Given reaction: $HI(g) \rightleftharpoons \frac{1}{2} H_2(g) + \frac{1}{2} I_2(g)$; $K_1 = 8.0$

If we reverse the reaction, the equilibrium constant becomes the reciprocal of the original: $\frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \rightleftharpoons HI(g)$; $K' = \frac{1}{K_1} = \frac{1}{8.0}$

Now, if we multiply this reaction by $2$, the equilibrium constant is squared: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$; $K_2 = (K')^2 = \left(\frac{1}{8.0}\right)^2 = \frac{1}{64}$