## delta G=0

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

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

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

Emily Glaser 1F
Posts: 156
Joined: Thu Jul 27, 2017 3:01 am

### delta G=0

When delta G=0, does that mean delta H and delta S are both zero too? Or they have numerical values that cancel?

Katie Lam 1B
Posts: 52
Joined: Fri Sep 29, 2017 7:06 am

### Re: delta G=0

Either delta H and delta S are 0, indicating that no reaction occurred, or delta H = T delta S and they cancel out.

John Huang 1G
Posts: 46
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### Re: delta G=0

Delta G=0 Does not necessarily mean delta H and delta S are zero as well. Delta G=0 means that the process is at equilibrium.

Sarkis Sislyan 1D
Posts: 31
Joined: Thu Jul 27, 2017 3:00 am

### Re: delta G=0

Delta G=delta H - TdeltaS so if delta G=0 then either both delta H and delta S are zero or delta H equals T times delta S

Emma Miltenberger 2I
Posts: 51
Joined: Thu Jul 27, 2017 3:00 am

### Re: delta G=0

It does not necessarily mean that delta H an delta S are zero. When delta G =0, the system is at equilibrium. Therefore, either delta H or delta S is zero or delta S is equal to delta H.

AtreyiMitra2L
Posts: 169
Joined: Fri Sep 29, 2017 7:03 am
Been upvoted: 1 time

### Re: delta G=0

When delta g is 0, it just means that it is at equilibrium. It is not enough to assume the other two quantities to alse be at 0.

Amy Zheng 2l
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

### Re: delta G=0

Delta G=equilbrium