At a certain temperature,the equilibrium cons | vedclass

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(C) Initial concentration of each gas $= \frac{1 \ mol}{10 \ L} = 0.1 \ M$.Reaction quotient $Q_c = \frac{[C_2][D_2]}{[A_2][B_2]} = \frac{0.1 	imes 0

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$0.133$

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(C) Initial concentration of each gas $= \frac{1 \ mol}{10 \ L} = 0.1 \ M$.Reaction quotient $Q_c = \frac{[C_2][D_2]}{[A_2][B_2]} = \frac{0.1 	imes 0.1}{0.1 	imes 0.1} = 1$.Since $Q_c > K_c$ $(1 > 0.25)$,the reaction proceeds in the backward direction.Let $x$ be the change in concentration at equilibrium.$A_{2(g)} + B_{2(g)} ightleftharpoons C_{2(g)} + D_{2(g)}$Equilibrium concentrations: $[A_2] = 0.1 + x, [B_2] = 0.1 + x, [C_2] = 0.1 - x, [D_2] = 0.1 - x$.$K_c = \frac{(0.1 - x)^2}{(0.1 + x)^2} = 0.25$.Taking the square root on both sides: $\frac{0.1 - x}{0.1 + x} = \sqrt{0.25} = 0.5$.$0.1 - x = 0.05 + 0.5x$ $\Rightarrow 1.5x = 0.05$ $\Rightarrow x = \frac{0.05}{1.5} = 0.0333 \ M$.Equilibrium concentration of $[A_2] = 0.1 + 0.0333 = 0.1333 \ M$.

At a certain temperature,the equilibrium constant $K_c$ is $0.25$ for the reaction:
$A_{2(g)} + B_{2(g)} \rightleftharpoons C_{2(g)} + D_{2(g)}$
If we take $1 \ mol$ of each of the four gases in a $10 \ L$ container,what would be the equilibrium concentration of $A_{2(g)}$ (in $M$)?

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At a certain temperature,the equilibrium constant $K_c$ is $0.25$ for the reaction:$A_{2(g)} + B_{2(g)} \rightleftharpoons C_{2(g)} + D_{2(g)}$If we take $1 \ mol$ of each of the four gases in a $10 \ L$ container,what would be the equilibrium concentration of $A_{2(g)}$ (in $M$)?