## deltaG at equilibrium

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

105311039
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### deltaG at equilibrium

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

Jacob Motawakel
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### Re: deltaG at equilibrium

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

Ryan Yee 1J
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### Re: deltaG at equilibrium

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
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### Re: deltaG at equilibrium

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
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### Re: deltaG at equilibrium

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?

Diana A 2L
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### Re: deltaG at equilibrium

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.

Ian Morris 3C
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### Re: deltaG at equilibrium

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?

Diana A 2L
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### Re: deltaG at equilibrium

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?
Last edited by Diana A 2L on Fri Mar 13, 2020 2:09 pm, edited 1 time in total.

Ian Morris 3C
Posts: 101
Joined: Wed Sep 18, 2019 12:18 am

### Re: deltaG at equilibrium

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

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