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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: khwcm on August 20, 2010, 01:46:13 AM

Title: about free energy
Post by: khwcm on August 20, 2010, 01:46:13 AM
The equation ΔG = -RT ln K relates the value of Kp, not Kc, to the change in standard free energy for a reaction in the gas phase, why.
Title: Re: about free energy
Post by: khwcm on August 21, 2010, 12:56:39 AM
really no one can help?
please...
Title: Re: about free energy
Post by: zeoblade on August 21, 2010, 10:36:49 AM
Kc being the equilibrium constant? yes it does

dG = -nFE = dH - TdS = G* - RT/(nF) ln(K)
Title: Re: about free energy
Post by: khwcm on August 21, 2010, 12:50:32 PM
this was a question copied from <Chemistry-the science in context>.
i also think that Kc can use instead of Kp....just dont understand it
Title: Re: about free energy
Post by: zeoblade on August 21, 2010, 06:28:52 PM
what do you mean by the term Kc and Kp?

if Kc is equilibrium constant Keq, and Kp is the solubility product constant Ksp, then both are interchangeable.

Ksp = 1/Kc

so dG = -nFE = -RT ln Kc = G* - RT/(nF) ln Kc
Title: Re: about free energy
Post by: khwcm on August 22, 2010, 05:01:04 AM
nono.. i mean Kc is equilibrium constant in terms of concentration while Kp is the equilibrium constant in terms of pressure
Title: Re: about free energy
Post by: zeoblade on August 22, 2010, 05:03:42 AM
either way, you can interconvert them
Title: Re: about free energy
Post by: MrTeo on August 22, 2010, 08:07:15 AM
This is something I thought about for a long time too: ok, the two constants (Kp and Kc) have more or less the same meaning so there are no real differences between them except for the units (that's why we use the well-known relation Kp=KcRT∆n)... But, if we substitute these two values in the free Gibbs energy formula we get different ∆G values: how can it be?
Title: Re: about free energy
Post by: khwcm on August 22, 2010, 09:51:12 AM
different value implies different units....seems nothing wrong..

btw...i just wonder that why it only works for Kp but not Kc ?
Title: Re: about free energy
Post by: MrTeo on August 22, 2010, 12:45:50 PM
Quote from: mcwhk on August 22, 2010, 09:51:12 AM
different value implies different units....seems nothing wrong..

Yes, but the matter is that the thermodinamical equilibrium constant is:

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B%5Calpha+A+%2B%5Cbeta+B+%5Crightleftharpoons+%5Cdelta+C+%2B%5Cgamma+D+%5C%5C%3Cbr+%2F%3EK_c%3D%5Cfrac%7B%5Cleft%5BC%5Cright%5D%5E%7B%5Cdelta%7D+%5Ccdot+%5Cleft%5BD%5Cright%5D%5E%7B%5Cgamma%7D%7D%7B%5Cleft%5BA%5Cright%5D%5E%7B%5Calpha%7D+%5Ccdot+%5Cleft%5BB%5Cright%5D%5E%7B%5Cbeta%7D%7D+%5Ccdot+%5Cleft%28%5Cfrac%7B1%7D%7BC_0%7D%5Cright%29%5E%7B%5Csigma%7D+%7D&hash=34a23adb8145aedf5fddf9db2d1c1e1d461d95e8)

where

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5Csigma%3D%5Cleft%28%5Cdelta%2B%5Cgamma+%5Cright%29+-+%5Cleft%28%5Calpha+%2B%5Cbeta+%5Cright%29+%7D&hash=bc21cee3df4d77673d02d3832c92d518b7d5cb8d)

and C0=1 M
So it's adimensional, due to the second term I wrote... (and this makes sense with the fact that it is the argument of a logarithm)
And Kp works the same way, with P0=1 atm instead of C0.
Title: Re: about free energy
Post by: demoninatutu on September 02, 2010, 09:04:32 PM
Quote
The equation ΔG = -RT ln K relates the value of Kp, not Kc, to the change in standard free energy for a reaction in the gas phase, why.

A second look at that question suggests that the point is that in the gas phase you have to use the Kp for the reaction and not the Kc of the solution-phase reaction.

For example, you couldn't use the Kc value of H2O + CO2 <—> H2CO3 in water to calculate the free energy change of the gas phase reaction.

The 'why' goes from the fairly obvious to the fairly complex, depending on how much detail you want to include, but basically heat of hydration will affect the free energy of the reaction in solution.