Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Arrhenius equation

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    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.

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

    3. 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?
  2. jcsd
  3. Mar 2, 2008 #2
    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.
  4. Mar 2, 2008 #3
    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?
  5. Mar 2, 2008 #4
    The equation you are using is incorrect.

    http://alt1.mathlinks.ro/Forum/latexrender/pictures/1/b/a/1bacaba62a260d46b7600cc644b6bba500f54b3b.gif [Broken]
    Last edited by a moderator: May 3, 2017
  6. Mar 2, 2008 #5
    I'm using the same equation as that....I'm just changing the locations of the corresponding k1/T1 and k2/T2 values.
  7. Mar 3, 2008 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Nice derivation, Roco. Rewrite the expression and solve for Ea. Use "T1+10" in place of "T2".
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook