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?

$4$ moles of $A$ are mixed with $4$ moles of $B$. At equilibrium for the reaction $A + B \rightleftharpoons C + D$,$2$ moles of $C$ and $D$ are formed. The equilibrium constant for the reaction will be

The equilibrium constant $(K_c)$ for the reaction,$CaSO_4 \cdot 5H_2O_{(s)} \rightleftharpoons CaSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)}$ is equal to

If for ${H_2(g)} + \frac{1}{2}{S_2(s)} \rightleftharpoons {H_2S(g)}$ and ${H_2(g)} + {Br_2(g)} \rightleftharpoons 2{HBr(g)}$ the equilibrium constants are $K_1$ and $K_2$ respectively,the reaction ${Br_2(g)} + {H_2S(g)} \rightleftharpoons 2{HBr(g)} + \frac{1}{2}{S_2(s)}$ would have equilibrium constant

$A$ quantity of $PCl_5$ was heated in a $10 \ L$ vessel at $250 \ ^oC$; $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. At equilibrium,the vessel contains $0.1 \ mol$ of $PCl_5$,$0.20 \ mol$ of $PCl_3$,and $0.2 \ mol$ of $Cl_2$. The equilibrium constant $(K_c)$ of the reaction is:

What is the equilibrium expression for the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$?