The value of equilibrium constant of the reaction, <br> $HI(g)\rightleftharpoons \frac{1}{2}H_2(g)+\frac{1}{2}I_2(g)$ is 8.0. <br> The equilibrium constant of the reaction, <br> $H_2(g)+I_2(g)\rightleftharpoons2HI(g)$ will be -

Question

( A ) $\frac{1}{16}$
( B ) $\frac{1}{8}$
( C ) $\frac{1}{64}$
( D ) 16