Ok, so the equilibrium constant in terms of pressure is the ratio of partial pressures at equilibrium of gaseous products to reactants (all raised to the powers equal to the stoichiometric coefficients). But if I have multiple equilibriums taking place simultaneously, where, let's say, there is one (gaseous) product common to both reactions, then that means the partial pressure of that substance will actually be equal to the sum of partial pressures of the moles of that substance created from both reactions.

For example:

$$\ce{X(s)} \ce{<=>} \ce{Y(g)} + \ce{Z(g)} $$

$$\ce{R(s)} \ce{<=>} \ce{Y(g)} + \ce{S(g)} $$

Here, let's say at equilibrium the partial pressure of Y and Z (the amount of them created from reaction 1) are $P_1$, and that of Y and S created in reaction 2 is $P_2$.

  1. My teacher said that the partial pressure of Y in the entire reaction mixture would be $P_1 + P_2$, which makes sense. But if we write the expression for $K_1$ and $K_2$, my teacher says we will take the partial pressure of Y to be $P_1 + P_2$. I'm having a hard time understanding why we won't just take the partial pressures of Y created in those specific reactions?

  2. Also, is it not true that the equilibrium constant of each of these reactions would change if we put them in a mixture together, since the concentration of $Y$ would change because that is present as a product in both reactions? Or would it just be that the equilibrium will move in the backward direction since more product is added?

  3. If the answer to (2) is that $K$ will not change, then that means if I only had reaction 1 taking place, equilibrium constant for that would also take pressure of Y to be $P_1 + P_2$, which seems obviously wrong to me- it should only be $P_1$.

Overall, stuff is not adding up in my mind for simultaneous equilibriums, if anyone could clear my concepts I would appreciate it!

  • 3
    $\begingroup$ Chemical equilibria are dynamical. This means that even in the equilibrium, the reactions are still happening but in both directions with the same reaction rate. Now think about a molecule of $\ce{Y}$. Once it is formed, it has no memory of where it came from. When it is formed, it is part of the whole collection of $\ce{Y}$ molecules in the container, and all of them together exert a what we call the partial pressure $P(\ce{Y})$, and act equally in both reactions. Even if a molecule was formed in the second reaction, it can then be used in the first reaction. Therefore, you cannot ... $\endgroup$
    – Domen
    Commented Jan 9, 2023 at 14:00
  • 2
    $\begingroup$ ... separate the two partial pressures of $\ce{Y}$ coming from the first or the second reaction because all $\ce{Y}$ molecules are indistinguishable. I hope this helps clearing up the confusion :) $\endgroup$
    – Domen
    Commented Jan 9, 2023 at 14:00
  • $\begingroup$ @Domen Yes, thank you! :) $\endgroup$
    – AVS
    Commented Jan 9, 2023 at 14:59
  • 3
    $\begingroup$ The equilibrium constant $K$ is just that, i..e constant at a fixed temperature so (2) is not correct. If now the amount of Y (or any other species) is changed then the amounts of the all species change until equilibrium is re-established; $K$ remains unchanged. $\endgroup$
    – porphyrin
    Commented Jan 10, 2023 at 8:05
  • $\begingroup$ Related: chemistry.stackexchange.com/q/111041 $\endgroup$
    – Karsten
    Commented Jan 10, 2023 at 13:01


Your Answer

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

Browse other questions tagged or ask your own question.