$5 \ \text{moles}$ of $SO_2$ and $5 \ \text{moles}$ of $O_2$ are allowed to react to form $SO_3$ in a closed vessel. At the equilibrium stage,$60\%$ of $SO_2$ is used up. The total number of moles of $SO_2$,$O_2$,and $SO_3$ in the vessel now is:

Match List-$I$ (Equations) with List-$II$ (Type of processes) and select the correct option.

List-$I$ (Equations)	List-$II$ (Type of processes)
$A. K_p > Q$	$(i)$ Non-spontaneous
$B. \Delta G^\circ < RT \ln Q$	$(ii)$ Equilibrium
$C. K_p = Q$	$(iii)$ Spontaneous and endothermic
$D. T > \frac{\Delta H}{\Delta S}$	$(iv)$ Spontaneous

The values of $K_{P_1}$ and $K_{P_2}$ for the reactions
$X \rightleftharpoons Y + Z$ --- $(1)$
$A \rightleftharpoons 2B$ --- $(2)$
are in the ratio of $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the total pressure at equilibrium for $(1)$ and $(2)$ are in the ratio:

For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,what is the relationship between $\Delta H$ and $\Delta E$?

$3.00 \ mol$ of $PCl_5$ kept in $1 \ L$ closed reaction vessel was allowed to attain equilibrium at $380 \ K$. If $1.59 \ mol$ of reactant was converted into the product at equilibrium,then $K_c$ is:

$CH_3COCH_{3(g)} \rightleftharpoons C_2H_{6(g)} + CO_{(g)}$. The initial pressure of $CH_3COCH_3$ is $100 \ mm$. When equilibrium is set up,the mole fraction of $CO_{(g)}$ is $\frac{1}{4}$. Hence,the partial pressure of $CO$ is: