In a 500 ,mL flask,the degree of dissociation | vedclass

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(B) The chemical equation for the dissociation is: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ Initial moles: $PCl_5 = 5$,$PCl_3 = 0$,$Cl_2 = 0$ Degree of

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$2.66$

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(B) The chemical equation for the dissociation is: $PCl_5 ightleftharpoons PCl_3 + Cl_2$ Initial moles: $PCl_5 = 5$,$PCl_3 = 0$,$Cl_2 = 0$ Degree of dissociation $\alpha = 0.4$ Moles at equilibrium: $PCl_5 = 5(1 - 0.4) = 3$,$PCl_3 = 5 	imes 0.4 = 2$,$Cl_2 = 5 	imes 0.4 = 2$ Volume of flask $V = 500 \,mL = 0.5 \,L$ Concentrations at equilibrium: $[PCl_5] = 3 / 0.5 = 6 \,mol/L$,$[PCl_3] = 2 / 0.5 = 4 \,mol/L$,$[Cl_2] = 2 / 0.5 = 4 \,mol/L$ Equilibrium constant $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{4 	imes 4}{6} = \frac{16}{6} = 2.66 \,mol/L$

In a $500 \,mL$ flask,the degree of dissociation of $PCl_5$ at equilibrium is $40 \%$ and the initial amount is $5 \,moles$. The value of equilibrium constant in $mol \,L^{-1}$ for the decomposition of $PCl_5$ is

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