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 $(K_c)$.

For the reaction $2 H_{2(g)} + 2 NO_{(g)} \rightarrow N_{2(g)} + 2 H_2O_{(g)}$,the observed rate expression is $rate = k_f [NO]^2 [H_2]$. The rate expression of the reverse reaction is:

At $780 \ K$ and $10 \ atm$ pressure,the equilibrium constant for the reaction $2 \ A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)}$ is $3.52$. At the same temperature and $7.04 \ atm$ pressure,the equilibrium constant for the same reaction is:

At $T \ K$,the equilibrium constants for the following two reactions are given below:
$2 A_{(g)} \rightleftharpoons B_{(g)} + C_{(g)} ; K_1 = 16$
$2 B_{(g)} + C_{(g)} \rightleftharpoons 2 D_{(g)} ; K_2 = 25$
What is the value of equilibrium constant $(K)$ for the reaction given below at $T \ K$?
$A_{(g)} + \frac{1}{2} B_{(g)} \rightleftharpoons D_{(g)}$

For the reaction $CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}$,the equilibrium constant $K$ is given by:

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: