Domenica DiStasio March 18, 2019Experiment 3: Chemical Kinetics-Determining the Rate EquationObjective:The objective of this lab was to use chemical kinetics in order to measure the speed (rate) of a chemical reaction. The reaction we were interested in was the reaction of potassium permanganate with oxalic acid. For our specific trials, we looked at how the concentration of different solutions (KMnO4, H2C2O4 and H2O) affected the rate of the reaction. Data Collection:Rate equation= k[KMnO4]x[H2C2O4]yRate = -D[KMnO4]/ Dt = -{[KMnO4]final - [KMnO4]init.]/(tfinal - tinitial)Volume Table KMnO4H2C2O4H2O Tria1 1 TimeTrial 2 TimeTrial 3 TimeAverageTimeExp 110 mL 20 mL 0 mL 175 s 174 s 178 s 175.6 sExp 25 mL 20 mL 5 mL 245 s 275 s 260 s 260 sExp 310 mL 10 mL 10 mL 309 s 299 s 304 s 304 sCalculations:M of KMnO4 = .021 mol/LM of H2C2O4 = .5 mol/LExperiment 1[KMnO4]= .1L*x (.021/.03L)=.007 MRate 1= .007M/175.6 s = 3.98E-5Msec-1[H2C2O4]= .02L x (.5M/.03L) =.333MRate 1= .333M/175.6s=1.9E-3Msec-1Experiment 2[KMnO4]= .005L*x (.021/.03L)=.035 MRate 2= .035M/260 s = 1.3E-4[H2C2O4]= .02L x (.5M/.03L) =.333MRate 2= .333M/260 s=1.3E-3Msec-1 Experiment 3[KMnO4]= (.01L x .021M) / .03L =.007 MRate 3= .007M/304 s = 2.3E-5[H2C2O4]= .01L x (.5M/.03L) =.166MRate 3= .166M/304 s=5.5E-4Msec-1 a) Finding x [KMnO4]Rate 1 / Rate 2 = 3.98E-5 / 1.3E-4 = (0.007 / 0.035)x 0.306 = 0.2x Log (0.306) / log (0.2) = xX = 0.736 ≈1b) Finding y [H2C2O4]Rate 1 / Rate 3 = 3.98E-5 / 2.3E-5 = (0.333 / 0.166)yLog (1.73) / log (2.01) = yY = .78 ≈1c) Overall reaction order = 1+1=2Rate = k[KMnO4]1[H2C2O4]1d) Solving for k using experiment oneRate = k[KMnO4]1[H2C2O4]13.98E-5 = k[0.007][0.333](3.98E-5 / 0.0023) = k (0.0023 / 0.0023)k = 0.0173Lmol-1sec-1e) Rate of reactionKMnO4 = 0.015L x (0.021M / 0.05L) = 0.0063MH2C2O4 = 0.03L x (0.5M / 0.05L) = 0.3MRate = k[KMnO4]1[H2C2O4]1Rate = (0.0173)(0.0063)(0.3) = 3.27E-5 M-1sec-1Conclusion:The order of the reaction with respect to [KMnO4] was 1, the order of the reaction with respect to[H2C2O4] was 1, so therefore the overall order of the reaction was 2. We found that k is equal to 0.0173Lmol-1sec-1and our rate is 3.27E-5 M-1sec-1 . If you increase the concentration, the rate of the reaction also increases. In this lab, there were many areas where there could have been an error in the data. If the amounts of each solution were not exact, then the time for each trial would have been thrown off which would then throw off the rate of our reaction. Human error would have attributed to the percent error of this lab (which we did not calculate but I imagine it would not be
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