## Question about Gibbs Free Energy

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

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

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

Nathan Sanchez 1F
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### Question about Gibbs Free Energy

What exactly is Gibbs Free Energy? I understand its equation, but what is the conceptual relationship between enthalpy, entropy, and temperature to give what we know as Gibbs Free Energy?

Chem_Mod
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### Re: Question about Gibbs Free Energy

The equation is at the heart of the relationship. Try to understand what would happen if you were to block out one term, and see how Gibbs free energy changes when there is only one variable. If this doesn't help you can always come to office hours!

Gibbs free energy is the measure of the maximum amount of work that a molecule could do. If entropy and temperature were not a factor, solely enthalpy, if enthalpy is positive, endothermic, the Gibbs free energy (the potential for work) of the species will go up. If the enthalpy of the reaction is negative, exothermic, the potential of a species will go down. Temperature is always positive, so it only acts as a scalar for the entropic value. That means at lower temperatures, the entropic effect on the potential of the species is small, and at a high temperature, the entropic effect on the potential of the species is large.

With purely entropic reactions, (no enthalpy), we can see that it is inversely proportional to Gibbs free energy. That means the more states there are (+ entropy), there is less potential for the system to do work, and a highly ordered system actually has more potential for work. This is because of the Second Law of Thermodynamics, which states that the universe tends towards disorder.