Chapter 17. Kinetics

Activation Energy and the Arrhenius Equation

Jessie A. Key

Learning Objectives

  • To gain an understanding of activation energy.
  • To determine activation energy graphically or algebraically.
Middle-aged man with a moustache and wearing a bowtie.
Figure 17.11 “Svante Arrhenius.” Swedish scientist Svante Arrhenius (1859–1927).

Earlier in the chapter, reactions were discussed in terms of effective collision frequency and molecule energy levels. In 1889, a Swedish scientist named Svante Arrhenius proposed an equation that relates these concepts with the rate constant:

k=Ae^{\frac{-E_a}{RT}}

where k represents the rate constant, Ea is the activation energy, R is the gas constant \left(\dfrac{8.3145\text{ J}}{\text{K mol}}\right), and T is the temperature expressed in Kelvin. A is known as the frequency factor, having units of L mol−1 s−1, and takes into account the frequency of reactions and likelihood of correct molecular orientation.

The Arrhenius equation allows us to calculate activation energies if the rate constant is known, or vice versa. As well, it mathematically expresses the relationships we established earlier: as activation energy term Ea increases, the rate constant k decreases and therefore the rate of reaction decreases.

Determining the Activation Energy

Graphically

We can graphically determine the activation energy by manipulating the Arrhenius equation to put it into the form of a straight line. Taking the natural logarithm of both sides gives us:

\ln k=-\dfrac{E_a}{RT}+\ln A

A slight rearrangement of this equation then gives us a straight line plot (y = mx + b) for ln k versus \dfrac{1}{T}, where the slope is -\dfrac{E_a}{R}:

\ln k=-\dfrac{E_a}{R}\left(\dfrac{1}{t}\right)+\ln A

Example 17.7

Using the data from the following table, determine the activation energy of the reaction:

Temperature (K) Rate Constant, k (s−1)
375 1.68 × 10−5
400 3.5 × 10−5
500 4.2 × 10−4
600 2.11 × 10−3

Solution

We can obtain the activation energy by plotting ln k versus \dfrac{1}{T}, knowing that the slope will be equal to -\dfrac{E_a}{R}.

First determine the values of ln k and \dfrac{1}{T}, and plot them in a graph:

\frac{1}{T} ln k
0.002667 −10.9941
0.0025 −10.2602
0.002 −7.77526
0.001667 −6.16107
Graphical determination of Ea example plot
Graphical determination of Ea example plot

    \begin{align*} \text{Slope}&=-\dfrac{E_a}{R} \\ \\ -4865\text{ K}&=-\dfrac{E_a}{8.3145\text{ J K}^{-1}\text{ mol}^{-1}} \\ \\ E_a&=4.0\times 10^4 \text{ J/mol} \end{align*}

Algebraically

The activation energy can also be calculated algebraically if k is known at two different temperatures:

    \begin{align*} \text{At temperature 1: }\ln k_1&=-\dfrac{E_a}{RT_1}+\ln A \\ \\ \text{At temperature 2: }\ln k_2&=-\dfrac{E_a}{RT_2}+\ln A \end{align*}

We can subtract one of these equations from the other:

\ln k_1-\ln k_2=\left(-\dfrac{E_a}{RT_1}+\ln A\right)-\left(-\dfrac{E_a}{RT_2}+\ln A\right)

This equation can then be further simplified to:

\ln \dfrac{k_1}{k_2}=\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)

Example 17.8

Determine the value of Ea given the following values of k at the temperatures indicated:

    \begin{align*} 600\text{ K: }k&=2.75\times 10^{-8}\text{ L mol}^{-1}\text{s}^{-1} \\ 800\text{ K: }k&=1.95\times 10^{-7}\text{ L mol}^{-1}\text{s}^{-1} \end{align*}

Solution

Substitute the values stated into the algebraic method equation:

    \begin{align*} \ln \dfrac{k_1}{k_2}&=\dfrac{E_a}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right) \\ \\ \ln \dfrac{2.75\times 10^{-8}\text{ L mol}^{-1}\text{s}^{-1}}{1.95\times10^{-7}\text{ L mol}^{-1}\text{s}^{-1}}&=\dfrac{E_a}{8.3145\text{ J K}^{-1}\text{mol}^{-1}}\left(\dfrac{1}{800\text{ K}}-\dfrac{1}{600\text{ K}}\right) \\ \\ 1.96&=\dfrac{E_a}{8.3145\text{ J K}^{-1}\text{mol}^{-1}}\left(-4.16\times10^{-4}\text{ K}^{-1}\right) \\ \\ 4.704\times 10^{-3}\text{ K}^{-1}&=\dfrac{E_a}{8.3145\text{ J K}^{-1}\text{mol}^{-1}} \\ \\ E_a&=3.92\times 10^4\text{ J/mol} \end{align*}

Key Takeaways

  • The activation energy can be graphically determined by manipulating the Arrhenius equation.
  • The activation energy can also be calculated algebraically if k is known at two different temperatures.

Media Attributions

License

Icon for the Creative Commons Attribution 4.0 International License

Activation Energy and the Arrhenius Equation Copyright © 2014 by Jessie A. Key is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book