In the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,initially $1 \text{ mole}$ each of $PCl_5$ and $PCl_3$ are present. At equilibrium,$x \text{ moles}$ of $PCl_5$ remain. What is the total number of moles at equilibrium (in $- x$)?

The dissociation of $CO_2$ is represented as $2CO_2(g) \rightleftharpoons 2CO(g) + O_2(g)$. If $2 \ mol$ of $CO_2$ are taken initially and $40\%$ of $CO_2$ dissociates,what will be the total number of moles at equilibrium?

For the reaction $2NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$ at $300 \ K$,the value of $K_p$ is $2 \ atm^{-1}$. The total pressure at equilibrium is $10 \ atm$. If the volume of the container becomes two times its original volume,what will be its equilibrium pressure at $300 \ K$ (in $atm$)?

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:
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$; $K_C = 0.0414\,mol\,L^{-1}$

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 $SO_2$ and $O_2$ at $5 \, atm$ pressure reacts $30\%$ until equilibrium is reached. Determine the total pressure of the equilibrium mixture in $atm$.
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$