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

For the following equilibrium `K_(c) = 6.3 xx 10^(14)` at `1000 K`
`NO(g) + O_(2)(g) hArr NO_(2)(g) + O_(2)(g)`
Both the forward the reverse reactions in the equilibrium are elementary bimolecular reactions. What is `K_(c)`, for the reverse reaction?

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For the reverse reactions ,
`K_(c_(("reverse"))) = (1)/(K_(c("forward"))) = (1)/(6.3 xx 10^(-14)) = 1.59 xx 10 ^(-15)`
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