IBChemistryHLRevision NotesHow Much, How Fast & How Far?How Fast? The Rate of Chemical ChangeDetermining Activation Energy & the Arrhenius Factor (HL)

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Determining Activation Energy & the Arrhenius Factor (HL) (HL IB Chemistry)

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Determining Activation Energy & the Arrhenius Factor

How to use the Arrhenius equation graph

  • Once the rate constant at different temperatures for a reaction has been determined experimentally, the results can be used to determine the activation energy, Ea, and the Arrhenius factor, A
  • This is best shown through a worked example

Worked example

The data in the table below was collected for a reaction.

Temperature / K bold 1 over bold T bold space bold divided by bold space bold K to the power of bold minus bold 1 end exponent 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. Complete the table
  2. Plot a graph of ln k against 1/T on the graph below
  3. Use this to calculate:
    1. the activation energy, Ea, in kJ mol-1
    2. the Arrhenius factor, A, of the reaction

 A blank graph with the y axis labelled ln k and the x axis labelled 1/T

Answer 1:

Temperature / K bold 1 over bold T bold space bold divided by bold space bold K to the power of bold minus bold 1 end exponent 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

Answer 2:

  • Choose a suitable scale for the axes
    • The scale does not need to start from (0,0)
    • The plotted points should fill as much of the graph provided as possible

Plot of ln k against 1/T

Answer 3a:

arrhenius-plot-to-calculate-ea

  • To calculate Ea:
    • Gradient = begin mathsize 14px style fraction numerator negative E subscript straight a over denominator R end fraction end style = –3666.6
    • Ea = –(–3666.6 x 8.31) = 30,469 J mol-1
  • Convert to kJ mol-1:
    • Ea = 30.5 kJ mol-1

Answer 3b:

arrhenius-plot-to-calculate-a

  • Choose a point on the graph:
    • (2.60 x 10-3, –7)
  • Use the logarithmic form of the Arrhenius equation
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
  • From the point chosen from the graph:
    • ln k = –7.0
    • begin mathsize 14px style 1 over T end style = 2.60 x 10-3
    • begin mathsize 14px style fraction numerator negative E subscript straight a over denominator R end fraction end style = fraction numerator negative 1.1 over denominator 0.3 space cross times space 10 to the power of negative 3 end exponent end fraction = –3666.6
  • Substituting these values:
    • –7.0 = (–3666.6 x 2.60 x 10-3) + ln A
    • –7.0 = –9.53 + ln A
  • Rearranging gives:
    • ln A = –7.0 + 9.53 = 2.53
    • A = e2.53
    • A = 12.55

Exam Tip

  • You are not required to learn these equations as they are given in the Data Booklet
  • 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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Author: Caroline

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.