The hydrolysis of СІ (CH2)6C CH3 in 80% ethanol follows the first-order rate equation. The values of the specific reaction rate constants are as follows: t°C                0                  25                 35                    45 k/s-1 : 1.06 X 10-5    3.19 x 10-4    9.86 x 10-4    2.92 x 10-3 (a) Plot log k against 1/T. (b) Calculate the activation energy, (c) Calculate the pre-exponential factor

General Chemistry - Standalone book (MindTap Course List)
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
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Chapter13: Rates Of Reaction
Section: Chapter Questions
Problem 13.22QP
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The hydrolysis of СІ (CH2)6C CH3 in 80% ethanol follows the first-order rate equation. The values of the specific reaction rate constants are as follows:

t°C                0                  25                 35                    45

k/s-1 : 1.06 X 10-5    3.19 x 10-4    9.86 x 10-4    2.92 x 10-3

(a) Plot log k against 1/T. (b) Calculate the activation energy, (c) Calculate the pre-exponential factor.

18.26 The hydrolysis of
Cl
(CH2)6C,
CH3
in 80% ethanol follows the first-order rate equation. The values
of the specific reaction rate constants are as follows:
25
35
45
t/°C
k/s-1
1.06 X 10-5 3.19 × 10¬4 9.86 × 10-4 2.92 × 10-3
(a) Plot log k against 1/T. (b) Calculate the activation energy.
(c) Calculate the pre-exponential factor.
Transcribed Image Text:18.26 The hydrolysis of Cl (CH2)6C, CH3 in 80% ethanol follows the first-order rate equation. The values of the specific reaction rate constants are as follows: 25 35 45 t/°C k/s-1 1.06 X 10-5 3.19 × 10¬4 9.86 × 10-4 2.92 × 10-3 (a) Plot log k against 1/T. (b) Calculate the activation energy. (c) Calculate the pre-exponential factor.
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