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I recently learned about the standard Gibbs free energy change of reaction, ΔG=ΔH-TΔS, and how its sign indicates whether the conversion of (ALL) reactants and products is spontaneous or not.

I then learned that the equilibrium constant of a reversible reaction depends on ΔG for some reason.

ΔG=-RTlnK

The equation shows that whether a reaction goes is (noticeably) reversible or effectively goes to completion depends on how -ve ΔG is.

Given the relationship between ΔG and K, is there any explanation of why reactions are reversible in the first place?

If a reaction is exergonic and the 100% conversion of reactants to products would increase total entropy, why shouldn't it go to completion?

If a reaction is endergonic and the 100% conversion of reactans to products would decrease total entropy, why should it occur to any extent at all?

Vulgar Mechanick
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    It's because of the entropy of mixing. If you take a look at the picture in my answer here https://chemistry.stackexchange.com/questions/127674/are-all-exothermic-reactions-spontaneous/127676#127676 , you'll see that the free energy of the system isn't a straight line from reactants to products as the reaction progresses. For the example shown there, if it were, then the equilibrium point would be all reactants. Instead, there's a dip, such that the minimum free energy, and thus the equilibrium, lies at a point somewhere between pure reactants and pure products. – theorist Jun 21 '21 at 05:23
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    For the simplest case, $\ce{A <-> B}$, the entropy of mixing is a maximum when you have half A and half B. In an ideal system, the free energy of mixing is T times the negative of the entropy of mixing, i.e., $\Delta G_{mix} = -T \Delta S_{mix}$. So the free energy of mixing is a minimum at the half way point. To get the actual free energy of the reaction mixture, you need to add the free energy of mixing which, being negative, causes a dip. In non-ideal systems, you need to add the enthalpy of mixing (assuming constant p), which can be either positive or negative. – theorist Jun 21 '21 at 05:27
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    Finally, in first-order phase transitions (these are the commonplace ones, like ice to liquid water), you have no entropy of mixing, because they're different phases. That's why, outside of the coexistence point (where the free energies of the pure phases are equal), phase transitions do go to completion, i.e., you have either all of one phase or all of another; and the phase you have is whichever one has the lowest free energy under the current conditions. – theorist Jun 21 '21 at 05:33
  • Ok so would it be correct to say that endergonic 'reversible' reactions attain equilibrium because the increase in free-energy due to some conversion of reactants into products is more than compensated by the decrease in free energy due to mixing, making the overall free energy change negative? – Vulgar Mechanick Jun 21 '21 at 06:59
  • Also, from ΔG=ΔG0 +RTLnQ, could i conclude that the entropy of mixing is RlnQ? – Vulgar Mechanick Jun 21 '21 at 07:02
  • Can you describe a reaction that is both exergonic and that doesn't increase total entropy (with 100% conversion or otherwise)? Exergonic refers to $\Delta G<0$ which under constant T and p is equivalent to saying that for the process $\Delta S_\textrm{total}>0$ – Buck Thorn Jun 21 '21 at 07:18
  • Βy total entropy I meant entropy of the universe. And apparently, as theorist points out, there's also free energy of mixing that must be taken into account, so I guess 100% conversion never maximizes entropy(of universe)? – Vulgar Mechanick Jun 21 '21 at 07:22

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All reactions are reversible and all reactions are spontaneous. If they told you differently, then they are too stupid to realize that saying that will just confuse you. If you start out with just pure reactants, combine them, and let them equilibrate, then some products will be present in the end. And if you start out with just pure products, combine them, and let them equilibrate, then some reactants will be present in the end. So, as I said, all reactions are reversible and spontaneous. The question is not "yes" or "no," it is only a question of to what extents.

A reaction being spontaneous "if its standard change in Gibbs free energy is negative" is only a very crude rule of thumb. All it means is that the equilibrium constant is greater than zero. But a positive $\Delta G^0$ does not mean that reactants will not form products. It just means that the reaction is less prone to approach completion. Note that I did not say that if $\Delta G^0$ is negative, the reaction will go to completion; it will just be more prone to approach completion.

Chet Miller
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  • But if all reagents and products are at constant activity then are you not guaranteed complete conversion (or no reaction)? – Buck Thorn Jun 21 '21 at 16:26
  • I am aware that all reactions are reversible to some extent. What I want to know is why? Someone mentioned a gibbs free energy of mixing due to increase in entropy when reactants and products mix. Could you elaborate on that? – Vulgar Mechanick Jun 21 '21 at 17:18
  • Not really. If the Standard Gibbs free energy change is carried out reversibly using compressors, semipermeable membranes, expanders, and a Van't Hoff equilibrium box, then there is no significant mixing involved. – Chet Miller Jun 21 '21 at 17:50
  • @BuckThorn I don't understand your question, or how it relates to what I said. – Chet Miller Jun 21 '21 at 17:51
  • Is it impossible to envision a situation in which $\Delta _r G$ does not depend on composition (that is, where it remains constant during the entire progress of a reaction)? – Buck Thorn Jun 21 '21 at 18:09
  • @BuckThorn Yes, in a Van't Hoff equilibrium box where reactants are added at the same chemical potentials as the mixture in the equilibrium box, and reactants are removed at the same chemical potential as in the mixture in the equilibrium box. In that case, $\Delta G_r=0$. Of course, in this case, the reactants and products are not in their standard states at 1 bar. – Chet Miller Jun 21 '21 at 19:06
  • In the previous comment, I meant that products are removed, not reactants. – Chet Miller Jun 22 '21 at 15:57
  • But what about a scenario where reagents and products are solids, each at constant activity, and assuming the substances to be otherwise immiscible. I can take an arbitrary amount of these through a cycle (say annealing them and cooling to obtain products) and in principle convert 100%. – Buck Thorn Jun 22 '21 at 16:05
  • @Buck Thorn Is there such a real world situation? If they are immiscible, does the reaction not take place in the gas phase? If so, is there not an equilibrium constant for the gas phase reaction? – Chet Miller Jun 22 '21 at 19:21
  • Well, I meant that the solid reactants reacted in a melt, upon heating, and otherwise formed separate solid phases. – Buck Thorn Jun 22 '21 at 19:46
  • @Buck Thorn So there is a melt phase reaction that occurs? If so, is there an equilibrium constant and standard free energy change associated with this melt phase reaction? – Chet Miller Jun 22 '21 at 20:06
  • I am just looking for a ploy to work around the immiscibility, something like the van't Hoff box. Gas phase would work too I guess. If you ignore the mechanism (how to mix the reagents) and focus on the relative stability of reagents and products, assuming these are constant (as in the van't Hoff box experiment) I can't see how not to conclude that you should not drive the reaction to completion in one direction. In such a situation $\Delta _r G $ is constant until completion of the reaction. Sorry about the double negatives. – Buck Thorn Jun 22 '21 at 20:20
  • Let us continue this discussion in chat. – Chet Miller Jun 22 '21 at 20:48