Gibbs free energy

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

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

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

Eileen Si 1G
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Gibbs free energy

Why does Gibbs free energy depend on the equilibrium constant and pressure? Also, is Gibbs free energy a state function?

Fiona Latifi 1A
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Re: Gibbs free energy

Gibbs free energy is a state function because it is determined by enthalpy and entropy, which are also state functions.

Kristina Rizo 2K
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Re: Gibbs free energy

Yes it is a state function.

Veronica_Lubera_2A
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Re: Gibbs free energy

At equilibrium, DeltaG=0 because the forward and backward reaction rates are equal. So the formula for nonstandard conditions (including not at equilibrium) is Delta G= DeltaG* + RT lnQ, where Q is the reaction quotient ([Products]/[Reactants]). At equilibrium, the Q variable would become K.

Callum Guo 1H
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Re: Gibbs free energy

yes, gibbs free energy is a state function

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Re: Gibbs free energy

Eileen Si 1G wrote:Why does Gibbs free energy depend on the equilibrium constant and pressure? Also, is Gibbs free energy a state function?

Yes! Gibbs free energy, G, is a state function. It it determined by its current state. This allows us to calculate the chnage based upon the final and inital values-- this is similar to entropy, S, and enthalpy, H.