Consider the reaction, $2 N_{2} O_{5} \to 4 N O_{2} + O_{2}$

In the reaction $N O_{2}$ is being formed at the rate of $0.0125 m o l L^{- 1} s^{- 1}$ . What is the rate of reaction at this time?

0.0018 m o l L^{- 1} s^{- 1}

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0.0031 m o l L^{- 1} s^{- 1}

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0.0041 m o l L^{- 1} s^{- 1}

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0.050 m o l L^{- 1} s^{- 1}

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Solution

The correct option is B $0.0031 m o l L^{- 1} s^{- 1}$
$R a t e = \frac{1}{4} \frac{d [N O_{2}]}{d t}$
$= \frac{1}{4} \times 0.0125 = 0.0031 m o l L^{- 1} s^{- 1}$ .

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Q. For the reaction,

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

, the rate of disappearance of

O_{2}

2 \times 10^{- 4} m o l L^{- 1} s^{- 1}

. The rate of appearance of

S O_{3}

is:

For the reaction, $N_{2} O_{5} (g) \to N O_{2} (g) + \frac{1}{2} O_{2} (g)$ The value of rate of disappearance of $N_{2} O_{5}$ is given as $6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1} .$ The rate of formation of $N O_{2}$ and $O_{2}$ is given respectively as: