A) \[10.0\]
B) \[8.5\]
C) \[10.5\]
D) \[3.9\]
Correct Answer: B
Solution :
\[2S{{O}_{2}}+{{O}_{2}}\rightleftharpoons 2S{{O}_{3}}\] At equilibrium 60% of \[\text{S}{{\text{O}}_{\text{2}}}\]is used up 5 moles of \[\text{S}{{\text{O}}_{\text{2}}}\]and 5 moles of \[{{\text{O}}_{2}}\] react to form \[\text{S}{{\text{O}}_{\text{3}}}\] Hence,\[\underset{5}{\mathop{2S{{O}_{2}}}}\,+\underset{5}{\mathop{{{O}_{2}}}}\,\rightleftharpoons \underset{-}{\mathop{2S{{O}_{3}}}}\,\] 60% of 5 moles \[=5\,\text{moles}\] \[\begin{align} & \text{initial}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,2x \\ & \text{used}\,\text{up}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,1.5\,\,\,\,\,\,\,\,2\times 1.5 \\ & \text{concentration}\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,3.5\,\,\,\,\,\,\,3 \\ & \text{at}\,\text{equilibrium}\,\,\,\, \\ \end{align}\] \[\therefore \]Total number of molecules \[=2+3.5+3=8.5\]You need to login to perform this action.
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