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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?

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    $\begingroup$ It's because of the entropy of mixing. If you take a look at the picture in my answer here chemistry.stackexchange.com/questions/127674/… , 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. $\endgroup$
    – theorist
    Commented Jun 21, 2021 at 5:23
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    $\begingroup$ 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. $\endgroup$
    – theorist
    Commented Jun 21, 2021 at 5:27
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    $\begingroup$ 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. $\endgroup$
    – theorist
    Commented Jun 21, 2021 at 5:33
  • $\begingroup$ 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? $\endgroup$
    – Vulgar Mechanick
    Commented Jun 21, 2021 at 6:59
  • $\begingroup$ Also, from ΔG=ΔG0 +RTLnQ, could i conclude that the entropy of mixing is RlnQ? $\endgroup$
    – Vulgar Mechanick
    Commented Jun 21, 2021 at 7:02

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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.

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  • $\begingroup$ But if all reagents and products are at constant activity then are you not guaranteed complete conversion (or no reaction)? $\endgroup$
    – Buck Thorn ♦
    Commented Jun 21, 2021 at 16:26
  • $\begingroup$ 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? $\endgroup$
    – Vulgar Mechanick
    Commented Jun 21, 2021 at 17:18
  • $\begingroup$ 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. $\endgroup$
    – Chet Miller
    Commented Jun 21, 2021 at 17:50
  • $\begingroup$ @BuckThorn I don't understand your question, or how it relates to what I said. $\endgroup$
    – Chet Miller
    Commented Jun 21, 2021 at 17:51
  • $\begingroup$ 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)? $\endgroup$
    – Buck Thorn ♦
    Commented Jun 21, 2021 at 18:09

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