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Math Help - Integrating the Arrhenius Equation

  1. #1
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    Integrating the Arrhenius Equation

    Hi everyone,

    I'm a bit rusty on my calc, and hope someone here can help me. In reviewing a particular scientific paper, the reaction rate equation is expressed as:

    \frac{dR}{dt}=Ae^{\frac{-E}{kT}

    Where \frac{dR}{dt} is the reduction in property with respect to time
    A is the steric constant
    k is the gas constant or Bolzmann constant (depending on units)
    T is absolute temperature
    and
    E is the activation energy of the reaction

    The paper then says "Integration of the rate equation, followed by the taking of logarithms, results in an equation of the form"
    \ln t = (\frac{E}{k})\frac{1}{T} + B

    I understand this to be true, but I am having a hard time getting from the first equation to the second equation on my own. If anyone could break it down step by step for me, I would be most appreciative.
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  2. #2
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    Treat everything on the RHS as a constant in time.

    First step:

    \displaystyle{\int dR=A e^{-\frac{E}{k T}}\int dt.}

    You're gonna perform the integrals, which is quite straight-forward, and then you'll have to define B such that you get the equation you need. Can you do the next step?
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  3. #3
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    Thanks for the quick response...OK, So I get
     (R+C_1)e^{\frac{E}{k T}}=A(t+C_2)

    and if I solve for t I get:

    t=(\frac{R+C_1}{A})e^{\frac{E}{k T}}-C_2

    I guess I need to assume R is a constant. If R is a constant, then R+C1/A is a constant I can call D

    t=De^{\frac{E}{k T}}-C_2

    Am I on the right track?

    I get stuck If I take the natural log of both sides of this

    \ln t=ln(De^{\frac{E}{k T}}-C_2)
    Because of the C2 term.

    If I ignore the C2 term, then it seems to work out

    \ln t=ln(De^{\frac{E}{k T}})

    \ln t=lnD + ln(e^{\frac{E}{k T}})

    \ln t=lnD + \frac{E}{k T}

    \ln t=lnD + (\frac{E}{k})\frac{1}{T}

    If I call lnD=B, then I get my desired form

    \ln t=B+(\frac{E}{k})\frac{1}{T}

    So, please help me see where I am going wrong. How can I ignore the constant created by the integration of dt? Or am I missing something?

    Thank you again.
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  4. #4
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    You don't need two constants of integration - one will suffice. I did it this way:

    \displaystyle{R=A\,e^{-\frac{E}{kT}}\,t+C.}

    I moved the constant of integration over to the other side, and then took the logarithm, etc.

    R is most definitely not a constant. If it were, its derivative would be zero, not the expression we have for dR/dt. When you define your B, it's obviously going to have R in it somewhere. Why does B have to be constant? Does the paper say B is constant?
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  5. #5
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    OK...I think I'm starting to see the light. You are right, R is not a constant, but a function of T. Certainly it needs to be accounted for. And how quickly I forgot that we are treating the initial RHS as a constant, and a constant times a constant is...duh.

    And the paper doesn't say explicitly that B is a constant, but the formula we arrive at looks like the general form for a line if we substitute m=E/k and, x=1/T, and y=lnt, so I was thinking B would be constant as an intercept.
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  6. #6
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    R is a function of t, not T, I think you meant to say. In this problem, that distinction is important!

    If the paper doesn't treat the second equation like a straight line, there's no need to regard it as the equation of a straight line.
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  7. #7
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    You are right, R is a function of t (time). dR/dt is a function of T (Temperature).

    But the paper does treat the equation like a straight line, as it then says "A plot of lnt versus 1/T is then a straight line of slope E/k, and is known as the Arrhenius plot."
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  8. #8
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    That must be at one instant in time. Is that the case?
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  9. #9
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    Perhaps...

    The plots being generated in this paper plot experimental data of lifetime (t) for multiple Temperatures (T). ln t is used for the y axis and 1/T is used for the x axis, then a best fit line is drawn, and the slope of that line is used to calculate E (Since k is a constant)

    But the value of R used to determine lifetime is being defined in the experiment...Ah Ha! So if we are saying that we are going to plot these points for values of t associated with a specific value of R, we will, for the purposes of the plot, be holding R constant. R varies with time, but the value of time we are using for any given T corresponds with a set, consistent value of R. So, in essence, we are allowing t to vary for a given T to get a specific value of R. It appears that this is why we get a straight line.

    Thank you so much for helping me to work through this.
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