Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of ·0.0047Ms-1: 2N2O (g) →2N2 (g) + O2 (g) Suppose a 2.0L ⁢ flask is charged under these conditions with 500. mmol of dinitrogen monoxide. After how much time is there only 250. mmol left? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

Question

Maddox

Chemistry

29 July, 00:46

Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of ·0.0047Ms-1: 2N2O (g) →2N2 (g) + O2 (g) Suppose a 2.0L ⁢ flask is charged under these conditions with 500. mmol of dinitrogen monoxide. After how much time is there only 250. mmol left? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

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Answers (1)

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Bennett Gray · Answer 1

250 mmol are left after 26.60 secs ≅ 26.6 seconds

Explanation:

zero order kinetics formula is:

[A] = [A₀] - kt

where [A] = amount left; [A₀] = amount remaining; k = rate constant; t = time

concentration [A₀] = mole/volume = 0.5 mole/2.0L = 0.25

[A] = 0.25 mole/2.0L = 0.125 M

k = 0.0047 Ms⁻¹

t = {[A]-[A₀]}/-k = (0.125 - 0.25) / (-0.0047) = 26.60 seconds = 26.6 seconds

the amount is reduced by half after 26.6 seconds