#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.
