Question

Consider the following equation 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 statement holds true regarding the $K_p$ and degree of dissociation '$\alpha$'.

• 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

Consider the following equation in a closed container
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

At eqm. 1-$\alpha$ 2$\alpha$
at a fixed temperature, the volume of the reaction container is halved, for this change, $K_p$ does not change but $\alpha$ changes.
The value of the equilibrium constant depends only on temperature. It is independent of volume of the reaction vessel, the concentrations (or partial pressures) of reactants and products and presence of catalyst or inert gas.
As pressure is increased due to lowering the volume, degree of dissociation will change to maintain the $K_p$.

Hence, the correct option is D.