The equilibrium composition for the reaction | vedclass

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(A) The initial equilibrium constant is $K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.40}{0.2 	imes 0.1} = 20$.When $0.2 \, mol$ of $Cl_2$ is added

Vedclass · Accepted Answer

$49$

Vedclass · Answer

(A) The initial equilibrium constant is $K_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.40}{0.2 	imes 0.1} = 20$.When $0.2 \, mol$ of $Cl_2$ is added to $1 \, L$ volume,the new concentration of $Cl_2$ becomes $0.1 + 0.2 = 0.3 \, M$.Let $x$ be the amount of $PCl_3$ and $Cl_2$ consumed to reach the new equilibrium.The new equilibrium concentrations are: $[PCl_3] = 0.2 - x$,$[Cl_2] = 0.3 - x$,and $[PCl_5] = 0.4 + x$.Substituting into the $K_c$ expression: $\frac{0.4 + x}{(0.2 - x)(0.3 - x)} = 20$.$0.4 + x = 20(0.06 - 0.5x + x^2) = 1.2 - 10x + 20x^2$.$20x^2 - 11x + 0.8 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x \approx 0.086$.Thus,$[PCl_5]_{eq} = 0.4 + 0.086 = 0.486 \, M = 48.6 	imes 10^{-2} \, mol \, L^{-1}$.Rounding to the nearest integer,the value is $49$.

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