Eight moles of a gas $AB_3$ attained equilibrium in a closed container of volume $1 \, dm^3$. The reaction is $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$. If at equilibrium $2 \, moles$ of $A_2$ are present,then the equilibrium constant is ...... $mol^2 \, L^{-2}$.

The equilibrium constants for the following reactions are $K_1$ and $K_2$,respectively.
$2 P_{(g)} + 3 Cl_{2(g)} \rightleftharpoons 2 PCl_{3(g)}$
$PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$
Then,the equilibrium constant for the reaction,$2 P_{(g)} + 5 Cl_{2(g)} \rightleftharpoons 2 PCl_{5(g)}$ is

$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
$56 \ g$ of nitrogen and $8 \ g$ of hydrogen gas are heated in a closed vessel. At equilibrium,$34 \ g$ of ammonia are present. The equilibrium number of moles of nitrogen,hydrogen and ammonia are respectively:

For the reaction $N_2 + O_2 \rightleftharpoons 2NO$ at $300 \, ^\circ C$,the value of $K_c$ is $9 \times 10^{-4}$. If equivalent amounts of $N_2$ and $O_2$ are used,what is the concentration of $NO$ at equilibrium (in terms of $a$) (in $, a$)?

One mole of $H_2O$ and one mole of $CO$ are taken in a $10 \ L$ vessel and heated to $725 \ K$. At equilibrium,$40 \%$ of water (by mass) reacts with $CO$ according to the equation:
$H_2O_{(g)} + CO_{(g)} \longleftrightarrow H_{2(g)} + CO_{2(g)}$
Calculate the equilibrium constant $(K_c)$ for the reaction.

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: