Question

Dinitrogen pentoxide decomposes to form nitrogen dioxide oxygen, following the equation 2N2O3 - 4NO, + 0,. At

a certain timepoint, N2O5 is being consumed at a rate of 0.1 M/s. What's the rate of production of NO, and O, at the
same timepoint?
A. NO, is produced at 0.2 M/s, and O, is produced at 0.05 M/s.
B. NO, is produced at 0.2 M/s, and O, is produced at 0.2 M/s.
C. NO, is produced at 0.05 M/s, and O2 is produced at 0.2 M/s.
D. NO, is produced at 0.1 M/s, and O, is produced at 0.05 M/s.​

Answer 1

NO2 is produced at 0.2 M/s, and O2 is produced at 0.05 M/s.

Step-by-step explanation:

Answer 2

A. N₂O₅ is produced at 0.2 mol·L⁻¹s⁻¹, and O₂ is produced at 0.05 mol·L⁻¹s⁻¹.

Step-by-step explanation:

2N₂O₅ ⟶ 4NO₂ + O₂

For every two molecules of N₂O₅ that disappear, we get four molecules of NO₂ and one of O₂.

NO₂ is forming twice as fast, and O₂ is forming half as fast, as N₂O₅ is decomposing.

N₂O₅ is decomposing at a rate of 0.1 mol·L⁻¹s⁻¹.

So, N₂O₅ is produced at 0.2 mol·L⁻¹s⁻¹, and O₂ is produced at 0.05 mol·L⁻¹s⁻¹.