At constant temperature, if the pressure is changed at equilibrium of a gaseous reaction, then will the values of `K_(p),K_(c) and K_(x)` change?

Explanation

We know, `DeltaG^(0)=-RTlnK_(p) or, K_(p)=e^((DeltaG^(0))/(RT))` where, `DeltaG^(0)`=standard free energy change of the reaction.
(1) The value of `DeltaG^(0)` depends only on temperature. Its value is independent of pressure. So, the value of `K_p` is independent of pressure.
(2) We know, `K_(p)=K_(c)(RT)^(Deltan)`. since value of `K_(p)` does not depend on pressure, according to this relationship, the vlaue of `K_(c)` is also independent of pressure.
(3) Again, we know, `K_(p)=K_(x)(P)^(Deltan) or, K_(x)=(K_(p))/((P)^(Deltan))`.
As `K_(p)` does not depend on pressure, accoding to this relation, the value of `K_(x)` depend on pressure. however if `Deltan=0`, then pressure will have no effect on `K_(x)`.