The equilibrium concentration of $x$,$y$ and $yx_2$ are $4$,$2$ and $2$ respectively for the equilibrium $2x + y \rightleftharpoons yx_2$. The value of equilibrium constant,$K_C$ is

If the equilibrium constant for the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $K$,what is the equilibrium constant of $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$?

The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:

$A$ reaction is $A + B \rightleftharpoons C + D$. Initially,we start with equal concentrations of $A$ and $B$. At equilibrium,the number of moles of $C$ is two times that of $A$. What is the equilibrium constant $(K_c)$ of the reaction?

One mole of a compound $AB$ reacts with one mole of a compound $CD$ according to the equation $AB + CD \rightleftharpoons AD + CB$. When equilibrium had been established,it was found that $\frac{3}{4} \ mol$ each of reactant $AB$ and $CD$ had been converted to $AD$ and $CB$. There is no change in volume. The equilibrium constant for the reaction is:

$PCl_5$,$PCl_3$,and $Cl_2$ are at equilibrium at $500 \ K$ with concentrations $[PCl_3] = 1.59 \ M$,$[Cl_2] = 1.59 \ M$,and $[PCl_5] = 1.41 \ M$. Calculate $K_c$ for the reaction:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$