$9.2 \ g$ of $N_2O_{4(g)}$ is taken in a $1 \ L$ closed vessel and heated until the following equilibrium is attained:
${N_2}{O_{4(g)}} \rightleftharpoons 2N{O_{2(g)}}$
If $50\%$ of $N_2O_{4(g)}$ dissociates at equilibrium,what will be the equilibrium constant (in $mol \ L^{-1}$)? (Mol. wt. of $N_2O_4 = 92$)

Equilibrium constants for the following reactions at $1200 \ K$ are given:
$2 \ H_2O_{(g)} \rightleftharpoons 2 \ H_{2(g)} + O_{2(g)}$
$K_1 = 6.4 \times 10^{-8}$
$2 \ CO_{2(g)} \rightleftharpoons 2 \ CO_{(g)} + O_{2(g)}$
$K_2 = 1.6 \times 10^{-6}$
The equilibrium constant for the reaction: $H_{2(g)} + CO_{2(g)} \rightleftharpoons CO_{(g)} + H_2O_{(g)}$
at $1200 \ K$ will be

For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_p = 41$ at $400 \ K$. Calculate $K_c$ for the following reactions at $400 \ K$:
$(a)$ $2N_{2(g)} + 6H_{2(g)} \rightleftharpoons 4NH_{3(g)}$
$(b)$ $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$(c)$ $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$

$A$ schematic plot of $\ln K_{eq}$ versus inverse of temperature $(1/T)$ for a reaction is shown below. The reaction must be

For the reaction,$2A_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$,the value of $K_c$ will be equal to:

The equilibrium constant for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$ at the same temperature is