If the equilibrium constant for the reaction,$2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $64$ at $500 \ K$,then the equilibrium constant for the reaction $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ at the same temperature is

The equilibrium constant for the reaction,$N_2 + 3H_2 \rightleftharpoons 2NH_3$ at $400 \ K$ is $41$. The equilibrium constant for the reaction,$\frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3$ at the same temperature will be closest to

In which of the following reactions will the concentration of the product be greater than the concentration of the reactant at equilibrium? ($k$ = equilibrium constant)

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

Two moles of $NH_3$ when put into a previously evacuated vessel $(1 \ L)$,partially dissociate into $N_2$ and $H_2$. If at equilibrium one mole of $NH_3$ is present,the equilibrium constant is

For the reactions $A \rightleftharpoons B; K_c = 2$,$B \rightleftharpoons C; K_c = 4$,and $C \rightleftharpoons D; K_c = 6$,the value of $K_c$ for the reaction $A \rightleftharpoons D$ is: