At T(K),K_{c} value for AO_{2(g)} + BO_{2(g)} | vedclass

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(C) The given reaction is $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$ with $K_c = 16$.Initial concentrations in $1 \ L$ flask are

Vedclass · Accepted Answer

$1.6$

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(C) The given reaction is $AO_{2(g)} + BO_{2(g)} ightleftharpoons AO_{3(g)} + BO_{(g)}$ with $K_c = 16$.Initial concentrations in $1 \ L$ flask are $[AO_2] = 1 \ M$,$[BO_2] = 1 \ M$,$[AO_3] = 1 \ M$,and $[BO] = 1 \ M$.The reaction quotient $Q_c = \frac{[AO_3][BO]}{[AO_2][BO_2]} = \frac{1 	imes 1}{1 	imes 1} = 1$.Since $Q_c Let $x$ be the amount of $AO_2$ and $BO_2$ consumed at equilibrium.Equilibrium concentrations: $[AO_2] = 1-x$,$[BO_2] = 1-x$,$[AO_3] = 1+x$,$[BO] = 1+x$.$K_c = \frac{(1+x)(1+x)}{(1-x)(1-x)} = \left(\frac{1+x}{1-x}ight)^2 = 16$.Taking the square root: $\frac{1+x}{1-x} = 4$.$1+x = 4 - 4x \implies 5x = 3 \implies x = 0.6$.Equilibrium concentration of $AO_3 = 1 + x = 1 + 0.6 = 1.6 \ mol \ L^{-1}$.

At $T(K)$,$K_{c}$ value for $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$ is $16$. In a closed $1 \ L$ flask,one mole each of $AO_2, BO_2, AO_3$ and $BO$ are taken and heated to $T(K)$. What is the concentration (in $mol \ L^{-1}$) of $AO_3$ at equilibrium?

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