Consider the following reversible chemical reactions:

The relation between $K_1$ and $K_2$ is

$A_2\left(g\right)+B_2\left(g\right)\stackrel{K_1}{\rightleftharpoons }2AB\left(g\right)\_\_\_\_\_\_\left(1\right)$

$6AB\left(g\right)\stackrel{K_2}{\rightleftharpoons }3A_2\left(g\right)+3B_2\left(g\right)\_\_\_\_\_\_\left(2\right)$

Reaction(2) = - 3  × reaction (1)

$\therefore K_2={\left(\frac{1}{K_1}\right)}^3\Rightarrow K_2=K_1^{-3}$