Equilibrium Constant and Gibbs Energy

[laser1]

Homework Statement

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Relevant Equations

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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?

 

[Chestermiller]

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 $\Delta n$.

 

[laser1]

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 $K_p=K_c(RT)^{\Delta n}$.

 

[Chestermiller]

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 $K_P$

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.

 

[DrDu]

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#