The equilibrium $NH_4HS_{(s)} \rightleftharpoons NH_{3(g)} + H_2S_{(g)}$ is set up at $127 \,^oC$ in a closed vessel. The total pressure at equilibrium was $20 \,atm$. The $K_C$ for the reaction is: (in $,M^2$)

The equilibrium concentrations of $N_2, H_2$ and $NH_3$ in the formation of $NH_3$ at $500 \ K$ are $1.25 \times 10^{-2} \ M, 4.0 \times 10^{-2} \ M$ and $1.6 \times 10^{-2} \ M$ respectively. The equilibrium constant $K_{p}$ at the same temperature is

$(1) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \ ; \ K_1$
$(2) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)} \ ; \ K_2$
$(3) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons H_2O_{(g)} \ ; \ K_3$
The equation for the equilibrium constant of the reaction
$2NH_{3(g)} + \frac{5}{2}O_{2(g)} \rightleftharpoons 2NO_{(g)} + 3H_2O_{(g)}$
$(K_4)$ in terms of $K_1$,$K_2$,and $K_3$ is

$K_{c}$ for the reaction,$A_{2(g)} \rightleftarrows B_{2(g)}$ is $99.0$. In a $1 \ L$ closed flask,two moles of $B_{2(g)}$ are heated to $T(K)$. What is the concentration of $B_{2(g)}$ (in $mol \ L^{-1}$) at equilibrium?

$2N_2O_{4(g)} \rightleftharpoons 4NO_{2(g)}$ has $K_p = 0.15 \, atm$ at $298 \, K$. Calculate $K_p$ in $torr$ and $K_c$ in $mol \, L^{-1}$. ($1 \, atm = 760 \, torr$; $R = 0.082 \, L \, atm \, mol^{-1} \, K^{-1}$).

For the reaction,$PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$,the value of $K_c$ at $250 \ ^oC$ is $26$. The value of $K_p$ at this temperature will be: