energy was invested in them (without it being actually invested). Wouldn't that mean that for that fraction of the reactants, in the high end of the energy distribution around some average, the reaction would occur (while not occurring for the rest and vast majority of the reactants who fail by themselves to attain the same energy level they would have had if ΔG energy were invested in them)? Is there a flaw in this understanding?

I guess the question here is whether the idea that $\Delta G > 0$ reactions are non-spontaneous is a statement of probability that says that such reactions are just very unlikely or is it an hard and fast rule like the conservation of energy?


Yes; you are on the right path in your second paragraph, but distinguish between activation energy and free energy as they are not connected. I'm sorry but I think that your third paragraph is not all correct as it stands. If the free energy change is positive the reaction is not spontaneous by definition. Perhaps you mean, if the overall enthalpy change is positive can the free energy now be negative if the overall entropy change is also positive? The answer is yes.
The thermodynamic quantities only give us information about initial and final states at equilibrium. The activation energy is obtained from kinetic, that is rate constant measurements usually vs, temperature and not from thermodynamic quantities.
In the vast majority of reactions the activation energy is greater than thermal energy (approx kBT, where kB is Boltzmann's constant). These reactions are generally slow because, by the decreasing exponential nature of the Boltzmann distribution with energy, not many reactants have enough energy to react in any given time period, say 1 sec. (This is lucky otherwise we would all have reacted eons ago!) In some special types of unimolecular reactions, e.g. electron transfer, bond dissociation & isomerization, with zero or very small activation energy then clearly these will be very fast possibly with rate constants $ >10^{12}\ /sec $.
In the case where the products are of higher energy than reactants, reaction still occurs but the equilibrium lies on the side of the reactants. Again by the Boltzmann distribution argument products react faster to reactants than the other way round.

  • $\begingroup$ Thank you for your answer. I wonder, though, if, for example, a reaction needs 50kJ/mol of work done on it to occur (ΔG = +50), then wouldn't a tiny fraction of the reactants achieve that level of energy even while the average reactant would have 50kJ/mol <i>less</i> energy than what's needed for the reaction? And if so, it seems that that a tiny fraction would be able to react even though the 50kJ of energy weren't invested in it. If this scenario is possible in principle, isn't that a case of a non-spontaneous reaction occurring spontaneously? $\endgroup$ – Daniel Jun 17 '16 at 21:56
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    $\begingroup$ You ask what happens when products are at a higher energy than reactants. In this case some reactants do still go to products but by the Boltzmann distribution it is more likely that products return to reactants so the equilibrium is on the side of the reactants. This is because in this case the activation barrier is smaller for products. At equilibrium, products and reactants are always interconverting. If you completely removed product the reaction would proceed until all reactant was used up. Thermodynamics tells us nothing about how long this might take. $\endgroup$ – porphyrin Jun 18 '16 at 6:36

I think you are confusing $\Delta G$ and $\Delta G^\circ$, where $\Delta G = \Delta G^\circ + RT \ln (Q)$. At $\Delta G$ = 0, $Q$ is equal to the equilibrium constant $K$, and $\Delta G^\circ = -RT \ln (K) $.

But $\Delta G$ and spontaneity and all that is about being away from equilibrium and getting back to equilibrium: $\Delta G = RT \ln (Q/K)$. So if you're not at equilibrium $Q$ < $K$ then $\Delta G $ < 0 and more product will be made spontaneously to get back to equilibrium whether the sign of $ \Delta G^\circ$ is positive or negative.


Thanks for your answers.

I wanted to write a comment, but it was kind of long, so I put it here:

My question had two parts:

1) Can a reactant, whose reaction is spontaneous, but whose Ea is higher than its current energy level, proceed with the reaction without an energy input?

Porphyrin'a answer fully answered the question in the affirmative. To paraphrase his\her answer: the reaction is going to occur cause it's spontaneous until equilibrium is reached, but it may be slow as only a small fraction of the reactants are going to reach the activation energy level at this temperature per unit time.

Porphyrin's answer had one more good insight: if the activation energy is high, and the kinetic rate constant is in turn low, equilibrium will tend to be on the side of the reactants (assuming the reverse reaction is faster).

2) I had second question, which in truth should have been a separate question as putting them together was probably confusing.

Here's what I meant for the example of water: let's say that water liquid and it's vapor are in equilibrium. As Jan Jensen pointed out, when at equilibrium ΔG = 0 (and since standard conditions of 1atm\25C are probably not equilibrium conditions, ΔGzero could be positive or negative - in the example of water, ΔGzero is slightly positive for evaporation under standard conditions).

Ok, so let's say we're at equilibrium and any additional condensation\evaporation is ΔG > 0.

As water is already at equilibrium, to evaporate a bit more you need to invest some positive energy (let's say 10kJ/mol). The idea of my second question was whether some small fraction of water molecules could possibly by themselves have kinetic energy that's higher by 10kJ/mol than the average for the rest of the water, and thus evaporate even though I didn't supply that energy from an external source? After all, the the water molecules have a distribution of kinetic energy around a mean...

It's hard for me to see why a bit of water could not in principle attain that energy level, but the answers above made me see one thing - even if a bit of water were to attain enough energy and evaporate against a positive ΔG, the vapor pressure would then be briefly higher than equilibrium, and soon thereafter some vapor will condense and equilibrium will be restored. (For a brief interlude condensation would have a ΔG < 0 until equilibrium is restored where ΔG = 0 again).

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    $\begingroup$ At equilibrium you have a distribution of KE (en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution) so fast molecules in the liquid will evaporate and slow molecules in the vapour will condense. If the rate of condensation = rate of evaporation then you have (a dynamic) equilibrium. $\endgroup$ – Jan Jensen Jun 23 '16 at 7:46
  • $\begingroup$ I agree with this last comment by Jan Jensen. To expand and clarify a little, any molecule could evaporate or condense but given the Boltzmann distribution the chance of more energetic molecules evaporating is greater that for slower ones, and vice-versa for condensation. $\endgroup$ – porphyrin Jul 1 '16 at 18:19

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