7.6 For the following equilibrium, Kc= 6.3 × 1014 at 1000 K

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

It is given that ${K}_{c}$ for the forward reaction is $6.3×{10}^{14}.$
Then, ${K}_{c}$ for the reverse reaction will be,
$K{\text{'}}_{c}=\frac{1}{{K}_{c}}\phantom{\rule{0ex}{0ex}}=\frac{1}{6.3×{10}^{14}}\phantom{\rule{0ex}{0ex}}=1.59×{0}^{-15}$