Question
Show that as long as the extent of the reaction, $\alpha,$ for the disassociation reaction $\mathrm{X}_{2} \rightleftharpoons 2 \mathrm{X}$ is smaller than one, $\alpha$ is given by $\alpha=\sqrt{\frac{K_{P}}{4+K_{P}}}$
Step 1
At equilibrium, let \(\alpha\) be the extent of the reaction, which means \(\alpha n_0\) moles of \(\mathrm{X}_{2}\) have dissociated. Show more…
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Show that as long as the extent of the reaction, $\alpha$ for the dissociation reaction $\mathrm{X}_{2} \rightleftharpoons 2 \mathrm{X}$ is smaller than $1, \alpha$ is given by $$ \alpha=\sqrt{\frac{K_{P}}{4+K_{P}}} $$
For the reaction $2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})$ if the degree of dissociation is $\alpha$ and $\mathrm{K}_{p}$ is the equilibrium constant, then (a) $\alpha=\frac{\sqrt{2 \mathrm{~K}_{\mathrm{p}}}}{1+\sqrt{2 \mathrm{k}_{\mathrm{p}}}}$ (b) $\alpha=\frac{2 \sqrt{\mathrm{K}_{\mathrm{P}}}}{1+2 \sqrt{\mathrm{K}_{\mathrm{P}}}}$ (c) $\sqrt{2 K_{\mathrm{P}}}=\left(\frac{\alpha}{1-\alpha}\right)$ (d) $2 \sqrt{K_{P}}=\left(\frac{\alpha}{1-\alpha}\right)$
For the reaction $\mathrm{AB}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{A}(\mathrm{g})+\mathrm{B}_{2}(\mathrm{~g})$ The degree of dissociation ' $\alpha$ ' is negligible as compared to 1 (unity); the degree of dissociation may be expressed as: (a) $\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}$ (b) $\alpha \propto \sqrt{\mathrm{V}}$ (c) $\alpha \propto \frac{1}{P}$ (d) $\alpha \propto \frac{1}{\mathrm{~V}}$
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