∆G and ∆G°

[Jasmine 2C]

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]

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

[Alice Chang 2H]

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]

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]

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]

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]

∆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]

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.