For the reactions:
$A \rightleftharpoons B$ $K_C = 2$
$B \rightleftharpoons C$ $K_C = 3$
$C \rightleftharpoons D + E$ $K_C = 5$
$K_C$ for the reaction $A \rightleftharpoons D + E$ is:

At $400 \ K$ in a closed vessel,the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ takes place. At equilibrium,the concentration of $H_2$ is $0.6 \ mol \ L^{-1}$,the concentration of $I_2$ is $0.8 \ mol \ L^{-1}$,and the concentration of $HI$ is $0.14 \ mol \ L^{-1}$. Calculate the equilibrium constant $(K_c)$.

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \, M$?
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$

At $490\,^{\circ}C$,the equilibrium constant for the synthesis of $HI$ is $50$. The value of $K$ for the dissociation of $HI$ will be:

Write the applications of equilibrium constants.

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: