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deltaG at equilibrium

Posted: Tue Mar 10, 2020 12:39 am
by 105311039
At equilibrium does deltaG=0? If so why? Thank you!

Re: deltaG at equilibrium

Posted: Tue Mar 10, 2020 12:58 am
by Jacob Motawakel
yes because there is no more potential energy for the reaction to move in any direction.

Re: deltaG at equilibrium

Posted: Wed Mar 11, 2020 8:41 pm
by 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

Re: deltaG at equilibrium

Posted: Wed Mar 11, 2020 9:37 pm
by 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.

Re: deltaG at equilibrium

Posted: Thu Mar 12, 2020 9:37 am
by Mariah
Ryan Yee 1J wrote: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?

Re: deltaG at equilibrium

Posted: Fri Mar 13, 2020 11:18 am
by Diana A 2L
Mariah wrote:
Ryan Yee 1J wrote: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.

Re: deltaG at equilibrium

Posted: Fri Mar 13, 2020 1:42 pm
by Ian Morris 3C
Diana A 2L wrote:
Mariah wrote:
Ryan Yee 1J wrote: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?

Re: deltaG at equilibrium

Posted: Fri Mar 13, 2020 2:00 pm
by Diana A 2L
Ian Morris 3C wrote:
Diana A 2L wrote:
Mariah wrote:
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?

Re: deltaG at equilibrium

Posted: Fri Mar 13, 2020 2:03 pm
by Ian Morris 3C
Diana A 2L wrote:
Ian Morris 3C wrote:
Diana A 2L wrote:
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