Consider the following reaction equilibrium:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Initially,$1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$ are taken in a $2 \ L$ flask. At equilibrium,if the number of moles of $N_2$ is $0.6$,what is the total number of moles of all gases present in the flask?

Equivalent amounts of $H_2$ and $I_2$ are heated in a closed vessel till equilibrium is obtained. If $80\%$ of the hydrogen is converted to $HI$,the $K_c$ at this temperature is

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:

For the reaction,$A \rightleftharpoons n B$,the concentration of $A$ decreases from $0.06 \ mol \ L^{-1}$ to $0.03 \ mol \ L^{-1}$ and that of $B$ rises from $0$ to $0.06 \ mol \ L^{-1}$ at equilibrium. The values of $n$ and the equilibrium constant for the reaction,respectively,are

$K_{c}$ 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 $mol \ L^{-1}$?

$56 \, g$ 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: