[upllop]

At equilibrium, if pressure is increased, then the equilibrium will shift to favour the reaction producing the least amount of molecules. e.g 3A    2B. In this reaction, if pressure is increased, then the reaction will shift to the right to produce more of B.

Would this mechanism result in pressure decreasing until it reaches its initial levels?

Also how can K (equilibrium constant), remain a constant? because pressure changes will alter concentrations of the substances, hence K will change based on pressure.

[Hunter2]

K is a constant. It will not change, if T = Constant.

Kp = p^2(B)/p^3(A)

Let say pressure p(B) = 10 bar and p(A) = 10 bar

The you get  K= (10 bar)^2/ (10 bar)^3 = 0,1 bar^-1

p = nRT/V. R, T = constant, V the volume of the vessel is also constant, then p ~ n.

If pressure for pB rises to 20 bar, then pA will be 15,874 bar.

[Borek]

At equilibrium, if pressure is increased, then the equilibrium will shift to favour the reaction producing the least amount of molecules. e.g 3A    2B. In this reaction, if pressure is increased, then the reaction will shift to the right to produce more of B.

Would this mechanism result in pressure decreasing until it reaches its initial levels?

No, it will change till K is satisfied, doesn't mean pressure will be identical to initial.

K will change based on pressure

It will not. Reaction quotient will initially change, then the reaction will proceed till Q=K.