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If a reaction A <-> B has a value of dG°>0, then I know that A -> B is endergonic while A <- B is exergonic.

Now if I start with 100% B I could see how equilibrium is reached, B reacts to A until it reaches the stable situation (for whatever temperature/pressure we're in).

If I start with 100% A however, I would say nothing would happen, but if that's the case then this situation is already stable, so how can the equilibrium still be the same as the previous situation (since concentrations don't alter K)?

And even then, I know that that's not how it works (with only the exergonic direction working until equilibrium), because the equilibrium is reached when both reaction rates are equal, and that means that the endergonic reaction does indeed work at all time. Even in the first case in fact there's A reacting to B all the time, even though it's endergonic.

So I guess my question ultimately is: where does the energy come from?

(Silly thought: it doesn't take the energy from the exergonic way to "power" the endergonic way right? That would make every reaction be a net 0, and would also block the situation with 100% A at the start)

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  • $\begingroup$ “If I start with 100% A however, I would say nothing would happen” - This is not correct. A will react to form B until the same equilibrium is reached. $\endgroup$
    – orthocresol
    Commented Jun 24, 2021 at 10:38
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    $\begingroup$ This is explained in any textbook on physical chemistry or thermodynamics; but what you need to look up is the relationship between $\Delta_r G$ and $\Delta_r G^\circ$. The former indicates the direction in which any particular system will go (if it is positive then the reaction goes backwards; negative then forwards; and zero indicates equilibrium); it is applicable to any system whether in chemical equilibrium or not. The latter only indicates the final position of equilibrium; it does not tell you anything about systems that are not in equilibrium, or how they will react. $\endgroup$
    – orthocresol
    Commented Jun 24, 2021 at 10:44
  • $\begingroup$ So the value of dG° doesn't have any "intuitive" meaning right? Looking at dG, if I'm at 100% A then dG will be negative and the reaction goes A->B, correct? Still, I'm left with the question of how the backwards reaction happens if it's unfavourable, even if A goes to B with negative dG, even at 99% A there's still some B going to A, until the rates of reaction are the same. How can that happen? $\endgroup$
    – wojif
    Commented Jun 24, 2021 at 10:58
  • $\begingroup$ $\Delta_r G$ doesn’t say anything about rates of reaction or whether they both occur; it only tells you about the net change. Both reactions will always occur, the equilibrium is not dictated by which one occurs, but rather which one is faster. Note that the rates are also a function of concentration so if there is very little B to begin with, then the reverse reaction will have a very tiny rate, and the forward reaction will predominate. This precisely corresponds to the case of $\Delta_r G < 0$ which predicts that the net reaction will be in the forward direction. $\endgroup$
    – orthocresol
    Commented Jun 24, 2021 at 11:04
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    $\begingroup$ With a single molecule its either A or B, so you would need to count the number of times / sec either its A or its B to get the equilibrium constant. Thermodynamics works only for ensembles of molecules but if you measure a single molecule many times over you can approximate this. $\endgroup$
    – porphyrin
    Commented Jun 24, 2021 at 12:13

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