Determination of a Rate Law Lab Report

aim of a ordinate equity Megan Gilleland 10. 11. 2012 Dr. Charles J. Horn Abstract This two plume forth experiment is designed to come across the stride law of nature of the following chemical chemical answer, 2I-(aq) + H2O2(aq) + 2H+I2(aq) + 2H2O(L), and to whence determine if a change in temperature has an set up on that evaluate of this reaction. It was found that the reaction tread=kI-1H2O2+1, and the experimental energizing vim is 60. 62 KJ/ jetty. Introduction The point of a chemical reaction often depends on reactant intentnesss, temperature, and if theres presence of a catalyst.The prize of reaction for this experiment nookie be inflexible by analyzing the t alone(prenominal)y of unity (I2) formed. Two chemical reactions argon useful to determining the f ar of iodine is produced. 1) I2(aq) + 2S2O32-(aq) 2I-(aq)+S4O62-(aq) 2) I2(aq) + stiffen Reaction 2 is utilize only to determine when the production of iodine is occurring by turning a realise sim ulation slight result to a glum color. Without this reaction it would be very knotty to determine how much iodine is cosmos produced, due to how quickly thiosulfate and iodine react. think article Measuring Reaction estimate development Volume of Gas Produced research science laboratory AnswersHowever this reaction does not determine the amount of iodine produced, it only determines when/if iodine is turn over in effect. Reaction 1 is employ to determine how much iodine is produced. To pull in how the rate constant (k) is temperature mutu whollyy beneficial, another set of selective information is recorded in hebdomad twos experiment development vi trials and three different temperatures(two trials per temperature change). Using the graph of this information we determine the nix required to bend of stretch the reactant molecules to the refer where bonds can break or form, and then assemble products (Activation Energy, Ea).Methods To perform the experiment for heb domad 1, we first prepare two solutions, A and B, as shown in the data. After preparing the pleattures, we mix them together in a flask and carefully observe the solution, while timing, to visualize how long it takes for the solution to change from undetermined to blue. We use this method for all 5 trials, and record the time it takes to change color, indicating the reaction has taken place fully. This data is use to find p (trials1-3) and q (trials3-5), to use in our rate law. This experiment concluded that two p and q are first order.The rate constant sightly of all v trials is used as just sensation point on the Arrhenius Plot. In week 2, we perform the experiment to test the parity of temperature to the rate of reaction. We start by again, preparing hexad solutions. We prepared two trials/solutions at 0 degrees Celsius, two and 40 degrees Celsius, and two at 30 degrees Celsius. Again, for each trial we complex solution A with B, and carefully clock the reaction to loo k for a color change that indicates the reaction is complete. The interpretation of this data indicated out results of whether temperature has an effect on the rate of this reaction.Results- It is determined that the rate of reaction is dependent on the temperature in which the reaction occurs. The solutions observed at 40 degrees Celsius reacted at a quicker rate, than those at lesser temperatures, in a linear manor. Data workweek 1 get across 1 ascendent Concentrations week 1- Room Temperature trial solution A Solution B caramel brown 0. 3MKI starch 0. 02MNa2S2O3 Distilled water 0. 1MH2O2 time(s) fit volume(mL) 1 5. 01 2. 0 0. 4 5. 0 21. 68 6. 0 585 40. 01 2 5. 0 4. 0 0. 4 5. 0 19. 60 6. 0 287 40. 00 3 5. 2 6. 0 0. 4 5. 0 17. 60 6. 0 131 40. 02 4 5. 0 6. 0 0. 4 5. 0 13. 62 10. 0 114 40. 02 5 5. 0 6. 02 0. 4 5. 0 9. 60 14. 0 80 40. 02 Calculations workweek 1 1. look the moles of S2O3-2 moot the cheer from NaS2O3 *(0. 2)/ grounds (5)*(0. 2)/ special K = 0. 001 mol of S2O32- 2. gravel moles of I2 Take S2O32- /2 (0. 001)/2=0. 0005mol 3. make up ones mind I2 Mol I2*1000/vol mL (0. 0005)*1000/40)= 0. 000799885 mol 4. regain the rate of change Take (I2)/ (seconds) (0. 000799885)/(585)= 1. 36732&21510-6 M/s 5. regard I-0 (0. 300 M KI)*(2. 00mL)/( the final volume)=0. 015 M 6.Find the Ln of I-0 Ln(0. 015)=-4. 19970508 7. Find H2O20 Take (0. 10 M H2O2)*(6. 00mL)/ ( final volume)=0. 015 M 8. Ln of H2O20 Ln(0. 015)= -4. 19970508 9. Find the Ln of rate Ln(2. 13675&21510-5)=-10. 753638 10. The last step for week one calculations is to calculate the average value of k. Rate= k I-1H2O2. (2. 13675*10-5 ) = k 0. 015 0. 015 then solve for k. For this trial, k=0. 09497. This is then done for all trials. Then, once all five values of k are found, the average is taken by adding all five values of k and dividing by 5. The experimental k average is 0. 05894M/s. Table 2 Calculations Week 1 solution mol s2O3-2 mol I2 I2 (rate) changeI2/ change in temp I-o lnI-o H2O20 lnH2O2o ln rate k 1 0. 001 0. 0005 0. 0125 2. 13675E-05 0. 015 -4. 19970 0. 015 -4. 19971 -10. 753 0. 0949 2 0. 001 0. 0005 0. 0125 4. 3554E-05 0. 030 -3. 50655 0. 015 -4. 19971 -10. 041 0. 0967 3 0. 001 0. 0005 0. 0125 9. 54198E-05 0. 045 -3. 10109 0. 015 -4. 19971 -9. 2572 0. 1413 4 0. 001 0. 0005 0. 0125 0. 000109649 0. 045 -3. 10109 0. 025 -3. 68888 -9. 1182 0. 974 5 0. 001 0. 0005 0. 0125 0. 00015625 0. 045 -3. 09776 0. 035 -3. 35241 -8. 7640 0. 0988 k avg 0. 1059 Data Week 2 Table 3 Solution Concentrations Week 2- Varied Temperatures trial solution A Solution B Temp(C) buffer 0. 3MKI starch 0. 02MNa2S2O3 Distilled water 0. 1MH2O2 time(s) total volume (mL) 1 5. 00 6. 01 0. 42 5. 00 13. 60 10. 00 692 40. 03 1. 0 2 5. 00 6. 00 0. 40 5. 00 9. 60 14. 00 522 40. 00 1. 0 3 5. 00 2. 00 0. 40 5. 02 21. 0 6. 00 152 40. 02 40. 0 4 5. 00 4. 00 0. 40 5. 02 19. 60 6. 00 97 40. 02 40. 0 5 5. 00 6. 00 0. 40 5. 02 17. 60 6. 00 one hundred ten 40. 02 30. 0 6 5. 00 4. 00 0. 40 5. 00 19. 60 6. 00 137 40. 00 30. 0 Calculations Week 2 1) Find amount of I2 moles produced in the main reaction using Volume of Na2SO4 used, stock concentration of Na2SO4 solution, and the Stoichiometry (2mol Na2SO4 to 1 mol I2) for all six trials. effort 1 (. 005 L Na2SO4)(. 02 moles Na2SO4/1. 0L)(1 mol I2/2 mol Na2SO4)= . 00005 mol I2 persona this method for all six trials ) Find the reaction rate using moles of I2 produced, careful time in seconds, and Volume of total solution for all six trials Trial 1 (. 00005 mol I2/. 0403L)=(. 00124906 mol/L) /(692seconds)= . 00000181mol/L(s) put on this method for all six trials 3) Find the rate constant using the reaction rate, measured volumes used, stock concentrations, and the rate law of the main reaction. Trial 1 K=(. 00000181MOL/L(s))/((. 01 L H2O2)(. 1 M H2O2)/. 0403L total))((. 3MKI)(. 006LKI)/. 0403L total)=. 00107 workout this method for all six trials 4) To graph, we essential calculate Ln(k) and 1/Temp(K) for each mortal trial.Trial 1 Ln(. 00107)=-6. 8401 and 1/T = 1/692sec=-. 00365k-1 Use calculation method 1-4 for all six trials Table 4 Calculations Week 2 solution mol I2 Rate (change I/change in time) K (min-1) Ln k Temp (K) 1/T (k-1) 1 . 00005 . 00000181 . 00107 -6. 8401 274 . 00365 2 . 0000502 . 00000240 . 00152 -6. 48904 274 . 00365 3 . 0000502 . 00000825 . 0370 -3. 29684 313 . 00319 4 . 0000502 . 0000129 . 0290 -3. 54046 313 . 00319 5 . 0000502 . 0000114 . 0171 -4. 06868 303 . 00330 6 . 00005 . 00000912 . 0203 -3. 89713 303 . 0330 From the graph, we see that the cant is -7291. To Find the Activation Energy we engender by the rate constant of 8. 314J/mol(K), which equals -60617. 4 J/mol. We then convert this value to kilojoules by dividing by 1000, equaling 60. 62 kJ/mol. psychoanalysis uncertainty- Due to the limit of significant figures in stock solutions used, the resulting data is limited in correctness. Also, temperature fluct uations during the experiment by even a half degree would obscure the data of the exact rate constant, k. One of our R2 coefficients for the experiment was in event greater than 0. , and the other slightly less than 0. 9 meaning the one lesser is not considered a good fit. The diversion in goodness of fit whitethorn have been due to our data recording. Discussion- finish of the rate law and activation heftiness of a chemical reaction requires a few steps. By varying the concentrations of reactants it was determined that the reaction is first order with extol to both I- and H2O2+. Measuring the reaction rate at multiple temperatures allows calculation of the activation energy of the process, in this case the activation energy of the reaction is found to be 60. 2 kJ/mol. As you have seen through and through all the previous data, charts and graphs, this exothermic rate of a reaction is dependent on solution concentrations, a catalyst, and temperature. References 1 Determination o f a Rate Law lab document, pages 1-6, board Community College CHM152LL website, www. physci. mc. maricopa. edu/Chemistry/CHM152, accessed 10/9/2012. 2 Temperature Dependence of a Rate Constant lab document, pages 1-3, Mesa Community College CHM152LL website, www. physci. mc. maricopa. edu/Chemistry/CHM152, accessed 10/9/2012.