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Arrhenius equation plot

  1. Apr 2, 2012 #1
    1. The problem statement, all variables and given/known data
    This isn't a homework question but this thing is disturbing from a long time.
    I attended a class on Arrhenius equation and its plots. One of the plots is disturbing me from long which is the plot between k vs T.
    Arrhenius equation:-
    [tex]k=Ae^{-\frac{E_a}{RT}}[/tex]
    k->Rate constant
    Ea->Activation Energy
    R->Gas constant
    T->Temperature

    2. Relevant equations



    3. The attempt at a solution
    Since the equation is of the form y=e-1/x, the graph should be like this one:-
    Click Here
    But my teacher made a completely different plot. I constantly said that it shouldn't be like that. Finally, when i checked wikipedia, to my surprise, the teacher was right. Here's the wiki article:- http://en.wikipedia.org/wiki/Arrhenius_plot
    I don't understand why the graph is different from that of y=e-1/x? Can somebody tell me where i am going wrong?

    Thanks!
     
  2. jcsd
  3. Apr 2, 2012 #2

    jedishrfu

    Staff: Mentor

    arent the wiki graphs log based?
     
  4. Apr 2, 2012 #3
    I am talking about this graph from the wiki page:-
    625px-NO2_Arrhenius_k_against_T.svg.png

    No, i don't think this one is log based.
     
  5. Apr 2, 2012 #4

    jedishrfu

    Staff: Mentor

    okay so the bottom chart is an arrhenius plot plotting 1/T vs ln(k) as defined in the wiki page and the upper chart is the same data but plotting T vs k.

    The only thing I can think of is that Ea changes with the square of T causing it to curve upward.
     
  6. Apr 2, 2012 #5

    I like Serena

    User Avatar
    Homework Helper

    Hey Pranav-Arora! ;)

    Here's another one from WolframAlpha.


    http://www4c.wolframalpha.com/Calculate/MSP/MSP82681a10gc2589cc12cc00000dh1e4827i034cf8?MSPStoreType=image/gif&s=20&w=300&h=183&cdf=RangeControl [Broken]

    plot[ y=e^(-1/x), {x,0.2,0.5} ]


    Doesn't it kind of look like your graph?
     
    Last edited by a moderator: May 5, 2017
  7. Apr 2, 2012 #6
    Hello ILS! :)

    Oh yeah, that's look like something which i am looking for. You made the graph for very small values of x but i don't think that this is correct because in the wiki graph, temperature is increasing to high numbers but it isn't turning like to be that of y=e^(-1/x).

    Ea changes with the square of T? I will have to check that out.
     
    Last edited by a moderator: May 5, 2017
  8. Apr 2, 2012 #7

    I like Serena

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    Well, the formula is undoubtedly based on the shape of the graph when plotting ln k versus 1/T.
    Then you apparently get the graph:

    500px-NO2_Arrhenius_lnk_against_T%5E-1.svg.png

    This is approximated by a straight line.

    With the relationship ##\ln k = c_1 - c_2 \cdot \frac 1 T##, you can deduce that ##k=A e^{-\frac B T}##.

    When I make an approximation of this, I get the following graph:

    http://www3.wolframalpha.com/Calculate/MSP/MSP65161a10geiaaagg40a900000h0fa40fba06a13c?MSPStoreType=image/gif&s=30&w=300&h=191&cdf=RangeControl [Broken]
    plot[ k=18e9 * e^(-13000/t), {t,590,660} ]


    Doesn't this kind of fit the expected graph?
     
    Last edited by a moderator: May 5, 2017
  9. Apr 3, 2012 #8
    How did you get this relation? :uhh:
     
  10. Apr 3, 2012 #9

    I like Serena

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    A straight line is defined by the equation:
    $$y = a x + b \qquad\qquad (1)$$
    where ##a## is the slope of the line, and ##b## is the y-coordinate where the line intersects the y-axis (the y-intercept).


    When we fit a straight line to a set of points, we try to find out what ##a## and ##b## have to be.


    In this case we have ##\ln k## on the y-axis, and ##\frac 1 T## on the x-axis.
    So we replace ##y## in equation (1) by ##\ln k##, and we replace ##x## by ##\frac 1 T##.

    The result is:
    $$\ln k = a \cdot \frac 1 T + b$$

    Since we already know that the line slopes down, I have taken the liberty of putting in a minus sign, and rewriting the equation with different constants ##c_1## and ##c_2## as:
    $$\ln k = c_1 - c_2 \cdot \frac 1 T$$
     
  11. Apr 3, 2012 #10
    I already knew about the straight line stuff, i was getting confused with those constant c1 and c2.
    Thank you for the help! :smile:
     
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