question_answer 1)

At\[450K,{{K}_{p}}=2\times {{10}^{10}}ba{{r}^{-1}}\]for the given reaction at equilibrium, \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)\] What is \[{{K}_{c}}\] at this temperature?

Answer:

Information shadow: The reaction is: \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)\] \[\Delta {{n}_{g}}=2-3=-1\] \[{{K}_{p}}=2\times {{10}^{10}}ba{{r}^{-1}}\] Problem solving strategy: \[{{K}_{p}}={{K}_{c}}\times {{(RT)}^{\Delta {{n}_{g}}}}\] \[{{K}_{c}}={{K}_{p}}/{{(RT)}^{\Delta {{n}_{g}}}}\] Working it out (i) \[{{K}_{c}}=\frac{2\times {{10}^{10}}}{{{(0.083\times 450)}^{-1}}}\] \[=7.47\times {{10}^{11}}mo{{l}^{-1}}L\]