A LevelChemistryAQARevision NotesAdvanced Physical ChemistryRate EquationsFinding Activation Energy

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

Exam code: 7405

Stewart Hird

Last updated

Finding Activation Energy

Finding the Activation Energy

  • Very often, the Arrhenius Equation is used to calculate the activation energy of a reaction

  • Either a question will give sufficient information for the Arrhenius equation to be used, or a graph can be plotted and the calculation done from the plot

Using the 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

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 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 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 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:

5.2.5 using Arrhenius plot to calculate Ea - plotted graph (WE)_2, downloadable AS & A Level Chemistry revision notes

  1. To calculate the activation energy, Ea:

    5.2.5 WE Arrhenius plot calculate Ea 1_1, downloadable AS & A Level Chemistry revision notes

    • 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)

    5.2.5 WE Arrhenius plot - calculate A (part 1)_1, downloadable AS & A Level Chemistry revision notes

    • 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 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 is helpful in understanding the effects of temperature on the rate constant.

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    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.