# Equilibrium constant always valid?

1. Nov 1, 2004

### arwelbath

DG = -RTln(K) always valid?

Hi,
For a reaction at equilibrium, A<==>B, then the free energy can be defined as
DG = -RTln(K), where K = /[A].

This is fine for an isolated equilibrium. But suppose that the system is not isolated so that A<==>B--->C for example, or if is part some even more complicated kinetic scheme. Does the relationship for DG still hold?
Ta.

Last edited: Nov 1, 2004
2. Nov 1, 2004

### mathlete

Yes, K is the same always - unless temperature changes. What you have is an application of Le Chatelier's Principle, whereby the concentration of B will shift so and [A] may be different but yield the same K.

3. Nov 1, 2004

### arwelbath

Thanks, but didn't frame my question quite right. What I mean is that is the relationship (DG = -RTln(K)) always true even if it isn't an isolated equilibrium? Is it okay for closed and open systems too?

4. Nov 1, 2004

### Bystander

irreversible as written, therefore, NO.

The expression is a valid description of the relationship between the initial and final states of a reversible process (or reaction) --- open, closed, riding a float in Mardi Gras, running for public office, running from the law --- in other words "ALWAYS" --- tack on an irreversible "sink" for the "final" state (C in your example), and the process is no longer reversible, the expression no longer includes the final "final" state.

Initial, final, and reversible.

Last edited: Nov 1, 2004
5. Nov 2, 2004

### arwelbath

Thanks Bystander.
The reason I ask is that this very often seems to be done. I could ref a number of DSC papers (for example) where people study coupled reversible and irreversible processes, and interpret the thermodynamics of the reversible process using that expression for the free energy in terms of K.

For example, the A<==>B---->C mechanism is ususaly called the Lumry-Eyring mechanism. The prcedure seems to be that as long as A<==>B is kinetically fast compared to B--->C, then they assume that the AB transition is approximately at equilibrium and can be analysed using classical reversible thermodynamics. Exactly where this approximation breaks down in terms of the relative reaction velocities seems to be not discussed however.

So my next question would be where to begin in terms of properly analysing thermodynamics of a coupled reversible / irreversible system like this? I can do the kinetics and get Arrhenius type activation energies for each process. But, the really interesting info would be DeltaG A-B and DeltaG B-C. Which brings me back to a question I've posted on here before. Is DeltaG defined for the irreversible process, and if so, how would it be calculated?

6. Nov 2, 2004

### Bystander

Irreversible process? del G? Yes. del G nought (the std. state that shows up in del G nought = -RTlnK)? Not in a way that can be determined from equilibrium measurements of reactant and product activities --- "irreversible" means that the system is never at equilibrium, del G is never zero, and del G nought can never be equated to RTlnK.