$2$ moles of $N_2$ are mixed with $6$ moles of $H_2$ in a closed vessel of $1 \ L$ capacity. If $50\%$ of $N_2$ is converted into $NH_3$ at equilibrium,find the value of $K_c$ for the reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

Solid $KClO_3$ is taken in a container maintained at a constant pressure of $1 \ atm$. Upon heating,the following equilibrium is obtained:
$2KClO_{3(s)} \rightleftharpoons 2KCl_{(s)} + 3O_{2(g)}$
If $\Delta H^o = 25 \ kcal/mol$ and $\Delta S^o = 50 \ cal/K$,at what temperature will equilibrium be established in the container? (Ignore variation of $\Delta H^o$ and $\Delta S^o$ with temperature.) $......K$

At $298 \ K$,
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), K_1 = 4 \times 10^5$
$N_2(g) + O_2(g) \rightleftharpoons 2NO(g), K_2 = 1.6 \times 10^{12}$
$H_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons H_2O(g), K_3 = 1.0 \times 10^{-13}$
Based on the above equilibria,the equilibrium constant of the reaction,
$2NH_3(g) + \frac{5}{2}O_2(g) \rightleftharpoons 2NO(g) + 3H_2O(g)$
is $.......... \times 10^{-33}$ (Nearest integer).

In a closed vessel of $1 \ L$ capacity,$2 \ mol$ of $N_2$ and $6 \ mol$ of $H_2$ are mixed. If at equilibrium $50\% \ N_2$ is converted into $NH_3$,then the value of $K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$ will be:

The values of $K_p$ for the reactions,
$X \rightleftharpoons Y + Z$ $...(i)$
$A \rightleftharpoons 2B$ $...(ii)$
are in the ratio $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the ratio of total pressure at equilibrium for $(i)$ and $(ii)$ is:

$5 \ moles$ of $SO_2$ and $5 \ moles$ of $O_2$ are allowed to react. At equilibrium,it was found that $60\%$ of $SO_2$ is used up. If the partial pressure of the equilibrium mixture is $1 \ atm$,the partial pressure of $O_2$ is (in $atm$)