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 holds true regarding the equilibrium constant $K_p$ 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 changes
• D. $K_p$ does not change, but $\alpha$ changes

Answer 1

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

At constnnt temperature $K_p$ or $K_c$ remains constant.
For the equilibria:
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
$K_p=K_c\times (RT{)}^{△n}$
$K_p=K_c$ because here $△n=1$
Since temperature is constant so $K_p$ or $K_c$ will remain constant. Further since volume is halved, the pressure will be doubled so $\alpha$ will decrease so as to maintain the constancy of $K_p$ or $K_c$.
$\begin{array}{cccc}& N_2{O}_4& \rightleftharpoons & 2N{O}_2\\ \text{initial}& 1& & 0\\ \text{at equilibrium}& (1-\alpha )& & 2\alpha \end{array}$
$\therefore \text{Total mole}=1-\alpha +2\alpha =1+\alpha$
Let total pressure = P
$\therefore p_{N_2{O}_4}=\frac{1-\alpha }{1+\alpha }.P$
$p_{N{O}_2}=\frac{2\alpha }{1-\alpha }P$
$K_p=\frac{p_{N{O}_2}^2}{p_{N_2{O}_4}}=\frac{4{\alpha }^2\times P}{(1-\alpha )(1+\alpha )}=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$
Since $K_p=$ constant, so $\alpha \propto \frac{1}{\sqrt{P}}$
So when volume is halved, pressure gets doubled and thus $\alpha$ will decrease.