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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium1 Mark
Question
For the following equilibrium, $\text{K}_{\text{c}}=6.3\times10^{14}\text{ at }1000\text{K}$
$\text{NO(g) + O}_3\text{(g)}\rightleftharpoons\text{NO}_2\text{(g) + O}_2\text{(g)}$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $\text{K}_{\text{c}}$, for the reverse reaction?
✓

Answer

For the reverse reaction $\text{K}_{\text{c}}=\frac{1}{\text{K}_{\text{c}}}=\frac{1}{6.3\times10^{14}}=1.59\times10^{-15.}$

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