In the reaction,$H_2 + I_2 \rightleftharpoons 2HI$. In a $2 \ L$ flask,$0.4 \ mol$ of each $H_2$ and $I_2$ are taken. At equilibrium,$0.5 \ mol$ of $HI$ are formed. What will be the value of equilibrium constant,$K_c$?

$K_p$ for the following reaction is $3.0$ at $1000 \ K$.
$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$
What will be the value of $K_c$ for the reaction at the same temperature?
(Given: $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)

$\Delta n$,the change in the number of moles for the reaction,$C_{12}H_{22}O_{11(s)} + 12O_{2(g)} \rightleftharpoons 12CO_{2(g)} + 11H_2O_{(l)}$ at $25 \ ^\circ C$ is

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

At $-20^{\circ} C$ and $1 \ atm$ pressure,a cylinder is filled with equal number of $H_2$,$I_2$ and $HI$ molecules for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$. The $K_p$ for the process is $x \times 10^{-1}$. Find the value of $x$. [Given: $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$]

For the following gas phase equilibrium reaction at constant temperature,$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$. If the total pressure is $\sqrt{3} \ atm$ and the pressure equilibrium constant $(K_p)$ is $9 \ atm$,then the degree of dissociation is given as $(x \times 10^{-2})^{-1/2}$. The value of $x$ is . . . . . . (Nearest integer)