$9.2 \ g$ of $N_2O_{4(g)}$ is taken in a closed $1 \ L$ vessel and heated until the following equilibrium is reached: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,$50\%$ of $N_2O_{4(g)}$ is dissociated. What is the equilibrium constant $K_c$ (in $mol \ L^{-1}$)? (Molecular weight of $N_2O_4 = 92$)

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:

The gas phase reaction $2A_{(g)} \rightleftharpoons A_{2(g)}$ at $400 \ K$ has $\Delta G^{\circ} = +25.2 \ kJ \ mol^{-1}$. The equilibrium constant $K_{C}$ for this reaction is $...... \times 10^{-2}$. (Round off to the nearest integer) $[$Use: $R = 8.3 \ J \ mol^{-1} \ K^{-1}$,$\ln 10 = 2.3$,$\log_{10} 2 = 0.30$,$1 \ atm = 1 \ bar]$ $[$antilog $(-3.3) = 5.01 \times 10^{-4}]$

In which of the following reactions is $K_p > K_c$ at $10 \ K$ temperature?

When $3 \ mol$ of $A$ and $1 \ mol$ of $B$ are mixed in a $1 \ L$ vessel,the following reaction takes place: $A_{(g)} + B_{(g)} \rightleftharpoons 2C_{(g)}$. If $1.5 \ mol$ of $C$ are formed at equilibrium,the equilibrium constant $(K_c)$ for the reaction is:

Consider the equilibrium,$H_2 + I_2 \rightleftharpoons 2 HI$. Calculate the equilibrium constant of the reverse reaction when the equilibrium concentrations of $H_2$,$I_2$,and $HI$ are $1.14 \times 10^{-2} \ mol \ L^{-1}$,$0.12 \times 10^{-2} \ mol \ L^{-1}$,and $2.52 \times 10^{-2} \ mol \ L^{-1}$,respectively.