The equilibrium concentrations of $HI$,$I_2$,and $H_2$ are $0.7 \ M$,$0.1 \ M$,and $0.1 \ M$ respectively. The equilibrium constant for the reaction $I_2 + H_2 \rightleftharpoons 2HI$ is:

Equilibrium constant,$K_{c}$ for the reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ at $500 \, K$ is $0.061$. At a particular time,the analysis shows that the composition of the reaction mixture is $[N_{2}] = 3.0 \, mol \, L^{-1}$,$[H_{2}] = 2.0 \, mol \, L^{-1}$,and $[NH_{3}] = 0.5 \, mol \, L^{-1}$. Is the reaction at equilibrium? If not,in which direction does the reaction tend to proceed to reach equilibrium?

For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the correct expression of equilibrium constant $K$ is

For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the value of $K_p$ changes with:

For the reaction $2 H_{2(g)} + 2 NO_{(g)} \rightarrow N_{2(g)} + 2 H_2O_{(g)}$,the observed rate expression is $rate = k_f [NO]^2 [H_2]$. The rate expression of the reverse reaction is:

Write the applications of equilibrium constants.