## Delta G and delta G naught

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

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

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

Jose_Arambulo_2I
Posts: 35
Joined: Wed Sep 21, 2016 2:59 pm

### Delta G and delta G naught

Hello!
I'm just a little confused about these two things. From my understanding, the naught refers to standard conditions, making me think that the only difference between the two values are that delta G naught is the change in free energy in 1 atm and 25 degrees Celsius and delta G is just the change in free energy in any other condition. If my assumption is correct, why did Dr. Lavelle emphasize that equilibrium takes place when delta G equals 0 and not when delta G naught equals zero?

Hannah_El-Sabrout_2K
Posts: 20
Joined: Wed Sep 21, 2016 2:58 pm

### Re: Delta G and delta G naught

You are right, the difference between the two is that delta G naught is at standard conditions. The reason Professor Lavelle emphasized it is because delta G naught is always the same because it is referring to when the reactants/products are at standard temperature/pressure. As the rxn goes towards equilibrium, delta G (without the naught) changes because the rxn is proceeding. So as the chemical rxn approaches equilibrium, delta G (without the naught) approaches zero. However, delta G naught remains the same because it is still referring to when the rxn is at standard conditions. I hope that helped!

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