The half-life of forward and reverse reactions are $400 \, sec$ and $100 \, sec$ respectively. If these half-lives are independent of the concentration of the reactant,find the equilibrium constant of the reaction.

In the reversible reaction $A + B \rightleftharpoons C + D$,the concentration of each $C$ and $D$ at equilibrium was $0.8 \ mol/L$. If the initial concentration of $A$ and $B$ was $1 \ mol/L$ each,then the equilibrium constant $K_c$ will be:

For the gaseous reaction $aA + bB \rightleftharpoons cC + dD$,the relation between $K_p$ and $K_c$ is:

At $600\ ^{\circ}C$,$NH_4COONH_2\ (s) \rightleftharpoons 2NH_3\ (g) + CO_2\ (g)$ has an equilibrium constant $K_p = 3.2 \times 10^2\ bar^3$. Calculate $K_c$. $\left( R = 0.082\ L\ atm\ K^{-1}\ mol^{-1} \right)$

$A + B \rightleftharpoons C + D$. If the final equilibrium concentrations of $A$ and $B$ are equal,and the equilibrium concentration of $D$ is twice that of $A$,what is the equilibrium constant $(K_c)$ of the reaction?

$1.1 \ mol$ of $A$ are mixed with $2.2 \ mol$ of $B$ and the mixture is kept in a $1 \ L$ vessel till the equilibrium is established in the reaction. If the molar concentration of $C$ at the equilibrium point is $0.2 \ M$,then find the value of the equilibrium constant $(K_c)$ for the reaction: $A + 2B \rightleftharpoons 2C + D$