[khwcm]

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.

[khwcm]

really no one can help?
please...

[zeoblade]

Kc being the equilibrium constant? yes it does

dG = -nFE = dH - TdS = G* - RT/(nF) ln(K)

[khwcm]

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

[zeoblade]

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

[khwcm]

nono.. i mean Kc is equilibrium constant in terms of concentration while Kp is the equilibrium constant in terms of pressure

[zeoblade]

either way, you can interconvert them

[MrTeo]

This is something I thought about for a long time too: ok, the two constants (K_p and K_c) 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 K_p=K_cRT^∆n)... But, if we substitute these two values in the free Gibbs energy formula we get different ∆G values: how can it be?

[khwcm]

different value implies different units....seems nothing wrong..

btw...i just wonder that why it only works for Kp but not Kc ?

[MrTeo]

different value implies different units....seems nothing wrong..

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



where



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 K_p works the same way, with P₀=1 atm instead of C₀.

[demoninatutu]

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 K_p for the reaction and not the K_c of the solution-phase reaction.

For example, you couldn't use the K_c 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.