$\Delta n$,the change in the number of moles for the reaction,$C_{12}H_{22}O_{11(s)} + 12O_{2(g)} \rightleftharpoons 12CO_{2(g)} + 11H_2O_{(l)}$ at $25 \ ^\circ C$ is

For the reaction $N_2O_{4_{(g)}} \rightleftharpoons 2NO_{2_{(g)}}$ at $298 \ K$,the equilibrium constant $K_p$ is $0.14 \ atm$. Calculate the value of $K_c$. $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

For the reaction $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$,if the total pressure in the reaction vessel at $105 \ ^\circ C$ is $1.12 \ atm$,then the $K_p$ for this equilibrium will be .........

At $1000 \ K$,if the equilibrium constant $K_p$ for the reaction $2 \ NOCl_{(g)} \rightleftharpoons 2 \ NO_{(g)} + Cl_{2(g)}$ is $4.157 \times 10^{-4} \ bar$,the $K_c$ (in $mol \ L^{-1}$) is $(R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1})$

For the reaction at $25\,^{\circ}C$,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,if $\Delta G^{\circ}_f$ for $N_2O_4$ and $NO_2$ are $23.49 \, kcal$ and $12.39 \, kcal$ respectively,then $K_p$ for the reaction is:

$9.2 \ g$ of $N_2O_{4(g)}$ is taken in a $1 \ L$ closed vessel and heated until the following equilibrium is attained:
${N_2}{O_{4(g)}} \rightleftharpoons 2N{O_{2(g)}}$
If $50\%$ of $N_2O_{4(g)}$ dissociates at equilibrium,what will be the equilibrium constant (in $mol \ L^{-1}$)? (Mol. wt. of $N_2O_4 = 92$)