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

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Question

(B) For the equilibrium: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$Initial moles: $1, 0, 0$Moles at equilibrium: $(1-\alpha), \alpha, \alpha$Total moles

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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: $1, 0, 0$Moles at equilibrium: $(1-\alpha), \alpha, \alpha$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{(\frac{\alpha}{1+\alpha} P)(\frac{\alpha}{1+\alpha} P)}{\frac{1-\alpha}{1+\alpha} P} = \frac{\alpha^2 P}{1-\alpha^2}$Since $\alpha$ is very small,$1-\alpha^2 \approx 1$,so $K_p \approx \alpha^2 P$Therefore,$\alpha^2 \propto \frac{1}{P}$ or $\alpha \propto \frac{1}{\sqrt{P}}$.

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

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