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Determining activation energy

  1. Jul 6, 2013 #1
    1. The problem statement, all variables and given/known data
    At room temperature (20°C) orange juice gets spoilt in about 64 hours. In a refrigerator at 3°C juice can be stored three times as long before it gets spoilt. Estimate (a) the activation energy of the reaction that causes the spoiling of juice. (b) How long should it take for juice to get spoilt at 40°C?

    (Answer: (a)43.46 kJ mol^(-1) (b) 20.47 hour)

    2. Relevant equations



    3. The attempt at a solution
    I guess I have to use the Arrhenius equation here but I don't have the rate constants at the two temperatures. How am I supposed to solve this?
     
  2. jcsd
  3. Jul 6, 2013 #2

    mfb

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    You have a ratio between the rate constants at two different temperatures. That is sufficient to calculate the activation energy, even if you do not know the other parameters.
     
  4. Jul 6, 2013 #3
    If ##k_1## is the rate constant at 20°C and ##k_2## at 3°C, does that mean ##k_1/k_2=3##?
     
  5. Jul 7, 2013 #4

    mfb

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    Sure.
     
  6. Jul 8, 2013 #5
    I tried that but I end up with a wrong answer.

    From Arrhenius equation,
    [tex]\ln\frac{k_1}{k_2}=-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)[/tex]
    Plugging in the values and solving for ##E_a##, I get a wrong answer. Here's the calculation:
    Wolfram|Alpha
     
  7. Jul 8, 2013 #6

    mfb

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    Just a calculation error in the fridge temperature.
    fixed
     
  8. Jul 8, 2013 #7
    :tongue:

    Thank you!
     
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