For the reaction 2A_{(g)} rightleftharpoons B | vedclass

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(B) The reaction is $2A_{(g)} \rightleftharpoons B_{(g)} + 3C_{(g)}$.At equilibrium,the number of moles are $n_A = 2 \ mol$,$n_B = 2 \ mol$,and $n_C =

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$\frac{1}{2} \ L$

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(B) The reaction is $2A_{(g)} ightleftharpoons B_{(g)} + 3C_{(g)}$.At equilibrium,the number of moles are $n_A = 2 \ mol$,$n_B = 2 \ mol$,and $n_C = 2 \ mol$.Let the volume of the flask be $V \ L$.The concentrations are $[A] = \frac{2}{V}$,$[B] = \frac{2}{V}$,and $[C] = \frac{2}{V}$.The equilibrium constant expression is $K_c = \frac{[B][C]^3}{[A]^2}$.Substituting the values: $16 = \frac{(\frac{2}{V})(\frac{2}{V})^3}{(\frac{2}{V})^2}$.Simplifying the expression: $16 = \frac{(\frac{2}{V}) 	imes \frac{8}{V^3}}{\frac{4}{V^2}} = \frac{16}{V^4} 	imes \frac{V^2}{4} = \frac{4}{V^2}$.Thus,$V^2 = \frac{4}{16} = \frac{1}{4}$.Therefore,$V = \sqrt{\frac{1}{4}} = \frac{1}{2} \ L$.

For the reaction $2A_{(g)} \rightleftharpoons B_{(g)} + 3C_{(g)}$,at a given temperature $K_c = 16$,what must be the volume of the flask if a mixture of $2 \ mol$ each of $A, B, C$ exists at equilibrium?

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