The unit of the equilibrium constant for the reversible reaction $H_2 + I_2 \rightleftharpoons 2HI$ is:

Observe the following equilibrium in a $1 \text{ L}$ flask. $A_{(g)} \rightleftharpoons B_{(g)}$. At $T \text{ K}$,the equilibrium concentrations of $A$ and $B$ are $0.5 \text{ M}$ and $0.375 \text{ M}$ respectively. $0.1 \text{ moles}$ of $A$ is added into the flask and heated to $T \text{ K}$ to establish the equilibrium again. The new equilibrium concentrations (in $\text{M}$) of $A$ and $B$ are respectively.

$3.1 \ mol$ of $FeCl_3$ and $3.2 \ mol$ of $NH_4SCN$ are added to $1 \ L$ of water. At equilibrium,$3.0 \ mol$ of $FeSCN^{2+}$ is formed. The equilibrium constant $K_c$ for the reaction is:
$Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$

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 $HI_{(g)} \rightleftharpoons \frac{1}{2} H_{2(g)} + \frac{1}{2} I_{2(g)}$?

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 constants for three reactions are given as:
$(I) \, CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2_{(g)}} + H_{2_{(g)}} ; \, k_1$
$(II) \, CH_{4_{(g)}} + H_2O_{(g)} \rightleftharpoons CO_{(g)} + 3H_{2_{(g)}} ; \, k_2$
$(III) \, CH_{4_{(g)}} + 2H_2O_{(g)} \rightleftharpoons CO_{2_{(g)}} + 4H_{2_{(g)}} ; \, k_3$
The correct relationship between their equilibrium constants is: