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Determine the rate of reaction at 150 K, k150K, if the rate constant, was determined to be k298K = 2.3x10³M ¹s¹ AH-4.12 kJ/mol This reaction takes place in two steps: S1 (fast): S2 (slow): CH₂CH₂COOH(aq) + H:O:(aq) → CH₂CH:COOOH(aq) Propionic Acid HP PPA a k150k 2.3x10³M¹s¹ с Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. bk150K 5.2x10-M¹s¹ d = CH-CH:COOH(aq) + H(aq) → CH-CH:COOH: (aq) CH,CH:COOH: (aq) + H:O:(aq) → CH-CH:COOH(aq) + H*(aq) k150K 1.9x10¹M¹S¹ = K. 298 K-2.05 k150K 9.4x10-2M-¹5-¹ E. (slow step)-44.2 kJ

Determine the rate of reaction at 150 K, k150K, if the rate constant, was determined to be k298K = 2.3x10³M ¹s¹ AH-4.12 kJ/mol This reaction takes place in two steps: S1 (fast): S2 (slow): CH₂CH₂COOH(aq) + H:O:(aq) → CH₂CH:COOOH(aq) Propionic Acid HP PPA a k150k 2.3x10³M¹s¹ с Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. bk150K 5.2x10-M¹s¹ d = CH-CH:COOH(aq) + H(aq) → CH-CH:COOH: (aq) CH,CH:COOH: (aq) + H:O:(aq) → CH-CH:COOH(aq) + H*(aq) k150K 1.9x10¹M¹S¹ = K. 298 K-2.05 k150K 9.4x10-2M-¹5-¹ E. (slow step)-44.2 kJ

Chemistry
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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5 Determine the rate of reaction at 150 K, k150K, if the rate constant, was determined to be k298K = 2.3x10³M-¹-¹ Q Search AURA 17 16 This reaction takes place in two steps: S1 (fast): S2 (slow): 11 b AHᵒrn-4.12 kJ/mol Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a k150k 2.3x10³M ¹s¹ с CH₂CH₂COOH(aq) + H₂O:(aq) → CH3CH₂COOOH(aq) Propionic Acid HP Ke 298 K= 2.05 d 6 k150k CH3CH₂COOH(aq) + H*(aq) → CH3CH₂COOH: (aq) CHICH:COOH: (aq) + H₂O2(aq) → CH3CH:COOH(aq) + H*(aq) k150K 5.2x10-5M¹S-¹ = 1.9x10¹M¹S-1 k150K 9.4x10-2M-¹5-¹ PPA 8 Ea (slow step)-44.2 kJ 88 O/0 ما 9000 L 1". 1
Transcribed Image Text:5 Determine the rate of reaction at 150 K, k150K, if the rate constant, was determined to be k298K = 2.3x10³M-¹-¹ Q Search AURA 17 16 This reaction takes place in two steps: S1 (fast): S2 (slow): 11 b AHᵒrn-4.12 kJ/mol Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a k150k 2.3x10³M ¹s¹ с CH₂CH₂COOH(aq) + H₂O:(aq) → CH3CH₂COOOH(aq) Propionic Acid HP Ke 298 K= 2.05 d 6 k150k CH3CH₂COOH(aq) + H*(aq) → CH3CH₂COOH: (aq) CHICH:COOH: (aq) + H₂O2(aq) → CH3CH:COOH(aq) + H*(aq) k150K 5.2x10-5M¹S-¹ = 1.9x10¹M¹S-1 k150K 9.4x10-2M-¹5-¹ PPA 8 Ea (slow step)-44.2 kJ 88 O/0 ما 9000 L 1". 1
Expert Solution
Step 1: Arrhenius equation

Answer:

ln straight k subscript 2 over straight k subscript 1 equals straight E subscript straight a over straight R open parentheses 1 over straight T subscript 1 minus 1 over straight T subscript 2 close parentheses

Here:

straight E subscript straight a equals Activation space energy straight R equals universal space gas space constant

straight k subscript 2 space and space straight k subscript 1 are the rate constants at temperatures T2 and T1 respectively.

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