Is $ Q $ dependent on the ϵ which is chosen in the expression for Gibbs free energy of a chemical reaction?

Question

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.

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*: 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.

Answer 1

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.

Answer 2

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).

Answer 3

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.