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

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(B) The decomposition reaction is: $PCl_5 \rightleftharpoons PCl_3 + Cl_2$Initial moles: $5, 0, 0$Moles at equilibrium: $5(1-\alpha), 5\alpha, 5\alpha

Vedclass · Accepted Answer

$2.66$

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(B) The decomposition reaction is: $PCl_5 ightleftharpoons PCl_3 + Cl_2$Initial moles: $5, 0, 0$Moles at equilibrium: $5(1-\alpha), 5\alpha, 5\alpha$Given $\alpha = 40 \% = 0.4$,so moles at equilibrium are: $5(1-0.4) = 3$,$5(0.4) = 2$,$5(0.4) = 2$Volume of flask $V = 500 \ mL = 0.5 \ L$Concentrations at equilibrium: $[PCl_5] = \frac{3}{0.5} = 6 \ M$,$[PCl_3] = \frac{2}{0.5} = 4 \ M$,$[Cl_2] = \frac{2}{0.5} = 4 \ M$$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{4 	imes 4}{6} = \frac{16}{6} = 2.66 \ mol \ L^{-1}$

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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