8 mol of AB_3 is taken in a 1.0 dm^3 contai | vedclass

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(C) The reaction is: $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$Initial moles: $8, 0, 0$At equilibrium,$n(A_2) = 2 \ mol$. Since the stoichio

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$27$

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(C) The reaction is: $2AB_{3(g)} ightleftharpoons A_{2(g)} + 3B_{2(g)}$Initial moles: $8, 0, 0$At equilibrium,$n(A_2) = 2 \ mol$. Since the stoichiometry of $A_2$ is $1$,the amount of $AB_3$ reacted is $2 	imes 2 = 4 \ mol$.Remaining moles at equilibrium:$n(AB_3) = 8 - 4 = 4 \ mol$$n(A_2) = 2 \ mol$$n(B_2) = 3 	imes 2 = 6 \ mol$Since the volume is $1.0 \ L$,the concentrations are $[AB_3] = 4 \ M, [A_2] = 2 \ M, [B_2] = 6 \ M$.$K_c = \frac{[A_2][B_2]^3}{[AB_3]^2} = \frac{2 	imes (6)^3}{(4)^2} = \frac{2 	imes 216}{16} = \frac{432}{16} = 27$.

$8 \ mol$ of $AB_3$ is taken in a $1.0 \ dm^3$ container. It dissociates according to the reaction $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$. If $2 \ mol$ of $A_2$ are present at equilibrium,the equilibrium constant $K_c$ of the reaction is ....... $mol^2 \ L^{-2}$.

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