The rate of disappearance of N2O5 is 3.6 x 10−4 M/s in the reaction below. What is the rate of appearance of NO2?

2 N2O5 → 4 NO2 + O2

Question

The rate of disappearance of N2O5 is 3.6 x 10−4 M/s in the reaction below. What is the rate of appearance of NO2?

2 N2O5 → 4 NO2 + O2

BlackTom AI · Accepted Answer

The question asks us to relate the rate of disappearance of N2O5 to the rate of appearance of NO2 using the balanced equation: 2 N2O5 → 4 NO2 + O2.
First, note the stoichiometric coefficients: 2 moles of N2O5 produce 4 moles of NO2. This means the rate at which NO2 appears is twice the rate at which N2O5 disappears.
Now evaluate each option:
- Option A: 3.6 x 10−4 M/s. This value matches the rate of N2O5 disappearance, not the NO2 appearance, so it would be incorrect because NO2 is produced, not consumed.
- Option B: 1.4 x 10−3 M/s. This is four times 3.6 x 10−4 (since 4 × 3.6e-4 = 1.44e-3) which would be appropriate if the stoichiometry were 1 N2O5 producing 4 NO2, but our ratio is 2:4, so this overestimates the rate by a factor of 2.
- Option C: 1.8 x 10−4 M/s. This is half of the given N2O5 disappearance rate, which would correspond to a 1:1 or incorrect interpretation of stoichiometry; it does not reflect the 2:4 ratio.
- Option D: 7.2 x 10−4 M/s. This value is twice the rate of N2O5 disappearance (2 × 3.6 x 10−4), aligning with the 2 N2O5 → 4 NO2 stoichiometry that says NO2 appears at twice the rate of N2O5 disappearance.
In summary, the correct interpretation hinges on the 2:4 ratio, which simplifies to a 1:2 relationship between N2O5 consumed and NO2 produced; doubling 3.6 x 10−4 M/s yields 7.2 x 10−4 M/s, the rate of NO2 appearance.

CHEM 1220 SP2025 (18215) Final Exam- Requires Respondus LockDown Browser

The rate of disappearance of N2O5 is 3.6 x 10−4 M/s in the reaction below. What is the rate of appearance of NO2? 2 N2O5 → 4 NO2 + O2

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