Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Question

The activation energy of a certain reaction is 41.8 kJ/mol . At 29∘C , the rate constant is 0.0180s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

Solve the following equation for T2:

ln ⁡(k2/k1) = Ea/R (1/T1−1/T2)

The temperatures must be in kelvins and the activation energy must be in J/mol. Use the following value for the gas constant: R=8.314 J/mol⋅K.

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