$A$ vessel at $1000 \ K$ contains $CO_2$ with a pressure of $0.5 \ atm$. Some of the $CO_2$ is converted into $CO$ on the addition of graphite. If the total pressure at equilibrium is $0.8 \ atm$,the value of $K_p$ is $..... \ atm$.

$2N_2O_{4(g)} \rightleftharpoons 4NO_{2(g)}$ has $K_p = 0.15 \, atm$ at $298 \, K$. Calculate $K_p$ in $torr$ and $K_c$ in $mol \, L^{-1}$. ($1 \, atm = 760 \, torr$; $R = 0.082 \, L \, atm \, mol^{-1} \, K^{-1}$).

Partial pressures of $A$,$B$,$C$,and $D$ for the gaseous system $A + 2B \rightleftharpoons C + 3D$ are $A = 0.20 \ atm$,$B = 0.10 \ atm$,$C = 0.30 \ atm$,and $D = 0.50 \ atm$. The numerical value of the equilibrium constant $(K_p)$ is:

At $527 \ ^oC$,the reaction given below has $K_c = 4$.
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$
What is the $K_P$ for the following reaction?
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$
The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

At $800 \ K$ in a closed vessel,the molar concentrations of $N_2, O_2$ and $NO$ at equilibrium are $3.2 \times 10^{-3} \ M, 4.2 \times 10^{-3} \ M$ and $2.8 \times 10^{-3} \ M$ respectively. The approximate values of $K_{c}$ and $\frac{1}{K_{c}}$ for the following reaction are respectively: $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$