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At Equilibrium G

Posted: Fri Feb 15, 2019 2:08 pm
by 305174946
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?)

Re: At Equilibrium G

Posted: Fri Feb 15, 2019 2:18 pm
by Samantha Kwock 1D
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.

Re: At Equilibrium G

Posted: Fri Feb 15, 2019 7:52 pm
by Parth Mungra
If K=0, then the reaction is neither spontaneous nor nonspontaneous

Re: At Equilibrium G

Posted: Sat Feb 16, 2019 5:42 am
by Dayna Pham 1I
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?

Re: At Equilibrium G

Posted: Sun Feb 17, 2019 10:38 am
by Dong Hyun Lee 4E
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.

Re: At Equilibrium G

Posted: Sun Feb 17, 2019 11:34 am
by 904936893
If K = 0, is the reaction at equilibrium?

Re: At Equilibrium G

Posted: Sun Feb 17, 2019 12:58 pm
by bonnie_schmitz_1F
904936893 wrote:If K = 0, is the reaction at equilibrium?


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