The gas phase decomposition of dinitrogen pentoxide is represented by this equation,
$2 N_{2} O_{5} (g) ⟶ 4 {NO}_{2} (g) + O_{2} (g)$
What is the rate of formation of oxygen gas (in $mol \cdot L \cdot s^{- 1})$ in an experiment where 0.080 mol of $N_{2} O_{5}$ is consumed in a 4.0 L container every 0.20 seconds?

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0.020

0.050

0.10

0.20

answer is B.

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Detailed Solution

The rate of formation of oxygen gas

$\frac{d [O_{2}]}{d t} = \frac{1}{2} \times \frac{0.08}{4} \times \frac{1}{0.20} \frac{d [O_{2}]}{d t} = 0.020 mol \cdot L \cdot s^{- 1}$

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