# Changing pressure does not change equilibrium constant?

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#### chaser0

##### Full Member
10+ Year Member
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

##### 1K member
10+ Year Member
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.

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

##### Membership Revoked
Removed
10+ Year Member
Shifting the equilibrium does not change the equilibrium constant.

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Last edited:

#### brandonh4

##### Full Member
10+ Year Member
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)