Question

Consider 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 held true regarding the equilibrium constant $\left(K_p\right)$ and degree of dissociation $\left(\alpha \right)$ :-

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

Answer 1

Answer: (D)

$K_p$ does not change, but $\alpha$ -changes

At constant temperature, the value of the equilibrium constant $K_p$ remains constant.

As pressure is increased, the equilibrium will shift in backward direction (with decrease in the number of moles of gaseous species) to nullify the effect of increased pressure. This decreases the degree of dissociation $\alpha$.