Question

Consider the decomposition of N₂O₅ according to the following equation:

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

If the rate of decomposition of N₂O₅ at a particular instant in a reaction vessel is 4.2 x 10⁻⁷ M/s, what is the rate of appearance of:
a) NO₂
b) O₂

Answer 1

The rate of appearance of NO₂ is 8.4 x 10⁻⁷ M/s and for O₂, it is 2.1 x 10⁻⁷ M/s, based on the stoichiometry of the decomposition of N₂O₅.

Step-by-step explanation:

To calculate the rate of appearance of NO₂ and O₂ based on the rate of decomposition of N₂O₅, we must consider the stoichiometry of the balanced chemical equation:

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

From this equation, we see the molar ratios are 2:4 for N₂O₅ to NO₂ and 2:1 for N₂O₅ to O₂. Therefore, for every 1 mole of N₂O₅ that decomposes, 2 moles of NO₂ are produced, and 0.5 moles of O₂ are produced.

The rate of decomposition of N₂O₅ is given as 4.2 x 10⁻⁷ M/s, so to find the rates of appearance for NO₂ and O₂, we use the molar ratios:

For NO₂:

• Rate of appearance = (4.2 x 10⁻⁷ M/s) × (4 mol NO₂ / 2 mol N₂O₅)
• Rate of appearance = 8.4 x 10⁻⁷ M/s

For O₂:

• Rate of appearance = (4.2 x 10⁻⁷ M/s) × (1 mol O₂ / 2 mol N₂O₅)
• Rate of appearance = 2.1 x 10⁻⁷ M/s