For the system $A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)}$,the equilibrium concentrations are $[A] = 0.06 \ mol/L$,$[B] = 0.12 \ mol/L$,and $[C] = 0.216 \ mol/L$. The $K_{eq}$ for the reaction is:

For the reactions and their equilibrium constants given below,
$CuCl_4^{2-} + Br^{-} \rightleftharpoons CuCl_3Br^{2-} + Cl^{-}$; $K_1$
$CuCl_3Br^{2-} + Br^{-} \rightleftharpoons CuCl_2Br_2^{2-} + Cl^{-}$; $K_2$
$CuCl_2Br_2^{2-} + Br^{-} \rightleftharpoons CuClBr_3^{2-} + Cl^{-}$; $K_3$
$CuClBr_3^{2-} + Br^{-} \rightleftharpoons CuBr_4^{2-} + Cl^{-}$; $K_4$
The equilibrium constant,$K$ for the reaction $CuCl_4^{2-} + 3Br^{-} \rightleftharpoons CuClBr_3^{2-} + 3Cl^{-}$,is

Before any reaction takes place,the value of the reaction quotient $(Q_c)$ is ......

In a closed vessel,$PCl_{5(g)}$ is obtained by the chemical reaction between $PCl_{3(g)}$ and $Cl_{2(g)}$. If the equilibrium concentrations in this vessel of $PCl_3$,$Cl_2$,and $PCl_5$ at $500 \ K$ are $1.59 \ M$,$1.59 \ M$,and $1.41 \ M$ respectively,then find the equilibrium constant $K_c$ for the reaction: $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$

For the chemical equilibrium $A + B \rightleftharpoons C + D$,when one mole of each reactant is mixed,$0.4 \ mol$ of each product is formed. The equilibrium constant $K_c$ is:

For the system $2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)}$,the expression for equilibrium constant $K$ is