$4.5 \ mol$ 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:

In the reaction $AB_{(g)} + CD_{(g)} \rightleftharpoons AD_{(g)} + CB_{(g)}$,one mole of $AB$ reacts with one mole of $CD$. When equilibrium is established,$3/4$ mole of each $AB$ and $CD$ are converted into $AD$ and $CB$. If the volume remains constant,the equilibrium constant for the reaction will be........

Equilibrium constant $(K_c)$ of $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$ is $5 \times 10^{3}$. What is the equilibrium concentration of $HI$,if equilibrium concentrations of $H_{2_{(g)}}$ and $I_{2_{(g)}}$ respectively are $2.2 \times 10^{-2} \ M$ and $2.2 \times 10^{-4} \ M$?

Two moles of $NH_3$ when put into a previously evacuated vessel $(1 \ L)$,partially dissociate into $N_2$ and $H_2$. If at equilibrium one mole of $NH_3$ is present,the equilibrium constant is

For the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the value of $K_p$ changes with:

For the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$,the equilibrium constant expression $K_c$ is: