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Arrhenius Equation problem

  1. Jan 25, 2010 #1
    1. The problem statement, all variables and given/known data
    At what temperature does K = A?

    2. Relevant equations
    [tex]k= Ae^{-Ea/RT}[/tex]

    3. The attempt at a solution
    In order for K to equal A, then [tex]e=0[/tex]. But at what temperature can [tex]e=0[/tex]? The temperature can't be 0, because that would be impossible. Any help is appreciated?
  2. jcsd
  3. Jan 25, 2010 #2


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    Do you mean [itex]e^{-Ea/RT}=1[/itex]? Why would that imply [itex]T=0[/itex]?
  4. Jan 25, 2010 #3
    I did not say that T should equal 0. I was asking what T must be if [tex]-Ea/RT= 0[/tex].
  5. Jan 25, 2010 #4
    Surely, the activiation energy has to be zero, it looks to be the only way to get zero in the equation, for what you're interested in. Since activation energy is dependent on temperature, make Ea zero, perhaps; although that seems highly questionable...

    From Wiki, concerning the equation for Ea: "While this equation suggests that the activation energy is dependent on temperature, in regimes in which the Arrhenius equation is valid this is cancelled by the temperature dependence of k. Thus Ea can be evaluated from the reaction rate coefficient at any temperature (within the validity of the Arrhenius equation)."
    Kind of goes round in a circle when I tried to find an answer. I would hazard a wild and unfounded guess that k=A is not possible. Do you know if it is plausible? Has it been given as an excercise?

    I'm interested to know now...
    Post if you find the answer! Good luck!
    Last edited: Jan 25, 2010
  6. Jan 25, 2010 #5
    I have found out that temperature must appraoch infinity for K to equal A. In [tex]k= Ae^{-Ea/RT}[/tex], as T approaches infinity, [tex]-Ea/RT[/tex] approaches 0. Basically, this is calculus more than anything.
  7. Jan 25, 2010 #6
    It seems so obvious now that you've pointed it out... I never considered approximating it to equal one...
    If you have a very small activation energy, and a very high temperature, I suppose you could argue that k is approximately A. Taking the limit as T tends to infinity doesn't seem to reflect any plausible, real situation, though.
    Once again, many thanks.
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