The second-order rate constant for the decomposition of NO2 (to NO and O2) at 573 K is 0.54 L/(mol · s). What is the time needed for an initial NO2 concentration of 0.5 mol/L to decrease to 1/3 of its initial concentration?
Please explain
Chemistry integrated law equation homework.?
2018-04-03 6:49 pm
回答 (1)
2018-04-03 7:22 pm
✔ 最佳答案
Rate constant, t = 0.54 L/(mol s)Initial NO₂ concentration, [NO₂]ₒ = 0.5 mol/L
Final NO₂ concentration, [NO₂] = 0.5/3 mol/L
Integrated rate equation for second-order reaction:
1/[NO₂] = kt + (1/[NO₂]ₒ)
1/(0.5/3) = 0.54t + (1/0.5)
0.54t = 2/0.5
Time taken, t = 2/(0.5*0.54) s = 7.4 s
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