At $783 \, K$ in the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,the molar concentrations $(mol \, L^{-1})$ of $H_2, I_2$ and $HI$ at some instant of time are $0.1, 0.2$ and $0.4$,respectively. If the equilibrium constant is $46$ at the same temperature,then as the reaction proceeds:

The equilibrium constant for the following reaction is $1.6 \times 10^{5}$ at $1024 \, K$:
$H_{2(g)} + Br_{2(g)} \longleftrightarrow 2HBr_{(g)}$
Find the equilibrium pressure of all gases if $10.0 \, bar$ of $HBr$ is introduced into a sealed container at $1024 \, K$.

$K_{c}$ for the following reaction is $99.0$: $A_{2(g)} \rightleftharpoons B_{2(g)}$. In a $1 \ L$ flask,$2 \ moles$ of $A_{2}$ were heated to $T(K)$ and equilibrium was reached. The concentrations at equilibrium of $A_{2}$ and $B_{2}$ are $C_{1}(A_{2})$ and $C_{2}(B_{2})$ respectively. Now,$1 \ mole$ of $A_{2}$ was added to the flask and heated to $T(K)$ to establish equilibrium again. The concentrations of $A_{2}$ and $B_{2}$ are $C_{3}(A_{2})$ and $C_{4}(B_{2})$ respectively. What is the value of $C_{3}(A_{2})$ in $mol \ L^{-1}$?

The amount of $PCl_5$ (in moles) that needs to be added to a $1\,L$ vessel at $250\,^oC$ in order to obtain $0.1\,mol$ of $Cl_2$ for the given reaction is:
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$; $K_C = 0.0414\,mol\,L^{-1}$

If the pressure in a reaction vessel for the following reaction is increased by decreasing the volume,what will happen to the concentrations of $CO$ and $CO_2$ ?
$H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)} + \text{Heat}$

At $T \ K$,$K_{c}$ for the reaction $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$ is $16$. If initially one mole each of all the four gases are taken in a $1 \ L$ vessel,the equilibrium concentrations of $SO_{3(g)}$ and $SO_{2(g)}$ in $mol \ L^{-1}$ respectively are: