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5.2.14Additional lessons

Determining Activation Energy and the Arrhenius Factor (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 5 parts Lesson 108 of 172

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Part 1 of 5Graphical Method

5.2.14 Determining Activation Energy and the Arrhenius Factor (HL)

Using a graph reduces the effect of random error compared with a calculation from only two points.

Linear form
  • ln k = −(Ea/R)(1/T) + ln A, which has the form y = mx + c.
Axes
  • Plot ln k on the vertical axis against 1/T on the horizontal axis, with T in kelvin.
Gradient
  • The gradient of the line is −Ea/R, so Ea = −gradient × R.
Intercept
  • The vertical intercept at 1/T = 0 equals ln A, so A = e^intercept.

Part 2 of 5Procedure

1
Measure the rate constant k at several different temperatures, keeping other conditions the same.
2
Convert each temperature to kelvin and calculate 1/T, then calculate ln k.
3
Plot ln k against 1/T and draw the line of best fit through the points.
4
Find the gradient using two points far apart on the line, taking care with the negative sign.
5
Multiply the gradient by −R to obtain Ea in J mol⁻¹, then convert to kJ mol⁻¹ if needed.

Part 3 of 5Interpreting the Graph

Interpreting the Graph
Steep line
  • A large negative gradient means high Ea, so the rate is very sensitive to temperature.
Shallow line
  • A small gradient corresponds to low Ea, so temperature has less effect on k.
Scatter
  • Points scattered around the line indicate experimental uncertainty.
Extrapolation
  • The intercept is found by extending the line, so its uncertainty is generally large.

Part 4 of 5Worked Example

Question

  • A plot of ln k against 1/T has gradient −6.02 × 10³ K and intercept +25.0.
  • Calculate Ea and A.
  • (R = 8.31 J K⁻¹ mol⁻¹)

  1. Ea = −gradient × R = 6.02 × 10³ × 8.31 = 5.00 × 10⁴ J mol⁻¹ = 50.0 kJ mol⁻¹.
  2. ln A = 25.0, so A = e25.0.
  3. A = 7.2 × 10¹⁰ (in the same units as k).

Answer Ea = 50.0 kJ mol⁻¹; A = 7.2 × 10¹⁰

Part 5 of 5Exam Checklist & Tips

5.2.14 – Key Points for the Exam

  • Gradient The slope of a graph, calculated as change in y divided by change in x.
  • Intercept The value of y where the line crosses the vertical axis.
  • Line of best fit A line drawn to represent the overall trend of data points.
  • Natural logarithm Logarithm to base e, written ln.

Common Mistakes

  • Plotting ln k against T instead of 1/T.
  • Forgetting the negative sign in the gradient, so Ea appears negative.
  • Reporting Ea in J mol⁻¹ when the answer is requested in kJ mol⁻¹.
  • Using Celsius temperatures to calculate 1/T.

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