## At Equilibrium G

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

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

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

305174946
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

### At Equilibrium G

For the equation G=-RTlnK
when K<1 then G is positive which means R>P right? (non-spontaneous?)
and when K>1 then G is negative which means R<P right? (spontaneous?)

Samantha Kwock 1D
Posts: 61
Joined: Fri Sep 28, 2018 12:24 am

### Re: At Equilibrium G

Yes, for K < 1, lnK will be negative, cancelling out the negative in -RTlnK. This means that G is positive. For K > 1, lnK will be positive and G will be negative. When G is negative, the reaction is spontaneous.

Parth Mungra
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### Re: At Equilibrium G

If K=0, then the reaction is neither spontaneous nor nonspontaneous

Dayna Pham 1I
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### Re: At Equilibrium G

Parth Mungra wrote:If K=0, then the reaction is neither spontaneous nor nonspontaneous

This sounds interesting! I’ve never heard of a case where K=0, can someone elaborate on this?

Dong Hyun Lee 4E
Posts: 68
Joined: Fri Sep 28, 2018 12:23 am

### Re: At Equilibrium G

An example of where K=0 would be the example he went over in class where when T=333K, BR2 was in both liquid and gas phase. This was called the boiling point. And when T was above 333K, the reaction favored foward and when T<333K a reverse reaction was favored.

904936893
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### Re: At Equilibrium G

If K = 0, is the reaction at equilibrium?

bonnie_schmitz_1F
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### Re: At Equilibrium G

904936893 wrote:If K = 0, is the reaction at equilibrium?

I think so since the reaction is no longer releasing or requiring energy.