$HI$ was heated in a closed tube at $440\,^{\circ}C$ until equilibrium was obtained. At this temperature,$22\%$ of $HI$ was dissociated. The equilibrium constant for this dissociation will be:

At equilibrium,the concentrations of $N_{2} = 3.0 \times 10^{-3} \, M$,$O_{2} = 4.2 \times 10^{-3} \, M$ and $NO = 2.8 \times 10^{-3} \, M$ in a sealed vessel at $800 \, K$. What will be $K_{c}$ for the reaction
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$

For the chemical equilibrium $A + B \rightleftharpoons C + D$,when one mole of each reactant is mixed,$0.4 \ mol$ of each product is formed. The equilibrium constant $K_c$ is:

The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:

At $T(K)$,the equilibrium constant of $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $49$. If $[H_2]$ and $[I_2]$ at equilibrium at the same temperature are $2.0 \times 10^{-2} \ M$ and $8.0 \times 10^{-2} \ M$ respectively,the $[HI]$ at equilibrium in $mol \ L^{-1}$ is:

Write the equilibrium constant $K$ for the reaction $CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}$