Animated Solution for Chemistry - Chemical Equilibrium: In which one of the following equilibria, $K_p\ne K_c$?

Bridging $K_p$ and $K_c$

When dealing with chemical equilibrium, we often express the equilibrium constant in two ways: $K_c$ (based on molar concentrations) and $K_p$ (based on partial pressures). But how are they connected? The bridge between them is a beautifully simple equation derived from the ideal gas law:

Here, $R$ is the universal gas constant, $T$ is the absolute temperature, and $\Delta n_g$ is the star of our show. If we want to find out when $K_p$ is not equal to $K_c$, we just need to look at the exponent, $\Delta n_g$. If $\Delta n_g=0$, then $(RT{)}^0=1$, making $K_p=K_c$. Therefore, for $K_p eq K_c$, we must find a reaction where $\Delta n_g eq 0$.

The Gas Only Club

What exactly is $\Delta n_g$? It stands for the change in the number of moles of gaseous substances during the reaction.

This is where most students make a silly mistake! You must strictly ignore any species in the solid $(s)$ or liquid $(l)$ state. Their active masses are considered constant and do not contribute to the partial pressure changes in the same way gases do. They are not invited to the $\Delta n_g$ club.

Let's put our detective glasses on and evaluate each option one by one.

Option (a): $2\text{C}(s)+{\text{O}}_2(g)\rightleftharpoons 2\text{CO}(g)$ Look closely at the states. Carbon is a solid! We ignore it. The product has $2$ moles of gas ($\text{CO}$), and the reactant side has $1$ mole of gas (${\text{O}}_2$). $\Delta n_g=2-1=1$. Since $\Delta n_g eq 0$, this means $K_p eq K_c$. We might have found our winner, but let's verify the others just to be sure.

Option (b): $2\text{HI}(g)\rightleftharpoons {\text{H}}_2(g)+{\text{I}}_2(g)$ Everything is a gas here. Products have $1+1=2$ moles. Reactants have $2$ moles. $\Delta n_g=2-2=0$. Here, $K_p=K_c$.

Option (c): ${\text{NO}}_2(g)+{\text{SO}}_2(g)\rightleftharpoons \text{NO}(g)+{\text{SO}}_3(g)$ Again, an all-gas party. Products have $1+1=2$ moles. Reactants have $1+1=2$ moles. $\Delta n_g=2-2=0$. Here too, $K_p=K_c$.

Option (d): $2\text{NO}(g)\rightleftharpoons {\text{N}}_2(g)+{\text{O}}_2(g)$ Products have $1+1=2$ moles. Reactants have $2$ moles. $\Delta n_g=2-2=0$. Once again, $K_p=K_c$.

After carefully analyzing all the options, it is crystal clear that only the first reaction has a non-zero $\Delta n_g$. The presence of solid carbon was the trap, but by sticking to the rule of counting only gases, we easily navigated through it.

Therefore, the correct answer is (a).