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Homework Help: Activation energy

  1. Mar 12, 2008 #1
    Please I need some help. I'm working on a chemistry question and have to determine the activation energy for a reverse reaction. How do you do that? I've looked all over my text and through my notes and cannot find a clear way on how to solve it.
  2. jcsd
  3. Mar 12, 2008 #2
    If the reaction is A --> B and this has a DeltaG(A-->B) and an activation energy E1, then the activation energy E2 of the reverse reaction B --> A is E2 = DeltaG(A-->B) - E1.
    Last edited: Mar 13, 2008
  4. Mar 12, 2008 #3
    Is DeltaG the heat of reaction?
  5. Mar 12, 2008 #4
    deltaG is called the Free Gibbs Energy.
    but I don't think we need deltaG here.
  6. Mar 12, 2008 #5
    The issue is not simple. See, e.g., Atkins - Physical Chemistry, par. 27.6.
  7. Mar 13, 2008 #6
    i'm still not getting it..... so here's the question:

    For the reaction CO + NO2 --> CO2 + NO the activation energy for the forward reaction is 135 kJ/mol of CO reacted.
    a) Determine the heat of reaction.
    For this I got -379.1 kJ/mol by using the heats of formation from my text and reversed the signs given to the reactants. I'm assuming this is right because I can't find another way to do this.

    b) From the data given. and the Delta Hr for the reaction, determine the activation energy (Ea) for the reverse reaction.
    This is where I'm stuck. I added the heat of reaction to the 135 kJ/mol given to me and got another negative number. Is this possible?

    Thanks for your help!
  8. Mar 13, 2008 #7
    You are right, I wrote "+" thinking "-". I'm sorry!
    The correct equation is:
    E2 = DeltaG(A-->B) - E1
    I have corrected my previous post.

    For the reaction:

    [tex]CO + NO_2\ \rightarrow\ CO_2 + NO[/tex]

    I get:

    [tex]\Delta H = -225.91\ kJmol^{-1}\ (Heat\ of\ reaction)[/tex]

    [tex]\Delta G = -221.95\ kJmol^{-1}[/tex]

    So, for the reversed reaction:

    [tex]CO_2 + NO \rightarrow\ CO + NO_2[/tex]

    you would have: 135 + 221.95 = 356.95 kJ/mol as Gibbs free energy of activation, assuming that 135 kJ/mol was the Gibbs free energy of activation of the reversed reaction.
    Last edited: Mar 13, 2008
  9. Mar 13, 2008 #8
    thanks alot really appreciate it
  10. Jun 18, 2008 #9
    would you possibly be able to show me the steps as to how you recieved -225.91 as your answer, I tried looking through multiply text books and internet sites but I can't seem to find where you got your numbers, greatly appreciate it if you could
  11. Jun 18, 2008 #10
    P.W. Atkins - Physical Chemistry - Fifth Edition:

    [tex] \Delta H^0_f(CO_2)_g = -393.51\ kJ mol^{-1} [/tex]

    [tex] \Delta H^0_f(NO)_g = 90.25\ kJ mol^{-1} [/tex]

    [tex] \Delta H^0_f(CO)_g = -110.53\ kJ mol^{-1} [/tex]

    [tex] \Delta H^0_f(NO_2)_g = 33.18\ kJ mol^{-1} [/tex]

    So, for the reaction

    [tex]CO + NO_2\ \rightarrow\ CO_2 + NO[/tex]

    you have:

    [tex]\Delta H^0_{reaction}\ =\ (-393.51\ +\ 90.25)\ -\ (-110.53\ +\ 33.18)\ =\ -225.91\ kJ mol^{-1}[/tex]
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