8 mol of AB_{3(g)} are introduced into a 1.0 | vedclass

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(C) The balanced chemical equation is: $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$Initial moles: $AB_3 = 8, A_2 = 0, B_2 = 0$Let $x$ be the e

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$27$

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(C) The balanced chemical equation is: $2AB_{3(g)} ightleftharpoons A_{2(g)} + 3B_{2(g)}$Initial moles: $AB_3 = 8, A_2 = 0, B_2 = 0$Let $x$ be the extent of reaction. At equilibrium,moles are: $AB_3 = 8 - 2x, A_2 = x, B_2 = 3x$Given that at equilibrium,$A_2 = 2 \ mol$,so $x = 2$.Equilibrium moles: $AB_3 = 8 - 2(2) = 4 \ mol$,$A_2 = 2 \ mol$,$B_2 = 3(2) = 6 \ mol$.Since the volume is $1.0 \ dm^3$,the concentrations are: $[AB_3] = 4 \ M, [A_2] = 2 \ M, [B_2] = 6 \ M$.The equilibrium constant $K_c$ is given by: $K_c = \frac{[A_2][B_2]^3}{[AB_3]^2}$$K_c = \frac{(2)(6)^3}{(4)^2} = \frac{2 	imes 216}{16} = \frac{432}{16} = 27$.

$8 \ mol$ of $AB_{3(g)}$ are introduced into a $1.0 \ dm^3$ vessel. If it dissociates as $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$. At equilibrium,$2 \ mol$ of $A_2$ are found to be present. The equilibrium constant of this reaction is

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