$2$ 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: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

Consider the following gas phase reaction being carried out in a closed vessel at $25^\circ\text{C}$: $2A(g) \rightarrow 4B(g) + C(g)$. The table provides the total pressure of the system at different time intervals. Calculate the pressure of $C(g)$ at $30$ minutes time interval.

Time (min)	Total pressure (mm Hg)
$30$	$300$
$\infty$	$600$

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:

$A$ gaseous reaction $A_{2(g)} \to B_{(g)} + \frac{1}{2} C_{(g)}$ shows an increase in pressure from $100 \ mm$ of $Hg$ to $120 \ mm$ of $Hg$ in $5 \ min$. What will be the rate of disappearance of $A_2$ in $mm$ of $Hg/min$?

The incorrect expression among the following is:

The reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2_{(g)}} + S_{2_{(g)}}$ is in equilibrium. If $0.5 \ mol$ of $H_2S$,$0.10 \ mol$ of $H_2$,and $0.4 \ mol$ of $S_2$ are taken in a $1 \ L$ vessel,the value of the equilibrium constant $(K)$ is .... $mol \ L^{-1}$.