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For the following equilibrium, Kc = 6.3 ...

For the following equilibrium, `K_c = 6.3 xx 10^14` at 1000 K. `NO_((g)) + O_(3(g)) = 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

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(`K_c` of reverse reaction of any reaction )= `1/(K_c "of forward reaction")`
`therefore K._c=1/(6.3xx10^14)=1.587xx10^(-15)`
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