For the reaction $P + Q \rightleftharpoons R + C$,the equilibrium constant $K_c$ is $10^{-2}$ and the forward rate constant $K_f$ is $10^{-1}$. The rate constant for the backward reaction $(K_b)$ will be:

For the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2}O_{2(g)}$,the equilibrium constant $K_C = 4.9 \times 10^{-2}$. Calculate the $K_C$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

The equilibrium concentrations of $X$,$Y$,and $YX_2$ are $4 \ mol/L$,$2 \ mol/L$,and $2 \ mol/L$ respectively for the equilibrium $2X + Y \rightleftharpoons YX_2$. The value of $K_c$ is:

One mole of $N_2O_4$ in a $1 \ L$ flask decomposes to attain the equilibrium $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At the equilibrium the mole fraction of $NO_2$ is $1/2$. Hence $K_C$ will be:

For the reaction $A + B \rightleftharpoons C + D$,the initial concentrations of $A$ and $B$ are equal. If the equilibrium concentration of $D$ is twice the equilibrium concentration of $A$,then the equilibrium constant $K_c$ is:

For the reversible reaction, $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$. When the partial pressure is measured in atmosphere, the value of $K_p$ at $500^{\circ}\text{C}$ is $1.44 \times 10^{-5}$. The value of $K_c$ when the concentration is expressed in $\text{mol L}^{-1}$ is: