At $T \ K$,the $K_C$ value for the reaction $\frac{1}{3} N_{2(g)} + H_{2(g)} \rightleftharpoons \frac{2}{3} NH_{3(g)}$ is $50$. The $K_C$ value for the reaction $2 NH_{3(g)} \rightleftharpoons N_{2(g)} + 3 H_{2(g)}$ at the same temperature is:

The equilibrium constant $K_C$ of the reaction,$2 A \rightleftharpoons B + C$ is $0.5$ at $25^{\circ} C$. The reaction will proceed in the backward direction,when concentrations $[A], [B]$ and $[C]$ are,respectively:

$8 \, \text{mol}$ of $AB_3$ is added to a $1.0 \, \text{dm}^3$ container. If it dissociates according to the reaction $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$ and $2 \, \text{mol}$ of $A_2$ are present at equilibrium,what is the equilibrium constant $(K_c)$ for this reaction?

How is the direction of a reaction predicted?

Dihydrogen gas is obtained from water gas by the following equation:
$\underbrace{CO_{(g)} + H_2O_{(g)}}_{water\,gas} \rightleftharpoons_{500^{\circ}C} CO_{2_{(g)}} + H_{2_{(g)}}$
At $733 \ K$,the concentrations of $CO$,$H_2O$,$CO_2$,and $H_2$ in water gas are $0.18$,$0.0412$,$0.15$,and $0.2 \ mol \ L^{-1}$ respectively. Find $K_c$.

For the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$,the equilibrium mixture is given. If $1 \ mol$ of $H_2S$,$0.2 \ mol$ of $H_2$,and $0.8 \ mol$ of $S_2$ are taken in a $2 \ L$ vessel,find the value of $K_c$.