# Thread: Integrating the Arrhenius Equation

1. ## 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.

2. 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?

3. 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.

4. 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?

5. 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.

6. 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.

7. 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."

8. That must be at one instant in time. Is that the case?

9. 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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# ahenius equation intergration

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