[disillusioned19]

Why is the Gibbs free energy change for a system = 0 at equilibrium? Surely there is a definite entropy factor which is non-zero, even at equilibrium?

Thanks in advance.

[Schrödinger]

But then there is also enthalpy, which should be considered, at a certain temperature...

[disillusioned19]

The only way for it to be zero would be if ΔH = TΔS, however I'm not sure why this would be the particular case at equilibrium.

[Bryan Sanctuary]

Recall at constant T and P that the Gibbs energy is a minimum at equilibrium.  So if you have a balanced chemical equation,

2A + B    Z    ΔG = 0

then at equilibrium there is no change.  The system is at a minimum, so there is no driving force, no chemical potential, so no energy is released.

You can write

ΔG = ΔG_o + RT ln([Z]/([A]²{B}))=ΔG_o + RT lnQ

where Q is the quotient product.  That is the concentrations are not the equilibrium values.  Hence there is a driving force which at constant T and P means that ΔG is negative if the process is spontaneous. 

At equilibrium Q  K_equil  and nothing changes.  Hence ΔG = 0 and what is left over is

ΔG_o =- RT lnK_equil

This is a form you likely know.

[Jorriss]

Why is the Gibbs free energy change for a system = 0 at equilibrium? Surely there is a definite entropy factor which is non-zero, even at equilibrium?

Thanks in advance.

At constant T and P, a system moves to a minimum gibbs free energy. From calculus, one knows that minimums mean that the derivative is zero so dG=0. At equilibrium, there is still a free energy but it's the smallest value the system can take.

http://sst-web.tees.ac.uk/external/U0000504/Notes/enzkin/Equilibria/Image77.gif

That graph is useful.

[Yggdrasil]

I like to think of it this way: at equilibrium, the forward reaction and reverse reaction are equally favorable, allowing changes to occur reversibly.  For the forward and reverse reactions to be equally favorable, they must have the same ΔG values:

ΔG_forward = ΔG_reverse

But we also know that the Gibbs free energy is a state function and so:

ΔG_forward = -ΔG_reverse

The only values of ΔG that satisfy these two equations are ΔG_forward = ΔG_reverse = 0.

[Kemi]

I like to think of it this way: at equilibrium, the forward reaction and reverse reaction are equally favorable, allowing changes to occur reversibly.  For the forward and reverse reactions to be equally favorable, they must have the same ΔG values:

ΔG_forward = ΔG_reverse

But we also know that the Gibbs free energy is a state function and so:

ΔG_forward = -ΔG_reverse

The only values of ΔG that satisfy these two equations are ΔG_forward = ΔG_reverse = 0.

Does ΔG_forward or ΔG_reverse really mean anything at equilibrium? There is a dynamic equilibrium, but thermodynamically the system has a constant value of Gibbs energy, and G_constant - G_constant = 0.

The ΔG of this topic is the reaction Gibbs energy. It is actually a derivative and zero at equilibrium, as Jorriss points out.

[Yggdrasil]

Here I consider ΔG_forward and ΔG_reverse to be free energy gained/lost when one molecule of reactants are converted to products and vice versa.  These values depend on both the identities of the products/reactants as well as the relative concentrations of products/reactants.

[Kemi]

Can we then write ΔG_forward = G_product,c - G_reactant,c, where c denotes the concentration in question? Likewise, ΔG_reverse = G_reactant,c - G_product,c, which would mean that ΔG_forward = ΔG_reverse only if G_reactant,c = G_product,c.

I would say the forward and reverse reaction rates are equal at equilibrium, but the reactions are not equally favorable (the position of equilibrium usually lies towards the products or the reactants, not halfway).

[beruniy]

Why is the Gibbs free energy change for a system = 0 at equilibrium? Surely there is a definite entropy factor which is non-zero, even at equilibrium?

I think that ΔS = 0 at equilibrium only in isolated system where is no energy and no matter exchange with surroundings.

If the system is closed (energy exchange but no matter exchange) we can use
Q = ΔS/T, so Q (=ΔH) - TΔS = 0; ΔG = 0 - this is the condition of equlibrium.

[juanrga]

Why is the Gibbs free energy change for a system = 0 at equilibrium? Surely there is a definite entropy factor which is non-zero, even at equilibrium?

I think that ΔS = 0 at equilibrium only in isolated system where is no energy and no matter exchange with surroundings.

If the system is closed (energy exchange but no matter exchange) we can use
Q = ΔS/T, so Q (=ΔH) - TΔS = 0; ΔG = 0 - this is the condition of equlibrium.

ΔS = 0 for any system at equilibrium, was it open, closed, or isolated.

[Jack Bauer]

I think you are forgetting to consider that the free energy of a reaction includes the free energy of mixing.  deltaG = 0 at some equilibrium point due to a balance between the Gibbs free energy change from going reaction to products (which is normally consisted to be negative) and the free energy of mixing reactants and products, which has a minimum point.  You will find this diagram in most good physical chemistry books.

Apologies if this is slightly incorrect, I have not reread this for a while.

I have not read this so cant say if it is all correct but it seems to cover it

http://pages.csam.montclair.edu/~kasner/Archives/Physicall1/Mixing1.pdf