Find out the value of $K_C$ for the following reaction from the value of $K_P$:
$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$
Given: $K_P = 8 \times 10^{12} \ atm$ at $500 \ K$,use $R = 0.08 \ L \ atm \ mol^{-1} \ K^{-1}$.

For the reactions $(1)$ and $(2)$ :
$A \rightleftharpoons B + C \dots (1)$
$D \rightleftharpoons 2E \dots (2)$
Given $K_{P_1} : K_{P_2} = 9 : 1$.
If the degree of dissociation of $A$ and $D$ is the same,then the total pressure at equilibria $(1)$ and $(2)$ are in the ratio (Assume reactions are started with equal number of moles of $A$ and $D$). (in $: 1$)

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.

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 $\left(R=0.083 \ L \ bar \ K^{-1} \ mol^{-1}\right)$

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 $m = -\frac{\Delta H^\circ}{2.303 R}$)

In the equilibrium system $AB_{(s)} \rightleftharpoons A_{(g)} + B_{(g)}$,if the equilibrium concentration of $A$ is doubled,the equilibrium concentration of $B$ will be .....