Please help!! I'm very confused about what equation I should be using. Not sure if I should be using the Arrhenius equation or not.?

2H2O2(aq)---> 2H2O(l) + O2(g)

The activation energy for this reaction is 75 kJxmol^-1. In the presence of a metal catalyst the activation energy is lowered to 49 kJxmol^-1.

At what temperature would the non-catalyzed reaction need to be run to have a rate equal to that of the metal catalyzed reaction at 25 degrees C?

2 Answers

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  • 3 years ago

    Arrhenius Equation [1]

    Rate = Aexp-Ea1/RT

    Ea = Activation energy in J mol^-1; R = 8.314 J mol^-1 K^-1; T in K

    Rate(cat) = Aexp-(49000/(8.314×298.2) = Aexp-(19.76) = A(2.609×10^-9)

    Rate(ucat) = A exp-(75000/(8.314×T)

    (2.609×10^-9) = exp-(75000/(8.314×T) (A’s cancel)

    ln of both sides

    -19.76 = -75000/(8.314×T)

    T = 75000/(19.76×8.314) = 456.4 – 273.2 = 183°C

    [1] http://www.chemguide.co.uk/physical/basicrates/arr...

    http://www.science.uwaterloo.ca/~cchieh/cact/c123/...

  • Al
    Lv 7
    3 years ago

    This has NOTHING to do with Arrhenius.

    • Melanie3 years agoReport

      According to the hint on the assignment it does... And if you say it doesn't the least you could do is explain why, or even better explain how to do it.

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