# How do we measure gibbs free energy for irreversible processes?

1. Sep 8, 2014

### hongiddong

I know that gibbs free energy for say a body will be equal to = Gibbs free standard energy at 1M and Ph7(-rtlnkeq) (Where k is the concentration of product/ concentration of reactant at equilibrium)+rtlnk.

How can we use the standard gibbs free for irreversible spontaneous processes? Is it that the product concentration is wayyyy higher than the reactant, therefore, irreversible chemical processes can actually be reversible, given we put more product into the body system so that le chatlier's principal may occur?

For another example for the standard free energy of keq, if the product concentration is way higher than reactant, is there even product moving to the reactant side and reactant moving to the product side at equal rates? Maybe it takes a longgg time but the rates occur. Or maybe when nothing moves across and no rates for both sides is an equal rate, omols/sec.

Last edited: Sep 8, 2014
2. Sep 9, 2014

### Andy Resnick

I don't understand your question- it sounds similar to an analysis of ATP hydrolysis- is that what you are referring to?

3. Sep 9, 2014

### hongiddong

I guess what I mean is that, we measure standard gibbs free energy based on a chemical reaction that is in equilibrium. We then use this to find the gibbs free energy of other systems with the same chemicals.
But for some reactions they are not reversible, meaning a crazy negative gibbs free energy change for the reactant to product.

G standard = -rtlnkeq (keq is equal to concentration of products/ concentration of reactants at equilibrium.

Perhaps equilibrium for this measured reaction takes a longggg time, or the reaction has equal rates because nothing goes back and forth?

4. Sep 9, 2014

### Andy Resnick

I guess I still don't understand what you mean. For example, within the cytoplasm ΔG for ATP hydrolysis = -57 kJ/mol. This is because at equilibrium, [ATP]/[ADP] = 0.0000001, while within your cells, [ATP]/[ADP] = 1000. This extreme departure from equilibrium is what provides the energy needed to keep you alive.

5. Sep 9, 2014

### Staff: Mentor

I think I understand your question. You are asking, "If we use the standard gibbs free energy to determine the equilibrium constants for reactions, how does this reconcile with reactions that are irreversible?" In this a correct interpretation of your question?

If it is a correct interpretation, then my answer is that there is no such thing as a perfectly irreversible reaction. All reactions are reversible to some extent. Reactions that we consider irreversible just have an extremely large equilibrium constant. I think this is what you were alluding to (correctly) when you were talking about crazy negative standard gibbs free energy change.

Chet

6. Sep 9, 2014

### hongiddong

THANK YOU CHET! YESSS! I am enlightened!

7. Jul 14, 2016

### Daniel Florez-Orrego

There are some theoretical approaches such as the Onsager Reciprocal relations in which it is attempted to match the fluxes and forces, i mean, the reaction rates and the -deltaG/T forces, for chemical reactions. This allows calculating the rate of the entropy production. Irreversible thermodynamics, however, still requires the approximation of the Local Thermodynamic Equilibrium hypothesis... I could recommend the works of Denbigh, De Donder, and lately, Kjelstrup, Hinderink and others. It is pretty interesting such topic. In my case, the reaction is ammonia, and it is expected that as an exothermic , equilibrium limited reaction at high temperatures, the equilibrium be not attained and rather, a refrigeration system shift the reacting mixture away from equilibrium to increase the ammonia yield... that is, even though equilibrium is not achieved at the exit of the reactor, it is required the assumption of this local equilibrium to perform the properties calculation : ).