Jennie Burgess - Experiment 3 Lab Report

Jennie Burgess Chem 112 March 3, 2024 99LN Experiment 3: Chemical Kinetics Determining the Rate Equation The overall goal of this experiment was to determine the effect of different concentrations of KMnO 4 , H 2 C 2 O 4 , and water on the rate constant, K, of the reaction. Procedure The experimental procedure was straightforward with setting up three burets and filling them with KMnO 4 , H 2 C 2 O 4 , and water. Each experiment called for a different amount of each substance. While timing and swirling, the solution was measured to see when it turned brown. Once the solution turned brown the timer was stopped and the time it took was recorded. Each experiment was repeated twice. After calculations were made for the rate of each experiment and the overall rate order. Data Figure 1. Results from Experiments KMnO 4 [A] H 2 C 2 O 4 [B] H 2 O Trial 1 Time (s) Trial 2 Time (s) Avg. Time (s) [A] Initial [B] Initial Exp 1 10.0 ml 20.0 ml 0.0 ml 5.20 m= 320s 5.30m= 330s 325s 0.007 mol/L 0.33 mol/L Exp 2 5.00 ml 20.0 ml 5.0 ml 6.51 m= 411s 5.48m = 348s 380s 0.0035 mol/L 0.33 mol/L Exp 3 10.0 ml 10.0 ml 10.0 ml 10.40m= 641s 10.15m= 615s 628s 0.007 mol/L 0.167 mol/L Figure 1. Describes the results from each experiment's two trials and the average time it took to complete each experiment as well as the initial concentrations of KMnO 4 and H 2 C 2 O 4 Calculations Time conversions Experiment 1: Time 1; (5 minutes x (60 seconds / 1 minute)] + 20 seconds= 320s Experiment 1: Time 2; (5 minutes x (60 seconds / 1 minute)]+ 30 seconds= 330s Average time= 320 seconds + 330 seconds / 2 = 325 s Experiment 2: Time 1; (6 minutes x (60 seconds / 1 minute)] + 51 seconds= 411s Experiment 2: Time 2; (5 minutes x (60 seconds / 1 minute)] + 48 seconds= 348 s Average time= 411 seconds + 348 seconds / 2= 380 s

Experiment 2: Time 1; (10 minutes x (60 seconds / 1 minutes)] + 40 seconds= 641s Experiment 2: Time 2; (10 minutes x (60 seconds / 1 minute)] + 15 seconds= 615s Average time= 614 seconds + 615 seconds / 2= 628s Total Volumes Experiment 1: 10 mL (KMnO 4 ) + 20 mL (H 2 C 2 O 4 ) + 0 mL (H 2 O) = 30 mL Experiment 2: 5 mL (KMnO 4 ) + 20 mL (H 2 C 2 O 4 ) + 5 mL (H 2 O) = 30 mL Experiment 3: 10 mL (KMnO 4 ) + 10 mL (H 2 C 2 O 4 ) + 10 mL (H 2 O) = 30 mL Initial Volumes of KMnO 4 and H 2 C 2 O 4 Molarity of KMnO 4 = 0.021 M Molarity of H 2 C 2 O 4 = 0.5 M Experiment 1 [A] i KMnO 4 initial= M KMnO 4 x Volume of KMnO 4 / Total Volume [A] i = 0.021M x 10ml / 30ml= 0.007 mol/L [B] i H 2 C 2 O 4 initial= M H 2 C 2 O 4 x Volume of H 2 C 2 O 4 / Total Volume [B] i = 0.5M x 20ml / 30ml= 0.33 mol/L Experiment 2 [A] i KMnO 4 initial= M KMnO 4 x Volume of KMnO 4 / Total Volume [A] i = 0.021M x 5.0 ml/ 30ml= 0.0035 mol/L [B] i H 2 C 2 O 4 initial= M H 2 C 2 O 4 x Volume of H 2 C 2 O 4 / Total Volume [B] i = 0.5 M x 20ml/ 30ml= 0.33 mol/L Experiment 3 [A] i KMnO 4 initial= M KMnO 4 x Volume of KMnO 4 / Total Volume [A] i = 0.021M x 10ml /30ml= 0.007mol/ L [B] i H 2 C 2 O 4 initial= M H 2 C 2 O 4 x Volume of H 2 C 2 O 4 / Total Volume [B] i = 0.5M x 10ml/ 30ml= 0.167mL Order of the Reaction in Respect KMnO 4 (x) 1. Using average times from experiments 1 and 2, to measure the effect on rate with different volumes of KMnO 4 Rate 1 [A] 1 / T1 = (0.007 M) / 325s = 2.15E-5 Ms -1 Rate 2 [A] 2 / T2= (0.0035 M)/ 380s= 9.21 E-6 Ms -1 Avg. Rate 1 / Rate 2 = 2.15E-5 / 9.21E-6= 2.33 Ms -1 Solve for X to complete overall reaction order equation

Finding the Rate of the Reaction for Hypothetical Experiment 4 Figure 2. The Volume of Experiment 4 Volume of KMnO 4 Volume of H 2 C 2 O 4 Volume of H 2 O Total Volume 15ml 30ml 5ml 50ml The figure above contains the volumes of the hypothetical experiment 4 Rate= k[KMnO 4 ] 1.22 [H 2 C 2 O 4 ] 1.98 [KMnO 4 ]= (0.021M) x 15ml / 50ml= 0.006M [H 2 C 2 O 4 ]= (0.5M) x 30ml / 50ml= 0.3M Rate= (0.08)(0.006) 1.22 (0.3) 1.98 Rate= 1.43E-5 Ms -1 Discussion The data and calculations yielded an overall reaction order of 3.20. The expected reaction order was to be 3. Despite the total volume changing in experiment 4 the concentrations of the reagents were not affected because of how the volume of each reagent was changed. In other words the ratio of reactants were the same. Experiment 4 would produce similar timing results to experiment 1 if it was completed. Conclusion This experiment had room for human error because it was easier to make mistakes when doing each trial. Even the slightest mistake could produce a result that did not match the goal. Buret readings could have been off, which would calculate an incorrect final. Precise measurement is crucial to accurate results. Since two separate groups did separate trials, swirling speed and timing could have been done differently as well. Overall the calculations support that the overall rate would be 3 for this type of reaction.