Introduction to Chemical Reactions
Chemical reactions entailing the movement of electrons from one species to the other are referred to as oxidation-reduction reactions. One such reaction occurs when oxalic acid reacts with potassium permanganate solution.
2 MnO−4 (aq) + 16 H+ (aq) + 5 C2O − 24(aq) → 2 Mn2+ (aq) + 8 H2O (l) + 10 CO2 (g)
KMnO4 has a deep purple colour, which turns to a brownish-yellow colour on reduction. KMnO4 is an oxidizing agent, whereas the oxalic acid is a reducing agent. The time needed to use up KMnO4 determines the rate of the reaction. The first rate of a reaction is the immediate change in concentration of a reactant as soon as the reaction commences. Activation energy, on the other hand, is the quantity of energy needed to kick off a chemical reaction (Elias 2002). This is usually related to the temperature in the system and can be evaluated using the Arrhenius equation
k=Ae-EaRT.
Activation energy can be calculated using the following modified version of the Arrhenius equation: “In (k2/k1) = Ea/R x (1/T1 – 1/T2) where Ea is the activation energy of the reaction in J/mol, R is the ideal gas constant (8.3145 J/K·mol), T1 and T2 are absolute temperatures, and k1 and k2 are the reaction rate constants at T1 and T2” (Activation energy example problem n.d.).
Aim
This experiment aimed at establishing the activation energy for the reaction between oxalic acid and potassium permanganate by first determining the rate equation for the reaction.
Method
The precise concentrations of oxalic acid and potassium permanganate solution provided were recorded. A bulb pipette was used to deliver 20 ml of 0.5 M oxalic acid into a clean, dry 100 ml conical flask for the first run. For the second run, 20 ml of oxalic acid and 5 ml of water were added to an empty conical flask, whereas 10 ml of acid and water were added to the third conical flask for the third run. Three test tubes were filled with 10 ml, 5 ml, and 10 ml of KMnO4.
For each run, the conical flask containing H2C2O4 and the test tube containing KMnO4 were placed in a beaker of water for approximately 5 minutes at room temperature after which the temperature of the water was measured. Potassium permanganate was then emptied into the acid in the beaker, and a stopwatch was started immediately. The mixture was thoroughly mixed by swirling, keeping the conical flask in the water until the potassium permanganate solution changed to a light yellowish-brown colour. The reaction time was then recorded. A second run was done, and the average of the two trials recorded. The whole procedure was replicated for the second and third runs.
The slowest run was reiterated at a higher temperature. This was to ascertain the activation energy for the reaction and was attained by running the experiment in a water bath at 45 oC.
Results
Potassium permanganate changed its colour from deep purple to a light brownish yellow colour as the reaction progressed. This change in colour signified the reduction of potassium permanganate by oxalic acid.
Table 1: Table of Results
The initial tempo of the reaction was obtained by dividing the concentration of KMnO4 by the time taken. Therefore, the initial rate for run 1 was given by 0.02/207= 9.67×10-5 M/s. The initial rate for run 2 was given by 0.02/120=1.67×10-4 M/s, whereas the initial rate for run 3 was 0.02/360=5.56×10-5M/s. In the final run (run 4), the initial rate was given by 0.02/33=6.06×10-4 M/s.
Comparing the initial velocities for runs 1 and 3 showed that increasing the concentration of oxalic acid doubled the initial reaction rate. Hence, the reaction was a first order regarding oxalic acid. Therefore, the value of a was 1. Comparing runs 1 and 2 revealed that decreasing the concentration of potassium permanganate almost doubled the initial reaction rate. Therefore, the value of b was also 1. Consequently, the rate law for the reaction was k (H2C2O4)1 (KMnO4)1.
The rate constant (k) for run 1 was 9.67×10-5 M/s=k (10)1(20)1
K=9.67×10-5 M/s/200
K=4.835×10-7M/s
The rate constant (k) for run 2 was 1.67×10-4 M/s=k (20)1(5)1
K=1.67×10-4 M/s/100
K=1.67×10-6M/s
The rate constant (k) for run 3 was 5.56×10-5 M/s=k (10)1(10)1
K=5.56×10-5 M/s/100
K=5.56×10-7M/s
The average value of k was given by (4.835×10-7+1.67×10-6+5.56×10-7)/3=8.882×10-7.
The rate constant for run 4 was 6.06×10-4M/s=k (10)1(10)1
K=6.06×10-4M/s/100
K=6.06×10-6M/s.
The activation energy was 76.4725 KJ/mol.
Discussion of the Reaction
The experimental activation energy for the reaction was 76.4725 KJ/mol. However, it was not possible to find literature values for comparison. The temperature of the reactants played a crucial role in determining the activation energy. Increasing the temperature accelerated the rate of the reaction because it lowered the activation energy and increased the collisions of the reacting species (Activation energy n.d.). That was why the reaction was extremely fast when the temperature was raised to 50oC.
Summary
The reaction rate increased with an increase in temperature because the rise in temperature lowered the activation energy.
Appendix: Calculation of Activation Energy
Calculation of activation energy used the formulae “In (k2/k1) = Ea/R x (1/T1 – 1/T2) where Ea was the activation energy of the reaction in J/mol, R was the ideal gas constant (8.3145 J/K·mol), T1 and T2 were absolute temperatures, and k1 and k2 were the reaction rate constants at T1 and T2” (Activation energy example problem n.d.).
T1= 25oC or 298K.
T2=50oC or 323K.
Therefore, In (6.06×10-6M/s ÷5.56×10-7M/s) =Ea÷8.3145J/K.mol× (1/298-1/323)
2.389=Ea/8.3145J/K.mol×2.597×10-4
2.389=Ea3.124×10-5
Ea=2.389/3.124×10-5
Ea=76,472.5 J/mol or 76.4725 KJ/mol.
References
Activation energy,n.d. Web.
Activation energy example problem, n.d. Web.
Elias, A. J. 2002, A collection of interesting general chemistry experiments, Universities Press (India) Private Limited, Hyderabad.