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Activation energy of a certain reaction

  1. Oct 23, 2007 #1
    The activation energy of a certain reaction is 45.8 kJ/mol. At 20 degrees celsius, the rate constant is 0.0130 s^(-1). At what temperature would this reaction go twice as fast?

    Arrhenius Equation:k = Ae^(-E_a/RT)

    ln(k_2 / k_1) = (E_a /R)(1/T_1 - 1/T_2)

    Okay, so I'm not quite sure how to approach this problem. I think you're supposed to solve for k_2 and then sub it in to find the activation energy and then find the temperature. But I don't know how to solve for k_2 if I'm not given a second temperature. Any help is appreciated!
  2. jcsd
  3. Oct 23, 2007 #2


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    You are given that k2/k1=2.
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