In a reaction $A + B \rightleftharpoons C + D$,the concentrations of $A$,$B$,$C$ and $D$ (in $mol/L$) are $0.5$,$0.8$,$0.4$ and $1.0$ respectively. The equilibrium constant is

For the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$,the equilibrium constant is $K$. For the reaction $2{N_2} + 6{H_2} \rightleftharpoons 4N{H_3}$,the equilibrium constant is $K'$. Then $K'$ is equal to:

The following concentrations were obtained for the formation of $NH_{3}$ from $N_{2}$ and $H_{2}$ at equilibrium at $500\, K$: $[N_{2}] = 1.5 \times 10^{-2}\, M$,$[H_{2}] = 3.0 \times 10^{-2}\, M$ and $[NH_{3}] = 1.2 \times 10^{-2}\, M$. Calculate the equilibrium constant.

If the equilibrium constant for the reaction,$2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $64$ at $500 \ K$,then the equilibrium constant for the reaction $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ at the same temperature is

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?

Explain the use of the equilibrium constant to predict the extent of a reaction with an example.