# The value of equilibrium constant of the reaction $HI (g) \rightleftharpoons \large\frac{1}{2} H_2(g) + \large\frac{1}{2} I_2$ is 8.0. The equilibrium constant of the reaction $H_2(g) +I_2(g) \rightleftharpoons 2HI(g)$ will be :

$\begin {array} {1 1} (1)\;\large\frac{1}{8} & \quad (2)\;\large\frac{1}{16} \\ (3)\;\large\frac{1}{64} & \quad (4)\;16 \end {array}$