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Okay, so I'm trying to derive an important formula which states that - ∆G(r) = ∆G° + RTln( RQ ) where ∆G(r) is the instantaneous rate of change of Gibbs energy with respect to moles of reagent consumed (am I right on this?) I'm hight confused because I have 'almost' derived it but due to some conceptual difficulty I'm not able to get the exact formula!

NOTE: (1) Please ignore any non expansion work done on the system and tell my mistake as simply as possible because I have just started thermodynamics.

(2) The line above G represent molar Gibbs energy of substance. The small letters on the side of the symbol of reactent are the stochiometric coefficients.

(3) Many times I see variations of this formula, instead of ∆G(r) they simply use ∆G, there is a difference between two. Is this formula valid if ∆G is used instead of ∆G(r) in left hand side of equation? Also don't see any 'sense' in using ∆G because if ∆G is there then we need the initial and final quantities of the substances in the reaction but in the reaction quotient we plug the instantaneous activities of reactents and products.

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I'm getting the above equation but a stochiometric coefficient is also included in this ! Obviously it is wrong. Kindly tell me my mistake.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – jonsca Apr 5 at 21:11
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∆G(r) is the instantaneous rate of change of Gibbs energy with respect to moles of reagent consumed (am I right on this?)

No, because then you would get different answers depending on which reactant you choose. Instead, it is with respect to change in amount divided by stoichiometric coefficient (sometimes called extent of reaction). This is similar to the way reaction rates are defined (change in concentration divided by stoichiometric coefficient).

If you update the definition of the Gibbs energy of reaction, you will get the expected equation in your answer.

Many times I see variations of this formula, instead of ∆G(r) they simply use ∆G, there is a difference between two. Is this formula valid if ∆G is used instead of ∆G(r) in left hand side of equation?

Textbooks and other sources are often fuzzy concerning the distinction of $\Delta G$ vs $\Delta_r G$. You are right in saying that the formula does not make sense if you are calculating the difference between two states with different concentration. The other thing that is fuzzy are the units, energy vs. energy per amount (e.g. kJ vs kJ/mol).

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  • $\begingroup$ Exactly the answer I was looking for! Interpreting the wrong meaning of ∆G(r) was my mistake. Thanks! $\endgroup$ – Shivansh J Apr 5 at 4:44
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I think you are using the first approach that I described. So, here is the analysis based on the first approach. Suppose you have the following numbers of moles of A, B, C, and D in the reaction mixture at some time: $n_A$, $n_B$, $n_C$, and $n_D$.

And the total pressure of the mixture (in bars) is held constant at P. So the partial pressure of species A is $$p_A=\frac{n_A}{n_A+n_B+n_C+n_D}P$$And the partial molar free energy of species A is $$\bar{G}_A=G^0_A+RT\ln{\left(\frac{n_A}{n_A+n_B+n_C+n_D}P\right)}$$where $G^0_A$ is the free energy of pure species A at the reference pressure of 1 bar. Similar equations can be written for the other species. The total free energy of the mixture at any point in time during the reaction is: $$G=n_A\bar{G}_A+n_B\bar{G}_B+n_C\bar{G}_C+n_D\bar{G}_D$$ Now, let the concentrations of the chemical species in the reaction mixture be expressed as follows: $$n_A=n_{A0}-a\xi$$ $$n_B=n_{B0}-b\xi$$ $$n_C=n_{C0}+c\xi$$ $$n_A=n_{D0}+d\xi$$ where the subscript 0 refers to the initial state and $\xi$ is the conversion. To find the equilibrium condition, you want to find the value of $ \xi $ that minimizes the free energy of the solution mixture. To do that, you have a calculus problem to solve.

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  • $\begingroup$ My understanding of the question is that the OP is trying to reconcile the different usages of $\Delta G$ and $\Delta_r G$, not trying to solve for the equilibrium concentrations. $\endgroup$ – Andrew Apr 5 at 0:24

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