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 \[S{{O}_{2}}\] is used up 5 moles of \[S{{O}_{2}}\] and 5 moles of \[{{O}_{2}}\] react to form \[S{{O}_{3}}\] Hence, \[\underset{5}{\mathop{2S{{O}_{2}}}}\,+\underset{5}{\mathop{{{O}_{2}}}}\,\rightleftharpoons \underset{-}{\mathop{2S{{O}_{3}}}}\,\] 60% of 5 moles =5 molesinitial | \[2x\] | \[x\] | \[2x\] |
used up concentration | \[3\] | \[1.5\] | \[2\times 1.5\] |
at equilibrium | \[2\] | \[3.5\] | \[3\] |
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