Number 4 4. The decomposition of N₂O5 to NO2 and O₂ is a first-order reaction with the rate constant, k = 1.2 x 10³ s¹ at 374K.a. Determine the half-life of this reaction.b. At 374 K, how much N₂O5 would remain in a 1.0 L flask 5.0 min after the introduction of 0.05 mol ofN₂O5?5. The reaction A➜ products is a second-order reaction and takes place at 32°C.a.Given that the half-life of this reaction is 4.22 x 104 s when the initial concentration of A is 0.2 M, what isthe rate constant for the reaction? Include units with your answer.

Answered: Image /qna-images/answer/ccb6a8ce-fa50-4506-893f-124d0aa44c98.jpg

4. The decomposition of N₂O5 to NO₂ and O₂ is a first-order reaction with the rate constant, k = 1.2 x 10³ s¹ at 374 K. a. Determine the half-life of this reaction. b. At 374 K, how much N₂O5 would remain in a 1.0 L flask 5.0 min after the introduction of 0.05 mol of N₂O5?

4. The decomposition of N₂O5 to NO₂ and O₂ is a first-order reaction with the rate constant, k = 1.2 x 10³ s¹ at 374 K. a. Determine the half-life of this reaction. b. At 374 K, how much N₂O5 would remain in a 1.0 L flask 5.0 min after the introduction of 0.05 mol of N₂O5?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Number 4

4. The decomposition of N₂O5 to NO2 and O₂ is a first-order reaction with the rate constant, k = 1.2 x 10³ s¹ at 374
K.
a. Determine the half-life of this reaction.
b. At 374 K, how much N₂O5 would remain in a 1.0 L flask 5.0 min after the introduction of 0.05 mol of
N₂O5?
5. The reaction A➜ products is a second-order reaction and takes place at 32°C.
a.
Given that the half-life of this reaction is 4.22 x 104 s when the initial concentration of A is 0.2 M, what is
the rate constant for the reaction? Include units with your answer.

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