# Standard Gibbs free energy equal to zero?

Kqwert
Homework Statement:
I have a question which asks:

What does it mean that deltaG_zero = 0?
Relevant Equations:
.
At equilibrium, we know that deltaG = 0. But what about deltaG_zero, i.e. the standard Gibbs free energy? When is deltaG_zero = 0?

In the solution manual it says that it means that K = 1, but by calculating an equilibrium constant you are already stating that we are at equilibrium? I.e. that deltaG = 0, and K = 1 implies deltaG_zero = 0. A bit confused by this.

Staff Emeritus
Problem Statement: I have a question which asks:

What does it mean that deltaG_zero = 0?
Relevant Equations: .

At equilibrium, we know that deltaG = 0. But what about deltaG_zero, i.e. the standard Gibbs free energy? When is deltaG_zero = 0?

In the solution manual it says that it means that K = 1, but by calculating an equilibrium constant you are already stating that we are at equilibrium? I.e. that deltaG = 0, and K = 1 implies deltaG_zero = 0. A bit confused by this.

You haven't said which textbook you're using, so it's not clear what the quantity ##K## is.

But also, when people use ##\Delta##, they usually mean the change in some quantity. So ##\Delta G_0 = 0## means no change in the value of ##G_0##. You're talking about change under what circumstances?

Mentor
At equilibrium, we know that deltaG = 0. But what about deltaG_zero, i.e. the standard Gibbs free energy? When is deltaG_zero = 0?
I'm assuming that by ##\Delta G^0 ## you mean the standard Gibbs free energy of a reaction (since it is that that is related to the equilibrium constant ##K##).

##\Delta G^0 = 0## when the Gibbs free energy of the products is the same as the Gibbs free energy of the reactants. For a reaction of the kind
A + B ##\rightleftharpoons## C + D
##\Delta G^0 = 0## means that the Gibbs free energy of A + B is the same as C + D, so no side of the equation is favored, and at equilibrium there will be as much A and B as C and D (hence ##K=1##).

In the solution manual it says that it means that K = 1, but by calculating an equilibrium constant you are already stating that we are at equilibrium? I.e. that deltaG = 0, and K = 1 implies deltaG_zero = 0. A bit confused by this.
##K## and ##\Delta G^0## are constants for a given reaction. They give information about what will happen at equilibrium. ##\Delta G## tells you about a particular system, and can tell you if it is at equilibrium or not.

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• stevendaryl
Mentor
For a reaction of the kind
A + B ##\rightleftharpoons## C + D
##\Delta G^0 = 0## means that the Gibbs free energy of A + B is the same as C + D, and at equilibrium there will be as much A and B as C and D (hence ##K=1##).
I don't think you really mean this. This will be the case only if you start out with equal amounts of A and B. For an ideal gas reaction, it means that the product of the partial pressures of A and B will be equal to the product of the partial pressures of C and D at equilibrium.

Mentor
I don't think you really mean this. This will be the case only if you start out with equal amounts of A and B. For an ideal gas reaction, it means that the product of the partial pressures of A and B will be equal to the product of the partial pressures of C and D at equilibrium.
You are of course correct. I had something in mind that I tried to express simply and it came out wrong Kqwert
With K I mean the equilibrium constant.

I got a bit confused here.

what exactly does it mean that ##\Delta G^0 = 0##? Apart from it meaning that the Gibbs free energy of the products are the same as the Gibbs free energy of the reactants. Is there something more we can deduce from it?

Mentor
With K I mean the equilibrium constant.

I got a bit confused here.

what exactly does it mean that ##\Delta G^0 = 0##? Apart from it meaning that the Gibbs free energy of the products are the same as the Gibbs free energy of the reactants. Is there something more we can deduce from it?
Are you asking for a physical interpretation (in terms of physical processing of the materials) for the standard change in free energy?

Kqwert
Are you asking for a physical interpretation (in terms of physical processing of the materials) for the standard change in free energy?
No, I was wondering what other "results" one could take from ##\Delta G^0 = 0##