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A reaction proceeds towards the direction of lesser Gibbs free energy (at constant $T$ (temperature) and $P$ (pressure)). So, we could say that Gibbs free energy at equilibrium is minimum.

On the other hand, we have $$\Delta G=\Delta G^\circ + RT\ln Q$$, where $Q$ is the reaction quotient. At equilibrium, $Q=K_\text{eq}$, and we already know that $\Delta G^\circ =-RT\ln K_\text{eq}$.

Substituting, we get $\Delta G=0$ at equilibrium. But, we know that $G$ minimized itself—thus there was a change in $G$ and $\Delta G < 0$.

What am I missing here?

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I think your question really arises from some confusion about what $\Delta G$ represents. In general, $\Delta X$ for a thermodynamic quantity $X$ is the change of $X$ along some process. You could make it clear by actually writing $\Delta G(\text{A}\rightarrow\text{B})$ where A and B are before and after states. (We'll note that, in the general case, $\Delta X$ depends on the path take from A to B, making this notation improper. If $X$ is a function of state, though, you're good to go.)

However, in the equation you quote:

$$\Delta G = \Delta G^0 + RT \ln Q$$

the $\Delta G$ is a free energy of reaction and should thus be denoted $\Delta_\mathrm r G$, with the correct equation being:

$$\Delta_\mathrm r G = \Delta_\mathrm r G^0 + RT \ln Q$$

The free energy of reaction is defined as $\Delta_\mathrm r G = G_{\text{products}} - G_{\text{reactants}}$.


Thus, this $\Delta_\mathrm r G$ is not the variation of $G$ over the entire reaction, which would be the $\Delta G$ of the system between the start of the reaction and the equilibrium.


PS: I think this link is the online resource I found with the clearer use and explanation of notations. Notations are important in thermodynamics!

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  • $\begingroup$ Hmm, but isn't G_prod-G_reactants negative as well? $\endgroup$ Apr 25 '12 at 19:06
  • $\begingroup$ @Manishearth At equilibrium, it'll be zero. $\endgroup$
    – F'x
    Apr 25 '12 at 19:07
  • $\begingroup$ Wait.. But doesn't a system go towards the lesser G? If G is nonzero elsewhere, this means that it had to increase at one point. $\endgroup$ Apr 25 '12 at 19:09
  • $\begingroup$ @Manishearth yes, the evolution of the system is such that it minimizes the free energy of the whole system. But what is your question? You have to be very clear about what system and what process/path you are talking about, otherwise it is confusing… $\endgroup$
    – F'x
    Apr 25 '12 at 19:12
  • $\begingroup$ I'm talking about a simple A-->B reaction._whole system_ is what I'd missed (which is a plus not a minus) thanks :) $\endgroup$ Apr 25 '12 at 19:17
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Gibbs free energy is a measure of how much "potential" a reaction has left to do a net "something." So if the free energy is zero, then the reaction is at equilibrium, an no more work can be done.

It may be easier to see this using an alternative form of the the Gibbs free energy, such as $\Delta G = -T\Delta S$.

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    $\begingroup$ But that's the point. $G$ is not zero by the formula, $\Delta G$ is. Or is my interpretation of the formula wrong? $\endgroup$ Apr 25 '12 at 18:47
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    $\begingroup$ Just because $G$ is at its minimum does not make it zero. (see here last part of introduction) $\endgroup$
    – soandos
    Apr 25 '12 at 18:51
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It might be helpful if we defined the changes in free energy in these equations in a more precise manner. Let's just consider an ideal gas reaction.

$\Delta G^\circ$ is the change in free energy for the transition between the following two thermodynamic equilibrium states:

State 1: Pure reactants (in separate containers) in stochiometric proportions at 1 atm pressure and temperature T

State 2: Pure products (in separate containers) in corresponding stochiometric proportions at 1 atm pressure and temperature T

To directly measure $\Delta G^\circ$, one would have to dream up a reversible path to transition from state 1 at state 2 and determine $\Delta G^\circ$ for that path. This path might involve the use of constant temperature reservoirs and semipermeable membranes.

$\Delta G$ is the change in free energy for the transition between the following two thermodynamic equilibrium states:

State 1: Pure reactants (in separate containers) in stochiometric proportions at specified pressures and temperature T

State 2: Pure products (in separate containers) in corresponding stochiometric proportions at specified pressures and temperature T

If the specified pressures just happen to correspond to the partial pressures of the gases in a reaction mixture at equilibrium, then $\Delta G = 0$. For small excursions of the partial pressures from the equilibrium values, $\Delta G$ will increase as the squares of the partial pressure increments. That is what we mean when we say that it is at a minimum at equilibrium.

To directly measure $\Delta G$, one would have to dream up a reversible path to transition from state 1 at state 2 and determine $\Delta G$ for that path. This path might involve the use of constant temperature reservoirs, small weights to be added or removed from a piston, and semipermeable membranes.

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  • $\begingroup$ Ignore my previous comment, sorry. Can you explain why $\Delta G = 0$ when moving from state 1 to state 2 provided that the pressures correspond to the partial pressures of the gas in equilibrium? I had always (mistakenly) assumed that $\Delta G = 0$ at equilibrium meant that $dG = 0$, i.e the slope $dG/d\xi$ where $\xi$ is the extent of reaction is zero, which is clear because $G$ of the whole system is minimum at equilibrium. Your answer gives a much more clear definition of what we mean by $\Delta G = 0$, but I can't come up with any reasoning as to why State 1 and State 2 have the same $G$ $\endgroup$ Oct 22 at 12:06
  • $\begingroup$ @Ashish Ahuja Are you familiar with the van't Hoff equilibrium box chemical reactor? Are you familiar with the open system (control volume) versions of the 1st and 2nd laws of thermodynamics? $\endgroup$ Oct 22 at 13:10
  • $\begingroup$ I was not aware of the van't Hoff reactor but I had a look at your answer here. Because in the reactor everything moves reversibly i.e theoretically the reactor could operate at an infinitely slow speed, the reaction must not be spontaneous in either direction and hence $\Delta G=0$; is this logic correct? I am not familiar with the open system versions of the 1st and 2nd law. I looked at the relevant wikipedia sections but not knowing what I was looking for it was hard for me to understand everything due to my current level of knowledge. $\endgroup$ Oct 23 at 7:10
  • $\begingroup$ @Axhish Ahuja The logic is correct. But I wish you had experience with the open system versions of the 1st and 2nd laws so that you could see how it is verified mathematically. For a better understanding of these, and how to apply them to the Van't Hoff equilibrium box, see Fundamentals of Engineering Thermodynamics by Moran et al. I think this book is available online. $\endgroup$ Oct 23 at 11:49
  • $\begingroup$ Could you let me know where the Van't Hoff equilibrium box is in the book, since I couldn't find it in the index at the back? $\endgroup$ Oct 23 at 12:21
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I suppose you are confused between $∆G$ and $G$. After equilibrium is achieved, $∆G$ is indeed zero, as no more reduction of free energy is possible (keeping conditions intact). What this also means is that $G$ is minimum.

Throughout the reaction, $∆G$ will have been negative until equilibrium is attained. This means the free energy will keep on reducing until no more reduction is possible. This indicates that $G$ is minimum, since $∆G$ is zero, i.e. the value of $G$ at equilibirum cannot be further diminished.

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    $\begingroup$ What $ΔG$ means in equilibrium? Doesn't $Δ$ refer to change between two states? $\endgroup$
    – ado sar
    Nov 21 '20 at 12:57
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    $\begingroup$ @adosar look at the framing of the question. The answer lies within the words of the question. Consider time t = zero when equilibrium has been achieved. Throughout the undergoing of the reaction before equilibrium, i.e. for all $t<0$, $∆G$ was negative. At $t=0$, $∆G$ is zero. What this means is $G$ minimized itself for the entire time t was less than zero. At $t=0$, it finally attains its minimum value and is not subject to any further reduction. Tl;dr - $∆G$ WAS negative for $t<0$ , $∆G$ FROM NOW ON is zero. $G$ WAS reducing itself, $G$ NOW cannot reduce anymore. $\endgroup$ Oct 23 at 15:13

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