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I've been having a lot of trouble trying to truly understand Gibbs free energy from a practical perspective. I have no background in physical chemistry, but I think I have a firm grasp on many of the concepts related to Gibbs energy. That being said, I'm going to try and reason using less rigorous mathematics. I'm having a hard time bridging the concepts of Gibbs free energy and the maximum useful non-PV work that can be extracted from an ideal chemical system.

My confusion may be due to some fundamental misunderstanding, but currently my confusion follows this line of thinking:

Most chemistry texts explain that $\Delta_\ G = wmax$.

The most common equations for calculating $\Delta_\ G$ are:

eq1. $\Delta_\ G = \Delta H - T\Delta S$

eq.2 $\Delta_\ G = \Delta G^\text{o} + RT\ln Q$

Both equations essentially make the same relationship, with the difference that $\Delta\Delta S_\mathrm{mixing}$ is baked in with $\Delta\ S^\text{0}$ in eq1.

Question 1.

This may just be a notation thing, but I want to confrim that $\Delta_\ G$ is just an alternative notation for $\Delta_\mathrm{r} G$ and $\Delta\ G^\text{o}$ is an alternative notation for $\Delta_\mathrm{r} G^\text{o}$. Considering I've seen both notations used and I think its safe to assume they are.

Question 2.

If $\Delta_\mathrm{r} G$ or alternatively, $\Delta\ G$ is the slope of the Gibbs energy curve at some non-standard state defined by the reaction quotient $Q$ how can this value be equal to the maximum useful non-PV work. I mean it would make sense if the system were at some steady-state such that the reaction continued to generate products with no net change in product/reactant concentrations. But in reality the value of $\Delta_\mathrm{r} G$ will continue to change until the reaction reaches its free energy minimum.

Maybe I'm trying to get these relations to do more than they meant to do. I fully appreciate that the slope of the Gibbs energy curve will allow for determination of conditions when $\Delta\ G = 0$ which is useful for predicting spontaneity. I just figured (probably incorrectly) that you could for example take some known mass of reactant and from there calculate the theoretical maximum non-PV work harvested from 100% reactants --> equilibrium.

Many Thanks!

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