If the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ at $720 \ K$ has $K = 48$,then for the reaction $2HI_{(g)} \rightleftharpoons H_{2(g)} + I_{2(g)}$,find its equilibrium constant.

If the equilibrium constant for $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ is $K,$ the equilibrium constant for $\frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{(g)}$ will be

In the reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$. At equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is:

If the equilibrium constants for the given reactions are $K_1$ and $K_2$ respectively,find the relationship between $K_2$ and $K_1$.
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$

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:

In the reaction $AB_{(g)} + CD_{(g)} \rightleftharpoons AD_{(g)} + CB_{(g)}$,one mole of $AB$ reacts with one mole of $CD$. When equilibrium is established,$3/4$ mole of each $AB$ and $CD$ are converted into $AD$ and $CB$. If the volume remains constant,the equilibrium constant for the reaction will be........