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Activation energy

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate Ea from given info:
    ln k = 0 t(sec-1) = 0.0033
    ln k = -0.2 t(sec-1) = 0.00335
    ln k = -0.4 t(sec-1) = 0.0034

    2. Relevant equations
    ln k = -(Ea/RT) + ln A

    3. The attempt at a solution
    I could find Ea if i had the temperature but i'm given the time. I think you can cancel out the ln A with the temperature can't you? But if you do that, Ea would be in J/K*mol, but Ea should be kJ/mol. This is all from a graph btw and the y intercept = -4515.7x + 14.749 and R2 = 0.9964. Am i suppose to treat this as y=mx + b?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 17, 2008 #2


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    Homework Helper

    Do you have any temperature data? You are going to need it.
  4. Nov 17, 2008 #3
    no temps are given. if they were, i would have been able to solve it.
  5. Nov 20, 2008 #4
    same here
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