# Why is the Arrhenius plot for the rate constant different from y=e-1/x?

• Saitama
In summary, the teacher made a different plot for y=e-1/x and it doesn't look like the one on wikipedia. I don't understand why the graph is different.
Saitama

## Homework Statement

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

## 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!

arent the wiki graphs log based?

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

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.

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

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

Doesn't it kind of look like your graph?

Last edited by a moderator:
I like Serena said:
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

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).

jedishrfu said:
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.

Last edited by a moderator:
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
plot[ k=18e9 * e^(-13000/t), {t,590,660} ]Doesn't this kind of fit the expected graph?

Last edited by a moderator:
I like Serena said:
With the relationship ##\ln k = c_1 - c_2 \cdot \frac 1 T##...

How did you get this relation?

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

I like Serena said:
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!

## 1. What is an Arrhenius equation plot?

An Arrhenius equation plot is a graphical representation of the relationship between the rate constant (k) of a chemical reaction and the temperature (T) at which the reaction occurs. It is based on the Arrhenius equation, which states that the rate constant is directly proportional to the activation energy (Ea) and inversely proportional to the temperature.

## 2. How is an Arrhenius equation plot used in chemistry?

An Arrhenius equation plot is used to determine the activation energy of a chemical reaction, which is an important factor in predicting the rate of a reaction. It can also be used to compare the rates of different reactions and to determine the temperature dependence of a reaction.

## 3. What does the slope of an Arrhenius equation plot represent?

The slope of an Arrhenius equation plot represents the activation energy (Ea) of the reaction. A steeper slope indicates a higher activation energy, while a shallower slope indicates a lower activation energy.

## 4. How is an Arrhenius equation plot affected by a change in temperature?

An increase in temperature causes the rate constant (k) to increase, resulting in a steeper slope on the Arrhenius equation plot. Conversely, a decrease in temperature causes the rate constant to decrease, resulting in a shallower slope on the plot.

## 5. What is the significance of the intersection point on an Arrhenius equation plot?

The intersection point on an Arrhenius equation plot represents the temperature at which the reaction rate is independent of temperature, also known as the Arrhenius temperature. This temperature is important in predicting the behavior of the reaction at different temperatures.

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