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 ofthe reaction container is halved. For this change, which of the following statements, holds false 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 $\alpha$ change
• C. $K_p$ changes but $\alpha$ does not change
• D. $K_p$ does not change but $\alpha$ changes

Answer 1

The correct option is D $K_p$ does not change but $\alpha$ changes

Degree of dissociation will decrease when volume is halved because of LeChatetier's Principle. When moles of product > moles of reactant for a gas phase reaction, decrease in volume will increase the pressure, causing less products to be formed.
Equilibrium constant depends only on temperature, so at a fixed temperature it will not change.