In which one of the following equilibria, K$$_p$$ ≠ K$$_c$$?

For every gaseous equilibrium we recall the relation

$$K_p=K_c\;(RT)^{\Delta n_g}$$

where $$\Delta n_g=\left(\text{total moles of gaseous products}\right)-\left(\text{total moles of gaseous reactants}\right).$$

If $$\Delta n_g=0$$, the factor $$(RT)^{\Delta n_g}=(RT)^0=1$$ and we immediately obtain $$K_p=K_c$$.

If $$\Delta n_g\neq 0$$, then $$(RT)^{\Delta n_g}\neq 1$$ and we have $$K_p\neq K_c$$. So our task reduces to evaluating $$\Delta n_g$$ for each given equilibrium.

Option A : $$2\,\text{HI}(g)\;\rightleftharpoons\;\text{H}_2(g)+\text{I}_2(g)$$

Total gaseous reactant moles  $$=2$$, product moles  $$=1+1=2$$. Hence

$$\Delta n_g = 2-2 = 0 \Longrightarrow K_p=K_c.$$

Option B : $$2\,\text{C}(s)+\text{O}_2(g)\;\rightleftharpoons\;2\,\text{CO}(g)$$

Only gases are counted, so solid carbon is ignored. Reactant gaseous moles  $$=1$$, product gaseous moles  $$=2$$. Therefore

$$\Delta n_g = 2-1 = +1 \neq 0 \Longrightarrow K_p\neq K_c.$$

Option C : $$2\,\text{NO}(g)\;\rightleftharpoons\;\text{N}_2(g)+\text{O}_2(g)$$

Reactant moles  $$=2$$, product moles  $$=1+1=2$$. Thus

$$\Delta n_g = 2-2 = 0 \Longrightarrow K_p=K_c.$$

Option D : $$\text{NO}_2(g)+\text{SO}_2(g)\;\rightleftharpoons\;\text{NO}(g)+\text{SO}_3(g)$$

Reactant moles  $$=1+1=2$$, product moles  $$=1+1=2$$. Hence

$$\Delta n_g = 2-2 = 0 \Longrightarrow K_p=K_c.$$

Among the four equilibria, only Option B gives $$\Delta n_g\neq 0$$ and therefore satisfies $$K_p\neq K_c$$.

Hence, the correct answer is Option B.