## Stability in Gibbs standard free energy of formation

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

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

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

Gurvardaan Bal1L
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### Stability in Gibbs standard free energy of formation

When determining how stable a compound is, does a more negative standard Gibbs free energy of formation refer to the more stable compound? Say if you had two compounds, one that is more negative than the other. Would we classify the more negative compound as "more stable"?

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### Re: Stability in Gibbs standard free energy of formation

I think that's true.

Phillip Winters 2F
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### Re: Stability in Gibbs standard free energy of formation

Yes, if the Gibbs standard free energy of one compound is more negative than another it is more stable because when it is in the form of its constituent elements it is less stable, meaning that when the reaction is spontaneous the formation of the product, which is more stable, is favored

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