November 30, 2020, 03:43:44 AM
Forum Rules: Read This Before Posting


Topic: How was ΔG = -RT ln(Keq) discovered?  (Read 33865 times)

0 Members and 1 Guest are viewing this topic.

Offline BioPhysicsDino

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
How was ΔG = -RT ln(Keq) discovered?
« on: August 24, 2009, 11:36:50 PM »
The relation, ΔG = -RT ln(Keq)
is ubiquitous.

It says that for a reaction, the change in Gibbs energy is proportional to the logarithm of the equilibrium constant.

But where does this come from?
I've been reading through many books and haven't yet found any that derive or indicate the history of this equation. Does anyone know?

Is this simply an empirical result? Is it derivable other than from statistical mechanics? I've found a couple places where this can be derived from statistical mechanical postulates, but is that the only way? Is that where this relation originally comes from? (I thought it was originally based in thermodynamics; am I wrong about that?)

Does anyone know the history of this equation? Who first wrote it down? In what publication?

If you can shed some light in where this relationship originally came from I would much appreciate it. Thank you. --Dino

Offline renge ishyo

  • Chemist
  • Full Member
  • *
  • Posts: 403
  • Mole Snacks: +67/-14
Re: How was ΔG = -RT ln(Keq) discovered?
« Reply #1 on: August 25, 2009, 12:48:15 PM »
The person who first derived the relation was Josiah Willard Gibbs. You can read a good background on his life on Wikipedia:

http://en.wikipedia.org/wiki/Josiah_Willard_Gibbs

and some info on the equation here:

http://en.wikipedia.org/wiki/Gibbs_free_energy

The Gibbs free energy can be derived in many ways so it is difficult to say which one is the most authentic. Gibbs work, like Einstein's, was ignored for many years before gaining widespread acceptance within the Physics community.

Offline BioPhysicsDino

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Re: How was ΔG = -RT ln(Keq) discovered?
« Reply #2 on: August 25, 2009, 09:08:43 PM »
Thanks.  I'd read both those pages before I posted my question.  Maybe I'll go back and read them again.  It just seems to me that there is some circular logic there.  He defines chemical potential which includes the number of molecules of each species, and then uses that in the definition for free energy so when you define an equilibrium constant it comes out related to the free energy.

I'm beginning to think that maybe ΔG = -RT ln(Keq) isn't something discovered or derived from a thermodynamic defnition of ΔG.  Rather Gibbs purposely defined free energy in a way that would relate it to the equilibrium constant, by definition!  I read in one place (can't remember where) that Gibbs actually set out from the start to try to define a "driver of reactions" in much the same way that a Force vector is a driver of mechanical motion.  Is that what's going on here? 

I had always thought that G = U + pV − TS was the whole definition.  If so, then how does ΔG = -RT ln(Keq) derive from that?

Where does this defintion come from:  dG = Vdp - SdT + ΣμidNi + ΣXidai

I can see once you have dNi in there it's not hard to bring in the equilibrium constant.
Is ΔG = -RT ln(Keq)  a definition rather than a derived or empirical relationship?

Offline renge ishyo

  • Chemist
  • Full Member
  • *
  • Posts: 403
  • Mole Snacks: +67/-14
Re: How was ΔG = -RT ln(Keq) discovered?
« Reply #3 on: August 25, 2009, 11:30:17 PM »
The best place to get the full details on how this relation came about is to take a course in Thermodynamics (or to get a good clear book to read on the subject such as Ira Levine's Physical Chemistry). It will take some time to go through all the different relations and to find out who developed them and when and to be satisfied by the explanations. Wikipedia will of course give only an overview, but relax there is no conspiracy...Gibbs really was the first one to describe the concept of free energy. Part of the problem is that the later derivations required the acceptance of such things as atoms which were confirmed a few years after Gibbs originally developed his theory together with the statistical view of thermodynamics (which Gibbs later played a role in developing as well). Both of these modern concepts are needed to really get to the more modern Delta G = - RTlnKeq in a satisfying way.

The term ΣμidNi is a term added to the equation to account for the changes in material equilibrium. When a chemical reaction occurs the material composition of a system changes (products get converted into reactants and so forth), so the thermodynamic relations (which were largely derived from systems that do not change their composition) needed an added term to account for the difference between the initial and final states when the chemical compositions of the two states are not the same. The above term accomplishes this where the N stands for the amount of substance (which would now probably be related to moles or molarity). It is this extra term that eventually got transformed into the more modern Delta G = -RTlnKeq.


Offline BioPhysicsDino

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Re: How was ΔG = -RT ln(Keq) discovered?
« Reply #4 on: August 26, 2009, 06:14:26 AM »
Thank you.  Your explanation of the ΣμidNi term was especially good and pretty much answers my question.  I also appreciate the recommendation of the good P.Chem book.  I've been debating with myself which one to buy.  Atkins (7th ed) appears very popular (although I hear the 8th edition is not as good).  I'll take a look at Levine.  I took P.Chem many years ago, but have been out of the field for a while, and don't have my old books, so trying to refresh.  Thanks again.

Offline zxt

  • Regular Member
  • ***
  • Posts: 60
  • Mole Snacks: +3/-5
Re: How was ΔG = -RT ln(Keq) discovered?
« Reply #5 on: August 26, 2009, 10:40:21 AM »
Quote
"driver of reactions"

That's difference in reactants and products' chemical potential. You can find the derivation of the equilibrium in any formal P.Chem.

Offline renge ishyo

  • Chemist
  • Full Member
  • *
  • Posts: 403
  • Mole Snacks: +67/-14
Re: How was ΔG = -RT ln(Keq) discovered?
« Reply #6 on: August 26, 2009, 12:14:09 PM »
The different textbooks for P.Chem excel at different things. Many people like Atkins the best because it is the most thorough when it comes to the math derivations and most people who study P.Chem are better at math than they are at English. Levine's book on the other hand excels because it is the most clearly explained book in English. He tells you what every single term means in all the equations he introduces. To someone like me that struggles with the derivations that approach is the best. It depends on who you are.

I sort of avoided the derivation of dG = -RTlnKeq because to do it for real would take too long, but so long as you plan to read up on it in a formal book I can give you a sneak peak of where it will come from. This isn't really a derivation because I am introducing things that I am not explaining but the book will cover the rest for me I think, and with that disclaimer:

dG = dH - TdS

Now to measure the difference in material equilibrium only keep pressure and temperature in the system constant for both the initial and final state. This is easy to do in real life: 1) Pressure is constant due to the atmosphere 2) You can keep the temperature constant by having the reaction start at room temp and cool down to room temp before measuring the final step. This makes the dH term equal to zero so that:

dG = 0 - TdS = -TdS

Now introducing the boltzman statistical definition of entropy which is klnS or RlnS (the mole version which is the one we want):

dG = - TR(LnS2 - lnS1) = -RTln(S2/S1)

Now, you can show (which I won't do...every P.Chem book shows this derivation) that S2/S1 can be related to the Volume of two gases expanding in the system as V2/V1.

dG = -RTln(V2/V1)

Now from the ideal gas law you can see that when T and P are constant that V is directly proportional to moles.

V = nRT/P = Cn (where C is a constant)

You can also arrange this in terms of molarity (which is what we want to ultimately do since we are using Keq):

n/V = RT/P = [M]

Making the substitutions into dG:

dG = -RTln(n2/n1) = -RTln([n2/V2]/[n1/V1]) = -RTln([M2]/[M1])

Finally if we define Keq as:

Keq = [M2]/[M1]

Then:

dG = - RTln(Keq)

A sloppy derivation in a few ways, but at least you can see that there is no magic involved.

Cheers.

Sponsored Links