For the reversible reaction $A + B \rightleftharpoons C + D$,the equilibrium concentrations of $C$ and $D$ are $0.8 \ mol/L$ each. If the initial concentrations of $A$ and $B$ were $1 \ mol/L$ each,calculate the equilibrium constant $K_c$.

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated at $473 \ K$ and $100 \ atm$ pressure. At equilibrium,the number of moles of $NH_3$ is $0.5 \ mol$. Calculate the equilibrium constant $K_p$ of the given reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

At $1000 \ K$,the equilibrium constant $K_C$ for the reaction $2 \ NOCl_{(g)} \rightleftharpoons 2 \ NO_{(g)} + Cl_{2(g)}$ is $4.0 \times 10^{-6} \ mol \ L^{-1}$. The $K_P$ (in bar) at the same temperature is $\left(R=0.083 \ L \ bar \ K^{-1} \ mol^{-1}\right)$

$1.1 \ mol$ of $A$ are mixed with $2.2 \ mol$ of $B$ and the mixture is kept in a $1 \ L$ vessel till the equilibrium is established in the reaction. If the molar concentration of $C$ at the equilibrium point is $0.2 \ M$,then find the value of the equilibrium constant $(K_c)$ for the reaction: $A + 2B \rightleftharpoons 2C + D$

For the reversible reaction in equilibrium:
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
Given $C_0 = C e^{-2.1 \times 10^{-3}t}$ for the forward reaction and $C'_0 = C' e^{-4.2 \times 10^{-4}t}$ for the backward reaction,calculate the equilibrium constant $K_c$ for the above reaction.

$A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$. The correct relationship between $K_P$,$\alpha$,and equilibrium pressure $P$ is: