## ∆G and ∆G°

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

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

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

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Jasmine 2C
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### ∆G and ∆G°

I know that if ∆G° is negative, that means the reaction is spontaneous (since K>1 and products are favored at equilibrium) and if ∆G° is positive, the reaction is not spontaneous. Does this apply for ∆G as well?

Sarah Zhari 1D
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### Re: ∆G and ∆G°

I would assume that these principles still apply for ∆G.

Alice Chang 2H
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Joined: Fri Aug 30, 2019 12:18 am

### Re: ∆G and ∆G°

The only difference between ∆G and ∆G° is that ∆G measures free energy for any conditions not standard while ∆G° measures free energy for standard conditions (1 atm and 25 degrees Celsius).

So yes, the same rules apply with the negative and positive ∆G and ∆G°.

RobertXu_2J
Posts: 104
Joined: Fri Aug 30, 2019 12:17 am

### Re: ∆G and ∆G°

Yes it does. Take note that ∆G° means that the reaction is spontaneous under STANDARD conditions, hence the "°". However, if it is proceeding under non standard conditions, you would need to use a formula for ∆G in order to find out if it is spontaneous or not.

Kayla Maldonado 1C
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### Re: ∆G and ∆G°

This also applies for ∆G. ∆G° is simply free energy measured under standard conditions at 1 atm and 25°C and ∆G is the free energy measured under a different pressure and temperature which must be calculated for a reaction.

AysiaB1I
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### Re: ∆G and ∆G°

I think that they are the same in terms of conditions for spontaneity, but ∆G° is for standard conditions. Which is 1 atm and 25°C and ∆G is just for any temperature and pressure.

Lauren Stack 1C
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### Re: ∆G and ∆G°

∆G° is for standard conditions, which ∆G is for non-standard conditions. Therefore, the principle still remains that a negative deltaG would mean that the reaction is spontaneous.

sarahsalama2E
Posts: 164
Joined: Fri Aug 30, 2019 12:16 am

### Re: ∆G and ∆G°

It is important to note the formulaic relationship between ∆G and ∆G°.
∆G = ∆G°+ RTlnQ. Make sure you are keeping this relationship in mind when seeing what direction the reaction would shift if the concentration of products or reactants changed or etc.

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