Calculate Ea: Activation Energy Solution

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    Activation Energy
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Discussion Overview

The discussion revolves around calculating the activation energy (Ea) using given data related to reaction rates (k) and time (t). Participants are attempting to apply the Arrhenius equation, but they express uncertainty due to the absence of temperature data.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant notes the relationship between ln k and Ea, referencing the equation ln k = -(Ea/RT) + ln A, but expresses confusion about how to proceed without temperature data.
  • Another participant emphasizes the necessity of temperature data to calculate Ea, indicating that it is essential for the solution.
  • Several participants agree that the lack of temperature information prevents them from solving the problem.

Areas of Agreement / Disagreement

Participants generally agree that the absence of temperature data is a significant barrier to calculating the activation energy, and no competing views are presented regarding the necessity of this data.

Contextual Notes

The discussion highlights the dependency on temperature for the calculation of activation energy, and participants acknowledge that without this information, the problem remains unsolvable.

magma_saber
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Homework Statement


Calculate Ea from given info:
ln k = 0 t(sec-1) = 0.0033
ln k = -0.2 t(sec-1) = 0.00335
ln k = -0.4 t(sec-1) = 0.0034

Homework Equations


ln k = -(Ea/RT) + ln A


The Attempt at a Solution


I could find Ea if i had the temperature but I'm given the time. I think you can cancel out the ln A with the temperature can't you? But if you do that, Ea would be in J/K*mol, but Ea should be kJ/mol. This is all from a graph btw and the y intercept = -4515.7x + 14.749 and R2 = 0.9964. Am i suppose to treat this as y=mx + b?
 
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Do you have any temperature data? You are going to need it.
 
no temps are given. if they were, i would have been able to solve it.
 
same here
 

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