If the equilibrium constant of the reaction $2HI \rightleftharpoons H_2 + I_2$ is $0.25$,then the equilibrium constant of the reaction $H_2 + I_2 \rightleftharpoons 2HI$ would be

The reaction,$2A_{(g)} + B_{(g)} \rightleftharpoons 3C_{(g)} + D_{(g)}$ is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00 \ M$. When equilibrium is reached,the concentration of $D$ is measured and found to be $0.25 \ M$. The value for the equilibrium constant for this reaction is given by the expression:

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

$4.5$ moles each of hydrogen and iodine are heated in a sealed $10 \ L$ vessel. At equilibrium,$3$ moles of $HI$ are found. The equilibrium constant for ${H_2}_{(g)} + {I_2}_{(g)} \rightleftharpoons 2HI_{(g)}$ is

$2$ moles of $PCl_5$ were heated in a closed vessel of $2 \ L$ capacity. At equilibrium,$40\%$ of $PCl_5$ is dissociated into $PCl_3$ and $Cl_2$. The value of equilibrium constant is

Consider the imaginary equilibrium $4A + 5B \rightleftharpoons 4X + 6Y$. The equilibrium constant $K_c$ has the unit: