The rate of formation of in the reaction 2SO2+O2→2SO3is 100g min-1. Hence rate of disappearance of O2 is

According to stoichiometric relation−dO2dt=+12dSO3dt−dO2dt=12×10080−dO2dt=58×32=20gmin−1

The rate of formation of in the reaction 2SO2+O2→2SO3is 100g min-1. Hence rate of disappearance of O2 is

According to stoichiometric relation−dO2dt=+12dSO3dt−dO2dt=12×10080−dO2dt=58×32=20gmin−1

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The rate of formation of in the reaction $2 {SO}_{2} + O_{2} \to 2 {SO}_{3}$ is 100g min^-1. Hence rate of disappearance of O₂ is

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$50 g \min^{- 1}$

$100 g \min^{- 1}$

$20 g \min^{- 1}$

$40 g \min^{- 1}$

answer is C.

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Detailed Solution

According to stoichiometric relation
$\begin{array}{l} \frac{- d [O_{2}]}{dt} = \frac{+ 1}{2} \frac{d [{SO}_{3}]}{dt} \\ \frac{- d [O_{2}]}{dt} = \frac{1}{2} \times \frac{100}{80} \\ \frac{- d [O_{2}]}{dt} = \frac{5}{8} \times 32 = 20 g \min^{- 1} \end{array}$