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I have a problem with the definition of the standard Gibbs energy and its connection to the equilibrium constants.

I think, that I've basically understood what the different equation mean but there is one thing, I'm unable to understand:

On the one hand:

One may describe a chemical reaction with $\Delta G=\Delta G^\circ + RT\ln{Q}$. In equilibrium $\Delta G = 0$ and the equation reads $\Delta G^\circ = -RT \ln{K}$.

On the other hand:

The definition of standard state is very clear: pressure = 1 bar and all reactants and products must have activity = 1.

If I consider these two aspects separately, everything seems to be fine. But these two concepts have to be valid at the same time, what leads to $\Delta G^\circ = 0$ (always), since $K=1$ (all activities are per definition = 1).

Therefore, $\Delta G^\circ$ would be always zero. I know that this isn't true, but I don't understand why.

Can anyone explain this to me?

Thanks!

user76122
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    "since $K=1$" not necessarily true; it is $Q = 1$. Nobody said that at standard state the system must be in equilibrium. – orthocresol Mar 24 '19 at 15:48
  • ok, but the formulae say: At standard state $\Delta G^\circ = -RT\ln{K}$ and K is the ratio of the activities of reactants and products in equilibrium, due to the standard state definition K = 1, because it says: each $a=1$. Doesn't this definition disagree with your comment? I find this all really confusing. – user76122 Mar 24 '19 at 16:14
  • I find that terribly confusing and wrong if it claims $K = 1$. – orthocresol Mar 24 '19 at 16:20
  • I know it is wrong, but I don't get why :( Did you understand my problem, that is the pure formula seems to contradict the general understanding... – user76122 Mar 24 '19 at 17:36
  • @user76122 Orthocresol is right. Your definition of standard state in the context of $\Delta G^{0}$ is wrong. The standard state of a pure material entails that its activity is 1. But in the context of the equation for $\Delta G$ you deal conceptually with mixtures and thus not pure materials. If you follow my derivation of the formula (see here) you can see which assumptions go into $\Delta G^{0}$. It is true that it is defined for standard pressure/concentration, but activity being equal to 1 is not presumed. – Philipp Mar 24 '19 at 18:24
  • @Philipp I read your linked answer. At the moment I think my problem arises, indeed, from a misunderstanding of $K$ and $Q$. Regarding the activities: In my understanding $a=1$ is automatically true in standard state. Since $a$ is a generalization of terms like $\frac{\varphi_i p_i}{p^0}$. If the system is in its standard state, i.e. $p=p^0$, $\varphi_i=1$ per definition and therefore $a=1$, isn't it? For example, in electrochemistry $a=1$ for every species is explicitly demanded in standard state. – user76122 Mar 25 '19 at 09:36
  • @user76122 I think you're still on the wrong track concerning activities and standard state. The activity being equal to 1 in standard state is only true for pure materials. Take the term $\frac{\varphi_{i} p_{i}}{p}$ for example. The $p_{i}$ is the partial pressure of substance $i$ in the mixture. When you have the mixture at standard conditions with $p=p^{0}$ and $\varphi = 1$ the term $\frac{\varphi_{i} p_{i}}{p^{0}} = \frac{p_{i}}{p^{0}}$ still isn't equal to 1 because $p = \sum_{j} p_{j}$. Only in the limit of enriching the mixture with $i$ until it's nearly pure, you get $p_i = p = p^0$. – Philipp Mar 25 '19 at 19:24
  • @Philipp I relooked at my textbook and you are right. So the "trick" is to know which standard state one should use for a real application? So far I found four standard states: infinit dilute solution, pure substance, ideal gas (standard pressure) and in electrochemistry all activities must equal 1. Are there even more? Is it just me or is this standard state topic rather confusing?!? – user76122 Mar 27 '19 at 12:21
  • @Philipp Hm strange, I just thought I got it and then I found the following video: https://de.coursera.org/lecture/advanced-chemistry/5-10-standard-versus-nonstandard-free-energy-change-vfbI5 at approx. 2:30 it is claimed, that in standard state the partial pressures are at 1 bar (std. pressure). That contradicts your comment, doesn't it? – user76122 Mar 27 '19 at 12:32
  • @user76122 In the answer of mine I linked you can find the equation $\sum_i \nu_{i} \mu_{i}^{0} = \Delta G^{0}$ with $\mu_{i}^{0} = \mu_{i}^{}(p^{0}, T)$ which is the definition of $\Delta G^{0}$, i.e. you can calculate the value of $\Delta G^{0}$ by adding up the chemical potentials of the pure* educts and products at the standard pressure of $1 , \mathrm{bar}$ (and at temperature $T$). So, you kind of start out conceptually with the unmixed educts in their standard states and I think this is what the video is referring to. – Philipp Mar 27 '19 at 20:49
  • @user76122 ...But when you mix the seperate educts - each gas being at standard pressure - and look at the way they mix, i.e. how the partial pressures turn out to be, you can derive the value of $\Delta G^{0}$ from this (via $K$). And this is what the equation $\ln K = -\frac{\Delta G^{0}}{RT}$ tells you. There is only one value of $\Delta G^{0}$ that will lead to this special set of partial pressures (represented by $K$) for the reaction at hand under standard conditions. – Philipp Mar 27 '19 at 20:55
  • @Philipp I really appreciate your patience, so I'm sorry to bother you again (maybe I'm just too stupid^^): Doesn't your explanation mean that there are two $\Delta G^\circ$? One for unmixed reactants and products in the standard state ($\sum_i\nu_i\mu_i^\circ=\Delta G^\circ$ with $\mu_i^\circ=\mu_i^*$) and one for the mixed state, where the latter can be calculated by $\Delta G^\circ = -RT\ln{K}$? I thought these two values must be identical? In this answer of https://chemistry.stackexchange.com/a/41864/76122 it is also mentioned, that all $a_i=1$, but maybe I'm missing something? – user76122 Mar 28 '19 at 10:15
  • @user76122 No, there aren't two values for $\Delta G^{0}$. It's just two ways of expressing the same quantity. One, $\sum_{i} \nu_i \mu^{0}{I} = \Delta G^0$ is the defining equation for $\Delta G^0$, the other $\Delta G^0 = - RT \mathrm{ln} K$ just follows from the equilibrium condition $\Delta G = 0$ when you express $\Delta G$ in terms of $\Delta G^0$. You can look at it like this: The chemical potentials $\mu{i}^{0}$ of educts and products which make up $\Delta G^0$ set up some kind of "force" that will push the equilibrium to a certain position, represented by $K$, once the components... – Philipp Mar 31 '19 at 22:13
  • @user76122 ... are mixed and equilibrate. A bit like an ideal spring with a characteristic spring constant that gets compressed by a certain force and upon release will oscillate at a characteristic frequency. – Philipp Mar 31 '19 at 22:20

2 Answers2

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As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$\Delta G=\Delta G^\circ+ RT\ln{1}=\Delta G^\circ$$ On the other hand at equilibrium $Q=K$ and so $$\Delta G=\Delta G^\circ + RT\ln{K}$$ This of course leads to $\Delta G^\circ = -RT\ln{K}$ since at equilibrium $\Delta G=0$.

So you might want to think of it as multiple statements:

  1. For the conversion of reactants to products in their standard states $Q=1$
  2. At equilibrium $\Delta G=0$
  3. At equilibrium $Q=K$
  4. $\Delta G^\circ$ is the free energy change for conversion of reactants to products in their standard states.

The first statement is consistent with the definition of standard states. The second and fourth statements follow from combination of the first and second laws of thermodynamics. The third statement is a definition of $K$.

$K$ and $\Delta G^\circ$ are very much connected, but $\Delta G^\circ$ does not describe the change in free energy from reactants to products at equilibrium.

Buck Thorn
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  • So, based on your answer and on the comments above, I think that my understanding increases. These are just two different point of views or two different points of the same equation: $\Delta G=\Delta G^\circ + RT\ln{Q}$. My misunderstanding arises from the equation: $\Delta G^\circ = - RT\ln{K}$, since it seems to connect standard ($\Delta G^\circ$) state with equilibrium ($K$) and that's not the meaning of this formula, correct? – user76122 Mar 25 '19 at 10:12
  • @user76122 $K$ and $\Delta G^\circ$ are very much connected. $\Delta G^\circ$ does not describe the change in free energy from reactants to products at equilibrium. I'll add this to my answer. – Buck Thorn Mar 25 '19 at 11:42
  • At equilibrium, there is no change in free energy neither a net conversion from reactants to products (or vice versa). In my understanding, if you consider the "hanging rope"-style curve of $G(\xi)$, $\Delta G^\circ$ and $K$ are just two different "points" on that curve, describing different things. The connection is, that these points lie on the same curve and therefore, can be calculated from each other. This is what is shown by the equations. Did you mean it that way (or at least in a similar way)? I'd like to emphasize that I'm trying to understand and not to correct you :) – user76122 Mar 25 '19 at 12:53
  • @user76122 I would not interpret $\Delta G^\circ$ and $K$ as points on the progress curve. Rather they are related to the slope of that curve. – Buck Thorn Mar 25 '19 at 19:25
  • Yes, of course. Please see the picture: https://i.stack.imgur.com/6MfNR.png What I meant was the following: If the system is in equilibrium, i.e. $\frac{\partial G}{\partial\xi}=\Delta G=0$, it corresponds to the red point on the curve. In standard state ($Q=1$) the system is at a point at which $\frac{\partial G}{\partial\xi}=\Delta G=\Delta G^\circ$. This corresponds to the green point. I'm not sure if the green point is always somewhere on the blue G curve or whether there are possibilities where the definition of a standard state is rather a hypothetical point? – user76122 Mar 27 '19 at 12:11
  • I guess you answer the question yourself: $\Delta G^\circ$ will be on the curve if Q=1 has an associated corresponding value of the progress coordinate on the curve. But I would consider that more of a coincidence than truly useful. But it doesn't hurt to continue digging deeper. – Buck Thorn Mar 27 '19 at 13:58
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    Thanks! By digging deeper, I found the following answer: https://chemistry.stackexchange.com/a/41864/76122 There it is claimed, that $Q=1$ has to lie on the curve, since $Q$ alters from $\infty$ to zero. – user76122 Mar 28 '19 at 09:42
  • @user76122 Yes, you are right. I'll have to eat my words for not thinking a little harder about this. I take back that last statement. – Buck Thorn Mar 28 '19 at 09:59
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What you enter into $K$ are not the activities of the pure reactants and pure products at standard state (if you did then, yes, $K$ would be 1). Rather, it is their activities at equilibrium (raised, of course, to the power of their respective stochiometric coefficients). And, at equilibrium, these activities are generally not equal to one.

theorist
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