# Questions regarding Gibbs energy changes

1. Apr 21, 2015

### davidbenari

In class we have been deducing the typical equations for chemical equilibrium. What bothers me was the term $\Delta Gº$ as opposed to $\Delta G$.

$\Delta G$ is fine with me because it takes into account the fact that during a reaction the multiple components find themselves in a mixture, and therefore their chemical potentials are different than $\mu º$. What bothers me about $\Delta Gº$ is that it doesn't seem that a "standard state reaction" is taking into consideration the mixing aspect.

$\Delta Gº=\sum \nu_p \mu_pº - \sum \nu_r \mu_rº$. With this difference I associate the following picture:

We have all components in different containers at first, mix them, get some products, and separate the products in different containers. $\Delta Gº$ is the difference between having products separated, and having reactants separated.

But it seems that this isn't what is implied by $\Delta Gº$. What is it really? Why is it useful?

I have one final question. In a mixture each component has its chemical potential $\mu$. Why is it valid to say that if I wanted to construct a system with $n_i$ moles of each component, I could just say that the Gibbs energy of the system is $G= \sum_i n_i \mu_i$. I know that the temptation might come because $\mu_i$ has units $J/mol$. that doesn't mean I can just multiply out moles. I have to integrate. Whats going on here?

You might say that were keeping relative concentration fixed, and integrating yields this result. But consider that at first we have an empty container and thus absolutely no relative concentrations.

You might say we establish this relative concentration after $dn_i$ of each component has been added. Sure okay, but why would the $\mu$ have the same value we are thinking about at first if such a small quantity has been added to the container and therefore we have a completely weird environment where each component is surrounded by an infinite volume. $\mu$ would surely change in such an environment?

Thanks and sorry for the lengthy question.

2. Apr 21, 2015

### Staff: Mentor

I had the same problems are you are having understanding this, until I became familiar with the van't hoff equilibrium box. Are you familiar with the van't hoff equilibrium box? If not, it will do you lots of good to learn about it. What you will find is that you don't mix the reactants first. First you either compress them or expand them from their initial pressure of 1 atm. until their free energies are equal to their partial molar free energies in the reaction mixture. At this point they are still in separate containers. I'm going to stop here and give you a chance to look over van't hoff equilibrium box.

Chet

3. Apr 22, 2015

### DrDu

There is also a mathematical explanation, which involves Euler's theorem on homogeneous functions:
Take in mind that G depends on T, P, and the $\{n_i\}$ and is itself an extensive function, while T and P are intensive. Increasing the size of the system can be done by scaling the n's (which are extensive quantities): $\{n_i\}\to \{\alpha n_i\}$. As G is extensive we have $G(T,P, \{\alpha n_i\})=\alpha G(T,P,\{n_i\})$. Differentiating the left side we get $dG/d\alpha =\sum_i (\partial G/\partial n_i) n_i=\sum_i \mu_i n_i$ while differentiating the right hand side simply get $dG/d\alpha=G(T,P,\{n_i\})$ which proves the statement.

Last edited: Apr 22, 2015
4. Apr 22, 2015

### Staff: Mentor

This is also worked out using a slightly different approach in Smith and Van Ness, which I recommended to David in recent post a few days ago (I think).

Chet

5. Apr 24, 2015

### davidbenari

I'm sorry it has taken me so long. I've just read about these boxes. The version I read was where there were pistons on one side with substance A at a concentration equal to the reaction mixture, and pistons on the other side with nothing at all. The pistons on one side were compressed slowly, and then the other ones were pulled out slowly. The work performed was calculated this way by checking the difference in free energy of the initial and final states.

**Note that the pistons on the other side were permeable to substance B only.

I still have the same doubts as before though (Except for the one which was proved with Eulers theorem). Any suggestions on what I should think about?

Thanks.

Last edited: Apr 24, 2015
6. Apr 24, 2015

### davidbenari

I guess my main doubt that still holds is the one were I'm concerned about Gº. How can this be the reaction change if its mathematical form is treating the components taken separately?

7. Apr 24, 2015

### Staff: Mentor

OK. I think I can help.

Let's confine attention to the case of reaction involving ideal gases. Why? If we can't do it for that case, we'll never be able to figure it out for non-ideal cases.

The equilibrium box is a chemical reactor in which all the reactants and products are in chemical equilibrium. Each reactant or product has a particular partial pressure, and the total pressure in the box is equal to the sum of the partial pressures. So no net reaction is taking place. The temperature is held constant at the reaction temperature using a constant temperature bath.

Now suppose we have the reactants in separate cylinders initially at 1 atm. Suppose that we expand or compress the gas in each cylinder of reactant isothermally to a pressure equal to the partial pressure of that reactant in the equilibrium box. We can calculate the change in free energy for each of these reactants by taking the integral of dG = VdP. When the pressures of the pure reactants match the partial pressures of the reactants in the equilibrium box, then their molar free energy is equal to the partial molar free energy (chemical potential) of the reactants in the equilibrium box.

The equilibrium box also has individual semi-permeable membranes through which each individual reactant can be injected into the box. Each reactant has its own semipermeable membrane. When the gas in a reactant cylinder is put into contact with the equilibrium box through its individual semi-permeable membrane, the pressure of the reactant species within the cylinder will be in equilibrium with the partial pressure of the same reactant in the equilibrium box, and there will be no tendency for the species to flow either into or out of the equilibrium box. Suppose we were to inject the reactant species in stoichiometric proportions very slowly through their individual semi-permeable membranes into the equilibrium box, while at the same time removing the product species through their individual semi-permeable membranes into individual cylinders at the same pressures as the partial pressures of the products in the equilibrium box. The change in free energy for this injection/removal operation would be zero, since the ΔG at equilibrium is zero. The concentrations of reactants and products in the equilibrium box has not changed, because stoichiometric proportions were injected, and stoichiometric proportions were removed.

Now we have products in their individual cylinders at the same pressures as the partial pressures of the reaction products in the equilibrium box. We next take the gases in each of the cylinders and compress or expand them reversibly to 1 atm. Here again, the change in free energy for each of the gases can be determined by using dG = VdP.

So the net effect of all this is to take the reactants at 1 atm in separate containers and produce the products in separate containers at 1 atm. The free energy change for this overall process is what we call ΔG0. Since the change in free energy for the injection/removal operation in the equilibrium box is zero, the ΔG0 must be equal to the changes in free energy required to change the pressures of the reactants and products from 1 atm to the partial pressures in the equilibrium box. That's how we determine the equation for the equilibrium constant.

I hope this makes sense.

Chet

8. Apr 24, 2015

### davidbenari

Thanks a lot Chet. I have one question though:

Many chemical reactions don't happen this way. For example, I don't see how $\Delta Gº$ for a chemical reaction happening inside a living cell is a good parameter for the free energy change: things aren't in their standard states in a cell. I don't understand how the equilibrium box is a good analogy for anything other than a very organised (maybe industrial) process?

Maybe $\Delta G$ refers to the precise system (e.g. what's happening inside the container at some moment in time), and $\Delta Gº$ refers to the process of wanting to produce certain products from certain reactants, and having to use something similar to the equilibrium box because thats the only way stuff happens?

Thanks again.

9. Apr 24, 2015

### Staff: Mentor

I don't quite get what you are saying here. The objective is to get the equilibrium constant, not necessarily the free energy change. This just gives you a well-defined structured method of determining the equilibrium constant in advance, without doing any experiments. You are aware the that $\Delta Gº$ for a reaction can be determined from tabulated values of the free energies of formation of the substances participating in the reaction, correct? What you are really interested in here is the relationship at equilibrium between the partial pressures of the substances that are in the equilibrium box (or anywhere else) where they are not in their standard states, but are in equilibrium. To get this, the only data you need are the free energies of formation of the different species.
$\Delta G$ can refer to what's happening inside the container at some moment of time (when the reaction is not yet at equilibrium) only if you accept the concept that it is possible to determine $\Delta G$ for a system that is not at equilibrium. Many people, including myself, accept this concept. Otherwise, $\Delta G$ just refers to the change in free energy from an initial thermodynamic equilibrium state in which the pure reactants are in separate containers to a final equilibrium state in which the pure products are in separate containers, all at pressures differing from 1 atm.

Chet

10. Apr 26, 2015

### davidbenari

What I mean is that I'm sometimes told to calculate the "Gibbs energy change for a reaction", and this is assumed to be $\Delta Gº$. I don't get this since, this parameter is calculating free energy changes when the components are separate, which seems to me very impractical.

Also, I've seen that during the derivations $Pº$ is assumed to be unchanging. This doesn't imply that the pressure of the system is unchanged right? But if the pressure is changing then $G$ shouldn't be used at all, because its useful to us when $T,P$ are held fixed.

11. Apr 26, 2015

### DrDu

2. The change of Delta G with pressure and temperature can be calculated from the molar volumes and reaction enthalpy, respectively, so you can take temperature and pressure into account easily.
1. Components are not taken separately, but in their standard states. For solutions, this is a solution containing 1 mol/ l (well, simplified). In equilibrium Delta G =0, hence you get from $\Delta G= \Delta G^0+RT \ln K$ that $\ln K=-\Delta G^0/RT$, i.e. the standard free energy is related to the equilibrium constant. That's why it is so important.

12. Apr 26, 2015

### davidbenari

I thought $\mu_Aº$ would be the chemical potential of pure A at some T,P. If they're not separate, then in any case they are not in the environment of the mixture, which is what matters isn't it?

A new question emerged though: Why am I allowed to use $Keq$ in the usual way its calculated when you haven't started with moles that are prescribed by the stoichiometric coefficients? E.g. You start with .5 moles in a reaction of A --> B.

Is this because the partial pressures are assumed to be the same whatever the quantity of matter? Like "equilibrium is equilibrium and will look the same if your system is big or small" or something like that?

Thanks.

13. Apr 26, 2015

### Staff: Mentor

The injection and removal from the equilibrium box through the semi-permeable membranes takes care of all the mixing, and and the reaction takes place within the equilibrium box. Like I said, determining $\Delta Gº$ is just a convenient way of determining the equilibrium constant. It is not used directly in actual equilibrium calculations in systems where the chemical species are intimately mixed, but the equilibrium constant determined from it is what applies to any equilibrium mixture, even at a different total pressure.
Right. The constant pressure only applies to the derivation of the equilibrium constant, using the equilibrium box as a simple vehicle for determining K. However, the value obtained for K from this calculation applies to a system at any arbitrary pressure. You need to get an appreciation of the cleverness of what they have done here.

As I said, for a system at any arbitrary value of P, the equilibrium constant determined from the equilibrium box derivation applies. This is very powerful stuff.

Chet

14. Apr 26, 2015

### Staff: Mentor

Yes, David. You are definitely correct about this. The derivation of $\Delta G^0$ starts with the pure reactants at 1 atm pressure and ends with the pure products at 1 atm pressure.

Chet

15. Apr 26, 2015

### Staff: Mentor

Where did I say that they are present in stoichiometric proportions within the equilibrium box? They should not necessarily be considered to be in stoichiometric proportions within the equilibrium box.

Chet

16. Apr 26, 2015

### Staff: Mentor

I should also mention that the total pressure inside the equilibrium box does not have to be 1 atm. So, within the equilibrium box, we truly can have an arbitrary equilibrium mixture at any pressure and any corresponding equilibrium composition. This mixture is representative of what we could have in any reactor or vessel at equilibrium.

Chet

17. Apr 26, 2015

### davidbenari

The reaction will occur in stoichiometric proportions and the actual number of moles participating in the reaction aren't necessarily the standard stoichiometric coefficients, which are used in the derivation of the $K_{eq}$. In my example of A-->B, if I start out with .5 of A then why am I allowed to use $K_{eq}$?

I've done the derivation for other mole quantities, but in the same proportions and have gotten that: $\Delta G =x \Delta Gº + RT Ln K^x$ where $x$ is the number that would multiply the entire chemical equation.

Now $K$ is the normal constant $\frac{ \prod (\frac{P_{products}}{Pº})^\nu_{products}}{\prod \Big(\frac{P_{reactants}}{Pº}\Big)^\nu_{reactants}}$. Which is equal to the one where the standard chemical coefficient are reacting only if the components are exerting the same partial pressures, which I guess is the case.

I have a feeling I'll be unclear in this paragraph, so please tell me so I can edit it, if that's the case.

note: By standard stoichiometric coefficient I am referring to the lowest possible integer valued numbers used.

Thanks.

Last edited: Apr 26, 2015
18. Apr 26, 2015

### Staff: Mentor

You're right. I'm pretty confused by this post. Maybe you can define a specific problem (precisely) and we can work our way through it. Please try to be precise about the situation we are considering. Let's assume that the reaction is aA+bB--->cC+dD, and we are dealing with ideal gas.

Chet

19. Apr 26, 2015

### davidbenari

Ok, so one problem (Which I've gotten out of my textbook) is :

-- In a 10L container 0.5 mols of H2 react with 0.5 moles of I2 at 448ºC.

H2(g)+I2(g) <--> 2HI(g)

If $K_{eq}=50$ calculate HI's partial pressure. ---

My specific doubt here is that one might want to construct $K_{eq}$ as $\frac{ (\frac{P_{HI}}{Pº})^2}{(\frac{P_{H2}}{Pº})(\frac{P_{I2}}{Pº})}$ because of the derivation of the equilibrium constant. The problem I have with this is that for me the exponents make no sense because we don't have ,e.g., 2 moles of HI. The value of $K_{eq}$ we were given is one that is considering a reaction where the moles participating are those prescribed by the coefficients.

After this I know I can manipulate the equation to yield $\frac{x_{hi}^2}{x_{H2}x_{I2}}=50$; use the ICE method, or whatever, and find the partial pressure.

But my specific doubt is with accepting $K{eq}$ for any number of moles participating. What's going on? If I take my consideration of the number of moles participating then you'll get the quotient I defined in the previous post with the factor of $x$, which was the number that multiplied a whole chemical equation such that the stoichiometric proportions are unaltered.

20. Apr 26, 2015

### davidbenari

I've seen the derivation starting from $\sum \nu_p(\mu_p +RT LnP_p) - \sum \nu_r(\mu_r + RT LnP_r)$ where $\nu$ indicates a stoichiometric coefficient. I interpret those coefficients as the actual moles participating. Maybe thats the source of my confusion.