For the following equilibrium,
Question:

For the following equilibrium, $K_{\mathrm{C}}=6.3 \times 10^{14}$ at $1000 \mathrm{~K}$

$\mathrm{NO}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{~g}) \longleftrightarrow \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$

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

Solution:

It is given that $K_{C}$ for the forward reaction is $6.3 \times 10^{14}$.

Then, $K_{C}$ for the reverse reaction will be,

$K_{C}^{\prime}=\frac{1}{K_{C}}$

$=\frac{1}{6.3 \times 10^{14}}$

$=1.59 \times 10^{-15}$