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Activation Energy type

  1. Jun 29, 2004 #1

    mrjeffy321

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    is there a way to tell the activation energy for a reaction mathmatically just based on what the reaction is, or what it contains? or do you just have to know what it is, based on looking off a table or some other way of finding that data.

    lets say I have a hypothetical reaction:
    X + Y -> XY

    is there any way I could figure out before hand what the activation energy would be, would the type of reaction make a diference?
    (for example synthesis, single/double replacement, combustion, decomposition,...)
     
  2. jcsd
  3. Jun 29, 2004 #2
    Arrhenius Equation

    Hi

    The most fundamental relationship involving the activation energy for a particular reaction is the Arrhenius Equation,

    [tex]
    k = Ae^{-\frac{E_{a}}{RT}}
    [/tex]

    So mathematically, if you know k (the rate constant) and the frequency factor (A) you should be able to extract the activation energy [tex]E_{a}[/tex] from this formula, at a particular temperature T.

    However, this does not completely answer your question...a reaction may be classified as a decomposition, double decomposition, combustion reaction but that only gives you a qualitative idea of how it proceeds. In fact often, you don't get even such an idea when dealing with apparently simple reactions.

    So as far as the mathematics is concerned, the above equation should do the job. A and T are the two parameters you need now.

    Cheers
    Vivek
     
    Last edited: Jun 29, 2004
  4. Jun 29, 2004 #3
    Hi

    Also, since k is a function of temperature, if [tex]k_{1}[/tex] and [tex]k_{2}[/tex] are rate constants measured at [tex]T_{1}[/tex] and [tex]T_{2}[/tex] we have from the Arrhenius Equation,

    [tex]
    \frac{k_{2}}{k_{1}} = e^{\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})}
    [/tex]

    So actually this eliminates the need to know the frequency factor A. All you need to know is the rate constants measured at two different temperatures or even their ratio [tex]\frac{k_{2}}{k_{1}}[/tex] and that is enough to get [tex]E_{a}[/tex], which turns out to be,

    [tex]
    E_{a} = R\frac{T_{1}T_{2}}{T_{2}-T_{1}}log_{e}\frac{k_{2}}{k_{1}}[/tex]

    Please make note of this correction in my previous post.
     
    Last edited: Jun 29, 2004
  5. Jun 30, 2004 #4

    Gokul43201

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    Given X + Y --> XY, it is as hard to know k and A as it is to know E(a). In fact, the intuitive path would be that a knowledge of A and E(a) can tell you what k is.

    E(a) is simply the energy of the highest point along the reaction co-ordinate. The exact value of E(a) can be determined though complex Qunatum Mechanical calculations...though there may be some pattern that is exhibited that depends on the type of reaction.

    Here are some of the factors that E(a) depends on :
    - geometry of the molecules
    - bonds strengths of all bonds
    - the number of bonds of each type that are broken/formed
    - medium/solvent
     
  6. Jun 30, 2004 #5

    mrjeffy321

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    so in other words, it will be hard, and even with those equations given above, I wont necesarily get an accurate answer.
     
  7. Jun 30, 2004 #6

    Gokul43201

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    Absolutely ! But if you knew the rate constants, you can calculate E(a) pretty accurately.

    Why do you want to know E(a), in any case ?
     
    Last edited: Jun 30, 2004
  8. Jun 30, 2004 #7
    Quantum Mechanical Calculations will be accurate anyway and will put everything else to rest. But at your level (I believe you are studying Chemical Kinetics from first principles or at least for the first time in this manner) would you be interested in delving into all that stuff immediately?

    Besides, since you mentioned tables in your first post, I mentioned these equations which are taught in a classical chemistry course and should be known to you already and at this level, you would be first exposed to tables containing values of the rate constant at different temperatures (the k1s and k2s), other experimental conditions remaining fixed.

    I understand that the equations I have mentioned are not accurate in a realworld situation but they use only the data you can easily get so they cannot be far from accurate in a controlled reaction experiment. In your lab for instance, you would carefully carry out X + Y -> XY and not simply mix the reagents.

    Also E(a) is known for such reasonably ideal conditions and yes, tables exist to find it out.

    Cheers
    Vivek
     
  9. Jun 30, 2004 #8
    Hey we're now entering a debate that is based on mathematical formulae and the physical principles. While I do not disagree with gokul, I would like to add that a simple intuitive approach is not a bad idea to understand things which cannot be exactly or easily expressed as mathematical equations. The Arrhenius Equation and its modification to eliminate the frequency factor are two things we were taught in school in our physical chemistry course and are mentioned in most books as fundamental relationships.

    How exactly you determine k is an experimental idea and yes there exist methods to do so without using activation energy. They are better ones indeed because in a particular situation you may not know the reaction coordinate in order to extract E(a) from it. So I agree that my equations aren't really as fundamental to find k, but if you're beginning an undergraduate course where you have little quantum mechanical experience they are good ways to get a fair idea.

    Cheers
    Vivek
     
  10. Jun 30, 2004 #9

    mrjeffy321

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    I am not looking for an exact amount for any reaction, I was just looking for more of a general idea, and the more exact that estimate can be, the better, so that is why I like the formula method of finding the energy, but I do agree that the actual energy required can very from reation to reaction.

    alot of this is very new to me, partly because of my inexperience, and also partly because I am having to go back and re-learn many concepts because my chemestry teacher was lets say not the best. so I think that going into quantum anything in order to find the answer might be a little to technical for me.

    now lets just say that we will use this equation:
    [tex]E_{a} = R\frac{T_{1}T_{2}}{T_{2}-T_{1}}log_{e}\frac{k_{2}}{k_{1}}[/tex]
    I will need to know these things:
    -the rate constants at the begining and ending temperatures?
    -the diferent temperatures?
    -R?
     
  11. Jul 1, 2004 #10

    Gokul43201

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    I was not suggesting that you try to calculate E(a) using Quantum Mechanics. That is a difficult problem for a molecular physicist.

    I was only trying to show you how diificult it would be to determine E(a) from first principles (theoretically). Of course, you can use experimental data like rate constants to deduce E(a).

    The rate constant is a function of temperature and is usually specified at some temperature.
    'R' is the ideal gas constant 8.314 J/K-mol .
     
  12. Jul 1, 2004 #11

    GCT

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    Some factors that will influence activation energy:

    entropy: yes, as you suggested whether a reaction is unimolecular, trimolecular, etc... will influence the activation energy. Activation energy, while in general chemistry is given as Ea, in organic you will find it as Ga (Gibbs free energy of activation). And as you may know, this variable is influenced by enthalpy as well as entropical factors.

    Hope this helps.


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    http://groups.msn.com/GeneralChemistryHomework
     
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