$A$ reaction is $A + B \rightleftharpoons C + D$. Initially,we start with equal concentrations of $A$ and $B$. At equilibrium,the number of moles of $C$ is two times that of $A$. What is the equilibrium constant $(K_c)$ of the reaction?

Consider the following reactions in which all the reactants and products are in gaseous state:
$2PQ \rightleftharpoons P_2 + Q_2\,;\,K_1 = 2.5 \times 10^5$
$PQ + \frac{1}{2}R_2 \rightleftharpoons PQR\,;\,K_2 = 5 \times 10^{-3}$
The value of the equilibrium constant for the reaction:
$\frac{1}{2}P_2 + \frac{1}{2}Q_2 + \frac{1}{2}R_2 \rightleftharpoons PQR$ is

In a reversible reaction $A \underset{k_2}{\overset{k_1}{\longleftrightarrow}} B$,the initial concentrations of $A$ and $B$ are $a$ and $b$ in moles per litre,and the equilibrium concentrations are $(a - x)$ and $(b + x)$ respectively. Express $x$ in terms of $k_1, k_2, a,$ and $b$.

Steam reacts with iron at high temperature to give hydrogen gas and $Fe_3O_{4(s)}$. The correct expression for the equilibrium constant is

For the following equilibrium,$K_{C} = 6.3 \times 10^{14}$ at $1000 \ K$:
$NO_{(g)} + O_{3(g)} \longleftrightarrow NO_{2(g)} + O_{2(g)}$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{C}$ for the reverse reaction?

The value of $K_{c}$ for the reaction $3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$ is $2.0 \times 10^{-50}$ at $25^{\circ} C$. If the equilibrium concentration of $O_{2}$ in air at $25^{\circ} C$ is $1.6 \times 10^{-2} \, M$,what is the concentration of $O_{3} ?$