At $T(K)$,$K_c$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $99.0$. Two moles of $A_{2(g)}$ were heated to $T(K)$ in a $1 \ L$ closed flask to reach the above equilibrium. What are the concentrations (in $mol \ L^{-1}$) of $A_{2(g)}$ and $B_{2(g)}$ respectively at equilibrium?

For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the correct expression of equilibrium constant $K$ is

If for ${H_2(g)} + \frac{1}{2}{S_2(s)} \rightleftharpoons {H_2S(g)}$ and ${H_2(g)} + {Br_2(g)} \rightleftharpoons 2{HBr(g)}$ the equilibrium constants are $K_1$ and $K_2$ respectively,the reaction ${Br_2(g)} + {H_2S(g)} \rightleftharpoons 2{HBr(g)} + \frac{1}{2}{S_2(s)}$ would have equilibrium constant

For the reaction,$A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)}$ at $25 \, ^oC$,$2 \, moles$ of $A$,$1 \, mole$ of $B$ and $1 \, mole$ of $C$ are present in a $1 \, L$ vessel. If $K_c$ for the reaction is $2$,then the reaction will proceed in:

At a given temperature,the equilibrium constant for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ is $2.4 \times 10^{-3}$. At the same temperature,the equilibrium constant for the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$ is:

Equilibrium constant,$K_{c}$ for the reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ at $500 \, K$ is $0.061$. At a particular time,the analysis shows that the composition of the reaction mixture is $[N_{2}] = 3.0 \, mol \, L^{-1}$,$[H_{2}] = 2.0 \, mol \, L^{-1}$,and $[NH_{3}] = 0.5 \, mol \, L^{-1}$. Is the reaction at equilibrium? If not,in which direction does the reaction tend to proceed to reach equilibrium?