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:

The suitable expression for the equilibrium constant of the reaction $2NO_{(g)} + Cl_{2(g)} \rightleftharpoons 2NOCl_{(g)}$ is

The following concentrations were observed at $500 \ K$ for the formation of $NH_3$ from $N_2$ and $H_2$. At equilibrium: $[N_2] = 2 \times 10^{-2} \ M$,$[H_2] = 3 \times 10^{-2} \ M$ and $[NH_3] = 1.5 \times 10^{-2} \ M$. The equilibrium constant for the reaction is:

$4.5 \ mol$ each of hydrogen and iodine are heated in a $10 \ L$ closed vessel. At equilibrium,$3 \ mol$ of $HI$ is formed. The equilibrium constant for the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ will be:

The decomposition of $N_2O_4$ to $NO_2$ is carried out at $280 \ K$ in chloroform. When equilibrium has been established,$0.2 \ mol$ of $N_2O_4$ and $2 \times 10^{-3} \ mol$ of $NO_2$ are present in $2 \ L$ solution. The equilibrium constant for the reaction $N_2O_4 \rightleftharpoons 2NO_2$ is:

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 $O_{3} ?$