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Calculating a reverse reaction

  1. Sep 10, 2006 #1
    Hello everyone,

    Here's my question.
    We have a reaction with an overal enthalpy change of -66KJ. The activation energy is 7KJ.
    We want the Ea for the reverse reaction.

    Formula:
    Ea (rev) = ∆E + Ea (fwd)
    So I wrote:
    Ea (rev) = -66 + 7 = 59 KJ.

    But the corrected exercise says: 66+7 = 73KJ.
    Why did we add 66, if it's a negative value?

    Thank you,

    J.
     
  2. jcsd
  3. Sep 10, 2006 #2

    GCT

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    Now try to see all of this practically.......the reaction itself is exothermic, the activation energy for the reverse reaction thus is greater then the forward assuming a single "transition state." Draw a free energy diagram, it should help greatly in understanding what's going on here.
     
  4. Sep 11, 2006 #3
    So since the reverse reaction is ENDOthermic, then the Ea should be greater than in the reaction that was exothermic... Would this be the correct way to think?

    ~J.
     
  5. Sep 11, 2006 #4

    Gokul43201

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    vet: Do you know how to draw the energy diagram? It takes just a couple of minutes to learn it, and once you do, problems like this become a piece of cake.

    Exothermic reaction, endothermic reaction - the positive x-direction is the direction of the forward reaction.
     
  6. Sep 11, 2006 #5
    I do, and I understand why the answer is 73, but I was just puzzled by the equation, since it seemed to lead to the wrong answer...

    Joanna.
     
  7. Sep 11, 2006 #6

    Bystander

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    Ea (rev) = ∆Erev + Ea (fwd)

    Remember to change the sign --- the reaction is proceeding in the reverse direction, and the sign of the energy change for the reaction is opposite that for the forward.
     
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