Solving Arrhenius Eqn: Double Reaction Rate at 10°C

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Discussion Overview

The discussion revolves around the Arrhenius equation and the conditions under which a temperature increase of 10 degrees Celsius can double the reaction rate. Participants are exploring the derivation of this relationship, focusing on the activation energy (Ea) and the implications for reaction rates.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks clarification on what to solve for when deriving the conditions for doubling the reaction rate, questioning the relevance of activation energy (Ea) in this context.
  • Another participant confirms that solving for Ea is necessary and suggests using arbitrary values for the calculations as long as they are consistent.
  • A participant expresses confusion about the specific value of Ea needed and how it relates to determining the reaction rate.
  • One participant challenges the correctness of the equation being used, implying there may be an error in the formulation.
  • Another participant indicates they are using the same equation but adjusting the placement of variables related to the rate constants and temperatures.
  • A later reply encourages rewriting the expression to solve for Ea, suggesting to use "T1+10" for the temperature increase.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the correct approach to the problem, with some questioning the equation's validity and others providing differing methods for solving for Ea.

Contextual Notes

There are unresolved questions regarding the specific values to use for the rate constants and temperatures, as well as the implications of the activation energy on the reaction rate.

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Homework Statement


A common rule of thumb is that temperature 10 degrees C for many reactions will double the reaction rate. Derive under what conditions this is true.

Homework Equations


ln(k) - ln(2k) = (-Ea/R) [(1/T1)-(1/T2)]

The Attempt at a Solution


I know I have to start off by plugging in values for T1 and T2, but what exactly am I trying to solve for? My teacher said to solve for Ea, but that is the measure of kinetic energy needed for reaction...what do I solve for when looking for reaction rate?

Also, can I leave the rate constants in or should I plug in actual values for those as well?
 
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I had the exact same question on a quiz, but instead we were given the room temperature. Indeed you do want to solve for Ea. You'll want to plug in values for ln k & temp. Think arbitrary values for what you want to plug in for the numbers, as long as they all relate to each other.
 
But what value of Ea am I looking for? If I'm plugging in all my values into that equation that I typed up there, of course I'll just get a value for Ea.
And why am I solving for Ea anyway? How will I know what the reaction rate is if I know Ea?
 
The equation you are using is incorrect.

http://alt1.mathlinks.ro/Forum/latexrender/pictures/1/b/a/1bacaba62a260d46b7600cc644b6bba500f54b3b.gif
 
Last edited by a moderator:
I'm using the same equation as that...I'm just changing the locations of the corresponding k1/T1 and k2/T2 values.
 
Nice derivation, Roco. Rewrite the expression and solve for Ea. Use "T1+10" in place of "T2".
 

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