For the reaction ${N_2}_{(g)} + 3{H_2}_{(g)} \rightleftharpoons 2{NH_3}_{(g)}$,the equilibrium constant ${K_P}$ is $5.8 \times 10^5$ at $298 \ K$. Calculate the value of the equilibrium constant ${K_C}$ (in $mol^{-2} \ L^2$) at the same temperature.

For the equilibrium,$2 \ NOCl \ (g) \rightleftharpoons 2 \ NO \ (g) + Cl_{2} \ (g)$,the value of the equilibrium constant,$K_{c}$ is $3.75 \times 10^{-6}$ at $1069 \ K$. Calculate the $K_{p}$ for the reaction at this temperature?

The reaction rate for the reaction $[PtCl_4]^{2-} + H_2O \rightleftharpoons [Pt(H_2O)Cl_3]^- + Cl^-$ was measured as a function of concentrations of different species. It was observed that $\frac{-d[[PtCl_4]^{2-}]}{dt} = 4.8 \times 10^{-5} [[PtCl_4]^{2-}] - 2.4 \times 10^{-3} [[Pt(H_2O)Cl_3]^-] [Cl^-]$,where square brackets are used to denote molar concentrations. The equilibrium constant $K_c = ...$. (Nearest integer)

At $413 \ K$ temperature and $100 \ atm$ pressure,$1 \ mol$ $N_2$ and $3 \ mol$ $H_2$ are heated in a closed vessel. At equilibrium,$0.50 \ mol$ $NH_3$ is present. Find $K_p$.

In a closed container of $1000\, cm^3$,$2\, mol$ of $PCl_5$,$2\, mol$ of $PCl_3$,and $3\, mol$ of $Cl_2$ are found to be at equilibrium at $27\, ^oC$. Then $K_P$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at $27\, ^oC$ is $.....$ $atm$.

$9.2 \ g$ of $N_2O_{4(g)}$ is taken in a closed $1 \ L$ vessel and heated until the following equilibrium is reached: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,$50\%$ of $N_2O_{4(g)}$ is dissociated. What is the equilibrium constant $K_c$ (in $mol \ L^{-1}$)? (Molecular weight of $N_2O_4 = 92$)