Relationship between Kp and Kc

Consider the general reversible gaseous reaction $aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g)$. Substituting $P_A=[A]RT$, $P_B=[B]RT$, $P_C=[C]RT$ and $P_D=[D]RT$ (Section 12.4.4) into the $K_p$ expression, $K_p=\dfrac{P_C^{\,c}P_D^{\,d}}{P_A^{\,a}P_B^{\,b}}$, gives $K_p=\dfrac{[C]^c(RT)^c\,[D]^d(RT)^d}{[A]^a(RT)^a\,[B]^b(RT)^b}=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}\times(RT)^{(c+d)-(a+b)}$. Since $K_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$ (Section 12.4.3), this simplifies to the key relation $K_p=K_c(RT)^{\Delta n}$, where $\Delta n=(c+d)-(a+b)$ = (total moles of gaseous products) $-$ (total moles of gaseous reactants) in the balanced equation, and $R=8.314$ (with pressure expressed in bar, since standard pressure is defined as 1 bar; recall $1\ \text{Pa}=1\ \text{N m}^{-2}$ and $1\ \text{bar}=10^5\ \text{Pa}$). Problem 12.2 applies this to the Haber equilibrium $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$, where $\Delta n=2-4=-2$, giving $K_p=K_c\,(RT)^{-2}$. Problem 12.3 applies it to $H_2(g)+I_2(g)\rightleftharpoo …