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For the following equilibrium,Kc=6.3xx10...

For the following equilibrium,`K_c=6.3xx10^14` at 1000 K
`NO(g)+O_3(g)iffNO_2(g)+O_2(g)`
What is `K_c` for the reverse reactions?

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`K_c` for the reverse reaction =`1/K_c=1/(6.3xx10^4)=0.159xx10^-14=1.59xx10^-15`
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