A pesticide decomposes at 25.0 °C in nature according to the following reaction 4XY(aq) ⇌ 2X2Y2(aq) , kc =1.0×10-6 If the reaction is found to be a 4th order reaction with a half-life of 2.5 years. If the initial concentration of XY is 0.100 M. The integrated rate law for the forth order reaction is given above. Calculate the rate constant for the decomposition of this pesticide and determine how long it will for 87.0 % of this pesticide to decompose? How long for it to reach equilibrium?
A pesticide decomposes at 25.0 °C in nature according to the following reaction 4XY(aq) ⇌ 2X2Y2(aq) , kc =1.0×10-6 If the reaction is found to be a 4th order reaction with a half-life of 2.5 years. If the initial concentration of XY is 0.100 M. The integrated rate law for the forth order reaction is given above. Calculate the rate constant for the decomposition of this pesticide and determine how long it will for 87.0 % of this pesticide to decompose? How long for it to reach equilibrium?
Chemistry
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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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A pesticide decomposes at 25.0 °C in nature according to the following reaction 4XY(aq) ⇌ 2X2Y2(aq) , kc =1.0×10-6
If the reaction is found to be a 4th order reaction with a half-life of 2.5 years. If the initial concentration of XY is 0.100 M.
The integrated rate law for the forth order reaction is given above.
Calculate the rate constant for the decomposition of this pesticide and determine how long it will for 87.0 % of this pesticide to decompose? How long for it to reach equilibrium?
![1
= kt +
0.3334
t1/2
3
3
3
3[A]:
3[A],°
k[A],°](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f32f609-10aa-427a-b18c-a27ffe69c67b%2F6bfeb87f-23f2-4a75-acc8-083eb9dbb9fe%2Fxodjkvq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1
= kt +
0.3334
t1/2
3
3
3
3[A]:
3[A],°
k[A],°
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