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In the case of activation energy diagrams in terms of potential energy, I understand that when we're talking about two individual molecules, the potential energy is highest in the transition state since it is the most unstable, and then falls.

When this activation energy diagram is expressed in terms of gibbs free energy:

enter image description here

As is pointed out by this answer,here, they indicate that for the individual molecule interpretation, macro properties such as quantities such as temperature and pressure are ill-defined.

However, if we treat this as representing the reaction macroscopically, that can't be the case also since that would violate the second law (gibbs free energy rising). You would instead expect something like this diagram:

enter image description here

xasthor
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  • In classical thermodynamics we always consider an ensemble, typically a mole of reactants. The activation energy 'free energy' is used as a way to include entropy changes for an ensemble of molecules when using transition state theory from a thermodynamic viewpoint. Even the activation energy is an ensemble measurement. – porphyrin Dec 16 '22 at 08:57
  • The x-axes in these plots is different. The overall $\Delta G$ plot uses the change in concentration. The plot with the activation energy uses a more abstract "reaction coordinate" which could be a spatial coordinate of the two interacting molecules, but also could be simply the time progression of the reaction of an individual molecule. So short answer - no it is not implying that the gibbs energy of the entire ensemble rises and falls in that manner – Andrew Dec 16 '22 at 14:07
  • in my previous comment by 'ensemble' I mean that the average value taken over numerous molecules is used. – porphyrin Dec 16 '22 at 15:36
  • @Andrew have a look at the thread I linked. The accepted answer claims it's for the ensemble which is part of what's confusing me – xasthor Dec 16 '22 at 16:54
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    It should be obvious from the time evolution of any reaction that the entire ensemble does not cross the transition state at the same time, not to mention that a macroscopic increase in G is a violation of the 2nd Law. It is true, however, that the value of $\Delta G^\ddagger$ is derived by treating the transition state as if there were a macroscopic population of that species. – Andrew Dec 16 '22 at 20:00
  • @Andrew That is what I thought as well. That the Gibbs free energy cannot rise, and I have also only ever seen Gibbs free energy used in a macroscopic context, which is why I was confused and asked this question – xasthor Dec 17 '22 at 04:38
  • Related: How exactly is activation energy defined? – Loong Dec 17 '22 at 10:02
  • I assume the activated complex phonomena of chemical reactions can be taken formally as 2 subsequent reactions, the first endergonic, the second exergonic, both without the respective own activated complex. – Poutnik Dec 17 '22 at 10:41
  • @xasthor Thanks for referencing my answer to a similar question. It is not an accepted answer though, and I'm not making a claim about the best way of labeling these diagrams. I did make a claim that some ways of labeling them don't make sense. Folks who are closer to or in the subdiscipline of physical chemistry probably can give a better answer to your question and to the linked question. – Karsten Jan 07 '23 at 14:43

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