$4.5$ moles each of hydrogen and iodine are heated in a sealed $10 \ L$ vessel. At equilibrium,$3$ moles of $HI$ are found. The equilibrium constant for ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$ is

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 $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$,the equilibrium mixture is given. If $1 \ mol$ of $H_2S$,$0.2 \ mol$ of $H_2$,and $0.8 \ mol$ of $S_2$ are taken in a $2 \ L$ vessel,find the value of $K_c$.

The equilibrium constant expression for a gas reaction is,
$K_{C} = \frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$
Write the balanced chemical equation corresponding to this expression.

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:

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: