If $K_C$ for the equilibrium reaction $2 ABC_{(g)} \rightleftharpoons 2 AB_{(g)} + C_{2(g)}$ is $X$ at $T \ K$,its $K_P$ at the same temperature is:

The equilibrium constant for the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ is $K.$ Then,the equilibrium constant for the equilibrium $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$ is

Calculate $K_{C}$ for the reversible process given below if $K_{P}=167$ and $T=800^{\circ}C$.
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$

For the following three equilibria $X, Y$,and $Z$ in the gaseous state at $300 \ K$,the increasing order of the ratio of $K_p$ to $K_c$ is:
$X: 2SO_2 + O_2 \rightleftharpoons 2SO_3$
$Y: PCl_5 \rightleftharpoons PCl_3 + Cl_2$
$Z: 2HI \rightleftharpoons H_2 + I_2$

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:

At $T(K)$,$K_{c}$ value for $AO_{2(g)} + BO_{2(g)} \rightleftharpoons AO_{3(g)} + BO_{(g)}$ is $16$. In a closed $1 \ L$ flask,one mole each of $AO_2, BO_2, AO_3$ and $BO$ are taken and heated to $T(K)$. What is the concentration (in $mol \ L^{-1}$) of $AO_3$ at equilibrium?