question_answer 1)

What is \[{{K}_{c}}\] for the following equilibrium when the equilibrium concentration of each substance is: \[[S{{O}_{2}}]=0.6M;[{{O}_{2}}]=0.82M;[S{{O}_{3}}]=1.90M?\] \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)\]

Answer:

Information shadow: The reaction is: \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)\] The equilibrium concentrations are: \[[S{{O}_{2}}]=0.6M;\,\,[{{O}_{2}}]=0.82M;\,\,\,\,[S{{O}_{3}}]=1.90M\] Problem solving strategy: The expression for equilibrium constant is: \[{{K}_{c}}\frac{{{[S{{O}_{3}}]}^{2}}}{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]}\,\,\,\,\,\,\,\,\,\,\,\,......(i)\] Working it out: Substituting the values in equation (i) we get, \[{{K}_{c}}=\frac{{{(1.90)}^{2}}}{{{(0.6)}^{2}}\times (0.82)}\] \[=12.228mo{{l}^{-1}}L\]