Rates - Activation Energy (OCR A Level Chemistry)

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

Temperature, T (K)	1 / T	Time, t (s)	1 / t	ln (1 / t)
297.0	0.003367	63	0.015873	-4.1431
310.3	0.003223	34	0.029412	-3.5264
316.9	0.003156	26	0.038462	-3.2581
323.6	0.003090	22	0.045455	-3.0910
335.3	0.002982	16	0.062500	-2.7726

• 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