$4.5 \ mol$ each of hydrogen and iodine are heated in a $10 \ L$ closed vessel. At equilibrium,$3 \ mol$ of $HI$ is formed. The equilibrium constant for the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ will be:

The compounds $A$ and $B$ are mixed in equimolar proportion to form the products,$A + B \rightleftharpoons C + D$. At equilibrium,one-third of $A$ and $B$ are consumed. The equilibrium constant for the reaction is:

In a chemical equilibrium $A + B \rightleftharpoons C + D$,when $1 \, mol$ each of two reactants are mixed,$0.5 \, mol$ each of the products are formed. The equilibrium constant is

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

$4.5 \text{ moles}$ each of hydrogen and iodine is heated in a sealed $10 \text{ litre}$ vessel. At equilibrium,$3 \text{ moles}$ of $HI$ were found. The equilibrium constant for $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is .......

If the equilibrium constant for $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $K$,then the equilibrium constant for $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ will be :