Calculating Q and K using Pressure and Concentration

[Glenda Marshall DIS 3M]

I remember from thermodynamics that you cannot use both concentrations and parietal pressures when calculating equilibrium constants which is why we always either calculated Kp or Kc. Why can we now use both of them when calculating K and Q for redox reactions?

For example, in homework #13.41 the solutions manual says that to calculate the Q value for 2H+(aq) + Cl- --> H2(g) + Cl2(g)

Q= (P H2)(P Cl2)
--------------
[H+]2 [Cl-]2

How is this different than before?
Thanks!

[Niharika Reddy 1D]

I had the same question, and it was answered on a previous post:

The exact reasons for this are quite technical and relate to the concept of standard states. In the end, if you want to use a "mixed" equilibrium expression, containing both concentrations and pressures, then the pressures must be in bars, and concentrations in molarity.

[Glenda Marshall DIS 3M]

Thanks Niharika! Sometimes I get a little lost in this big chemistry community world and don't see every post!

[Justin Le 2I]

I think part of the explanation is because ln ab = ln a + ln b. So it's like doing the concentrations and pressures separately and using log rules to combine them, which makes it easier on the calculator work.