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6.1.3Unit 6. Practical Skills in Chemistry II

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

5 min read 6 parts Lesson 7 of 14 Unit quiz

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Part 1 of 6Arrhenius Equation

6.1.3 Activation Energy

Rate constants increase with temperature in a predictable exponential way.

Form
  • k = Ae^(–Ea/RT), where A is a constant, Ea the activation energy, R the gas constant and T the temperature in kelvin.
Logarithmic form
  • ln k = –Ea/RT + ln A, which has the form y = mx + c.
Graph
  • Plotting ln k against 1/T gives a straight line with gradient –Ea/R and intercept ln A.
Units
  • R = 8.31 J mol⁻¹ K⁻¹, so Ea is found in J mol⁻¹ and then divided by 1000 for kJ mol⁻¹.

Part 2 of 6Practical Method

1
Choose a clock reaction, such as the iodine clock, that gives a clear colour change at a convenient time.
2
Place the reactants in a water bath and record the temperature of the mixture just before and after mixing.
3
Mix, start the stopwatch, and record the time for the end point, then repeat at about five temperatures from 10 °C to 50 °C.
4
Convert each temperature to kelvin by adding 273 and calculate 1/T.
5
Use 1/time as a measure of rate, calculate ln(1/time), and plot against 1/T.

Part 3 of 6Interpreting the Graph

Interpreting the Graph
Gradient
  • A steeper negative gradient means a larger activation energy.
Calculation
  • Ea = –gradient × R, so a gradient of –5200 K gives Ea = 5200 × 8.31 ≈ 43 200 J mol⁻¹.
Intercept
  • The intercept on the y-axis gives ln A, related to collision frequency and orientation.
Best-fit line
  • A line of best fit reduces the influence of random errors in individual results.

Part 4 of 6Accuracy Points

Temperature control
  • Small errors in T cause large errors in 1/T, so keep the mixture at constant temperature.
Units
  • Use kelvin in all Arrhenius calculations.
Range
  • A wider temperature range makes the gradient more accurate.
Approximation
  • Using 1/time as the rate is valid only if the extent of reaction at the end point is the same each time.

Part 5 of 6Worked Example

Question

  • A graph of ln(rate) against 1/T has gradient –5200 K.
  • Calculate Ea in kJ mol⁻¹ (R = 8.31 J mol⁻¹ K⁻¹).

  1. Gradient = –Ea ÷ R, so Ea = 5200 × 8.31.
  2. 5200 × 8.31 = 43 212 J mol⁻¹.
  3. Divide by 1000 to give kJ mol⁻¹.

Answer Ea = 43.2 kJ mol⁻¹

Part 6 of 6Exam Checklist & Tips

6.1.3 – Key Points for the Exam

  • Activation energy The minimum energy that colliding particles need for a reaction to occur.
  • Arrhenius equation The equation relating the rate constant to temperature and activation energy.
  • Gradient The slope of a line, found by dividing the change in y by the change in x.
  • Rate constant The proportionality constant k in the rate equation.

Common Mistakes

  • Using temperatures in °C for 1/T, which gives a wrong gradient.
  • Forgetting the minus sign in gradient = –Ea/R.
  • Quoting Ea in J mol⁻¹ with units of kJ mol⁻¹ without dividing by 1000.
  • Plotting ln k against T rather than 1/T.

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