Topic: equilibrium and K (Read 4924 times)

briteyellowness · « **on:** August 19, 2005, 10:18:11 AM »

at equilibrium, G=0, does K =1?
G=-RTlnK

0 = -RTln1

and ln1 equals 0...?
yes....

Blueshawk · « **Reply #1 on:** August 19, 2005, 11:25:52 AM »

Mathematically yes.. 0= -RTlnK divide out -R and T 0=lnK raise each side by e

e⁰ = e^lnK 1=K

Winga · « **Reply #2 on:** August 19, 2005, 11:33:07 AM »

Quote from: briteyellowness on August 19, 2005, 10:18:11 AM

at equilibrium, G=0, does K =1?
G=-RTlnK

0 = -RTln1

and ln1 equals 0...?
yes....

G = G' + RTlnK
at eqm, G = 0
G' = -RTlnK

Blueshawk · « **Reply #3 on:** August 19, 2005, 11:43:24 AM »

From winga's post I am wrong, but with the equation I was given my statement was true. But thank you Winga for clearing up the equation....its been awhile since I used it...so I answered it without realizing that G' is to be included.

briteyellowness · « **Reply #4 on:** August 19, 2005, 05:32:04 PM »

thanks mcat cramming has gotten my mind turned to mush, but blueshawk, you were right cuz i meant G'.

UGHHHHHHHHH

Topic: equilibrium and K (Read 4924 times)

briteyellowness

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