For the following equilibrium, `K_(c)=6.3xx10^(14)` at 1000 K: `NO(g)+O_(3)(g)hArr NO_(2)(g)+O_(2)(g)`. Both The forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is `K_(c)` for the reverse reaction ?

Explanation

For a reversible reaction if the values of `K_(c)` for the forward and reverse direction be `K_(c_(1))& K_(c_(2))` respectively, then `K_(c_(2))=(1)/(K_(c_(1)))`
Therefore, for the given reaction, `K_(c)` for the reverse direction`=(1)/(6.3xx10^(14))=1.587xx10^(-15)`.