Changing pressure does not change equilibrium constant?

[chaser0]

Advertisement - Members don't see this ad

I was wondering about the logic of this.

If you increase the pressure, the reaction moves in the direction with the least amount of moles.
How exactly is this not changing the equilibrium constant? The equilibrium ratio of R/P is changing....

 

[milski]

If you're changing pressure, you're changing it proportionally for all gasses. The side which has more moles will have the change at higher power and be more affected by the change.

For example, if you had:

A+2B<==>3C+4D

K=[C]^3*[D]^4/[A]^2

If you double the pressure, you'll double each of [A][C][D]. The quotient becomes:

Q=8[C]^3*16[D]^4/2[A]4^2=16K

To get back to the same K we'll need to decrease the partial pressures of the products and increase the partial pressures of the reactants.

 

[MedPR]

Shifting the equilibrium does not change the equilibrium constant.

 

[Class1P]

See above.

The only thing that will change the Keq is the temperature. This is because Keq is equal to (the rate constant forward) / (the rate constant reverse) and the rate constants are dependent on the arrhenius equation:

http://www.chemguide.co.uk/physical/basicrates/arrhenius.html

 

[brandonh4]

I think the relevant concept is that, while Q will change, K remains the same at constant temperature. The reaction shifts due to the stress (Q changes), but it will shift back to K (eqm constant)