question_answer 61)

The value of \[{{K}_{c}}\] for the reaction \[3{{O}_{2}}(g)\rightleftharpoons 2{{O}_{3}}(g)\]is\[2\times {{10}^{-50}}\]at\[{{250}^{{}^\circ }}C\]. If the equilibrium concentration of \[{{O}_{2}}\]in air at\[25{{\,}^{o}}C\,is1.6\text{ }\times \text{ }{{10}^{-2}}M\], what is the concentration of\[{{O}_{3}}?\]

Answer:

Information shadow: Reaction: \[3{{O}_{2}}(g)\rightleftharpoons 2{{O}_{3}}(g)\] \[{{K}_{c}}=2\times {{10}^{-50}}\] At equilibrium, \[[{{O}_{2}}]=1.6\times {{10}^{-2}}M\] \[[{{O}_{3}}]=?\] Problem solving strategy: Expression of \[{{K}_{c}}\] for the given reaction is: \[{{K}_{c}}=\frac{{{[{{O}_{3}}]}^{2}}}{{{[{{O}_{2}}]}^{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......(i)\] Substituting the values of \[{{K}_{c}}\]and \[[{{O}_{2}}]\] we can calculate the concentration of ozone, i.e.,\[[{{O}_{3}}]\] Working it out: \[2\times {{10}^{-50}}=\frac{{{[{{O}_{3}}]}^{2}}}{{{[1.6\times {{10}^{-2}}]}^{3}}}\] \[[{{O}_{3}}]=2.86\times {{10}^{-28}}M\]