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:
\[2\times
{{10}^{-50}}=\frac{{{[{{O}_{3}}]}^{2}}}{{{[1.6\times {{10}^{-2}}]}^{3}}}\]
\[[{{O}_{3}}]=2.86\times
{{10}^{-28}}M\]
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