A 1 L vessel contains 2 moles of PCl_5 init | vedclass

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(C) The reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Initially,moles: $PCl_5 = 2, PCl_3 = 0, Cl_2 = 0$At equilibrium,moles: $PCl

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$0.5$

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(C) The reaction is: $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Initially,moles: $PCl_5 = 2, PCl_3 = 0, Cl_2 = 0$At equilibrium,moles: $PCl_5 = 2-x, PCl_3 = x, Cl_2 = x$Since the volume is $1 \ L$,the concentrations are equal to the number of moles.$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{x \cdot x}{2-x} = 1$$x^2 = 2 - x \implies x^2 + x - 2 = 0$$(x+2)(x-1) = 0$Since $x$ cannot be negative,$x = 1$.Degree of dissociation $(\alpha)$ $= \frac{	ext{moles dissociated}}{	ext{initial moles}} = \frac{x}{2} = \frac{1}{2} = 0.5$.

$A$ $1 \ L$ vessel contains $2 \ moles$ of $PCl_5$ initially. If $K_c$ is found to be $1$,the degree of dissociation of $PCl_5$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ is:

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