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 $.0$)

Consider the following gaseous equilibria with equilibrium constants $K_{1}$ and $K_{2}$ respectively:
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$
$2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$
The equilibrium constants are related as:

If the equilibrium constant for the reaction $2AB \rightleftharpoons A_2 + B_2$ is $49$,then the equilibrium constant for the reaction $AB \rightleftharpoons \frac{1}{2}A_2 + \frac{1}{2}B_2$ will be:

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:

Consider the following equilibrium reaction in gaseous state at $T(K)$.
$A(g) + 2B(g) \rightleftharpoons 2C(g) + D(g)$
The initial concentration of $B$ is $1.5$ times that of $A$. At equilibrium,the concentrations of $A$ and $B$ are equal. The equilibrium constant for the reaction is

How is the direction of a reaction predicted?