Chemistry Equilibrium Constant and Gibbs Energy

AI Thread Summary
The discussion centers on the relationship between Gibbs free energy change (ΔG) and the equilibrium constants Kc and Kp. It clarifies that ΔG should be expressed as ΔG° to reflect standard states, and emphasizes that the correct equilibrium constant to use in the equation ΔG = -RT ln K is Kp, not Kc. The relationship Kp = Kc(RT)Δn is acknowledged as correct, but it's noted that K in the ΔG equation refers to activities rather than concentrations or partial pressures. The participants are encouraged to review the derivation connecting ΔG° and Kp for better understanding. Overall, the conversation highlights the importance of precision in thermodynamic equations.
laser1
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We have that ##\Delta G = -RT\ln K##. This is in my lecture notes. However, it does not specify whether ##K## is ##K_c## or ##K_p##. Fair enough, I assumed that it could be both. However, when writing out the definitions of ##K_p## and ##K_c##, and using the fact that ##P=CRT##, where ##C## is the concentration, defined as ##n/V##, I noted the fact that ##K_p=K_c(RT)^{\Delta n}##.

So let's say $$\Delta G = -RT\ln K_p = -RT\ln K_c$$ It is clear that this equation cannot be true, right? As you get an extra factor of ##-RT\Delta n \ln(RT)## on one side. Where am I going wrong?
 
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You are going wrong in a number of places. First of all, in your initial equation, you should have ##\Delta G^0##, the free energy change between the standard states of reactants and products. Secondly, in your equation, you should have ##K_P##, nor ##K_C##. Third, the expression for ##K_P## should involve only partial pressures (in bars) and not ##\Delta n##.
 
Chestermiller said:
You are going wrong in a number of places. First of all, in your initial equation, you should have ##\Delta G^0##, the free energy change between the standard states of reactants and products. Secondly, in your equation, you should have ##K_P##, nor ##K_C##. Third, the expression for ##K_P## should involve only partial pressures (in bars) and not ##\Delta n##.
Thanks for the reply.

1. Fair enough.
2. Okay, so the ##K## in the formula refers to ##K_p##, NOT ##K_c##. Can you provide some insight on this? Or is it just by definition.
3. Are you sure? book states that ##K_p=K_c(RT)^{\Delta n}##.
 
laser1 said:
Thanks for the reply.

1. Fair enough.
2. Okay, so the ##K## in the formula refers to ##K_p##, NOT ##K_c##. Can you provide some insight on this? Or is it just by definition.
You need to review the derivation of the relationship between ##\Delta G^0## and ##K_P##
laser1 said:
3. Are you sure? book states that ##K_p=K_c(RT)^{\Delta n}##.
Your relationship between Kp and Kc is correct, if R is expressed in Joules/mole-K.
 
Note that K in the Delta G equation involves the activities, not the concentrations or partial pressures. In ideal systems #c/c^o=p/p^o#
 
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