At $500 \ K$,for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the $K_p$ is $0.036 \ atm^{-2}$. What is its $K_C$ in $L^2 \ mol^{-2}$? $(R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1})$.

In the reversible reaction $A + B \rightleftharpoons C + D$,the concentration of each $C$ and $D$ at equilibrium was $0.8 \ mol/L$. If the initial concentration of $A$ and $B$ was $1 \ mol/L$ each,then the equilibrium constant $K_c$ will be:

For the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + 1/2 O_{2(g)}$,the equilibrium constant $K_c = 4.9 \times 10^{-2}$. What is the value of $K_c$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$?

The reaction between $N_2$ and $H_2$ to form ammonia has $K_c = 6 \times 10^{-2}$ at the temperature $500 \ ^oC$. The numerical value of $K_p$ for this reaction is

The value of $K_p$ for the following reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$ is $1.2 \times 10^{-2}$ at $106.5 ^oC$. The value of $K_c$ for this reaction is

For the reversible reaction in equilibrium
$N_{2(g)} + O_{2(g)} \underset{k_2}{\overset{k_1}{\longleftrightarrow}} 2NO_{(g)}$
If the rate constant for the forward reaction is $k_1 = 2.1 \times 10^{-3} \ s^{-1}$ and for the backward reaction is $k_2 = 4.2 \times 10^{-4} \ s^{-1}$,then the equilibrium constant $K_c$ for the above reaction is: