For the reaction A_{(g)} + 2B_{(g)} rightleft | vedclass

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(A) The reaction quotient $Q_c$ is given by the expression: $Q_c = \frac{[C]^2}{[A][B]^2}$.Given concentrations in a $10 \, L$ vessel:$[A] = \frac{2 \

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Forward direction

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(A) The reaction quotient $Q_c$ is given by the expression: $Q_c = \frac{[C]^2}{[A][B]^2}$.Given concentrations in a $10 \, L$ vessel:$[A] = \frac{2 \, 	ext{mol}}{10 \, L} = 0.2 \, M$$[B] = \frac{3 \, 	ext{mol}}{10 \, L} = 0.3 \, M$$[C] = \frac{1 \, 	ext{mol}}{10 \, L} = 0.1 \, M$Substituting these values into the expression for $Q_c$:$Q_c = \frac{(0.1)^2}{(0.2)(0.3)^2} = \frac{0.01}{0.2 	imes 0.09} = \frac{0.01}{0.018} = \frac{10}{18} \approx 0.556$.Since $Q_c (0.556) < K_c (3.6)$,the reaction will proceed in the forward direction.

For the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)} + D_{(s)}$,$2 \, \text{moles}$ of $A$,$3 \, \text{moles}$ of $B$ and $1 \, \text{mole}$ of $C$ are present in a $10 \, L$ vessel. If $K_c$ for the reaction is $3.6$,the reaction will proceed in:

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