In the formation of sulphur trioxide by the contact process,
$2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g)$
The rate of reaction is expressed as
$- \frac{d (O_{2})}{dt} = 2 .5 \times 10^{- 4} {molL}^{- 1} s^{- 1}$
The rate of disappearance of (SO₂) will be

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$50 .0 \times 10^{- 4} mol L^{- 1} s^{- 1}$

$- 2.25 \times 10^{- 4} mol L^{- 1} s^{- 1}$

$3.75 \times 10^{- 4} mol L^{- 1} s^{- 1}$

$5 .0 \times 10^{- 4} mol L^{- 1} s^{- 1}$

answer is A.

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Detailed Solution

$\begin{array}{l} 2 {SO}_{2} (g) + O_{2} (g) ⇌ 2 {SO}_{3} (g) \\ Rate = \frac{- d [{SO}_{2}]}{2 dt} = \frac{- d [O_{2}]}{dt} \\ ∴ \frac{- d [{SO}_{2}]}{dt} = 2 \times \frac{- d [O_{2}]}{dt} \\ = 2 \times 2 .5 \times 10^{- 4} \\ = 5 .0 \times 10^{- 4} {molL}^{- 1} s^{- 1} \end{array}$