The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written ask=Ae−Ea/RT where R is the gas constant (8.314 J/mol⋅K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction.However, a more practical form of this equation is lnk2k1=EaR(1T1−1T2) which is mathmatically equivalent tolnk1k2=EaR(1T2−1T1) where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). The activation energy of a certain reaction is 36.9 kJ/mol . At 27∘C , the rate constant is 0.0130s−1 . At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units.
The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written ask=Ae−Ea/RT where R is the gas constant (8.314 J/mol⋅K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction.However, a more practical form of this equation is lnk2k1=EaR(1T1−1T2) which is mathmatically equivalent tolnk1k2=EaR(1T2−1T1) where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). The activation energy of a certain reaction is 36.9 kJ/mol . At 27∘C , the rate constant is 0.0130s−1 . At what temperature in degrees Celsius would this reaction go twice as fast? Express your answer with the appropriate units.
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
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as
k=Ae−Ea/RT
where R is the gas constant (8.314 J/mol⋅K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction.
However, a more practical form of this equation is
lnk2k1=EaR(1T1−1T2)
which is mathmatically equivalent to
lnk1k2=EaR(1T2−1T1)
where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2).
The activation energy of a certain reaction is 36.9 kJ/mol . At 27∘C , the rate constant is 0.0130s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?
Express your answer with the appropriate units.
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