The Arrhenius Equation (Oxford AQA International A Level Chemistry)

Revision Note

Richard Boole

Written by: Richard Boole

Reviewed by: Stewart Hird

The Arrhenius Equation

  • The rate equation shows how each reactant affects the rate of the reaction

  • 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

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

bold italic k bold equals bold italic A bold italic e to the power of stretchy left parenthesis fraction numerator negative E subscript a over denominator RT end fraction stretchy right parenthesis end exponent

  • Terms in the Arrhenius Equation:

    • 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 that occur in solution

    • Reactions that occur on the surface of a catalyst

Activation Energy Calculations

  • The Arrhenius Equation is often used to calculate:

    • The activation energy, Ea, of a reaction

    • The Arrhenius constant, A, for a reaction

  • A question will give sufficient information for:

    • The Arrhenius equation to be used

    • Or, a graph to be plotted and the calculation done from the plot

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:

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

  • How temperature affects the rate constant and rate of reaction:

    • An increase in temperature gives a greater value of ln k

    • This results in a higher value for the rate constant, k

    • Since the rate of the reaction depends on the rate constant, k, an increase in k also means an increased rate of reaction

  • How activation energy affects the rate constant:

    • An increase in the activation energy, Ea, means a smaller proportion of molecules has the required activation energy

    • This results in a lower value for the rate constant, k

    • Since the rate of the reaction depends on the rate constant, k, a decrease in k also means a decreased rate of reaction

  • 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 Tips and Tricks

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:

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

    • bold italic E subscript bold a bold equals stretchy left parenthesis ln space k minus ln space A stretchy right parenthesis bold cross times bold italic R bold italic T

  • 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

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)

  • The equation shows that 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

Example graph of ln k over 1/T 

5.2.5 Arrhenius sketch of ln k against 1_T_2
The graph of ln k over 1/T is a straight line with gradient -Ea/R

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.

Last updated:

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?

Richard Boole

Author: Richard Boole

Expertise: Chemistry

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.

Stewart Hird

Author: Stewart Hird

Expertise: Chemistry Lead

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.