Activation energy
16.2 Activation energy (2 hours)
Pause for thought
The relationship between the value of k, the rate constant, and temperature is given by the Arrhenius equation:
k = A exp(–Ea/RT)
where k is the rate constant, A is another constant sometimes known as the Arrhenius constant but more usually as the frequency factor (or pre-exponential factor), Ea is the activation energy for the reaction, R is the gas constant and T is the temperature in Kelvin.
In theory students should know how to express this in logarithmic format but both the Arrhenius equation and its logarithmic form are provided in the Data booklet so they only really need to know how to use it. It is worth relating this to the Nature of Science (and to TOK) by stressing that this an example of an empirical relationship as it cannot be proven mathematically. Note that in the 'Pause for thought' on the page for sub-topic 6.1 Collision theory & rates of reaction I have used the Arrhenius equation to show that the rule of thumb which says that raising the temperature by ten degrees doubles the rate of reaction depends very much upon the value of the activation energy.
The equation is named after the Swedish chemist, Svante Arrhenius, who is perhaps more famous for being the first person to realise that electrolytes are dissociated into their positive and negative ions in aqueous solution and that the more dilute the solution the greater the degree of dissociation. There is a cautionary tale here – don’t just dismiss your student’s ideas out of hand. When Arrhenius submitted his work on electrolytes for his doctoral thesis at the University of Uppsala in 1884 the staff at the university thought his work was substandard and awarded him the lowest possible pass mark. Just 19 years later in 1903, after the brilliance of his revolutionary work had been fully recognized, he was awarded the Nobel Prize in Chemistry.
Nature of Science
Changing the temperature of a reaction has a much greater effect on the rate of reaction than can be explained by its effect on the rate of collisions. The Arrhenius equation, which proposes a quantitative model to explain the effect of temperature change on the rate of reaction, is a good example of how theories can be supported or falsified and replaced by new theories.
Learning outcomes
After studying this topic students should be able to:
Understand
- The Arrhenius equation is based on the temperature dependence of the rate constant and can be used to determine the activation energy.
- A plot of 1/T against ln k produces a straight line with the gradient equal to – Ea / R and the intercept equal to ln A.
- A, the frequency factor (or pre-exponential factor), is related to the frequency of collisions with the correct orientations.
Apply their knowledge to:
- Use the Arrhenius equation in its normal form
k = A exp(–Ea/RT). - Analyse graphical representation of the Arrhenius equation in its logarithmic (linear) form
ln k = –Ea/RT + ln A. - Describe the relationships between temperature and rate constant, k; and between frequency factor, A, and the complexity of molecular collisions.
- Determine and evaluate the values of activation energy and frequency factors from data.
Clarification notes
Illustrate multi-step reactions using energy level diagrams showing the rate determining step in the diagrams.
Consider various different data sources when using the logarithmic (linear) expression:
ln k = –Ea/RT + ln A.
The expression:
is given in Section 1 of the data booklet.
International-mindedness
Nothing is listed on the syllabus under this heading.
Teaching tipsThis is a relatively straightforward topic to teach - although some students may find the mathematics difficult. From the rate equation students can see that if the concentrations of the reactants remain constant then the rate constant must increase in value as the temperature (in Kelvin) is increased. The qualitative explanation for this can be related to the Maxwell-Boltzmann distribution as more reactant particles will possess the necessary activation energy. They will need to be given the Arrhenius equation. The reason for putting it into its logarithmic form is to get it into the form of y = mx + c. That is, a plot of ln k against 1/T will give a straight line. The gradient will be equal to – Ea/R and so the activation energy for the reaction can be determined. When the graph is extrapolated back to the ln k axis the intercept will be ln A and hence the value for the Arrhenius constant can be obtained.
In practice this is not easy to do as the values tend to be in a very small region of the graph and the axes need to be extended considerably. Once Ea has been found it is usually easier to obtain A by substituting back into the equation where the rate constant at a particular temperature is known as A is now the only unknown. |
Study guide
Page 50
Questions
For ten 'quiz' multiple choice questions with the answers explained see MCTest: Activation energy.
For short-answer questions on activation energy which can be set as an assignment for a test, homework or given for self study together with model answers see Activation energy questions.
Vocabulary list
frequency factor
Arrhenius equation
Arrhenius constant
ln (log to the base e)
IM, TOK, 'Utilization' etc.
See separate page which covers all of Topics 6 & 16
Practical work
Other resources
1. A video by Richard Thornley on the Arrhenius equation and how to determine the activation energy for a reaction.
2. A simulation from the Wolfram Demonstration Project (you will need to download the free Wolfram CDF player to use it)
Arrenhius equation for rate and velocity