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Relation between gibbs free energy and equilibrium constant

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HARI A
#1
Apr28-13, 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.
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Yanick
#2
Apr29-13, 09:41 AM
P: 382
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 pseudo-derivation applicable to a freshman level Gen. Chem. course.
morrobay
#3
May2-13, 08:53 PM
P: 380
In this reference , post #2 PhaseShifter states: q = Keq
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 ?

Borek
#4
May3-13, 04:44 AM
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Relation between gibbs free energy and equilibrium constant

Quote Quote by morrobay View Post
In this reference , post #2 PhaseShifter states: q = Keq
Can someone show how this is ?
Do you know what Q is? (To be honest using q is IMHO confusing).
morrobay
#5
May4-13, 02:28 AM
P: 380
Quote Quote by Borek View Post
Do you know what Q is? (To be honest using q is IMHO confusing).
Thanks, was looking at q as in heat. There is not a question on the equivalence of ΔG for
ΔG = -RTlnKeq and ΔG = ΔH - TΔS as Ill show for anyone who wants to see it
N2 + 3H2 --> 2NH3
Keq = (NH3)2/(N2)(H2)3 = 6.73 * 105 , 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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