## Gibbs Free Energy: State Function?

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

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

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

Grace Boyd 2F
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### Gibbs Free Energy: State Function?

I was wondering if someone could explain to me why Gibbs Free Energy is considered a state function. Thank you!

Chem_Mod
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### Re: Gibbs Free Energy: State Function?

Gibbs Free Energy is a state function because it depends only on the initial and final states of Gibbs Free Energy. We use the equation of $\Delta G^{o} = \sum \Delta G_{f}^{o}(products) - \sum \Delta G_{f}^{o}(reactants)$, so the difference between products and reactants only matters.

Lauren Seidl 1D
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### Re: Gibbs Free Energy: State Function?

Also, gibbs free energy is equal to enthalpy plus entropy, which are both state functions. Thus, Gibbs free energy must also be a state function.

Sarah Maraach 2K
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### Re: Gibbs Free Energy: State Function?

Generally, anything with a delta in front of it will be a state function, because the delta means 'change in'. So, to calculate the change, you must subtract the initial value from the final value, therefore the only relevant values are your initial and final, fitting the definition of a state function.