If the rate of appearance of O2 in the reaction: 2O3(g)-----3O2(g) is 0.250 M/s over the first 5.50 s, how much oxygen will form
during this time?
1 answer:
<u>Given:</u>
Rate of appearance of O2 = 0.250 M/s
Time period = 5.50 s
<u>To determine:</u>
The concentration of O2 formed
<u>Explanation:</u>
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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