## DG = 0 [ENDORSED]

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

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

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

Miranda 1J
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am

### DG = 0

Why is it that at equilibrium we say that DG is equal to zero? Do we have to know why or just that we use that in our calculations?

RyanS2J
Posts: 32
Joined: Thu Jul 27, 2017 3:00 am

### Re: DG = 0  [ENDORSED]

DeltaG is 0 at equilibrium because neither the forward nor the reverse reactions are favored. They occur at equal rates. If deltaG were to be negative and spontaneous for one reaction (either forward or reverse), that would mean the opposite reaction (either reverse or forward, respectively), would have to be positive and nonspontaneous.

Katelyn 2E
Posts: 35
Joined: Sat Jul 22, 2017 3:01 am

### Re: DG = 0

Adding on to what Ryan said, DeltaG at equilibrium means there's no net flow, or no tendency for the reaction to favor either direction.

Ryan Sydney Beyer 2B
Posts: 82
Joined: Fri Sep 29, 2017 7:07 am
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### Re: DG = 0

We can also say that an endergonic process has a positive delta g and is non spontaneous, an exergonic process has a negative delta g and is spontaneous so it makes sense that a process at equilibrium has a delta g of zero.