The Arrhenius Equation (HL) (DP IB Chemistry: HL): Revision Note

Caroline Carroll

Written by: Caroline Carroll

Reviewed by: Richard Boole

Updated on

The Arrhenius equation

  • Rate equations include the rate constant k

    • This reflects collision frequency and energy requirements, not just concentrations

  • k is constant when the only variable is concentration

    • k changes if temperature or catalysts change

  • Increasing the temperature increases the proportion of molecules with energy greater than the activation energy, Ea

  • As temperature increases, more molecules have enough energy to overcome the activation energy

  • This means that:

    • The rate of reaction increases

    • The rate constant increases

What is the Arrhenius equation?

  • The Arrhenius equation is a mathematical expression that describes the temperature dependence of reaction rates

    • It links the rate constant, temperature and activation energy

  • The Arrhenius equation describes reactions that:

    • Involve gases

    • Occur in solution

    • Occur on the surface of a catalyst

  • The Arrhenius equation can be found in the IB Chemistry data booklet (Section 1):

k space equals space A e to the power of fraction numerator italic minus E subscript a over denominator R T end fraction end exponent to the power of blank

What are the terms in the Arrhenius equation?

  • k is the rate constant

    • It determines how quickly the reaction proceeds

  • A is the Arrhenius factor

    • This is also known as the frequency factor or pre-exponential factor

    • It is a constant that takes into account the frequency of collisions with proper orientations

    • The value of A:

      • Changes slightly with temperature but is still considered constant

      • Is characteristic of a specific reaction

  • Ea is the activation energy

    • Activation energy is the minimum energy per mole of reactant particles required to start a chemical reaction

    • The common units of activation energy are J mol⁻¹ and kJ mol⁻¹

    • The value of Ea is characteristic of a specific reaction

  • R is the gas constant

    • It is a fundamental physical constant for all reactions

    • The value of R is 8.31 J K⁻¹ mol⁻¹

    • It can be found in the IB Chemistry data booklet (Section 2)

  • T is temperature

    • The unit for temperature in this equation is Kelvin, K

  • e is a mathematical constant

    • It can be found on your calculator

    • It has the approximate value of 2.718

  • k and T are the only variables in the Arrhenius equation

Using the Arrhenius equation

  • Taking the natural logarithm of the Arrhenius equation gives a straight-line

  • This makes it easier to:

    • Determine Ea and A from experimental data

    • Solve problems when two sets of k and T are known

  • Both forms of the Arrhenius equation are given in the IB Chemistry data booklet (Section 1)

    • The natural logarithmic form is:

ln k=EaRT+ln A 

  • This version of the equation shows how:

    • Temperature affects the rate constant, k:

      • Increasing temperature increases ln k

      • This gives a higher value of k, increasing the rate of reaction

    • Activation energy, Ea, affects the rate:

      • A higher Ea means fewer molecules have enough energy to react

      • This decreases the rate of reaction, giving a lower value of k

  • The values of k and T can be determined experimentally:

    • These can be used to calculate Ea for a reaction

    • This is the most common Arrhenius calculation

Examiner Tips and Tricks

In exams, you could be asked to calculate any part of the Arrhenius equation

  • Use the natural logarithmic form when solving for Ea, A, or T

    • It is easier to rearrange and substitute into

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:

  • Use the logarithmic form of the Arrhenius equation to calculate Ea:

  ln k=EaRT+ln A

  • Rearrange the Arrhenius equation for Ea:

  EaRT + ln k = ln AEaRT = ln A  ln kEa = (ln A  ln k) × RT

  • Insert values from the question:

Ea = [(ln 4.6×1013)(ln 6.25×104)]×(8.31×400)= [(31.4597)(7.3778)]×3324= 129,095.85 J

  • Convert Ea to kJ:

   Ea = 129,095.85 ÷ 1000= 129 kJ mol1

Graphing the Arrhenius equation

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

  • Plotting ln k against 1/T gives a straight line:

    • The slope of the line is EaR

    • The y-intercept of the line is ln A

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
  • The equation of a straight line is:

y = mx + c

  • This matches the natural logarithmic Arrhenius equation:

ln k=EaRT+ln Aln k=EaR×1T+ln A

  • The variables are:

    • y = ln k

    • x = 1T

    • m = EaR (the gradient)

    • c = ln A (the y-intercept)

Calculating the activation energy

  • The activation energy can be calculated from the slope:

Ea = - gradient x R

  • Since the slope is negative:

    • The minus signs cancel

    • Ea is a positive value

Calculating the Arrhenius factor

  • The Arrhenius factor, A, can be calculated from the y-intercept:

A = ey-intercept 

Examiner Tips and Tricks

  • 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.

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Caroline Carroll

Author: Caroline Carroll

Expertise: Head of Content Delivery

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 delivering high-quality resources to help students achieve their full potential.

Richard Boole

Reviewer: Richard Boole

Expertise: Curriculum Expert

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.