For the reaction $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$,the equilibrium constant $K_p$ is $16$. If the volume of the container is reduced to $\frac{1}{2}$ of its original value,what will be the value of $K_p$ at the same temperature?

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. If the equilibrium constant $(K_C)$ for the above reaction at $500 \ K$ is $1.79$ and the equilibrium concentrations of $PCl_5$ and $PCl_3$ are $1.41 \ M$ and $1.59 \ M$,respectively,then the concentration of $Cl_2$ is approximately: (in $M$)

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:

$8 \, \text{mol}$ of $AB_3$ is added to a $1.0 \, \text{dm}^3$ container. If it dissociates according to the reaction $2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$ and $2 \, \text{mol}$ of $A_2$ are present at equilibrium,what is the equilibrium constant $(K_c)$ for this reaction?

In a chemical equilibrium $A + B \rightleftharpoons C + D$,when $1 \, mol$ each of two reactants are mixed,$0.5 \, mol$ each of the products are formed. The equilibrium constant is

For the reaction equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$,the concentrations of $N_2O_4$ and $NO_2$ at equilibrium are $4.8 \times 10^{-2} \ mol \ L^{-1}$ and $1.2 \times 10^{-2} \ mol \ L^{-1}$ respectively. The value of $K_c$ for the reaction is: