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10. The hydrolysis of (CH2)6C(CI)CH; in 80% ethanol follows the first-order rate equation. The values of the specific reaction rate constants are as follows: T, °C 25 35 45 k, 104 s1 1.06 3.19 9.86 2.92 (a) Plot Ink against 1/T. (b) Calculate the activation energy. (c) Calculate the pre-exponential factor.

10. The hydrolysis of (CH2)6C(CI)CH; in 80% ethanol follows the first-order rate equation. The values of the specific reaction rate constants are as follows: T, °C 25 35 45 k, 104 s1 1.06 3.19 9.86 2.92 (a) Plot Ink against 1/T. (b) Calculate the activation energy. (c) Calculate the pre-exponential factor.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
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10. The hydrolysis of (CH2)6C(Cl)CH3 in 80% ethanol follows the first-order rate equation. The values of the specific reaction rate constants are as follows: Т, "С k, 104 s-1 25 35 45 1.06 3.19 9.86 2.92 (a) Plot Ink against 1/T. (b) Calculate the activation energy. (c) Calculate the pre-exponential factor.
Transcribed Image Text:10. The hydrolysis of (CH2)6C(Cl)CH3 in 80% ethanol follows the first-order rate equation. The values of the specific reaction rate constants are as follows: Т, "С k, 104 s-1 25 35 45 1.06 3.19 9.86 2.92 (a) Plot Ink against 1/T. (b) Calculate the activation energy. (c) Calculate the pre-exponential factor.
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