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I know that Gibbs free energy change represents the amount of the non-expansionary work that a reaction is capable of doing but what happens to this energy at equilibrium? Why is the system unable to do work? Why is the Gibbs free energy change equal to zero at equilibrium?

Edit: Please don't simply write the derivation that we use to conclude that ∆$G$ must be zero at equilibrium. Please try to explain the physical significance behind this fact.

Please also note that I have already searched this website for an answer to my question, found only the following question but the answers only explained the Math:

Gibbs free energy-zero or minimum

Thanks ever so much in advance :) Regards.

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  • $\begingroup$ What exactly are you looking for when you say "physical significance"? $\endgroup$ – getafix Aug 29 '16 at 4:09
  • $\begingroup$ At equilibrium, exactly what happens to the energy that had once been present, allowing the reaction to do work? What is the meaning of ∆$G$=0? $\endgroup$ – user33789 Aug 29 '16 at 4:12
  • $\begingroup$ It means the chemical potential of the reactants and products is the same $\endgroup$ – getafix Aug 29 '16 at 4:14
  • $\begingroup$ Can you please elaborate that point? I don't quite understand fully. $\endgroup$ – user33789 Aug 29 '16 at 4:15
  • $\begingroup$ You would have to allow the use of some mathematics then $\endgroup$ – getafix Aug 29 '16 at 4:23
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I detect some inaccuracy in for question. I am going to try to clarify a few things since I perceive genuine curiosity in your question. The answer can not be so short.

First some clarifications:

Why is the Gibbs free energy change equal to zero at equilibrium?

The very common phrase free energy change at equilibrium is inadequate. In thermodynamics the state of a simple system is specified with state parameters. In the matter at hand temperature ($T$) and pressure ($p$). As the system would be described by $T$ and $p$, any "change" that does not involve changes in $T$ or $p$ is outside of the scope of the theoretical treatment. That is, to say the system is at equilibrium is accurate within a theoretical framework, so change should refers to changes within the same framework. Of course they could be important in uncommon cases but they are not included in the model. So, let use the $^*$ symbol to refer to an equilibrium state. At equilibium $G = G(T^*,p^*)$. Any change that do not involve $T$ and $p$ will let the system in the equilibrium state (although in others way the system can look not the same). So, its change is zero just because is the same function with the same values for its arguments.

At equilibrium the system is in an stationary state. That is, the system (thermodynamic!) state do not change with time. So, if you inspect how $T$ and $p$ evolve in time they do not change anymore. So any spontaneous change won't change the value of $G$. To get a change of $G$ you must make a perturbation on the system by making a change in the environment or the walls that divide them up, but once it is done the system is not in equilibrium state anymore (although it can reach other equilibrium state in the future).

That is why at equilibrium there are not useful work available from the system, because it is in a stationary state in harmony with the environment. So changes won't take place. Notice that to get work from the system it is needed some kind of macroscopic movement (according to the definition of mechanical work).

You said:

I know that Gibbs free energy change represents the amount of the non-expansionary work that a reaction is capable...

It is not exactly true, it is not a requirement that the process involve a reaction.

In the ligh of getafix commentary:

Take in mind that physical significance most times, in the core, is just an inaccurate idea about a physical theory/model/law. Although it may be useful sometimes from a practical perspective (and many times leads to catastrophic conclusions), it has not an add value. There is the nature. There are good formal descriptions of it obtained through years of hard work from experienced people. There are pictorial interpretations of those formal descriptions used when formalism turns too hard/complex (physical significance). I noticed that it is fashion to say that one get the true understanding of phenomena when one get the physical significance. I think that one truly understand the phenomena when feels natural the formal description and is also aware that it is just a description.

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I'm really not getting why that post didn't answer OP's query.

Nevertheless, I'm posting the following argument as an answer.


Why is the Gibbs free energy change equal to zero at equilibrium?

Well, it's not 'Gibbs free energy change' $\Delta G$ but rather Reaction Gibbs Energy $\Delta _r G$ which is zero at equilibrium, if written accurately.

Reaction Gibbs Energy is defined as the rate of change of Gibbs Free Energy per unit change in composition i.e.,

$$\Delta_r G =\frac{\Delta G}{\Delta n} \tag 1$$

where $\Delta n$ is the change in composition of the system.

Now, OP must be aware of the first derivative test for finding local minima; that means for reaching the equilibrium position, the rate of change of Gibbs Free Energy per unit composition must be zero and the rate must change from negative in the neighbourhood of the extrema point of Gibbs Free Energy vs. Composition graph viz.

$$\Delta_r G= 0\;.$$

Now, question might arise as that of the physical implication of association of equilibrium position with $\Delta_r G= 0\;.$

The definition of equilibrium position sheds light on this:

A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain 'spontaneous in neither direction' .

This happens when the reaction Gibbs energy becomes zero viz. $\Delta_rG=0.$

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  • $\begingroup$ First of all, that post didn't answer my query because I was looking to get an intuitive feel for this concept and not the mathematical description. What is the definition of ∆$n$ for a system? Yes, I'm aware if the first derivative test, but I'm afraid I didn't understand much of what came after that in your answer. Okay, it's spontaneous in neither direction, but what does all this have to do with energy? Please try to understand what I am asking; the term is "Reaction Gibbs energy" (please elaborate on that too) so what does it have to do with energy and why is this energy becoming zero? $\endgroup$ – user33789 Aug 29 '16 at 4:42
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    $\begingroup$ @KaumudiHarikumar: The 'intuitive feel' of the query lies in that very phrase 'spontaneous in neither direction'. Do you know what spontaneous means? Try to correlate it with the Second Law in the light of Gibbs Free Energy. $\endgroup$ – user5764 Aug 29 '16 at 4:45
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I don't think I am putting out something truly novel here either; the answer by @MAFIA36790 and the discussion in the link you yourself post it seem quite satisfactory to me, yet I shall try to put a different spin on things so to speak.

Let me define the chemical potential, as the partial molar Gibbs energy of species "$j$":

$$\mu_j = \left ( \frac{\partial G}{\partial n_j} \right)_{p,T,n'}$$

and for a system of components A, B, etc. the fundamental equation of chemical thermodynamics gives us:

$$\mathrm dG= V~\mathrm dp-S ~\mathrm dT+\Sigma \mu_j~\mathrm dn_j$$

Now, I consider a simple equilibria $$R \rightleftharpoons P $$

and additionally, I introduce a parameter " $\xi$ " called the extent of the reaction (it has the dimensions of amount of substance)

for macroscopic changes, the amount of a component $j$ changes by $\nu_j\Delta\xi$ ($\nu_j$ are the "signed" stoichiometric coefficients; positive for products, and negative for reactants)

Now, if we defined $$\Delta _rG = \left ( \frac{\partial G}{\partial \xi} \right)_{p,T} $$

This $\Delta _r G$ still holds the meaning you might already be familiar with, namely:

$$\Delta _r G = \Delta _f G\textrm{ (products)} - \Delta _f G\textrm{ (reactants)}$$

and I have nearly extended it to mean the derivative defined above so that i can relate it to the extent of a reaction (measured in amounts of substance consumed/produced)

and at constant temperature and pressure, from the second equation in this post:

$$ \mathrm dG = (\mu_P - \mu_R)~ \mathrm d\xi$$

so, $$ \Delta _rG = \left ( \frac{\partial G}{\partial \xi} \right)_{p,T} = (\mu_P - \mu_R)$$

Now, as already mentioned, if we were to plot the reaction Gibbs energy as a function of the extent of the reaction, a minimum on the plot would correspond to the first derivative defined above being equal to zero. (or, equivalently in the chemical potential of reactants and products being exactly equal). This is the condition for equilibrium.

Perhaps, stooping down to layman terminology (caution), one can say $\Delta _r G $ is like a "force" that drives a reaction forward, and the system is in equilibrium when this driving force is zero (and moves in the reverse direction when it changes sign).

An appropriate mechanical analogy could perhaps be two weights on either end of a string wrapped across a pulley. If the heavier weight is higher, it would descend raising the lower weight. The system would come to rest when they are in balance.

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  • $\begingroup$ Thank you, but I have only just graduated high school and I'm not aware of some of the terms that you(and others in the post that I have linked in my question)have used. Nevertheless, I will try to understand what you have said as best I can. At my level, we aren't introduced to many important concepts, which poses a problem for understanding clearly. One last doubt; how can you use the analogy of a seesaw like static mechanism to help explain a dynamic equilibrium? When you say that the driving force is zero, I don't understand this because like I just said, isn't equilibrium dynamic? $\endgroup$ – user33789 Aug 29 '16 at 5:26
  • $\begingroup$ It's not a perfect analogy, I admit. Which is have included (a literal) word of caution. If you want to improve the mechanical analogy, then imagine perturbing it equilibrium perhaps? Also, maybe at your level then you could just take Gibbs energy = 0 as a definition of equilibrium? $\endgroup$ – getafix Aug 29 '16 at 5:29
  • $\begingroup$ Yeah, well, this is the part that is very annoying; when I try to understand more but a barrier of some advanced science props up. I only wanted to know why "the ability of the system to do non-expansional work"(for this is how it is defined for us)reduces to zero at equilibrium and in neither of your answers have I found anything useful. Sigh. I guess that I will have to accept the fact that there is no way for me to understand it without understanding the advanced terms and must resort to not knowing for the time being :( $\endgroup$ – user33789 Aug 29 '16 at 5:58

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