One mole of $N_2O_4$ in a $1 \ L$ flask decomposes to attain the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At the equilibrium the mole fraction of $NO_2$ is $1/2$. Hence $K_C$ will be:

If the equilibrium constant of a reaction is $20.0$ at equilibrium,and the rate constant of the forward reaction is $10.0$,then the rate constant for the backward reaction is:

For the reactions:
$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$
$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$
$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$
Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to:

For the reaction $2Ag_2O_{(s)} \rightleftharpoons 4Ag_{(s)} + O_{2(g)}$,the partial pressure of $O_2$ is given by:

For the reaction $AB_{(g)} \rightleftharpoons A_{(g)} + B_{(g)}$,$AB$ is $33.3\%$ dissociated at a total equilibrium pressure of $P$. Therefore,$P$ is correctly related to $K_p$ by which of the following options?

If $0.5 \ mol$ of $H_2$ and $0.5 \ mol$ of $I_2$ are reacted in a $10 \ L$ vessel at $444 \ ^\circ C$,and the equilibrium constant $K_c$ at the same temperature is $49$,then the ratio of $[HI]$ to $[I_2]$ is .......