For the reaction ${N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}}$,the reaction quotient is given by ${Q = [NH_3]^2 / ([N_2][H_2]^3)}$. The reaction will proceed from left to right when ........

Consider the following gaseous equilibria with equilibrium constants $K_{1}$ and $K_{2}$ respectively:
$SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$
$2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$
The equilibrium constants are related as:

In the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the equilibrium concentrations of $PCl_5$ and $PCl_3$ are $0.4 \ mol/L$ and $0.2 \ mol/L$ respectively. If the value of $K_c$ is $0.5$,what is the concentration of $Cl_2$ in $mol/L$?

The following equilibrium constants are given: $N_2 + 3 H_2 \rightleftharpoons 2 NH_3$ $(k_1)$,$N_2 + O_2 \rightleftharpoons 2 NO$ $(k_2)$,$H_2 + 1/2 O_2 \rightleftharpoons H_2 O$ $(k_3)$. The equilibrium constant for the oxidation of $1 \text{ mole } NH_3$ by oxygen to give $NO$ according to the reaction $NH_3 + 5/4 O_2 \rightleftharpoons NO + 3/2 H_2 O$ is:

$A$ reaction with reaction quotient $Q_C$ and equilibrium constant $K_C$ will proceed in the direction of the products when:

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. If the equilibrium constant $(K_C)$ for the above reaction at $500 \ K$ is $1.79$ and the equilibrium concentrations of $PCl_5$ and $PCl_3$ are $1.41 \ M$ and $1.59 \ M$,respectively,then the concentration of $Cl_2$ is approximately: (in $M$)