# Equilibrium constant magnitude for concentration vs. pressure

When introducing the concept of equilibrium constant, we often teach that the size of $$K$$ relative to $$1$$ indicates whether equilibrium lies toward the left (reactants) or right) products) of the balanced equation. However, we also know that there are different equilibrium constants depending on whether you express the numbers as ratios of concentrations ($$K_c$$) or pressures ($$K_P$$). So how should I think about the fact that the numerical value of $$K$$ depends on how I choose to express things?

I can illustrate what I mean with a simple example: Consider the reaction $$\ce{CO (g) + 2H2(g) <=> CH3OH(g)}$$ At $$220^\circ$$C, this reaction as $$K_c = 10.5$$ but $$K_P = 6\times10^{-3}$$. So, $$K_c$$ indicates an equilibrium lying far to the right and $$K_P$$ indicates one lying far to the left. How should I interpret this?

This is a classical scaling issue. For example, is a $$1$$ ton object lighter than a $$100$$ milligram object?
The rule that $$K>1$$ or $$K<1$$ should not be taught because it will lead to exactly the same fallacies as raised in the question. As you know very well that $$K_c$$ and $$K_p$$ are related by $$K_p=K_c(RT)^{\Delta n}.$$
Then, at least three scenarios can arise depending on the value of $${\Delta n}$$ as being equal to $$0,1, -1$$ in simple examples. This will lead to $$K_p=K_c,$$ $$K_p=RTK_c,$$ and $$K_p=\frac{K_c}{RT}.$$
• Thank you, this is really clear! But it brings up two questions: (1) If I had given a reaction with the same number of moles on each side, would $K_c$ have to equal $K_p$? And (2) if I convert partial pressures and concentrations to activities, do I get to the same value of $K$ in either case? Could you show how I might do that so that I can see the same numerical value results? Commented Dec 19, 2022 at 15:21