# Arrhenius equation plot

1. Apr 2, 2012

### Pranav-Arora

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:-
$$k=Ae^{-\frac{E_a}{RT}}$$
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:-
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. Apr 2, 2012

### Staff: Mentor

arent the wiki graphs log based?

3. Apr 2, 2012

### Pranav-Arora

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

4. Apr 2, 2012

### 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.

5. Apr 2, 2012

### I like Serena

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
6. Apr 2, 2012

### Pranav-Arora

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
7. Apr 2, 2012

### I like Serena

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:

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
8. Apr 3, 2012

### Pranav-Arora

How did you get this relation? :uhh:

9. Apr 3, 2012

### I like Serena

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:
$$\ln k = c_1 - c_2 \cdot \frac 1 T$$

10. Apr 3, 2012

### Pranav-Arora

I already knew about the straight line stuff, i was getting confused with those constant c1 and c2.
Thank you for the help!