Question

Consider the following equilibrium in a closed container,
$N_2{O}_{4\left(g\right)}\leftrightharpoons 2N{O}_{2\left(g\right)}$
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 ($K_p$) 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.

Answer:- (D) $K_P$ does not change, but $\alpha$ changes

For the equillibrium $\to N_2{O}_4\rightleftharpoons 2N{O}_2$

$K_P=K_C\times {\left(RT\right)}^{\Delta n}$

Since temperature is constant so $K_{C}or{K}_P$ will remain constant.

Since volume is halved, the pressure will be doubled, thus to maintain the constancy of $K_{C}or{K}_P,\alpha$ will decrease.