A sample of pure $\mathrm{PCl}_5$ was introduced into an evacuated vessel at $473 \mathrm{~K}$. After equilibrium was attained, concentration of $\mathrm{PCI}_5$ was found to be $0.5 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}$. If value of $K_{\mathrm{c}}$ is $8.3 \times 10^{-3}$, what are the concentrations of $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$ at equilibrium? $\mathbf{P C l}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$

Question

A sample of pure $\mathrm{PCl}_5$ was introduced into an evacuated vessel at $473 \mathrm{~K}$. After equilibrium was attained, concentration of $\mathrm{PCI}_5$ was found to be $0.5 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}$. If value of $K_{\mathrm{c}}$ is $8.3 \times 10^{-3}$, what are the concentrations of $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$ at equilibrium? $\mathbf{P C l}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$,

MARKS App · Accepted Answer

$\mathrm{K}_{\mathrm{c}}=\frac{x^2}{\left(0.5 \times 10^{-1}\right)}=8.3 \times 10^{-3}$ (Given) or $x^2=\left(8.3 \times 10^{-3}\right)\left(0.5 \times 10^{-1}\right)=4.15 \times 10^{-4}$ or $x=\sqrt{4.15 \times 10^{-4}}$ $=2.03 \times 10^{-2} \mathrm{M}=0.02 \mathrm{M}$

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