For the equilibrium $ \text{Liquid} \rightleftharpoons \text{Vapour} $,which of the following expressions is correct?

For the reaction $N_2 + O_2 \rightleftharpoons 2NO$,the equilibrium concentrations of $N_2$,$O_2$,and $NO$ are $0.25 \ mol/L$,$0.05 \ mol/L$,and $1.0 \ mol/L$ respectively. What are the initial concentrations of $N_2$ and $O_2$?

At $T \ K$,$K_{c}$ for the reaction $SO_{2(g)} + NO_{2(g)} \rightleftharpoons SO_{3(g)} + NO_{(g)}$ is $16$. If initially one mole each of all the four gases are taken in a $1 \ L$ vessel,the equilibrium concentrations of $SO_{3(g)}$ and $SO_{2(g)}$ in $mol \ L^{-1}$ respectively are:

$A$ vessel at $1000 \ K$ contains $CO_2$ with a pressure of $0.5 \ atm$. Some of $CO_2$ is converted into $CO$ on addition of graphite. If total pressure at equilibrium is $0.8 \ atm$,then $K_P$ is : (in $atm$)

One mole of $N_2O_4(g)$ is taken in a closed container at $1 \ atm$ and $300 \ K$. When it is heated to $600 \ K$,$20 \%$ of $N_2O_4(g)$ dissociates into $NO_2(g)$. The resulting pressure is .......... $atm$.

$X_{2(g)} + Y_{2(g)} \rightleftharpoons 2Z_{(g)}$
$X_{2(g)}$ and $Y_{2(g)}$ are added to a $1 \ L$ flask and it is found that the system attains the above equilibrium at $T \ K$ with the number of moles of $X_{2(g)}$, $Y_{2(g)}$ and $Z_{(g)}$ being $3$, $3$ and $9 \ mol$ respectively (equilibrium moles). Under these conditions of equilibrium, $10 \ mol$ of $Z_{(g)}$ is added to the flask and the temperature is maintained at $T \ K$. Then the number of moles of $Z_{(g)}$ in the flask when the new equilibrium is established is . . . . . . . (Nearest integer).