7.6     For the following equilibrium, Kc = 6.3 × 1014 at 1000 K               $NO_{(g)}+O_{3}_{(g)}\rightleftharpoons NO_{2}_{(g)}+O_{2}_{(g)}$Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

$K_c=6.3\times 10^{14}$
we know that $K'_c$ for the reverse reaction is the inverse of the forward equilibrium constant. Thus it can be calculated as:
$K'_c=\frac{1}{K_c}$
$=\frac{1}{6.3\times 10^{14}}$
$=1.58\times 10^{-15}$