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4.1.10Unit 4. Rates, Equilibria and Further Organic Chemistry

Activation Energy and Arrhenius — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 6 parts Lesson 10 of 56 Unit quiz

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Part 1 of 6The Arrhenius Equation

4.1.10 Activation Energy and Arrhenius

The equation links the rate constant to temperature and to the energy barrier.

Form
  • k = A e^(−Eₐ/RT), where A is a constant, Eₐ is the activation energy, R is the gas constant and T is the temperature in kelvin.
Exponential term
  • e^(−Eₐ/RT) represents the fraction of molecules with energy at least equal to Eₐ.
Effect of T
  • Raising T makes the exponent less negative, so k increases rapidly with temperature.
Effect of Eₐ
  • A larger activation energy gives a smaller k at the same temperature, so the reaction is slower.

Part 2 of 6Logarithmic Form

Logarithmic Form
Equation
  • Taking natural logs gives ln k = ln A − Eₐ/(RT).
Straight line
  • This has the form y = mx + c, where y is ln k and x is 1/T.
Gradient
  • The gradient of the line is −Eₐ/R, which is negative.
Intercept
  • The intercept on the y-axis is ln A.
QuantityValue or unit
R8.31 J K⁻¹ mol⁻¹
TKelvin, K
EₐJ mol⁻¹, then converted to kJ mol⁻¹
Plotln k against 1/T

Part 3 of 6Finding Eₐ Experimentally

1
Measure the rate constant k at several different temperatures with the other conditions kept the same.
2
Convert each temperature to kelvin and calculate 1/T, and also ln k.
3
Plot ln k on the y-axis against 1/T on the x-axis and draw the best-fit line.
4
Calculate the gradient of the line using two points far apart.
5
Eₐ = −gradient × R, converting from J mol⁻¹ to kJ mol⁻¹ by dividing by 1000.

Part 4 of 6Interpreting Temperature Effects

Rule of thumb
  • The rate roughly doubles for a 10 K rise in many reactions near room temperature, but this depends on Eₐ.
Comparison
  • The effect of temperature is larger for reactions with a higher activation energy.
Catalysts
  • A catalyst lowers Eₐ, so the exponential term and so k increase at the same temperature.
Units
  • If Eₐ is in kJ mol⁻¹ it must be converted to J mol⁻¹ before using R in J K⁻¹ mol⁻¹.

Part 5 of 6Worked Example

Question

  • A graph of ln k against 1/T passes through (0.00320 K⁻¹, −6.30) and (0.00300 K⁻¹, −4.50).
  • Calculate Eₐ.

  1. Gradient = (−4.50 − (−6.30)) ÷ (0.00300 − 0.00320) = 1.80 ÷ (−0.00020) = −9000 K.
  2. Eₐ = −gradient × R = 9000 × 8.31 = 74 790 J mol⁻¹.
  3. Eₐ = 74.8 kJ mol⁻¹ to three significant figures.

Answer Eₐ = 74.8 kJ mol⁻¹

Part 6 of 6Exam Checklist & Tips

4.1.10 – Key Points for the Exam

  • Arrhenius equation An equation relating the rate constant to temperature and activation energy.
  • Activation energy The minimum energy that colliding particles need to react.
  • Gas constant R = 8.31 J K⁻¹ mol⁻¹.
  • Natural logarithm A logarithm to base e, written ln.
  • Pre-exponential factor The constant A, related to collision frequency and orientation.

Common Mistakes

  • Using temperatures in °C instead of kelvin.
  • Forgetting to convert Eₐ from J mol⁻¹ to kJ mol⁻¹ at the end.
  • Losing the minus signthe gradient is −Eₐ/R, and so is negative.
  • Plotting ln k against T instead of against 1/T.

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