# Why are the conditions of ΔG° partial pressure fixed at 1bar

1. Jul 12, 2015

### sgstudent

Shouldn't the ΔG° partial pressure of the components be based on the K? Where ΔG°=-RTlnK such that K is based on the partial pressures of the gas involved?

If we were to set the partial pressure to be 1bar each then every reaction having the same stoichiometric proportion of reactants and products would also have the same K value which isn't true.

2. Jul 14, 2015

### DrDu

The full equation for the Gibbs enthalpy change is $\Delta G(,T, P, {P_i} )=\Delta G^\circ(T, P, {P_i^\circ=1 bar}) +RT \ln K$ where $K=\prod_i (P_i/P_i^\circ)^{\nu_i}$.
You can learn two things from this equation: 1. In equilibrium $\Delta G=0$ and you get your equation which links K to the standard enthalpy.
2. if all partial pressures equal the standard pressure $P^\circ$=1 bar, then K=1 and $\Delta G=\Delta G^\circ$, i.e. $\Delta G^\circ$ is the Gibbs enthalpy change when all components are under their standard pressure of 1 bar.

3. Jul 14, 2015

### Staff: Mentor

I would add to DrDu's response that $ΔG^0$ is easiest to think of as going from the pure reactants at 1 atm pressure and temperature T, and going to the pure products also at 1 atm pressure and temperature T. So you don't even have to think of $ΔG^0$ in terms of partial pressures, only in terms of pure species total pressures. This applies particularly well when you are dealing with non-ideal systems.

Chet

4. Jul 16, 2015

### sgstudent

Hmm if we had a reaction A↔B+C would the ΔG° be based on 1 bar of pure reactants to 1 bar of B and 1 bar of C?
If so shouldn't the Q here be 1x1/0? Since there aren't any products left? Why would the Q be 1x1/1?

5. Jul 16, 2015

### DrDu

$\Delta G= \sum_i \mu _i \nu_i$ where $\nu_i$ are the stochiometric coefficients (1 for B and C and -1 for A in your example) and $\mu_i$ are the chemical potentials of the components under standard conditions. For a gas $\mu_i (p^\circ) =G_i(p^\circ)/n_i$, i.e. the free enthalpy of one mole of the pure gas under standard pressure. That you calculate $\Delta G^\circ$ at standard pressure, does not mean that you have to take Q at standard pressure.
Just to make this clear: The "Delta" in $\Delta G$ does not imply a difference between initial and final states.

6. Jul 16, 2015

### Staff: Mentor

Just to be clear, ΔG0(T) means the change in Gibbs free energy in going from equilibrium state 1 to equilibrium state 2, where

Equilibrium State 1: Pure reactants in separate containers, each at temperature T and 1 atm pressure
Equilibrium State 2: Pure products in separate containers, each at temperature T and 1 atm pressure

So there is no Q associated with ΔG0, but, if you insist on having a Q value, what is the natural log of 1?

To get a better understanding of all this, it is very very helpful to study the Van't Hopf equilibrium box. Here you have a box with an equilibrium mixture of reactants and products in the box. You take pure reactants in separate containers at 1 atm pressure and isothermally reversibly drop the pressure of each to its partial pressure in the equilibrium box. You then inject the reactants into the equilibrium box in stoichiometric proportions, each through a separate semi-permeable membrane, while, at the same time removing the products in stoichiometric proportions at their partial pressures into separate containers through separate semi-permeable membranes. You then reversibly increase the pressures of each of the products to 1 atm pressure. The net effect of this process is convert stoichiometric quantites of pure reactants in equilibrium state 1 to stoichiometric quantities of pure products in equilibrium state 2.

Chet

7. Jul 16, 2015

### DrDu

While this is perfectly feasible to define e.g. $\Delta U$ or $\Delta H$ I don't think this is correct for $\Delta G$. The problem is that in state 1, the pressure of the products is zero and the log of 0 is infinite. The same holds true for the reactands in state 2.

8. Jul 16, 2015

### Staff: Mentor

Yes, but, in state 1, the number of moles of the products (in their separate containers) is zero, and, in state 2, the number of moles of the reactants (in their separate containers) is zero. So the free energy of the products is zero. Anyway, you can still take the pressures of the products as 1 atm even though they are not present.... basically, there are negligible quantities of the products at 1 atm.

You certainly can't be suggesting that if I have reactants in separate containers at 1 atm pressure, the free energy of such a system is infinite. Of course, it's not; in this state, the free energy of the system is just equal to the sum of the free energies of formation of the reactants. And, in state 2, the free energy of the system is just equal to the sum of the free energies of formation of the products.

Chet

9. Jul 16, 2015

### DrDu

Ok, thinking about it, you are completely right.

10. Jul 17, 2015

### sgstudent

I was a little confused when I read the first quote. If I were to set a Q for it it would be based on the pressures of the products and reactants at that point. But since by definition its the conversion of the reactants to products at 1 atm it doesn't really have a Q. Is this what you mean by Q not associating with ΔG°?

But after reading the second quote it sort of makes sense for it to have a Q? If we were to assume the reactant proceeding then initially we would have 1 atm of reactants in separate containers and negligible amounts of 1 atm of products in separate containers as well. And so the Q=1. Would this be the correct way to think about this?

Lastly, when using the this equation ΔG=ΔG°+RTlnQ wouldn't the ΔG° be different if the Q were in terms of Qp or Qc where Qp is based on the pressure ratio and Qc on the concentration ratio?

Thanks for the help

11. Jul 17, 2015

### DrDu

If you look at my post #2 you can see that K (or Q) depends on both $P_i$ and $P^\circ_i$. In fact the whole equation is not conceptually different from e.g. Gay Lussac's law for an ideal gas: $\frac{T (P_1)}{T(P_2)}=\frac {P_1}{P_2}$

12. Jul 17, 2015

### Staff: Mentor

Yes. Pretty much so.
There are more than one way of thinking about this. It depends on whether you are willing to accept the idea that the free energy of a mixture can be defined even if the mixture in not in chemical equilibrium. Many people in Physics Forums would not accept this idea. I for one do accept it. In such cases, you can, for any set of partial pressures of the gases, calculate a free energy for the mixture. So if you start off with the mixture in an initial non-equilibrium state, you can then track how the free energy of the mixture changes as the reaction proceeds towards equilibrium. Equilibrium will be achieved when the calculated free energy passes through a minimum. But, even though there is a ΔG involved, this is not the ΔG we are talking about. In this situation, if you solve for the minimum by setting the derivative of the free energy to zero with respect to the conversion, you will find that the condition for equilibrium to be obtained is when RTlnQ=-ΔG0, where the ΔG0 here is the one described in post #6. Try doing this derivation, starting, say, from pure reactants and see what you get.

I feel that it is simpler to understand the basics here by going from initial states where you have only reactants present in separate containers and going to final states in which you have only products present in separate containers. I think that this makes the analysis much cleaner. To bring about the reversible transition from the initial state to the final state in such situations, you need to have something like a van't hoff equilibrium cell to carry out the reactions in such a way that the partial pressures of the species within the cell do not change while you add reactants through semipermeable membranes, and remove products through semipermeable membranes. The key to such a set up is that, if the partial pressures of the reactants being injected into the cell through the semipermeable membranes and the partial pressures of the products being removed from the cell through the semipermeable membranes are equal to the pressures of the corresponding pure gases on the other side of the membrane, then the changes in their free energies will be zero. It would be very helpful if you could read more about the van't hoff equilibrium cell. In any event, using the conceptual model of the van't hoff equilibrium cell, you can determine for any arbitrary values of the pressures of the pure reactants and the pure products, the change in free energy from the initial to the final state (assuming stoichiometric conversion). If the pressures of the pure reactants and pure products happen to match those present in an equilibrium mixture, the ΔG that you would determine in this way would be zero. If the pressures of the pure reactants and pure products happen to all be 1 atm, the ΔG that you would determine in this way would be ΔG0. For arbitrary pressures of the pure reactants and products, the difference between the ΔG and the ΔG0 would be obtained by expanding or compressing the reactants and products reversibly between the actual pressure and 1 atm, using reversible compressors or expanders (e.g., turbines). This is where the Q would come in.
The ΔG0 between the states cannot be different, because it is determined only by the states and not by how you calculate things. However, if you are going to use Qc instead of Qp, then you need to include a factor in the Q term for the conversion from partial pressure to concentration, since p = cRT. That factor will not be present if there is no change in the number of moles in the reaction, but it will be present otherwise.

Chet

13. Jul 18, 2015

### sgstudent

Does this mean whenever i have a Qp I would have to convert it to a Qc by using the conversion Kp=Kc(RT)^n or Qp=Qc(RT)^n? The equation wouldn't work if I simply substituted Qp into the RTlnQ?

Hmm we divide by P° so if we used the units atm we would be dividing it by 1. By still wouldn't that value be different if concentrations were used? Because if were were to use Ci/C° as Qc where the units of C is M then wouldn't C° be 1M?

14. Jul 18, 2015

### DrDu

Of course

15. Jul 18, 2015

### sgstudent

Sorry do you mean of course to the first statement where we converted Qp to Qc for the RTlnQ equation?

16. Jul 18, 2015

### Staff: Mentor

The Q in the equation ΔG=ΔG0+RTlnQ is Qp. If you want to work in terms of concentrations, then you need to substitute the conversion from Qp to Qc.

Chet[/QUOTE]

17. Jul 18, 2015

### sgstudent

[/QUOTE]

But wouldn't the conversion from Qp to Qc change the value of the Q term? And as a result change the ΔG0 term?

Or am I misinterpreting this wrongly?

18. Jul 18, 2015

### Staff: Mentor

It's a math thing. The conversion from Qp to Qc would involve inclusion of the (RT)n factor that you had in your previous post. That would not affect ΔG0, because it would be in the ln term. Even if you then split the log term into the sum of the logs of the product, it still wouldn't affect ΔG0. Don't forget that dG at constant temperature is Vdp = RTlnp. So G is directly associated with p.

Chet

19. Jul 18, 2015

### DrDu

Also, the change of the standard state from 1 bar to 1 mol/l affects both $\Delta G^\circ$ and also K, as it depends then on $c_i/c_i^\circ$ instead of $P_i/P^\circ$.

20. Jul 18, 2015

### Staff: Mentor

Interesting. I've never heard of 1 mole/liter being taken as the standard state for G of a gas. Is there a reference on this? Even for pure liquids, the standard state is usually taken as the 1 atm. pressure.

Chet