The amount of $PCl_5$ (in moles) that needs to be added to a $1\,L$ vessel at $250\,^oC$ in order to obtain $0.1\,mol$ of $Cl_2$ for the given reaction is:
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$; $K_C = 0.0414\,mol\,L^{-1}$

One mole of $N_2O_4(g)$ is taken in a closed vessel at $300 \, K$ under $1 \, atm$ pressure. When it is heated to $600 \, K$,$20 \%$ of $N_2O_4(g)$ dissociates into $NO_2(g)$. What is the final pressure in $atm$?

In a one-litre flask,$6$ moles of $A$ undergoes the reaction $A_{(g)} \rightleftharpoons P_{(g)}$. The progress of product formation at two temperatures (in Kelvin),$T_1$ and $T_2$,is shown in the figure:
If $T_1=2 T_2$ and $(\Delta G_2^{\Theta}-\Delta G_1^{\Theta})=R T_2 \ln x$,then the value of $x$ is. . . . .
$[\Delta G_1^{\Theta}$ and $\Delta G_2^{\Theta}$ are standard Gibb's free energy change for the reaction at temperatures $T_1$ and $T_2$,respectively.]

$A$ mixture of $1.57 \ mol$ of $N_2$,$1.92 \ mol$ of $H_2$ and $8.13 \ mol$ of $NH_3$ is introduced into a $20 \ L$ reaction vessel at $500 \ K$. At this temperature,the equilibrium constant,$K_c$ for the reaction $N_{2(g)} + 3H_{2(g)} \longleftrightarrow 2NH_{3(g)}$ is $1.7 \times 10^2$. Is the reaction mixture at equilibrium? If not,what is the direction of the net reaction?

If the equilibrium constant $K_c = 0.04$,how many moles/liter of $PCl_5$ are required to obtain $0.1$ mole of $Cl_2$?

$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are taken in a $4 \ L$ closed vessel at a constant temperature. The reaction is $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$. If $0.25\%$ of $N_2$ is converted into ammonia,calculate $K_c$ for this reaction and the $K_c'$ for the reaction $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightleftharpoons NH_{3(g)}$.