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 hold 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
• D. $K_p$ does not change but $\alpha$ changes

Answer 1

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

Use Le-Chatelier principle
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
$\therefore$ Number of moles are increasing in forward reaction.
Such reactions are favoured at low pressure according to Le-Chatelier principle.
Given, Volume is halved.
$\therefore$ Pressure is doubled
$\therefore$ Equilibrium will shift backward and degree of dissociation $\left(\alpha \right)$ will decreases.
$K_p$ is constant at constant temperature.