For the reactions $A \rightleftharpoons B; K_c = 2$,$B \rightleftharpoons C; K_c = 4$,and $C \rightleftharpoons D; K_c = 6$,the value of $K_c$ for the reaction $A \rightleftharpoons D$ is:

The equilibrium constants for the following reactions are given at $25^{\circ} C$:
$2 A \rightleftharpoons B + C, K_{1} = 1.0$
$2 B \rightleftharpoons C + D, K_{2} = 16$
$2 C + D \rightleftharpoons 2 P, K_{3} = 25$
The equilibrium constant for the reaction $P \rightleftharpoons A + \frac{1}{2} B$ at $25^{\circ} C$ is

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 $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$,the value of $K_p$ changes with:

In a closed vessel,$PCl_{5(g)}$ is obtained by the chemical reaction between $PCl_{3(g)}$ and $Cl_{2(g)}$. If the equilibrium concentrations in this vessel of $PCl_3$,$Cl_2$,and $PCl_5$ at $500 \ K$ are $1.59 \ M$,$1.59 \ M$,and $1.41 \ M$ respectively,then find the equilibrium constant $K_c$ for the reaction: $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$

At $T \ K$,the equilibrium constant for the reaction $a A_{(g)} \rightleftharpoons b B_{(g)}$ is $K_c$. If the reaction takes place in the following form $2a A_{(g)} \rightleftharpoons 2b B_{(g)}$,its equilibrium constant is $K_c^{\prime}$. The correct relationship between $K_c$ and $K_c^{\prime}$ is