For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ at $25^o C$,the value of $K_c$ is $6.10 \times 10^{-3}$. What is the value of $K_c$ for the equilibrium reaction $NO_{2(g)} \rightleftharpoons 1/2 N_2O_{4(g)}$?

The value of $K_{C}$ for the equilibrium reaction: $CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ at $T \ K$ is $0.036$. If the equilibrium concentration of $CO_{2(g)}$ is $0.004 \ M$,the concentration of $CO_{(g)}$ in $mol \ L^{-1}$ is:

In the reaction,$H_2 + I_2 \rightleftharpoons 2HI$. In a $2 \ L$ flask,$0.4 \ mol$ of each $H_2$ and $I_2$ are taken. At equilibrium,$0.5 \ mol$ of $HI$ are formed. What will be the value of equilibrium constant,$K_c$?

$A + B \rightleftharpoons C + D$. 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?

In a chemical equilibrium,the rate constant of the backward reaction is $7.5 \times 10^{-4}$ and the equilibrium constant is $1.5$. The rate constant of the forward reaction is:

$4.0 \, mol$ of argon and $5.0 \, mol$ of $PCl_5$ are introduced into an evacuated flask of $100 \, L$ capacity at $610 \, K$. The system is allowed to equilibrate. At equilibrium,the total pressure of the mixture was found to be $6.0 \, atm$. The $K_p$ for the reaction is ...... [Given: $R = 0.082 \, L \, atm \, K^{-1} \, mol^{-1}$]