For the equilibrium $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$,the partial pressures of $SO_2$,$O_2$,and $SO_3$ are $0.662 \ atm$,$0.101 \ atm$,and $0.331 \ atm$ respectively. If the equilibrium concentrations of $SO_2$ and $SO_3$ are made equal,the partial pressure of $O_2$ will be ..... $atm$.

At $700 \, K$,the equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \longleftrightarrow 2 HI_{(g)}$ is $54.8$. If $0.5 \, mol \, L^{-1}$ of $HI_{(g)}$ is present at equilibrium at $700 \, K$,what are the concentrations of $H_{2(g)}$ and $I_{2(g)}$,assuming that we initially started with $HI_{(g)}$ and allowed it to reach equilibrium at $700 \, K$?

Observe the following equations:
$Ag^{+} + NH_3 \rightleftharpoons [Ag(NH_3)]^{+}$,$K_1 = 1.6 \times 10^3$
$[Ag(NH_3)]^{+} + NH_3 \rightleftharpoons [Ag(NH_3)_2]^{+}$,$K_2 = 6.8 \times 10^3$
The equilibrium constant for the following reaction,$Ag^{+} + 2 NH_3 \rightleftharpoons [Ag(NH_3)_2]^{+}$ is

$N_2O_{4(g)}$ at $300 \ K$ is kept in a closed container under $1 \ atm$. At equilibrium,$20\%$ of $N_2O_{4(g)}$ is converted to $NO_{2(g)}$.
$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
Hence,the resultant pressure is: (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:
$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$; $K_C = 0.0414\,mol\,L^{-1}$

$0.6 \ mol$ of $NH_3$ in a reaction vessel of $2 \ dm^3$ capacity was brought to equilibrium. The vessel was then found to contain $0.15 \ mol$ of $H_2$ formed by the reaction $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$. Which of the following statements is true?