$A$ mixture of $1.57 \ mol$ of $N_2$,$1.92 \ mol$ of $H_2$ and $8.13 \ mol$ of $NH_3$ is introduced into a $20 \ L$ reaction vessel at $500 \ K$. At this temperature,the equilibrium constant,$K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ is $1.7 \times 10^2$. Is the reaction mixture at equilibrium? If not,what is the direction of the net reaction?

(D) The given reaction is: $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$
The concentrations of the species are:
$[N_2] = \frac{1.57}{20} \ mol \ L^{-1} = 0.0785 \ mol \ L^{-1}$
$[H_2] = \frac{1.92}{20} \ mol \ L^{-1} = 0.096 \ mol \ L^{-1}$
$[NH_3] = \frac{8.13}{20} \ mol \ L^{-1} = 0.4065 \ mol \ L^{-1}$
The reaction quotient $Q_c$ is calculated as:
$Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.4065)^2}{(0.0785)(0.096)^3}$
$Q_c = \frac{0.16524}{0.0785 \times 0.0008847} \approx \frac{0.16524}{0.00006945} \approx 2.38 \times 10^3$
Since $Q_c \approx 2.38 \times 10^3$ and $K_c = 1.7 \times 10^2$,we observe that $Q_c > K_c$.
Therefore,the reaction mixture is not at equilibrium and will proceed in the reverse direction to reach equilibrium.