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In this lecture by MIT open courseware (*), Moungi Bawendi states that we can have some change in our chemical system containing reactants and products scaled by some $ \epsilon$ amount. Using this set up and the equation which gives chemical potential as a function of pressure (**) he derives the $\Delta G$ as a function of $ \epsilon$ as shown below:

$$ \mu_{i} (T , P_{tot} ) = u_{i}^{o} (T) + RT \ln \frac{P}{P_o}$$

and, with the expression for $ \Delta G $ in terms of chemical potentials:

$$ \Delta G = \sum_{i=prods} \mu_i \nu_i - \sum_{i= reacts} \mu_i \nu_i$$

He gets,

$$ \Delta G(\epsilon) = \epsilon [ \Delta G^{o} + RT \ln \frac{ (P_d)^{v_d} (P_c)^{v_c} }{ (P_b)^{v_b} (P_a)^{v_a} }]$$

Now, this expression I have a few questions about:

  • Is the pressure ratio term on the right hand side dependent on $\epsilon$? Like I think that as the reaction proceeds, would each pressure term fluctuate?

  • How does one correctly interpret the final equation ? Is it related to calculus of variations ? I ask this because it looks very similar ot the derivation of fundamental theorem of calculus of variations (found here)

  • Usually we assign a Gibbs free energy to a whole reaction, but here there is an aspect of this $\epsilon$ paremeter, so how does this idea of parameterizing chemical reactions relate to the regular values of Gibbs free energy we see in textbooks? The professor equates bracketed term to $\Delta G$ but the connection is not a 100% clear to me. Perhaps an explicit explanation would help.


*: Reference video

**: page-2 of these notes from mit ocw

Note: I had posted this question more than three months back on physics stack exchange without finding any explanatory answer. Hopefully the question finds more home here.

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    $\begingroup$ I'm not able to see the notes, and the video doesn't have anything involving $\epsilon$ as far as I can tell. Can you please provide a precise definition of what $\epsilon$ represents or a better reference? $\endgroup$ Aug 7 '21 at 23:48
  • $\begingroup$ $\epsilon$ is an infinitesimally small change in amounts defined to see whether the Gibbs energy change is positive or negative if starting with a certain set of partial pressures. This means the partial pressures do not change significantly.. $\endgroup$ Aug 8 '21 at 2:43
  • $\begingroup$ Pls check the edited link @ChetMiller $\endgroup$
    – Buraian
    Aug 8 '21 at 2:44
  • $\begingroup$ Check out this link. Page 623 on Chemical Potential and Thermodynamics of Spontaneous Chemical Change, has a pretty clear derivation and explanation. pearsoncanada.ca/media/highered-showcase/multi-product-showcase/… $\endgroup$
    – lDuong
    Aug 8 '21 at 3:51
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The other two answers interpret $\epsilon$ as the extent of reaction $\xi$ and make factually accurate statements under this assumption. I disagree that $\epsilon$ should be interpreted as $\xi$; instead, $\epsilon$ seems to be used as a (small) perturbative parameter that describes how the system will change under a small perturbation from its current state. (In general, this does not seem to be a useful way of analyzing chemical reactions, since parametrizing the extent of reaction by $\xi$ is more versatile and provides more information besides.)

Is the pressure ratio term on the right hand side dependent on ϵ? Like I think that as the reaction proceeds, would each pressure term fluctuate?

It will be approximately independent of $\epsilon$ because $\epsilon$ is taken to be small. However, it would not be independent of $\xi$ for the reasons you describe in your second question.

How does one correctly interpret the final equation ? Is it related to calculus of variations ? I ask this because it looks very similar ot the derivation of fundamental theorem of calculus of variations (found here)

The change in the Gibbs free energy due to this small perturbation is given, to first order, by $\Delta G(\epsilon)$. It is related to the calculus of variations in the sense that they both deal with infinitesimal analysis, but lots of things deal with infinitesimal analysis and there is no specific connection to be made, so this analogy is neither very precise nor very useful.

Usually we assign a Gibbs free energy to a whole reaction, but here there is an aspect of this ϵ paremeter, so how does this idea of parameterizing chemical reactions relate to the regular values of Gibbs free energy we see in textbooks? The professor equates bracketed term to ΔG but the connection is not a 100% clear to me. Perhaps an explicit explanation would help.

This question is really asking about $\xi$ rather than $\epsilon$. In the notes you link, the equation $\Delta G = \Delta G^\circ + RT\ln Q$ is derived by setting $\epsilon = 1$, which does not seem appropriate to me. Certainly the final result is correct, but there are clearer and more mathematically responsible derivations that you should be able to find in any standard physical chemistry textbook.

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    $\begingroup$ I wrote one of the other answers, and failed to clarify this. Good answer and nice catch of the misconception the OP is struggling with. $\endgroup$ Aug 17 '21 at 21:00
  • $\begingroup$ Could you suggest a source which gives a better derivation? $\endgroup$
    – Buraian
    Aug 18 '21 at 4:19
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    $\begingroup$ @Buraian, maybe Atkins' or McQuarrie's Physical Chemistry? $\endgroup$ Aug 18 '21 at 12:21
  • $\begingroup$ I had tried Atkins but in the edition I had, I was more confused because they used the other epsilon like looking symbol in it and I thought it was the same. I'll check it out again when I get time $\endgroup$
    – Buraian
    Aug 18 '21 at 12:22
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Is the pressure ratio term on the right hand side dependent on $\epsilon$? Like I think that as the reaction proceeds, would each pressure term fluctuate?

You choose to make $\epsilon$ very small so you can neglect changes in partial pressures, concentration, or activity.

How does one correctly interpret the final equation ?

The Gibbs energy is an extensive quantity. The more reacts, the bigger the change in Gibbs energy (e.g. you can do more work with a D battery compare to a AAA battery, even though they have the same voltage).

Usually we assign a Gibbs free energy to a whole reaction, but here there is an aspect of this $\epsilon$ paremeter, so how does this idea of parameterizing chemical reactions relate to the regular values of Gibbs free energy we see in textbooks?

For the Gibbs energy, you can state the change at a given set of concentrations, but you can't just assign a Gibbs energy to the entire reaction (you would have to integrate over the extent of reaction from initial to final state). In contrast for enthalpy, it is trivial to do the analogous calculation because the enthalpy is not concentration-dependent in the first approximation. It is clear that the Gibbs energy changes with concentrations, otherwise reactions would not attain equilibrium from either side of the equilibrium (e.g. starting with all reactants or all products).

The next step in this derivation would be to define the intrinsic quantity $\Delta_r G$, the Gibbs energy of reaction, as

$$ \Delta_r G = \frac{dG}{d\epsilon}$$

In textbooks, it is more common to use $\xi$ (xi) as symbol of the extent of reaction, i.e. how the concentrations changed. $\Delta_r G$ is dependent on concentration, but in the limit of an infinitesimal change in $\xi$, you can neglect the changes in concentration and just use the initial concentrations (or activities or partial pressures or fugacities).

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  • $\begingroup$ "You choose to make ϵ very small so you can neglect changes in partial pressures, concentration, or activity." I have a hard time believing this, because the professor showed how varying epsilon runs us through the whole gibbs free energy curve for the reaction in the video $\endgroup$
    – Buraian
    Aug 8 '21 at 15:24
  • $\begingroup$ You keep epsilon small and use different sets of partial pressures along the reaction path to explore the slope of Gibbs energy change at different stages in the reaction. @Buraian $\endgroup$ Aug 8 '21 at 19:31
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I can see what the professor is doing here.

First of all, he implicitly asks us to accept the idea that the free energy of an ideal gas mixture that is not at chemical equilibrium can still be calculated for the mixture using the usual expression for a gas mixture at equilibrium: $$G=n_a\mu_a+n_b\mu_b+n_c\mu_c+n_d\mu_d$$ I have no problem with his doing this if you don't.

He then looks at the effect of a small change in reaction conversion $d\epsilon$, such that $$dn_a=-\nu_ad\epsilon$$$$dn_b=-\nu_bd\epsilon$$$$dn_c=+\nu_cd\epsilon$$$$dn_d=+\nu_dd\epsilon$$This causes a change in G given by $$dG=-\mu_a\nu_ad\epsilon-\mu_b\nu_bd\epsilon+\mu_c\nu_cd\epsilon+\mu_d\nu_dd\epsilon=d\epsilon[-\mu_a\nu_a-\mu_b\nu_b+\mu_c\nu_c+\mu_d\nu_d]$$ The rest is straightforward algebra.

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  • $\begingroup$ How would you explain the point about pressure fluctuations? Would it be the same as @Karsten Theis $\endgroup$
    – Buraian
    Aug 9 '21 at 14:16
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    $\begingroup$ Not exactly. You are aware that, because G is an extensive property, the changes in the partial molar quantities such as chemical potentials must satisfy the Clausius Duhem equation, right? That would allow us to write the change in G without including terms for the changes in chemical potential (which would ordinarily follow from the product rule for differentiation). $\endgroup$ Aug 9 '21 at 14:29
  • $\begingroup$ You can work it out algebraically, and you will find that those pressure variation changes (as a result of changes in mole fraction) cancel out. $\endgroup$ Aug 9 '21 at 14:35
  • $\begingroup$ Clausius Duhem, I checked it up online, seems to be a result relating to Chemical Engineering and the math back ground for understanding the derivation directly seems a bit out of my reach. The algebra is indeed simple as you said, but the problem for me is fully comprehending the premise which the derivation begins with. Oh well, this turned out more complicated than I had initially expected. $\endgroup$
    – Buraian
    Aug 9 '21 at 14:36
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    $\begingroup$ Like I said, those cancel out as a result of the Clausius Duhem equation. Do the algebra, and you will see this for this particular problem... $\endgroup$ Aug 9 '21 at 16:19

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