On a given condition,the equilibrium concentrations of $HI$,$H_2$,and $I_2$ are $0.80 \ mol/L$,$0.10 \ mol/L$,and $0.10 \ mol/L$ respectively. The equilibrium constant for the reaction $H_2 + I_2 \rightleftharpoons 2HI$ will be:

$HI$ 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 $400 \ K$ in a closed vessel,the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ takes place. At equilibrium,the concentration of $H_2$ is $0.6 \ mol \ L^{-1}$,the concentration of $I_2$ is $0.8 \ mol \ L^{-1}$,and the concentration of $HI$ is $0.14 \ mol \ L^{-1}$. Calculate the equilibrium constant $(K_c)$.

In the figure shown below,reactant $A$ (represented by square) is in equilibrium with product $B$ (represented by circle). The equilibrium constant is:

If the equilibrium constant for the reaction $2AB \rightleftharpoons A_2 + B_2$ is $49$,then the equilibrium constant for the reaction $AB \rightleftharpoons \frac{1}{2}A_2 + \frac{1}{2}B_2$ will be:

State the law of chemical equilibrium for an equilibrium process.