Question

The following equilibrium in a closed container $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant $\left(K_p\right)$ and degree of dissociation?

• A. Neither $K_p$ and nor $\alpha$ changes.
• B. Both $K_p$ and $\alpha$ changes.
• C. $\alpha$ changes
• D. $K_p$ does not change

Answer 1

The correct option is B Both $K_p$ and $\alpha$ changes.

When the volume of the container is halved, the pressure is doubled. Due to this the equilibrium will shift in the backward direction which will decrease the number of moles and pressure. Hence the value of the equllibrium constant $K_p$ and the degre of dissociation $\alpha$ changes.