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Arrhenius equation plot 
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#1
Apr212, 12:36 PM

P: 3,806

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 E_{a}>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
Apr212, 12:43 PM

P: 2,812

arent the wiki graphs log based?



#3
Apr212, 12:59 PM

P: 3,806

I am talking about this graph from the wiki page:
No, i don't think this one is log based. 


#4
Apr212, 01:41 PM

P: 2,812

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. 


#5
Apr212, 01:50 PM

HW Helper
P: 6,187

Hey PranavArora! ;)
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? 


#6
Apr212, 02:04 PM

P: 3,806

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). 


#7
Apr212, 02:22 PM

HW Helper
P: 6,187

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? 


#8
Apr312, 09:22 AM

P: 3,806




#9
Apr312, 09:32 AM

HW Helper
P: 6,187

$$y = a x + b \qquad\qquad (1)$$ where ##a## is the slope of the line, and ##b## is the ycoordinate where the line intersects the yaxis (the yintercept). 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 yaxis, and ##\frac 1 T## on the xaxis. 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: $$\ln k = c_1  c_2 \cdot \frac 1 T$$ 


#10
Apr312, 10:31 AM

P: 3,806

Thank you for the help! 


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