## Why are standard free energies of formation zero for elements in their stablest form?

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

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

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

Harjas Sabharwal 1G
Posts: 42
Joined: Sat Jul 22, 2017 3:01 am

### Why are standard free energies of formation zero for elements in their stablest form?

I understand that ΔH⁰f is 0 for an element in its standard state.
But then by using the equation ΔG⁰f = ΔH⁰f - TΔS⁰f, shouldn't ΔG⁰f have some other value than 0? because elements in there standard states do have some ΔS⁰f value.
Ex: ΔG⁰f = 0 - (298K)ΔS⁰f ≠ 0

Michelle Chernyak 1J
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am

### Re: Why are standard free energies of formation zero for elements in their stablest form?

It is favorable for a substance to be at its most stable state, therefore when a substance reaches that, it is going to want to maintain that most stable state. This means that it wont spontaneously progress in either direction, it will just stay in its current form.

Hope that helps!

Austin Ho 1E
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

### Re: Why are standard free energies of formation zero for elements in their stablest form?

The principle is that ΔG°= 0 for elements because we think about the reaction of their formation. For an element like Carbon, its formation reaction would simply be:

C(s) -> C(s). Since both sides are identical (Carbon is identical to itself), they must both have the same ΔG°. Therefore, ΔG° for the reaction is 0 since there is literally no difference between the products and reactants. This is why ΔG°f for elements is 0.