At $400 \ K$ in a closed vessel,the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ takes place. At equilibrium,the concentration of $H_2$ is $0.6 \ mol \ L^{-1}$,the concentration of $I_2$ is $0.8 \ mol \ L^{-1}$,and the concentration of $HI$ is $0.14 \ mol \ L^{-1}$. Calculate the equilibrium constant $(K_c)$.

At $T(K)$,$K_c$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $99.0$. Two moles of $A_{2(g)}$ were heated to $T(K)$ in a $1 \ L$ closed flask to reach the above equilibrium. What are the concentrations (in $mol \ L^{-1}$) of $A_{2(g)}$ and $B_{2(g)}$ respectively at equilibrium?

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:

One mole of a compound $AB$ reacts with one mole of a compound $CD$ according to the equation $AB + CD \rightleftharpoons AD + CB$. When equilibrium had been established,it was found that $\frac{3}{4} \ mol$ each of reactant $AB$ and $CD$ had been converted to $AD$ and $CB$. There is no change in volume. The equilibrium constant for the reaction is:

Write the equilibrium constant (equilibrium law) for the following reactions:
$(i)$ $4NH_{3(g)} + 5O_{2(g)} \rightleftharpoons 4NO_{(g)} + 6H_{2}O_{(g)}$
$(ii)$ $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$
$(iii)$ $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

How is the direction of a reaction predicted?