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7.6     For the following equilibrium, Kc = 6.3 × 1014 at 1000 K

               NO_{(g)}+O_{3}_{(g)}\rightleftharpoons NO_{2}_{(g)}+O_{2}_{(g)}

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction?

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It is given that,
K_c=6.3\times 10^{14}
we know that K'_c for the reverse reaction is the inverse of the forward equilibrium constant. Thus it can be calculated as:

         =\frac{1}{6.3\times 10^{14}}
          =1.58\times 10^{-15}

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