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.
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.
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
Chapter12: Chemical Kinetics
Section: Chapter Questions
Problem 108CWP: Consider the hypothetical reaction A2(g) + B2(g) 2AB(g), where the rate law is: [A2]t=k[A2][B2] The...
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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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