In the data analysis section of this lab you will be preparing a graph of lnK vs 1/T for trials 1-4. This graph correspond to the Arrhenius equation in it's linear form. ln k = -Ea/R (1/T) + lnA (R=0.008314 kJ/mol K) If the slope of the graph you prepare is - 9537.3, What is the activation energy (Ea) for the reaction in kJ/mol (Answer to 4 sig figs)
In the data analysis section of this lab you will be preparing a graph of lnK vs 1/T for trials 1-4. This graph correspond to the Arrhenius equation in it's linear form. ln k = -Ea/R (1/T) + lnA (R=0.008314 kJ/mol K) If the slope of the graph you prepare is - 9537.3, What is the activation energy (Ea) for the reaction in kJ/mol (Answer to 4 sig figs)
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter14: Chemical Kinetics: The Rates Of Chemical Reactions
Section: Chapter Questions
Problem 31PS
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In the data analysis section of this lab you will be preparing a graph of lnK vs 1/T for trials 1-4. This graph correspond to the Arrhenius equation in it's linear form.
ln k = -Ea/R (1/T) + lnA
(R=0.008314 kJ/mol K)
If the slope of the graph you prepare is - 9537.3, What is the activation energy (Ea) for the reaction in kJ/mol
(Answer to 4 sig figs)
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