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 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

$\begin{array}{cccc}& N_2{O}_4\left(g\right)& \rightleftharpoons 2N{O}_2\left(g\right)& \left(Total\right)\\ & 1-\alpha & 2\alpha & 1+\alpha \\ pi:& \frac{1-\alpha }{1+\alpha }p& \frac{2\alpha }{1+\alpha }p& K_p=\frac{2\alpha }{1-{\alpha }^2}p\end{array}$
At constant temperature, having the volume changed will change both p and $\alpha$ but $K_p$ remains constant.