In this equation that relates Standard free energy and the equilibrium constant:

$\Delta Gº=-RT \ln K_{eq}$

My textbook (and my teacher) say both $K_p$ (constant related to pressure) and $K_c$ (constant related to concentration) can be used in the place of $K_{eq}$.

When using the equation that relates $K_p$ with $K_c$

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

We would get two different $\Delta Gº$ values, which shouldn't be possible.

My question is, why can we use both constants if their values are different?


1 Answer 1


The equilibrium constant is defined by activities:

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The activity of a species in a liquid solution can be approximated to: $a=c/c^0$ where c is the concentration of the species and $c^o$ is the standard concentration ($1M$). This approximation leads to $K_{eq}=K_c$. The activity of gasses is called fugacity and it can be approximated to: $f=p/p^0$ where p is the pressure of the gas and $p^0$ is the standard pressure ($1$ $bar$). This leads to $K_{eq}=K_p$. Therefore, if a reaction takes place in e.g. water, one should use $K_c$, but if the reaction takes place in the gaseous state, $K_p$ should be used when calculating $dG$.


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