If the rate of appearance of o2 in the reaction: 2o3((g) is 0.250 m/s over the first 5.50 s, how much oxygen will form during this time?

Given:

Rate of appearance of O2 = 0.250 M/s

Time period = 5.50 s

To determine:

The concentration of O2 formed

Explanation:

2O3 (g) ↔ 3O2 (g)

Rate of appearance of O2 = 1/3 * Δ[O2]/Δt

Based on the given data:

0.250 M/s = 1/3 * [O2]/5.50 s

[O2 ] = 0.250 Ms⁻¹ * 3 * 5.50 s = 4.125 M

Ans: Amount of oxygen formed is 4.13 M

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