Consider the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$. The equilibrium constant of the above reaction is $K_p$. If pure ammonia is left to dissociate,the partial pressure of ammonia at equilibrium is given by (Assume that $P_{NH_3} \ll P_{total}$ at equilibrium and $P$ is the total pressure).

In a closed vessel at $448^{\circ} C$,$0.5 \ mol$ of $H_2$ and $0.5 \ mol$ of $I_2$ react to form hydrogen iodide.
Reaction: $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,$K_c = 50$.
$(i)$ Calculate the moles of $I_2$ that remain unreacted at equilibrium.
$(ii)$ Calculate $K_p$.

In reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$,but at equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is:

Reaction between $N_{2}$ and $O_{2}$ takes place as follows:
$2 N_{2(g)} + O_{2(g)} \longleftrightarrow 2 N_{2}O_{(g)}$
If a mixture of $0.482 \ mol$ of $N_{2}$ and $0.933 \ mol$ of $O_{2}$ is placed in a $10 \ L$ reaction vessel and allowed to form $N_{2}O$ at a temperature for which $K_{c} = 2.0 \times 10^{-37}$,determine the composition of the equilibrium mixture.

The equilibrium constants of the following are
$N_2 + 3H_2 \rightleftharpoons 2NH_3 \,; \quad K_1$
$N_2 + O_2 \rightleftharpoons 2NO \,; \quad K_2$
$H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \,; \quad K_3$
The equilibrium constant $(K)$ of the reaction:
$2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O$ is:

$PCl_{5}$ dissociates as $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. $5 \, \text{moles}$ of $PCl_{5}$ are placed in a $200 \, L$ vessel which contains $2 \, \text{moles}$ of $N_{2}$ and is maintained at $600 \, K$. The equilibrium pressure is $2.46 \, atm$. The equilibrium constant $K_{p}$ for the dissociation of $PCl_{5}$ is $...... \times 10^{-3}$. (nearest integer) (Given: $R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1}$: Assume ideal gas behaviour)