At T K,K_c for the reaction AO_{2(g)} + BO_{ | vedclass

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(A) The reaction is $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$.Given $K_c = 16$ and initial concentrations $[AO_2] = [BO_2] = [AO

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$1.6$

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(A) The reaction is $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$.Given $K_c = 16$ and initial concentrations $[AO_2] = [BO_2] = [AO_3] = [BO] = 1.0 \ M$.Let the change in concentration be $x$.At equilibrium:$[AO_2] = 1 - x$$[BO_2] = 1 - x$$[AO_3] = 1 + x$$[BO] = 1 + x$$K_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = 16$Taking the square root of both sides:$\frac{1+x}{1-x} = 4$$1 + x = 4 - 4x$$5x = 3 \implies x = 0.6$Equilibrium concentration of $[BO] = 1 + x = 1 + 0.6 = 1.6 \ mol \ L^{-1}$.

At $T \ K$,$K_c$ for the reaction $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$ is $16$. One mole each of reactants and products are taken in a $1 \ L$ flask and heated to $T \ K$,and equilibrium is established. What is the equilibrium concentration of $BO$ (in $mol \ L^{-1}$)?

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