3
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

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 $$K_p=\frac{K_c}{RT}.$$

In short the simple rule of the equilibrium is favored if it is greater than unity is fallacious because it breaks down right here!

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Or rather, comparison K and 1 should be done only for true and unitless thermodynamic constants, expressed in relative terms of activities (or fugacities), referring themselves to their values at standard state. $\endgroup$
    – Poutnik
    Commented Dec 19, 2022 at 11:00
  • $\begingroup$ 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? $\endgroup$
    – pQ12branch
    Commented Dec 19, 2022 at 15:21
  • 1
    $\begingroup$ I agree with Poutnik. The implication of the conditions K>1 or K<1 should be taught but be properly explained. $\endgroup$
    – Buck Thorn ♦
    Commented Dec 19, 2022 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.