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Relation between gibbs free energy and equilibrium constant 
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#1
Apr2813, 09:15 PM

P: 4

I am familiar with the equation ΔG=ΔG°+RT ln(Q).But I can't derive it.We have to use the equation to derive nernst equation. So please help.



#2
Apr2913, 09:41 AM

P: 380

I don't remember the derivation of the above equation off the top of my head, but I'm sure you can google it or look in any intro to the Thermodynamics or Physical Chemistry text.
As far as deriving the Nernst equation from what you have: ΔG = nFE and ΔG° = nFE° plug those two into ΔG = ΔG° + RTlnQ and do some simple algebraic rearranging. This may not be useful to you if you are in an advanced class which requires derivations of the equations that I have taken for granted. In other words this is a pseudoderivation applicable to a freshman level Gen. Chem. course. 


#3
May213, 08:53 PM

P: 375

In this reference , post #2 PhaseShifter states: q = K_{eq}
Can someone show how this is ? http://www.physicsforums.com/showthread.php?t=332342 Also how does R the gas constant , Avogadro's number x Boltzamann's constant, apply to ΔG for solutions ? 


#4
May313, 04:44 AM

Admin
P: 23,374

Relation between gibbs free energy and equilibrium constant



#5
May413, 02:28 AM

P: 375

ΔG = RTlnK_{eq} and ΔG = ΔH  TΔS as Ill show for anyone who wants to see it N_{2} + 3H_{2} > 2NH_{3} K_{eq} = (NH_{3})^{2}/(N_{2})(H_{2})^{3} = 6.73 * 10^{5} , lnK=13.4 , RT= 2.473 kJ/m So ΔG = RTlnK = 33kJ/m In agreement with ΔG = ΔH  TΔS ΔH = 92kJ/m, T = 298K , ΔS = 198J/m 33kJ/m = 92kJ/m + 59kJ/m 


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