#sf(K_p)# is the equilibrium constant expressed in partial pressures.

#sf(K_c)# is the equilibrium constant expressed in concentrations.

For the general expression:

#sf(aA+bBrightleftharpoonscC+dD)#

We get:

#sf(K_p=(p_C^(c)xxp_D^(d))/(p_A^(a)xxp_B^(b)#

and

#sf(K_c=([C]^(c)[D]^(d))/([A]^(a)[B]^(b)))#

The Ideal Gas expression gives us:

#sf(PV=nRT)#

#:.##sf(P=n/V.RT)#

Since #sf(n/V)# is the concentration we can say that:

#sf(P=[gas].RT)#

From this it can be shown that the relationship between #sf(K_p)# and #sf(K_c)# is given by:

#sf(K_p=K_c(RT)^(Deltan))#

Where #sf(Deltan)# is the number of moles of product molecules - the number of moles of reactant molecules as they appear in the equation.

I'll give 3 examples that show the possibilities that I think you are asking for:

(1)

#sf(H_2+I_2rightleftharpoons2HI)#

#sf(Deltan=2-(1+1)=0)#

#:.##sf(K_p=K_(c)cancel((RT)^0)=K_c)#

Here you can see that #sf(K_p)# and #sf(K_c)# have the same numerical value and they are dimensionless quantities.

(2)

#sf(PCl_5rightleftharpoonsPCl_3+Cl_2)#

#sf(Deltan=(1+1)-1=1)#

#:.##sf(K_p=K_c(RT)^1=K_cRT)#

(3)

#sf(2SO_2+O_2rightleftharpoons2SO_3)#

#sf(Deltan=2-(2+1)=-1)#

#:.##sf(K_p=K_c(RT)^(-1)=K_c/(RT))#

In (2) and (3) it is not valid to compare the relative magnitudes of #sf(K_p)# and #sf(K_c)# as the numbers now have different dimensions.

Its like saying 10 kg is bigger than 2 km.