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 45k/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 ofCl(CH2)6C,CH3in 80% ethanol follows the first-order rate equation. The valuesof the specific reaction rate constants are as follows:253545t/°Ck/s-11.06 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.

T°C 1/T k (s-1) log (k) 0 0 0.0000106 -4.974694135 25 0.04 0.000319 -3.496209317 35…

Answered: The hydrolysis of СІ (CH2)6C CH3 in 80%…

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter12: Chemical Kinetics

Section: Chapter Questions

Problem 63AE: For the following reaction profile, indicate a. the positions of reactants and products. b. the...

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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.

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