Chemical and Ionic Equilibrium - Result Question 18
####18. Consider the following equilibrium in a closed container
$$ N _2 O _4(g) \rightleftharpoons 2 NO _2(g) $$
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 $\left(K _p\right)$ and degree of dissociation $(\alpha)$ ?
$(2002,3 M)$
(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
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Solution:
$$ \begin{aligned} & N _2 O _4(g) \rightleftharpoons \underset{2}{2} \rightleftharpoons \underset{2}{2} NO _2(g) \quad \begin{array}{l} \text { Total } \ 1+\alpha \end{array} \ p _i: \quad \frac{1-\alpha}{1+\alpha} p & \frac{2 \alpha}{1+\alpha} p \quad K _p=\frac{4 \alpha^{2}}{1-\alpha^{2}} p \end{aligned} $$
At constant temperature, halving the volume will change both $p$ and $\alpha$ but $K _p$ remains constant.
