$A$ quantity of $PCl_5$ was heated in a $10 \ L$ vessel at $250 \ ^oC$; $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. At equilibrium,the vessel contains $0.1 \ mol$ of $PCl_5$,$0.20 \ mol$ of $PCl_3$,and $0.2 \ mol$ of $Cl_2$. The equilibrium constant $(K_c)$ of the reaction is:

For the reactions $SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}$ and $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$,if the equilibrium constants at $298 \ K$ are $K_1$ and $K_2$ respectively,then the correct relationship between them is .......

$15$ moles of $H_2$ and $5.2$ moles of $I_2$ are mixed and allowed to attain equilibrium at $500 \, ^oC$. At equilibrium,the concentration of $HI$ is found to be $10$ moles. The equilibrium constant for the formation of $HI$ is

For the reactions and their equilibrium constants given below,
$CuCl_4^{2-} + Br^{-} \rightleftharpoons CuCl_3Br^{2-} + Cl^{-}$; $K_1$
$CuCl_3Br^{2-} + Br^{-} \rightleftharpoons CuCl_2Br_2^{2-} + Cl^{-}$; $K_2$
$CuCl_2Br_2^{2-} + Br^{-} \rightleftharpoons CuClBr_3^{2-} + Cl^{-}$; $K_3$
$CuClBr_3^{2-} + Br^{-} \rightleftharpoons CuBr_4^{2-} + Cl^{-}$; $K_4$
The equilibrium constant,$K$ for the reaction $CuCl_4^{2-} + 3Br^{-} \rightleftharpoons CuClBr_3^{2-} + 3Cl^{-}$,is

Consider the following reactions in which all the reactants and products are present in gaseous state:
$2xy \rightleftharpoons x_2 + y_2$ $K_1 = 2.5 \times 10^5$
$xy + \frac{1}{2}z_2 \rightleftharpoons xyz$ $K_2 = 5 \times 10^{-3}$
The value of $K_3$ for the equilibrium $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$ is:

If the equilibrium constant for the reaction $A + 3B \rightleftharpoons 2C$ is $K_1$,then the equilibrium constant $K_2$ for the reaction $2C \rightleftharpoons A + 3B$ should be: