One mole of $N_2O_4$ in a $1 \ L$ flask decomposes to attain the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At the equilibrium the mole fraction of $NO_2$ is $1/2$. Hence $K_C$ will be:

For the given equilibrium reaction,$2 A(g) \rightleftharpoons 2 B(g) + C(g)$,the equilibrium constant $(K_c)$ at $1000 \ K$ is $4 \times 10^{-4}$. Calculate $K_p$ for the reaction at $800 \ K$ temperature.

$NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$
The reaction was started with some amount of $NH_4HS$. The equilibrium pressure at $25^{\circ}C$ is $0.5 \ atm$. What is $K_p$ for the reaction (in $atm^2$)?

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

For the equilibrium,$2 \ NOCl \ (g) \rightleftharpoons 2 \ NO \ (g) + Cl_{2} \ (g)$,the value of the equilibrium constant,$K_{c}$ is $3.75 \times 10^{-6}$ at $1069 \ K$. Calculate the $K_{p}$ for the reaction at this temperature?

The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below: $A(g) \rightleftharpoons B(g) + C(g)$.

$1/T \text{ (K}^{-1})$	$\log_{10} K_p$
$0.05$	$3.5$
$0.06$	$2.5$
$0.07$	$1.5$

The magnitude of $\frac{\Delta H^\circ}{R}$ calculated from the above data is . . . . . . . (Note: The slope $m = -\frac{\Delta H^\circ}{2.303 R}$)