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 given statements holds 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 $\alpha$ 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

$K_p$ remains constant with $T$.

For the reaction,
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

$K_c=\frac{4a^2{\alpha }^2}{(a-a\alpha )\cdot V}$ ...(i)

where, $a$ is the initial mole of $N_2{O}_4$ present in $VL$ and $\alpha$ is its degree of dissociation.

Also, $K_p=K_c{\left(RT\right)}^{\Delta n}$

On reducing the volume of container to $\frac{V}{2}L$, initial concentration of $N_2{O}_4$ becomes $\frac{2a}{V}$. An increase in concentration leads to more dissociation of $N_2{O}_4$ in order to have $K_c$ constant, a characteristic constant for a given reaction at a temperature.

Hence, option D is correct.