Domenica DiStasio April 2, 2019Experiment 4: Chemical Kinetics-Effect of Temperature… Arrhenius EquationObjective: The objective of this lab is to determine the effect that temperature has on the speed of a chemical reaction. We are going to look at how the activation energy in the Arrhenius Equation changes as the temperature of H2C2O4 increases/decreases. Procedure:1. Using a buret place 20mL of ~0.5M H2C2O4 in an Erlenmeyer Flask and 10mL of ~0.02M KMnO4 in a 15cm test tube. 2. Place both vessels in your water bath and let them sit there for five minutes. 3. Set up your second trial at this temperature. 4. Mix the reactants and record the time it takes for the solution again to go yellow/brown. Swirl the contents regularly. 5. Repeat the procedure with another sample at this temperature. 6. Using the average time, determine k at this temperature. 7. Repeat this procedure for the other suggested temperatures. 8. When complete, determine the Activation Energy by plotting ln k v 1/T. (Remember, temperature must be in Kelvin). *Repeat this 4 times at temperatures 0, 10, 20, and 30 degrees.* Only do 1 trial for the temp at 0 degrees.Data Collection: Table 1: Temperature and concentration trials and their reaction timesKMnO4mLH2C2O4mLTemp.0CReaction Time Trial 1 sReaction Time Trial 2 sAverage Times 10 mL 20mL 0 769 s --------- 769 s10 mL 20 mL 10 410 s 470 s440 s10 mL 20 mL 20 320 s 299 s*temp was at 23 at end309.5 s10 mL 20 mL 30 113 s 105 s 109 sCalculations:1. Determining the Rate (Temperature at 0oC) - [KMnO4]/t ---- Rate/([KMnO4][H2C2O4])[KMnO4] = (.021 mol/ 1000 ml) * 10mL * (1000 mL/30mL) = .007 MR= k[KMnO4][ H2C2O4]R= [KMnO4]/t = .007 M/ 769 s = 9.1*10-6 mol/L/sec2. Calculation on Determining k (Temperature at 0oC)[KMnO4] = (.021 mol/ 1000 ml) * 10mL * (1000 mL/30mL) = .007 M[H2C2O4] = (.5 mol/ 1000 mL) * 20 mL * (1000 mL/ 30 mL) = .333 MR= k[KMnO4][ H2C2O4]K = (R/[KMnO4][ H2C2O4])= (9.1*10-6 mol/L/sec) / (.007 M * .333M) = .00391 M*secTable 2: Calculations of k for each temperature trial T oCT K1/T 1/Kk Ln k0 273.15 .00366 .00391 -5.5510 283.15 .00353 .00683 -4.9920 293.15 .00341 .00970 -4.6430 303.15 .00330 .0276 -3.59Discussion:The activation energy tells you the minimum amount of energy required for a reaction to occur. The lower the activation energy, the faster the reaction. In this experiment, the reaction’s activation should be lower than the typical bond energy because the temperature is increasing with each reaction, which means the activation energy should be decreasing. This makes sense regarding our graph because as the temperature increases, the activation energy is decreasing, hence the negative slope. Figure 1: Scatter plot with trend line showing the relationship between Ln k vs. 1/T (Kelvin). The slope is y= -5150.5x + 13.205.Ea Equation: Slope= -5150.5x + 13.205-Ea/R = -5150.5x Ea = -(5150.5)(8.314 JmolK)= 42821 / 1000= 42.821 kJmol Ea = 42.821
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