For the reaction: $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,$K_P = 2 \times 10^{12}$ at $27^{\circ} C$ and $1 \ atm$ pressure. The $K_C$ for the same reaction is $......... \times 10^{13}$. (Nearest integer)
(Given $R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$)

For the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant at temperature $T$ is $4 \times 10^{-4}$. Find the value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$.

$2 \ mol$ of $N_2$ is mixed with $6 \ mol$ of $H_2$ in a closed vessel of $1 \ L$ capacity. If $50\%$ of $N_2$ is converted into $NH_3$ at equilibrium,the value of $K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ 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:

In the synthesis of $HI$,the amounts of $H_{2(g)}$,$I_{2(g)}$,and $HI_{(g)}$ at equilibrium were found to be $0.8 \ mol$,$0.8 \ mol$,and $2.4 \ mol$ respectively in a $10 \ L$ vessel. Calculate the equilibrium constant $(K_c)$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ and the equilibrium constant for the reverse reaction.

Equilibrium constant for the reaction $H_2O_{(g)} + CO_{(g)} \rightleftharpoons H_{2(g)} + CO_{2(g)}$ is $81$. If the velocity constant of the forward reaction is $162 \ L \ mol^{-1} \ s^{-1}$,what is the velocity constant (in $L \ mol^{-1} \ s^{-1}$) for the backward reaction?