Equilibrium constant $(K_c)$ of $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$ is $5 \times 10^{3}$. What is the equilibrium concentration of $HI$,if equilibrium concentrations of $H_{2_{(g)}}$ and $I_{2_{(g)}}$ respectively are $2.2 \times 10^{-2} \ M$ and $2.2 \times 10^{-4} \ M$?

The value of $K_{c}$ for the reaction $3 O_{2(g)} \longleftrightarrow 2 O_{3(g)}$ is $2.0 \times 10^{-50}$ at $25^{\circ} C$. If the equilibrium concentration of $O_{2}$ in air at $25^{\circ} C$ is $1.6 \times 10^{-2} \, M$,what is the concentration of $O_{3} ?$

The reaction,$CO_{(g)} + 3H_{2(g)} \longleftrightarrow CH_{4(g)} + H_{2}O_{(g)}$ is at equilibrium at $1300 \, K$ in a $1 \, L$ flask. It also contains $0.30 \, mol$ of $CO$,$0.10 \, mol$ of $H_{2}$,and $0.02 \, mol$ of $H_{2}O$ and an unknown amount of $CH_{4}$ in the flask. Determine the concentration of $CH_{4}$ in the mixture. The equilibrium constant,$K_{c}$ for the reaction at the given temperature is $3.90$.

For the reaction ${N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}}$,the reaction quotient is given by ${Q = [NH_3]^2 / ([N_2][H_2]^3)}$. The reaction will proceed from left to right when ........

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 unit of the equilibrium constant for the reversible reaction $H_2 + I_2 \rightleftharpoons 2HI$ is: