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: $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$

Consider the following gaseous equilibrium reactions $(I)$,$(II)$ and $(III)$ with equilibrium constants $K_1$,$K_2$ and $K_3$ respectively:
$I$) $\frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3$
$II$) $2 NO \rightleftharpoons N_2 + O_2$
$III$) $H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2 O$
The correct expression for the equilibrium constant for the gaseous equilibrium reaction $2 NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2 NO + 3 H_2 O$ is

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 ........

One mole of a compound $AB$ reacts with one mole of a compound $CD$ according to the equation $AB + CD \rightleftharpoons AD + CB$. When equilibrium had been established,it was found that $\frac{3}{4} \ mol$ each of reactant $AB$ and $CD$ had been converted to $AD$ and $CB$. There is no change in volume. The equilibrium constant for the reaction is:

For the reaction $A_{(g)} + B_{(g)} \rightleftharpoons 2 C_{(g)}$; $K_{c} = 4$. If equilibrium concentration of $A_{(g)}$ and $B_{(g)}$ are found to be $0.1 \ M$ and $0.4 \ M$ respectively. Determine equilibrium concentration of $C_{(g)}$. (in $M$)

For the reactions $A \rightleftharpoons B; K_c = 2$,$B \rightleftharpoons C; K_c = 4$,and $C \rightleftharpoons D; K_c = 6$,the value of $K_c$ for the reaction $A \rightleftharpoons D$ is: