Calculating Q and K using Pressure and Concentration


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Glenda Marshall DIS 3M
Posts: 68
Joined: Sat Sep 27, 2014 3:00 am

Calculating Q and K using Pressure and Concentration

Postby Glenda Marshall DIS 3M » Sat Feb 07, 2015 6:50 pm

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
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

Re: Calculating Q and K using Pressure and Concentration

Postby Niharika Reddy 1D » Sat Feb 07, 2015 7:37 pm

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

Chem_Mod wrote: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
Posts: 68
Joined: Sat Sep 27, 2014 3:00 am

Re: Calculating Q and K using Pressure and Concentration

Postby Glenda Marshall DIS 3M » Sat Feb 07, 2015 8:28 pm

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

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

Re: Calculating Q and K using Pressure and Concentration

Postby Justin Le 2I » Mon Feb 09, 2015 11:41 pm

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.


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