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I have read that in a reversible process, entropy can be transferred but not generated. Such that the entropy of the universe stays constant. On the other side, there is the Gibbs Energy, which can be defined as $-TdS$, where $dS$ is the change of the entropy of the universe, meaning also an exergonic chemical reaction, for instance, increases the entropy of the universe. As far as I know, for Gibbs Energy Processes occur under the assumption that they are isothermal and isobaric, as well as proceeding in a reversible manner. Where do I have the mistake?

Here is the source from where I have the $-TdS$ definition

Urb
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Achilles Zehnter
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    Can you clarify what your question is? No real macroscale process is reversible; all increase total entropy. This is equivalent to minimizing the relevant potential, such as the Gibbs free energy. The use of the Gibbs free energy does not assume reversibility, and your definition of it is incorrect; it is $G\equiv H-TS$ or $dG=-S,dT+V,dP$ (for a simple closed system). – Chemomechanics Jun 17 '21 at 06:27
  • -TS=H-TS is derived from the Second Law.It has been corrected now, thanks. Hope my question now makes sense. And the Gibbs energy change is not assumed to be for reversible processes only? I have thought because of the fact that it represents the maximum amount of non-expansive work, it must be proceeding in a reversible manner. – Achilles Zehnter Jun 17 '21 at 06:34

2 Answers2

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In a reversible process, a system is in thermal equilibrium with its surroundings at every instant of change. As a result the morsel of energy it looses at a temp. $T$ is the morsel of energy surroundings gain at the same temp. $T$, hence the net entropy change in reversible processes is zero.

Note that in a reversible process, while the entropy of the universe is unchanged - of the system's isn't.

If you know about state variables skip to What this means below

While its 'easiest' to calculate entropy change for reversible processes, its numerically same for irreversible processes too that take the system between the same initial and final state. However, its not as easy to calculate.

Observables like these that are path independent and only dependent on the initial and final system state are called state variables. Gibbs free energy is one of them.

What this means is that

during exergonic reactions, where $\Delta G_{sys}<0$, occurs regardless of the path the system takes, as reasoned above. So reversibility is not a pre-requisite for spontaneous reactions. In fact for reversible processes, $\Delta G_{sys}=0$ as they are always in equilibrium thus technically, not spontaneous.

Hence, for spontaneous reactions that occur irreversibly, there can be a change in entropy of the system and the universe without any contradictions.

lineage
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  • Thank you for your answer! It clarified part of my question. The other part is how it would be even possible to go from state 1 to 2 in a reversible manner with respect to delta G. When -TdS, S for the entropy of the universe, is defined as the Gibbs Energy under isothermal and isobaric conditions as life in a human cell for chemical reaction, must mean, a negative change in Gibbs Energy corresponds to an increase in the entropy of the universe. Then this must be a contradiction to the statement that in a reversible process the entropy of the universe does not change? – Achilles Zehnter Jun 17 '21 at 08:39
  • @AchillesZehnter Can one not have heat transfer during a reversible process? remember enthalpy change also appears in $\Delta G$ – lineage Jun 17 '21 at 08:40
  • Mmh, let me ask it differently. If -TdS=dG, S the total entropy of the universe, than in a reversible process no entropy is generated. All the entropy is constant but can be transferred by heat, as far as I know. If a chemical reaction is spontaneous dG<0 and -TdS<0 and therefore dS>0. When dS>0 entropy has been generated, irreversibilites must have been occured. If they have occured it was not reversible. I understand that it can change in the system by heat transfer but then for a reversible process it must change by the same amount in order to keep the entropy of the universe the same? – Achilles Zehnter Jun 17 '21 at 08:53
  • In this source they say that is -TdS which would then be what you have said.https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/The_Four_Laws_of_Thermodynamics/Second_Law_of_Thermodynamics – Achilles Zehnter Jun 17 '21 at 09:04
  • @AchillesZehnter I think you are absolutely right...in reversible processes, the system is in equilibrium and for systems in equilibrium, $\Delta G=0$. – lineage Jun 17 '21 at 09:17
  • I think I got it. But first, I think more about it. I will answer then again, when I think I got it really. Huge thanks for your patiency and answers! – Achilles Zehnter Jun 17 '21 at 09:25
  • @AchillesZehnter so you doubt was: "1. reversible processes don't gen. entopy. 2. exergonic do 3. gibbs change is reversible - how to reconcile" the only flaw is to assume change in gibbs is reversible. As u stated $..\Delta S_{uni}$ that's definitely not-reversible – lineage Jun 17 '21 at 09:26
  • Therefore, if I started with 100 reactants and the reaction is exergonic, as well as in eqilibrium I would have a ratio of reactants to products of 78:22. Then the system fails to return to 100 reactants, meaning the transition from 100 to 78:22 generated entropy? – Achilles Zehnter Jun 17 '21 at 11:01
  • @AchillesZehnter u had posted an earlier comment - something along the lines of "all reactions that proceed must be irreversible as reversible ones are at equilibrium anyways"...something like that...may you plz re-post it? I would like to elaborate on that and make that part of the main answer – lineage Jun 17 '21 at 17:01
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For the typical case of a reaction process that is carried out reversibly, going from pure reactants at T and P to pure products at T and P, the $\Delta G$ refers to the change in Gibbs free energy of the system (i,e, the chemicals participating in the reaction), not the universe. The change in free energy of the universe in such a situation is zero, with $\Delta H_{system}=-\Delta H_{surroundings}$ and $\Delta S_{system}=-\Delta S_{surroundings}$.

Chet Miller
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  • Hi, thank you for your answer. But if a reaction goes to completion it is considered to be irreversible and spontaneous, respectively. – Achilles Zehnter Jun 17 '21 at 13:00
  • https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/19%3A_Equilibrium/19.09%3A_Nonreversible_Reactions – Achilles Zehnter Jun 17 '21 at 13:07
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    You are confusing reversible chemical reactions with reversible processes. You do know the difference, right? The word "reversible" has to entirely different contexts. Also, all chemical reactions are reversible to some extent, unless their standard Gibbs free energy change is minus infinity. Also, considering a reaction spontaneous if its Gibbs free energy is negative is just a crude rule of thumb. – Chet Miller Jun 17 '21 at 15:44
  • No, sorry. I considered them to be equal. I have some difficulty with the exact english translations for my language. I meant with reversible process the process when one starts with 100 reactants in an exergonic reaction, for example, and ends up with 78:22 ratio for reactants to products, when the system is at chemical equilibrium. The question was whether it is possible that this reaction proceeds reversibly to the chemical equilibrium and back to the initial state of 100 products in this case. – Achilles Zehnter Jun 17 '21 at 16:12
  • ...free energy of the system (i,e, the chemicals participating in the reaction), not the universe yep that's the immediate reaction I had too, but as stated in OP's reference, $\Delta G$ can also be defined as $-T\Delta S_{uni}$ - afet all it must have some correlation to $\Delta S_{uni}$ to be deemed the arbiter of spontaneity of a reaction. In light of this, the OP is right to bring into his arguments $\Delta S_{uni}$ – lineage Jun 17 '21 at 16:58
  • It is possible to carry out the process of going from 100% reactants to 78.22 ratio for reactants to products reversibly, but the setup and steps to do this are not straightforward, and, for example, might involve compressors, semi=permeabile membranes, a Van't Hoff, equilibrium box, and reversible expansions. – Chet Miller Jun 17 '21 at 18:11
  • @lineage It is possible to define a $\Delta G$ for the universe as $\Delta H_{uni}-T\Delta S_{uni}$ if all the heat transfer between system and surroundings is at temperature T, but, for a reversible process, this is zero. – Chet Miller Jun 17 '21 at 18:16
  • @lineage So,....? What is your point? – Chet Miller Jun 17 '21 at 19:56
  • @ChetMiller was merely justifying OP's use of $\Delta S_{uni}$. – lineage Jun 17 '21 at 19:58
  • @ChetMiller reversible $\implies \Delta G=0$..never disagreed...i hope that was clear – lineage Jun 17 '21 at 20:00