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Arrhenius equation plot

by Pranav-Arora
Tags: arrhenius, equation, plot
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Pranav-Arora
#1
Apr2-12, 12:36 PM
P: 3,807
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!
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jedishrfu
#2
Apr2-12, 12:43 PM
P: 2,982
arent the wiki graphs log based?
Pranav-Arora
#3
Apr2-12, 12:59 PM
P: 3,807
I am talking about this graph from the wiki page:-


No, i don't think this one is log based.

jedishrfu
#4
Apr2-12, 01:41 PM
P: 2,982
Arrhenius equation plot

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.
I like Serena
#5
Apr2-12, 01:50 PM
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Hey Pranav-Arora! ;)

Here's another one from WolframAlpha.




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


Doesn't it kind of look like your graph?
Pranav-Arora
#6
Apr2-12, 02:04 PM
P: 3,807
Quote Quote by I like Serena View Post
Hey Pranav-Arora! ;)

Here's another one from WolframAlpha.




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

Doesn't it kind of look like your graph?
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).

Quote Quote by jedishrfu View Post
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.
Ea changes with the square of T? I will have to check that out.
I like Serena
#7
Apr2-12, 02:22 PM
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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:



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:


plot[ k=18e9 * e^(-13000/t), {t,590,660} ]


Doesn't this kind of fit the expected graph?
Pranav-Arora
#8
Apr3-12, 09:22 AM
P: 3,807
Quote Quote by I like Serena View Post
With the relationship ##\ln k = c_1 - c_2 \cdot \frac 1 T##...
How did you get this relation?
I like Serena
#9
Apr3-12, 09:32 AM
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Quote Quote by Pranav-Arora View Post
How did you get this relation?
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:
$$\ln k = c_1 - c_2 \cdot \frac 1 T$$
Pranav-Arora
#10
Apr3-12, 10:31 AM
P: 3,807
Quote Quote by I like Serena View Post
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:
$$\ln k = c_1 - c_2 \cdot \frac 1 T$$
I already knew about the straight line stuff, i was getting confused with those constant c1 and c2.
Thank you for the help!


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