question_answer 34)

For the following equilibrium \[{{K}_{c}}=6.3\times {{10}^{14}}\]at 1000 K \[NO(g)+{{O}_{3}}(g)\rightleftharpoons N{{O}_{2}}(g)+{{O}_{2}}(g)\] Both forward and backward reactions in the equilibrium are elementary bimolecular reactions. What is the\[{{K}_{c}}\], for the reverse reaction?

Answer:

The given reaction is: \[NO(g)+{{O}_{3}}(g)\rightleftharpoons NO(g)+{{O}_{3}}(g)\] \[{{K}_{c}}\frac{[N{{O}_{2}}][{{O}_{2}}]}{[NO][{{O}_{3}}]}=6.3\times {{10}^{14}}\,\,\,\,\,....(i)\] For the reverse reaction, \[N{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons NO(g)+{{O}_{3}}(g)\] \[K_{c}^{'}=\frac{[NO][{{O}_{3}}]}{[N{{O}_{2}}][{{O}_{2}}]}=\frac{1}{{{K}_{c}}}\] \[=\frac{1}{6.3\times {{10}^{14}}}=1.587\times {{10}^{-15}}\]