Question

Consider the following equilibrium in a closed container $N_2 O_4 \, (g) \rightleftharpoons 2NO_2 \, ( g)$ 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 $( K_p )$ and degree of dissociation $( \alpha)$ ?

• 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

Answer: (D)

$K_p$ does not change but $\alpha$ changes

Answer 2

\begin{array} \ N_2 O_4 \ (g) \rightleftharpoons 2NO_2 \ ( g) \ \ \ \ \ \ \ \ \ Total \\\ 1 - \alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 + \alpha \\\ \end{array}
$P_1 = \frac{ 1 - \alpha }{ 1 + \alpha} \frac{ 2 \alpha}{ 1 + \alpha } p \ \ K_p = \frac{ 4 \alpha^2 }{ 1 - \alpha^2 } p$
At constant temperature, having the volume will change both p and $\alpha \ but \ K_p$ remains constant.