For the reaction $x \rightleftharpoons y$,which of the following factors will affect the value of $[\text{Product}] / [\text{Reactant}]^{-1}$ at equilibrium?

$HI$ was heated in a closed tube at $440\,^{\circ}C$ until equilibrium was obtained. At this temperature,$22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be:

If the equilibrium constant for $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $K$,then the equilibrium constant for $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ will be :

From equations $1$ and $2$,
$CO_2 \rightleftharpoons CO + \frac{1}{2} O_2 \, [K_{C_1} = 9.1 \times 10^{-12} \, \text{at} \, 1000^{\circ} C] \, \text{(Eq. } i\text{)}$
$H_2O \rightleftharpoons H_2 + \frac{1}{2} O_2 \, [K_{C_2} = 7.1 \times 10^{-12} \, \text{at} \, 1000^{\circ} C] \, \text{(Eq. } ii\text{)}$
The equilibrium constant for the reaction,$CO_2 + H_2 \rightleftharpoons CO + H_2O$ at the same temperature,is

The following equilibrium constants are given:
$N_{2} + 3 H_{2} \rightleftharpoons 2 NH_{3} ; K_{1}$
$N_{2} + O_{2} \rightleftharpoons 2 NO ; K_{2}$
$H_{2} + \frac{1}{2} O_{2} \rightleftharpoons H_{2} O ; K_{3}$
The equilibrium constant for the oxidation of $2 \text{ mole}$ of $NH_{3}$ to give $NO$ is

Equilibrium constant $(K_c)$ of $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$ is $5 \times 10^{3}$. What is the equilibrium concentration of $HI$,if equilibrium concentrations of $H_{2_{(g)}}$ and $I_{2_{(g)}}$ respectively are $2.2 \times 10^{-2} \ M$ and $2.2 \times 10^{-4} \ M$?