To use the Arrhenius equation to calculate the activation energy. As temperature rises, the average kinetic energy of molecules increases. In a chemical reaction, this means that a higher percentage of the molecules possess the required activation energy, and the reaction goes faster. This relationship is shown by the Arrhenius equation k=Ae−Ea/RT where k is the rate constant, A is the frequency factor, Ea is the activation energy, R = 8.3145 J/(K⋅mol) is the gas constant, and T is the Kelvin temperature. The following rearranged version of the equation is also useful: ln(k2k1)=(EaR)(1T1−1T2) where k1 is the rate constant at temperature T1, and k2 is the rate constant at temperature T2. The rate constant of a chemical reaction increased from 0.200 s−1 to 2.80 s−1 upon raising the temperature from 26.0 ∘C to 52.0 ∘C . Calculate the value of (1T1−1T2) where T1 is the initial temperature and T2 is the final temperature. Express your answer numerically.

Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter11: Chemical Kinetics: Rates Of Reactions
Section: Chapter Questions
Problem 11.ACP: (Section 11-5) A rule of thumb is that for a typical reaction, if concentrations are unchanged, a...
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To use the Arrhenius equation to calculate the activation energy.

As temperature rises, the average kinetic energy of molecules increases. In a chemical reaction, this means that a higher percentage of the molecules possess the required activation energy, and the reaction goes faster. This relationship is shown by the Arrhenius equation

k=Ae−Ea/RT

where k is the rate constant, A is the frequency factor, Ea is the activation energy, R = 8.3145 J/(K⋅mol) is the gas constant, and T is the Kelvin temperature. The following rearranged version of the equation is also useful:

ln(k2k1)=(EaR)(1T1−1T2)

where k1 is the rate constant at temperature T1, and k2 is the rate constant at temperature T2.

The rate constant of a chemical reaction increased from 0.200 s−1 to 2.80 s−1 upon raising the temperature from 26.0 ∘C to 52.0 ∘C .

Calculate the value of (1T1−1T2) where T1 is the initial temperature and T2 is the final temperature.
Express your answer numerically.
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