Therhelp modynamics and the equilibrium constant

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 4K views
mcato_O
Messages
3
Reaction score
0
Therhelp ! modynamics and the equilibrium constant

Thermodynamics and the equilibrium constant

Okay I have a question about the equation (* means change) (*() means stander condition) so [G=*(G)+RT ln (Keq)] or just equation *(G)= - RT ln(Keq)
So what dose the (Keq) stander for? Kc? Kp? Ksp?? Kf or Kd

Is it always Kc regardless the chemical reaction? (that’s what my prof told me)
And then use the equation Kp=Kc (RT)^*n to convert K value?

Or is does Keq depends on the chemical reaction Kc for solution Kp for Gas?(from the textbook)

And can someone please derived the equation [G=*(G)+RT ln (Keq)] for me?
Where is it come from and why does it makes sense?
 
Physics news on Phys.org
The derivation is a bit extensive, and you'll have to cover some ground in Physical Chemistry before you can make sense out of this particular topic.

The Kc,Kp,Ksp,Kf,Kd are the equilibrium constants with respect to the concentration Kc, and pressure Kp; the rest are specific forms uniquely suited to the reaction dynamics. That is, they be in terms of the concentration or pressure. Ksp is in reference to the solubility, Kf to complex ion formation, Kd to dissociation.