Rates - Activation Energy (OCR A Level Chemistry A): Revision Note

PAG 10.3: Rates – Activation Energy

The Ethanedioic Acid and Potassium Manganate(VII) Reaction

• A less common experiment to determine the activation energy of a reaction is the reaction between potassium manganate(VII) solution and ethanedioic acid

2MnO₄^– (aq) + 5H₂C₂O₄ (aq) + 6H⁺ (aq) → 2Mn²⁺ (aq) + 8H₂O (l) + 10CO₂ (g)

• In this reaction, potassium manganate(VII) with ethanedioic acid in the presence of sulfuric acid

• The key point in this experiment is the colour change of the manganese which changes from purple to brown 

Method

• Using a different / clean measuring cylinder for each chemical, transfer 10.0 cm³ of each of the following into a separate boiling tube

  • 0.1 mol dm^-3 potassium manganate(VII) solution, KMnO₄ (aq) 

  • 2.0 mol dm^-3 sulfuric acid, H₂SO₄ (aq)

  • 0.25 mol dm^-3 ethanedioic acid, H₂C₂O₄ (aq)

• Warm / heat all three boiling tubes in a water bath

• Place a test tube in a test tube rack

• Measure and record the temperature of all three solutions

• Add 0.5 cm³ of the potassium manganate(VII) solution to the test tube

• Add 0.5 cm³ of the sulfuric acid to the test tube

• Then add 1.0 cm³ of the ethanedioic acid to the test tube

• Immediately start the stopwatch and swirl the mixture

• Stop the stopwatch when the purple colour changes to pale brown and record the time

• Repeat the experiment for a minimum of four more / different temperatures

The Ethanedioic Acid and Potassium Manganate(VII) Reaction

Specimen Results

• The temperature of the reaction is converted to 1 / T by calculating the reciprocal value

• The rate is found by calculating the reciprocal of time, t

• This is then converted into the natural logarithm function to give ln(1 / t) or ln(rate)

• A graph of ln(rate) versus 1 / T is then plotted:

Analysis

An ln(rate) versus 1 / T graph for the Ethanedioic Acid and Potassium Manganate(VII) Reaction

• From this graph, we can see that the line of best fit is a straight line

  • The equation for a straight line is y = mx + c

  • The Arrhenius equation can be rearranged to ln k = ln A - E_a/RT

  • Therefore, the gradient of the line is -E_a / R 

• This means that we can use the gradient from the graph to calculate the activation energy for this reaction

  • Gradient =  can be rearranged to E_a = –gradient x R

  • E_a = –x 8.31 = 30616 J mol^-1

  • E_a = = 30.616 kJ mol^-1