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$2 {SO}_{2} + O_{2} ⇌ 2 {SO}_{3} .$ If $K_{C} = 100,$ the concentration of ${SO}_{3}$ and ${SO}_{2}$ are found to be equal at equillibrium, the concentration of $O_{2}$ will be

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0.01M

100M

0.001 M

0.1M

answer is C.

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Detailed Solution

.KC=[SO₃]² / [O₂][SO₃]² Given [SO₃]=[SO₂]=1

⇒100=1/[O₂]

⇒[O₂]=1/100=0.01M .

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