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$a$ moles of $\mathrm{PCl}_{5}$ is heated in a closed container to equilibriate $\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}$ at a pressure of $P$ atm. If $x$ moles of $\mathrm{PCl}_{5}$ dissociate at equilibrium, then
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The correct answer is:
$\frac{x}{a}=\left(\frac{K_{p}}{K_{p}+p}\right)^{1 / 2}$
$\begin{array}{l}
\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \\
\begin{array}{l}
1-\alpha \\
\alpha=\frac{x}{a} \quad \alpha \\
\text { Total } \text { moles }=1-\alpha+\alpha+\alpha=1+\alpha \\
K_{p}=\frac{P_{\mathrm{PCl}_{3}} \times P_{\mathrm{Cl}_{2}}}{P_{\mathrm{PCl}_{5}}}=\frac{\left[\frac{\alpha}{1+\alpha} \cdot p\right]\left[\frac{\alpha}{1+\alpha} \cdot p\right]}{\frac{1-\alpha}{1+\alpha} \cdot p} \\
=\frac{\alpha^{2} p}{1-\alpha^{2}} \Rightarrow \alpha=\left(\frac{K_{p}}{\mathrm{~K}_{p}+P}\right)^{1 / 2}
\end{array}
\end{array}$
\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})} \\
\begin{array}{l}
1-\alpha \\
\alpha=\frac{x}{a} \quad \alpha \\
\text { Total } \text { moles }=1-\alpha+\alpha+\alpha=1+\alpha \\
K_{p}=\frac{P_{\mathrm{PCl}_{3}} \times P_{\mathrm{Cl}_{2}}}{P_{\mathrm{PCl}_{5}}}=\frac{\left[\frac{\alpha}{1+\alpha} \cdot p\right]\left[\frac{\alpha}{1+\alpha} \cdot p\right]}{\frac{1-\alpha}{1+\alpha} \cdot p} \\
=\frac{\alpha^{2} p}{1-\alpha^{2}} \Rightarrow \alpha=\left(\frac{K_{p}}{\mathrm{~K}_{p}+P}\right)^{1 / 2}
\end{array}
\end{array}$
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