The degree of dissociation (alpha) of PCl_5 o | vedclass

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(B) For the equilibrium: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$Initial moles: $1 \quad 0 \quad 0$At equilibrium: $(1 - \alpha) \quad \alpha \quad \al

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$\alpha \propto \frac{1}{\sqrt{P}}$

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(B) For the equilibrium: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$Initial moles: $1 \quad 0 \quad 0$At equilibrium: $(1 - \alpha) \quad \alpha \quad \alpha$Total moles at equilibrium: $(1 - \alpha) + \alpha + \alpha = 1 + \alpha$Partial pressures: $p_{PCl_5} = \frac{1 - \alpha}{1 + \alpha} P$,$p_{PCl_3} = \frac{\alpha}{1 + \alpha} P$,$p_{Cl_2} = \frac{\alpha}{1 + \alpha} P$$K_P = \frac{p_{PCl_3} \cdot p_{Cl_2}}{p_{PCl_5}} = \frac{\left( \frac{\alpha}{1 + \alpha} P \right) \left( \frac{\alpha}{1 + \alpha} P \right)}{\left( \frac{1 - \alpha}{1 + \alpha} P \right)} = \frac{\alpha^2 P}{1 - \alpha^2}$For $\alpha Thus,$\alpha^2 \approx \frac{K_P}{P}$,which implies $\alpha \propto \frac{1}{\sqrt{P}}$.

The degree of dissociation $(\alpha)$ of $PCl_5$ obeying the equilibrium $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is related to the pressure at equilibrium by

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