For the given hypothetical reactions,the equilibrium constants are as follows:
$X \rightleftharpoons Y ; K_1=1.0$
$Y \rightleftharpoons Z ; K_2=2.0$
$Z \rightleftharpoons W ; K_3=4.0$
The equilibrium constant for the reaction $X \rightleftharpoons W$ is (in $.0$)

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 $(K_c)$.

$PCl_5$,$PCl_3$,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:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$

$2 NOCl_{(g)} \rightleftharpoons 2 NO_{(g)} + Cl_{2(g)}$
In an experiment,$2.0 \ mol$ of $NOCl$ was placed in a $1 \ L$ flask and the concentration of $NO$ after equilibrium was established,was found to be $0.4 \ mol/L$. The equilibrium constant at $30^{\circ} C$ is $....... \times 10^{-4}$.

For the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$,the equilibrium constant is $K$. For the reaction $2{N_2} + 6{H_2} \rightleftharpoons 4N{H_3}$,the equilibrium constant is $K'$. Then $K'$ is equal to:

The reaction,$2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$ is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00 \ M$. When equilibrium is reached,the concentration of $D$ is measured and found to be $0.25 \ M$. The value for the equilibrium constant for this reaction is given by the expression: