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Half life

Rate of forma...

Question

Rate of formation of ${S O}_{3}$ in the following reaction $2 {S O}_{2} + O_{2} ⟶ 2 {S O}_{3}$ is $100 g {m i n}^{- 1}$ . Hence disappearance of $O_{2}$ is:

50 g {m i n}^{- 1}

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40 g {m i n}^{- 1}

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200 g {m i n}^{- 1}

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20 g {m i n}^{- 1}

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Solution

The correct option is D $20 g {m i n}^{- 1}$
$2 {S O}_{2} + O_{2} ⟶ 2 {S O}_{3}$
From above reaction we can say that $2$ moles of ${S O}_{3}$ is produced from $1$ mole of $O_{2}$ .
$∴$ rate of disappearance of $O_{2} = \frac{r a t e o f f o r m a t i o n o f {S O}_{3}}{2}$
$= \frac{100 / 80}{2} \times 32 g m {m i n}^{- 1}$
$= 20 g$ ${m i n}^{- 1}$ [ Since molecular weight of ${S O}_{3}$ and $O_{2}$ are $100$ and $32$ respectively]

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