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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 6 of 73
Marks: +1, -0
For the following equilibrium, KcK_c = 6.3 × 101410^{14} at 1000 K.
NO(g)+O3(g)\mathrm{NO}_{(g)} + \mathrm{O}_{3(g)} ⇌ NO2(g)+O2(g)\mathrm{NO}_{2(g)} + \mathrm{O}_{2(g)} Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is KcK_c , for the reverse reaction?
Solution:  
NO(g)+O3(g)\mathrm{NO}_{(g)} + \mathrm{O}_{3(g)} ⇌ NO2(g)+O2(g)\mathrm{NO}_{2(g)} + \mathrm{O}_{2(g)}
KcK_c = [NO2][O2][NO][O3]\frac{[\mathrm{NO}_2][\mathrm{O}_2]}{[\mathrm{NO}][\mathrm{O}_3]} = 6.3 × 101410^{14}
For the reverse reaction, Kc′K_c' = [NO][O3][NO2][O2]\frac{[\mathrm{NO}][\mathrm{O}_3]}{[\mathrm{NO}_2][\mathrm{O}_2]} = 1Kc\frac{1}{K_c} = 16.3×1014\frac{1}{6.3 \times 10^{14}} = 1.587 × 10−1510^{-15}
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