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?

For the following equilibrium, `K_(c )=6.3xx10^(14) at 1000 K` `NO(g)+

1.	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?
Answer» For the reverse reaction `K_(c) = (1)/(K_(c)) =(1)/(6.3 xx 10^(14)) = 1.59 xx 10^(-15).`

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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?

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