Equilibrium constant,$K_{c}$ for the reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ at $500 \, K$ is $0.061$. At a particular time,the analysis shows that the composition of the reaction mixture is $[N_{2}] = 3.0 \, mol \, L^{-1}$,$[H_{2}] = 2.0 \, mol \, L^{-1}$,and $[NH_{3}] = 0.5 \, mol \, L^{-1}$. Is the reaction at equilibrium? If not,in which direction does the reaction tend to proceed to reach equilibrium?

(A) The given reaction is: $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$
At a particular time,the concentrations are: $[N_{2}] = 3.0 \, mol \, L^{-1}$,$[H_{2}] = 2.0 \, mol \, L^{-1}$,$[NH_{3}] = 0.5 \, mol \, L^{-1}$
The reaction quotient $Q_{c}$ is calculated as:
$Q_{c} = \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}$
Substituting the values:
$Q_{c} = \frac{(0.5)^{2}}{(3.0)(2.0)^{3}} = \frac{0.25}{3.0 \times 8} = \frac{0.25}{24} \approx 0.0104$
Given $K_{c} = 0.061$.
Since $Q_{c} \neq K_{c}$,the reaction is not at equilibrium.
Since $Q_{c} < K_{c}$,the reaction will proceed in the forward direction to reach equilibrium.