Activation energy experiments (Core Practical 10)

Collecting the data

• Use a reaction that can be timed (for example a clock reaction), with the same concentrations each time.
• Warm each reactant separately in a water bath until it reaches the set temperature, then mix and start timing.
• Measure the temperature of the mixture (for example at the start and end, and take the mean).
• Use at least five temperatures over a good range.

Processing

Rate is proportional to 1/t, and k is proportional to rate, so ln(1/t) = -Ea/R x (1/T) + constant.

1. Convert each temperature to kelvin: T = θ + 273.

2. Calculate 1/T (K^-1) and ln(1/t).

3. Plot ln(1/t) (y) against 1/T (x) and draw the best-fit straight line.

4. Gradient = -Ea/R, so Ea = -gradient x 8.31 J mol^-1, then convert to kJ mol^-1.

Errors and improvements

• The mixture cools (or warms) during the reaction at temperatures far from room temperature: use a water bath throughout and record the mixture's temperature.
• Judging the end point: use a light sensor.
• At high temperatures the time is short, so reaction time in starting and stopping the clock gives a large percentage error: use lower concentrations so times are longer.
• Too few temperatures make the gradient unreliable.

Worked example

Times for a clock reaction: 20 °C 180 s, 30 °C 89 s, 40 °C 46 s, 50 °C 25 s, 60 °C 14 s. Find Ea.

1. Convert to K and calculate 1/T and ln(1/t) for each point.
2. Best-fit gradient of ln(1/t) against 1/T = -6227 K
3. Ea = 6227 x 8.31 = 51744 J mol^-1

Answer: Ea = 52 kJ mol^-1

Check yourself

1. Calculate 1/T for a temperature of 45 °C.
   Show answer

   0.00314 K^-1. T = 45 + 273 = 318 K; 1/T = 3.14 x 10^-3 K^-1.

2. A graph of ln(1/t) against 1/T has a gradient of -7350 K. Calculate the activation energy in kJ mol^-1. (R = 8.31 J mol^-1 K^-1)
   Show answer

   61.1 kJ mol^-1. Ea = 7350 x 8.31 = 61079 J mol^-1 = 61.1 kJ mol^-1.

3. A reaction takes 46 s. Calculate ln(1/t).
   Show answer

   -3.83. ln(1/46) = -3.83.

4. Why are the reactant solutions warmed separately before mixing?
   • To stop them evaporating
   • To make the colour change sharper
   • So the reaction starts at the chosen temperature, not while warming up
   • To dissolve the starch
   Show answer

   So the reaction starts at the chosen temperature, not while warming up. Mixing first would let the reaction run at lower temperatures.

5. What is the sign of the gradient of a graph of ln(1/t) against 1/T?
   • Negative
   • Positive
   • Zero
   • It depends on the reaction
   Show answer

   Negative. Higher T (smaller 1/T) gives a faster rate (larger ln(1/t)).

6. In a run at 60 °C, the mixture cools to 52 °C by the end. Which improvement is best?
   • Use a larger stopwatch
   • Add more starch
   • Keep the reaction vessel in a 60 °C water bath and record the mixture's mean temperature
   • Use a measuring cylinder instead of a burette
   Show answer

   Keep the reaction vessel in a 60 °C water bath and record the mixture's mean temperature. The temperature must be known and constant during the run.

7. A student plots ln(1/t) against 1/θ with θ in °C. What is wrong?
   • The temperature must be in kelvin for the Arrhenius relationship
   • The y-axis should be t, not 1/t
   • The graph should be against θ, not 1/θ
   • Nothing is wrong
   Show answer

   The temperature must be in kelvin for the Arrhenius relationship. The Arrhenius equation uses absolute temperature.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson IAL Chemistry teacher practical guide
• Pearson examiners' report, WCH16 October 2024

Last updated 8 November 2026. Spotted a mistake? Tell us.