A sample of pure PCl_{5} was introduced into | vedclass

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(A) Let the concentrations of both $PCl_{3}$ and $Cl_{2}$ at equilibrium be $x \, mol \, L^{-1}$.The given reaction is:$PCl_{5(g)} \longleftrightarrow

Vedclass · Accepted Answer

$0.02 \, mol \, L^{-1}$

Vedclass · Answer

(A) Let the concentrations of both $PCl_{3}$ and $Cl_{2}$ at equilibrium be $x \, mol \, L^{-1}$.The given reaction is:$PCl_{5(g)} \longleftrightarrow PCl_{3(g)} + Cl_{2(g)}$At equilibrium,the concentrations are:$[PCl_{5}] = 0.5 	imes 10^{-1} \, mol \, L^{-1}$$[PCl_{3}] = x \, mol \, L^{-1}$$[Cl_{2}] = x \, mol \, L^{-1}$The equilibrium constant expression is:$K_{c} = \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}$Substituting the given values:$8.3 	imes 10^{-3} = \frac{x 	imes x}{0.5 	imes 10^{-1}}$$x^{2} = 8.3 	imes 10^{-3} 	imes 0.5 	imes 10^{-1}$$x^{2} = 4.15 	imes 10^{-4}$$x = \sqrt{4.15 	imes 10^{-4}} \approx 2.04 	imes 10^{-2} \, mol \, L^{-1}$Thus,$[PCl_{3}] = [Cl_{2}] \approx 0.02 \, mol \, L^{-1}$.

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