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Find the rate constant, given temperature and activation energy.

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    A reaction is found to have an activation energy of 38.0 kJ/mol. If the rate constant for this reaction is 1.60 × 102 M-1s-1 at 249 K, what is the rate constant at 436 K?

    2. Relevant equations

    [tex]ln\frac{K_{2}}{K_{1}}=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)[/tex]

    3. The attempt at a solution

    Given:

    R=8.314
    T1=249K
    T2=436K
    Ea=160

    [tex]ln(K_{2})=\frac{38}{8.314}\left(\frac{1}{249}-\frac{1}{436}\right)+ln(160)[/tex]

    Which equals 161.257 which is incorrect. Any clues where I went wrong would be greatly appreciated. Thanks in advance.

    Joe
     
  2. jcsd
  3. Sep 18, 2010 #2

    Borek

    User Avatar

    Staff: Mentor

    38000
     
  4. Sep 19, 2010 #3
    Ah ha! I should have noticed that being that R has units of J not Kj. Thank you very much Borek.

    Joe
     
  5. Sep 19, 2010 #4

    Borek

    User Avatar

    Staff: Mentor

    kJ, not Kj...
     
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