For the reaction:

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
Choose the correct option which describes the overall rate of reaction

r = - 2 \times \frac{d [S O_{3}]}{d t}

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r = + \frac{d [O_{2}]}{d t}

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r = - 2 \times \frac{d [S O_{2}]}{d t}

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r = \frac{- 1}{2} \times \frac{d [S O_{2}]}{d t}

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Solution

The correct option is D $r = \frac{- 1}{2} \times \frac{d [S O_{2}]}{d t}$
For the given reaction:
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$

Overall rate $r$ can be expressed as:

$r = \frac{- 1}{2} \times \frac{d [S O_{2}]}{d t} = - 1 \times \frac{d [O_{2}]}{d t} = \frac{+ 1}{2} \times \frac{d [S O_{3}]}{d t}$
where:

$- \frac{d [S O_{2}]}{d t} = Rate of disappearance of S O_{2} (g)$

$- \frac{d [O_{2}]}{d t} = Rate of disappearance of O_{2} (g)$

$+ \frac{d [S O_{3}]}{d t} = Rate of formation of S O_{3} (g)$

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How will the rate of reaction $2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$ change if the volume of the reaction vessel is halved :

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