# How do you plot the Arrhenius equation?

Mar 26, 2016

You draw a graph of $\ln k$ vs. $\frac{1}{T}$.

#### Explanation:

The Arrhenius equation is

$k = A {e}^{- {E}_{a} / R T}$

where

• $k$ is the rate constant
• $A$ is the pre-exponential factor
• ${E}_{a}$ is the activation energy
• $R$ is the universal gas constant
• $T$ is the temperature

If we take the logarithms of both sides, we get

lnk = lnA color(white)(l)–E_a/(RT)

or

$\ln k = - \left({E}_{a} / R\right) \frac{1}{T} + \ln A$

This is the equation of a straight line

$y = m x + b$

where $m$ is the slope and $b$ is the slope of a plot of $y$ vs $x$.

$\ln k = \left(\text{-} {E}_{a} / R\right) \frac{1}{T} + \ln A$
$\textcolor{w h i t e}{l l} y = \textcolor{w h i t e}{m m} m \textcolor{w h i t e}{m l l} x \textcolor{w h i t e}{l l} + \textcolor{w h i t e}{l l} b$

By comparing the equations, we see that the slope of the line is $\text{-} {E}_{a} / R$ and the $y$ intercept is $\ln A$

Hence, if we draw a graph of $\ln k$ vs. $1 / T$ at various temperatures, we should get a straight line.

Example

For a certain reaction, the data were:

$\boldsymbol{T} \boldsymbol{\text{/°C" color(white)(ml)bb"k/s"^"-1}}$
$\textcolor{w h i t e}{l l} 477 \textcolor{w h i t e}{m l l} 0.00018$
$\textcolor{w h i t e}{l l} 523 \textcolor{w h i t e}{m l l} 0.0027$
$\textcolor{w h i t e}{l l} 577 \textcolor{w h i t e}{m l l} 0.030$
$\textcolor{w h i t e}{l l} 623 \textcolor{w h i t e}{m l l} 0.26$

Plot the Arrhenius equation using the above values.

Solution

We have to plot $\ln k$ vs $1 / T$.

We must convert the temperatures to kelvins and take their reciprocals.

We must also take the natural logarithms of the rate constants.

Although we could do this by hand, I find it easier to do the calculations and the plot in Excel.

Here's are my results.