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For the following equilibrium, K(c )=6.3...

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?

Text Solution

Verified by Experts

The correct Answer is:
`1.587xx10^(-15)`
`K_c` (reverse reaction)
`=1/(6.3xx10^14)=1.587xx10^(-15)`
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