Syllabus Edition

First teaching 2023

First exams 2025

|

The Arrhenius Equation (HL) (HL IB Chemistry)

Revision Note

Caroline

Author

Caroline

Last updated

The Arrhenius Equation

  • The rate equation shows how each of the reactants in a reaction affects the rate of the reaction and it includes the rate constant, k
  • However, k only remains constant if the concentration of the reactants is the only factor which is changed
    • If the temperature is changed or a catalyst is used or changed, then the rate constant, k, changes
  • At higher temperatures, a greater proportion of molecules have energy greater than the activation energy
  • Since the rate constant and rate of reaction are directly proportional to the fraction of molecules with energy equal or greater than the activation energy, then at higher temperatures:
    • The rate of reaction increases
    • The rate constant increases

What is the Arrhenius equation?

  • The relationship between the rate constant, the temperature and also the activation energy is given by the Arrhenius equation:

k space equals space A straight e to the power of fraction numerator negative straight E subscript straight a over denominator RT end fraction end exponent to the power of blank

  • Where:
    • k = Rate constant
    • A = Arrhenius factor (also known as the frequency factor or pre-exponential factor) which is a constant that takes into account the frequency of collisions with proper orientations
    • Ea = Activation energy (J mol-1)
    • R = Gas constant (8.31 J K-1 mol-1)
    • T = Temperature (Kelvin, K)
    • e = Mathematical constant (can be found on your calculator - it has the approximate value of 2.718)
  • Ea and A are constants that are characteristic of a specific reaction
    • A does vary slightly with temperature but it can still be considered a constant
  • R is a fundamental physical constant for all reactions
  • k and T are the only variables in the Arrhenius equation
  • The Arrhenius equation is used to describe reactions that involve gases, reactions occurring in solution or reactions that occur on the surface of a catalyst

Using the Arrhenius Equation

  • The Arrhenius equation is easier to use if you take natural logarithms of each side of the equation, which results in the following equation:

ln space k space equals space ln space A space minus space fraction numerator E subscript straight a over denominator R T end fraction

  • The Arrhenius equation can be used to show the effect that a change in temperature has on the rate constant, k, and thus on the overall rate of the reaction
    • An increase in temperature (higher value of T) gives a greater value of ln k and therefore a higher value of k
    • Since the rate of the reaction depends on the rate constant, k, an increase in k also means an increased rate of reaction
  • The equation can also be used to show the effect of increasing the activation energy on the value of the rate constant, k
    • An increase in the activation energy, Ea, means that the proportion of molecules which possess at least the activation energy is less
    • This means that the rate of the reaction, and therefore the value of k, will decrease
  • The values of k and T for a reaction can be determined experimentally
    • These values of k and T can then be used to calculate the activation energy for a reaction
    • This is the most common type of calculation you will be asked to do on this topic

Examiner Tip

  • In the exam, you could be asked to calculate any part of the Arrhenius equation
    • Using the equation in its natural logarithm form makes this easier.

Worked example

Calculate the activation energy of a reaction which takes place at 400 K, where the rate constant of the reaction is 6.25 x 10-4 s-1.

A = 4.6 x 1013 and R = 8.31 J K-1 mol-1.

Answer:

  • Rearrange the Arrhenius equation for Ea:

  table row cell ln space k space end cell equals cell space ln space A space minus space fraction numerator E subscript straight a over denominator R T end fraction end cell row blank blank blank row cell fraction numerator E subscript straight a over denominator R T end fraction space plus space ln space k space end cell equals cell space ln italic space A end cell row blank blank blank row cell fraction numerator E subscript straight a over denominator R T end fraction space end cell equals cell space ln space A space minus space ln space k end cell row blank blank blank row cell E subscript straight a space end subscript end cell equals cell space open parentheses ln space A space minus space ln italic space k close parentheses space cross times space R T end cell end table

  • Insert values from the question:

table row cell E subscript straight a space end cell equals cell space open square brackets open parentheses ln space 4.6 cross times 10 to the power of 13 close parentheses minus open parentheses ln space 6.25 cross times 10 to the power of negative 4 end exponent close parentheses close square brackets cross times open parentheses 8.31 cross times 400 close parentheses end cell row blank blank blank row blank equals cell space open square brackets open parentheses 31.4597 close parentheses minus open parentheses negative 7.3778 close parentheses close square brackets cross times 3324 end cell row blank blank blank row blank equals cell space 129 comma 095.85 space straight J end cell end table

  • Convert Ea to kJ:

   table row cell E subscript straight a space end cell equals cell space 129 comma 095.85 space divided by space 1000 end cell row blank equals cell space bold 129 bold space bold kJ end cell end table

Graphing the Arrhenius Equation

Finding the activation energy and Arrhenius factor

  • A graph of experimental data can be used to determine the activation energy and the Arrhenius factor

Arrhenius equation graph

  • A graph of ln k against 1/T can be plotted, and then used to calculate Ea
    • This gives a line which follows the form y = mx + c

    Graph of ln k against 1/T

A sketch of ln k against 1/T shows a straight line graph with a negative gradient

The graph of ln k against 1/T is a straight line with gradient -Ea/R

  • From the graph, the equation is in the form of y = mx + c is as follows:
ln k = fraction numerator negative E subscript straight a over denominator R end fraction begin mathsize 14px style 1 over T end style + ln A
           
y = m    x + c
  • Where:
    • y = ln k
    • x = begin mathsize 11px style 1 over T end style
    • m = begin mathsize 14px style fraction numerator negative E subscript straight a over denominator R end fraction end style (the gradient)
    • c = ln A (the y-intercept)
  • Ea can be calculated by:
    • Ea = - gradient x R
  • As the slope has a negative gradient, the minus signs cancel giving Ea as a positive value
  • The Arrhenius factor, A, can be calculated by:
    • A = ey-intercept 

Examiner Tip

  • If the x-axis does not start at the origin, you cannot use the y-intercept to find A 
  • Instead, take a point from the graph and substitute the values of ln k, 1/T and the gradient into the logarithmic Arrhenius equation to find ln A, and use this to calculate A.

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Caroline

Author: Caroline

Expertise: Physics Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.