For the reaction $MgCO_{3(s)} \rightleftharpoons MgO_{(s)} + CO_{2(g)}$,the value of $K_p$ is:

At $1000 \ K$,a vessel contains $CO_2$ at a pressure of $0.5 \ atm$. Some $CO_2$ is converted into $CO$ by the addition of graphite. If the total pressure at equilibrium is $0.8 \ atm$,what is the value of $K_p$ in $atm$?

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:

Assertion : For reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the unit of $K_C$ is $L^2 \, mol^{-2}$.
Reason : For the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,the equilibrium constant $K_C = \frac{[NH_3]^2}{[N_2][H_2]^3}$.

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:

$A$ mixture of $0.3 \ mol$ of $H_2$ and $0.3 \ mol$ of $I_2$ is allowed to react in a $10 \ L$ evacuated flask at $500 \ ^oC$. The reaction is $H_2 + I_2 \rightleftharpoons 2HI$,and the equilibrium constant $K_c$ is found to be $64$. The amount of unreacted $I_2$ at equilibrium is $...... \ mol$.