deltaG at equilibrium

[105311039]

At equilibrium does deltaG=0? If so why? Thank you!

[Jacob Motawakel]

yes because there is no more potential energy for the reaction to move in any direction.

[Ryan Yee 1J]

Yes, because of the equation: delta(G) = delta(G0) + RTln(Q) and since we are at equilibrium, Q=K as well as delta(G0) = -RTln(K). So if you plug things back in, you get delta(G) = -RTln(K) + RTln(K) which equals 0

[Brooke Yasuda 2J]

Yes, also remember that the value of delta G can help to determine the direction of the reaction that is favored. For example, a more negative delta G means that it is spontaneous and favoring the forward reaction. As the reaction progresses and moves towards equilibrium, the concentrations of the reactants and products change and delta G moves towards zero. Just as when the Q reaches the K value, Delta G reaches 0, which indicates equilibrium.

[Mariah]

Yes, because of the equation: delta(G) = delta(G0) + RTln(Q) and since we are at equilibrium, Q=K as well as delta(G0) = -RTln(K). So if you plug things back in, you get delta(G) = -RTln(K) + RTln(K) which equals 0

I thought that delta G naught, was just at standard conditions not equilibrium?

[Diana A 2L]

Yes, because of the equation: delta(G) = delta(G0) + RTln(Q) and since we are at equilibrium, Q=K as well as delta(G0) = -RTln(K). So if you plug things back in, you get delta(G) = -RTln(K) + RTln(K) which equals 0

I thought that delta G naught, was just at standard conditions not equilibrium?

You’re right the difference is that delta G naught is at standard conditions. Delta G naught is always the same because it is referring to when the reactant/products are at standard conditions.

[Ian Morris 3C]

Yes, because of the equation: delta(G) = delta(G0) + RTln(Q) and since we are at equilibrium, Q=K as well as delta(G0) = -RTln(K). So if you plug things back in, you get delta(G) = -RTln(K) + RTln(K) which equals 0

I thought that delta G naught, was just at standard conditions not equilibrium?

You’re right the difference is that delta G naught is at standard conditions. Delta G naught is always the same because it is referring to when the reactant/products are at standard conditions.

Would that mean standard conditions are not always at equilibrium?

[Diana A 2L]

I thought that delta G naught, was just at standard conditions not equilibrium?

You’re right the difference is that delta G naught is at standard conditions. Delta G naught is always the same because it is referring to when the reactant/products are at standard conditions.

Would that mean standard conditions are not always at equilibrium?

YES! Exactly. Delta G naught DOES NOT describe the change in free energy from reactants or products at equilibrium. Standard state conditions lead to Q=1 not K=1. At standard conditions, it is required that the activities of pure reactants and pure products is equal to 1. These activities are different from the activities of the reactants and products at equilibrium. Does that make sense?

[Ian Morris 3C]

You’re right the difference is that delta G naught is at standard conditions. Delta G naught is always the same because it is referring to when the reactant/products are at standard conditions.

Would that mean standard conditions are not always at equilibrium?

YES! Exactly. Delta G naught DOES NOT describe the change in free energy from reactants you products at equilibrium. Standard state conditions lead to Q=1 not K=1. At standard conditions, it is required that the activities of pure reactants and pure products is equal to 1. These activities are different from the activities of the reactants and products at equilibrium. Does that make sense?

Yes, thank you so much! this will help me immensely on the final