The equilibrium of formation of phosgene is represented as:
$CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$
The reaction is carried out in a $500 \ mL$ flask. At equilibrium,$0.3 \ mol$ of phosgene,$0.1 \ mol$ of $CO,$ and $0.1 \ mol$ of $Cl_2$ are present. The equilibrium constant $(K_c)$ of the reaction is:

For the reaction $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,$K_c = 1.8 \times 10^{-6}$ at $185\,^{\circ}C$. At $185\,^{\circ}C$,the value of $K_c$ for the reaction $NO_{(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons NO_{2(g)}$ is

Dihydrogen gas is obtained from water gas by the following equation:
$\underbrace{CO_{(g)} + H_2O_{(g)}}_{water\,gas} \rightleftharpoons_{500^{\circ}C} CO_{2_{(g)}} + H_{2_{(g)}}$
At $733 \ K$,the concentrations of $CO$,$H_2O$,$CO_2$,and $H_2$ in water gas are $0.18$,$0.0412$,$0.15$,and $0.2 \ mol \ L^{-1}$ respectively. Find $K_c$.

The equilibrium constant for the equilibrium $2HX_{(g)} \rightleftharpoons H_{2(g)} + X_{2(g)}$ is $1 \times 10^{-5}$. What will be the equilibrium concentration of $HX$ if the equilibrium concentrations for $H_2$ and $X_2$ are $1.2 \times 10^{-3} \ M$ and $1.2 \times 10^{-4} \ M$ respectively?

What is the equilibrium expression for the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$?

From equations $1$ and $2$,
$CO_2 \rightleftharpoons CO + \frac{1}{2} O_2 \, [K_{C_1} = 9.1 \times 10^{-12} \, \text{at} \, 1000^{\circ} C] \, \text{(Eq. } i\text{)}$
$H_2O \rightleftharpoons H_2 + \frac{1}{2} O_2 \, [K_{C_2} = 7.1 \times 10^{-12} \, \text{at} \, 1000^{\circ} C] \, \text{(Eq. } ii\text{)}$
The equilibrium constant for the reaction,$CO_2 + H_2 \rightleftharpoons CO + H_2O$ at the same temperature,is