A LevelChemistryAQARevision NotesAdvanced Physical ChemistryRate EquationsFinding Activation Energy

Finding Activation Energy (AQA A Level Chemistry): Revision Note

Exam code: 7405

Stewart Hird

Written by: Stewart Hird

Reviewed by: Caroline Carroll

Updated on

Finding Activation Energy

  • The Arrhenius equation is commonly used to calculate the activation energy (Ea) of a reaction

  • The equation is:

k = A e−Ea/RT

  • Where

    • k is the rate constant,

    • A is the Arrhenius constant,

    • Ea is the activation energy,

    • R is the gas constant, and

    • T is the temperature in kelvin

  • In some cases, sufficient data are provided to substitute directly into the Arrhenius equation

    • Alternatively, a graph of ln k against 1/T can be plotted

    • The gradient of this straight-line graph is equal to −Ea/R, which can then be used to calculate the activation energy

Using the Arrhenius equation:

  • Remember, it is usually easier to use the version of the Arrhenius equation after natural logs of each side have been taken

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:

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

  • Rearrange the equation for Ea:

    • fraction numerator bold italic E subscript bold italic a over denominator bold italic R bold italic T end fraction bold plus bold ln bold space bold italic k bold equals bold ln bold space bold italic A

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

    • Error converting from MathML to accessible text.

  • Insert the values from the question:

    • Error converting from MathML to accessible text.

    • Error converting from MathML to accessible text.

    • Ea = 129095.85 J

  • Convert from J to kJ:

    • Ea = 129 kJ

Using an Arrhenius plot:

  • 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 plotting ln k against 1/T with a red line of best fit, indicating the gradient as -Ea/R and the intercept as ln A.
The graph of ln k against 1/T is a straight line with gradient -Ea/R

  • From the graph, the equation in the form of y = mx + c is:

bold ln bold space bold italic k bold equals fraction numerator bold minus bold italic E subscript bold a over denominator bold italic R end fraction bold space bold 1 over bold italic T bold plus bold ln bold space bold italic A

  • Where:

    • y = ln k 

    • x = begin mathsize 14px 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)

Worked Example

  1. Complete the following table.

Temperature
/ K

1/T
/ K-1

Time, t
/ s

Rate constant, k
/ s-1

ln k

310

3.23 x 10-3

57

-9.2

335

31

3.01 x 10-4

-8.1

360

2.78 x 10-3

19

5.37 x 10-4

-7.5

385

2.60 x 10-3

7

9.12 x 10-4

  1. Plot a graph of ln k against 1/T.

  2. Using the graph and the following equation, calculate the activation energy, Ea, for this reaction.

bold ln bold space bold italic k bold equals fraction numerator bold minus bold italic E subscript bold a over denominator bold italic R end fraction bold space bold 1 over bold italic T bold plus bold ln bold space bold italic A

  1. Calculate the Arrhenius constant, A, for this reaction.

Answers:

  1. The completed table is:

Temperature
/ K

1/T
/ K-1

Time, t
/ s

Rate constant, k
/ s-1

ln k

310

3.23 x 10-3

57

1.01 x 10-4

-9.2

335

2.99 x 10-3

31

3.01 x 10-4

-8.1

360

2.78 x 10-3

19

5.37 x 10-4

-7.5

385

2.60 x 10-3

7

9.12 x 10-4

-7.0

  1. The graph of ln k against 1/T is:

Graph showing a straight line through blue data points. X-axis: 1/T × 10^-3 (K^-1), Y-axis: ln k, range -5 to -10.

  1. To calculate the activation energy, Ea:

    Graph titled "Calculate the Activation Energy" with a plot of ln k versus 1/T. A linear line through points, gradient calculation shown as -3666.6.

    • Gradient = -Ea / R = -3666.6

    • Ea = -(-3666.6 x 8.31) = 30, 469 J mol-1

    • Ea = -(-3666.6 x 8.31) = 30.5 kJ mol-1

  2. To calculate the Arrhenius constant, A:

    • Choose a point on the graph, e.g (2.60 x 10-3, -7.0)

    Graph with ln k on the y-axis and 1/T x 10^-3 (K^-1) on the x-axis. Points are marked with crosses and annotated (2.60 x 10^-3, -7).

    • Substitute values into the equation bold ln bold space bold italic k bold equals fraction numerator bold minus bold italic E subscript bold a over denominator bold italic R end fraction bold space bold 1 over bold italic T bold plus bold ln bold space bold italic A

    • -7.0 = (-3666.6 x 2.60 x 10-3) + ln A

    • -7.0 = -9.53 + ln A

    • ln A = 2.53

    • A = e to the power of 2.53 end exponent

    • A = 12.55

Examiner Tips and Tricks

You are not required to learn these equations. However, you do need to be able to rearrange them, and knowing them helps you understand the effects of temperature on the rate constant.

Unlock more, it's free!

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

the (exam) results speak for themselves:

I would just like to say a massive thank you for putting together such a brilliant, easy to use website.I really think using this site helped me secure my top gradesin science and maths. You really did save my exams! Thank you.

Beth
IGCSE Student

This website is soooo useful and I can’t ever thank you enough for organising questions by topic like this. Furthermore, the name of the website could not have been more appropriate as it literally did SAVE MY EXAMS!

Fathima
A Level Student

Incredible! SO worth my money, the revision notes have everything I need to know and are so easy to understand. I actually enjoy revising! It makes me feel a lot more confident for my GCSEs in a few months.

Kate
GCSE Student

Absolutely brilliant, both my girls used it for A levels and GCSE. It's saves on paper copies, also beneficial exam questions ranked from easy to hard. It's removed a lot of stress from the exams.

Sameera
Parent

Just to say that your resources are the best I have seen and I have been teaching chemistry at different levels for about 40 years

Mark
Chemistry Teacher

Excellent

Our rating represented in Trustpilot stars
Read more reviews
Test yourself
Previous:The Arrhenius EquationNext:Concentration-Time Graphs
Author
Reviewer
Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Content Creator

Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Physics & Chemistry Subject 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.