$PCl_5$,$PCl_3$,and $Cl_2$ are at equilibrium at $500 \ K$ with concentrations $[PCl_3] = 1.59 \ M$,$[Cl_2] = 1.59 \ M$,and $[PCl_5] = 1.41 \ M$. Calculate $K_c$ for the reaction:
$PCl_5 \rightleftharpoons PCl_3 + Cl_2$

Given the following equilibria:
$K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$
$K_2 : N_2 + O_2 \rightleftharpoons 2NO$
$K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$
The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ is:

The equilibrium constant for the given reaction is $100$.
$N_{2(g)} + 2 O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$
What is the equilibrium constant for the reaction given below?
$NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$

What is the equilibrium constant $(K_C)$ for the given reaction?
$N_2 + O_2 \rightleftharpoons 2 NO$
Where the equilibrium concentrations of $N_2$,$O_2$ and $NO$ are found to be $4 \times 10^{-3} \ M$,$3 \times 10^{-3} \ M$ and $3 \times 10^{-3} \ M$ respectively.

At $400 \ K$ in a closed vessel,the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ takes place. At equilibrium,the concentration of $H_2$ is $0.6 \ mol \ L^{-1}$,the concentration of $I_2$ is $0.8 \ mol \ L^{-1}$,and the concentration of $HI$ is $0.14 \ mol \ L^{-1}$. Calculate the equilibrium constant $(K_c)$.

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