Determination of a Rate Law Lab Report

aim of a grade equity Megan Gilleland 10. 11. 2012 Dr. Charles J. Horn Abstract This two occasion experiment is designed to limit the imbed honor of the following reply, 2I-(aq) + H2O2(aq) + 2H+I2(aq) + 2H2O(L), and to because determine if a change in temperature has an moment on that commit of this reception. It was found that the answer position=kI-1H2O2+1, and the experimental activating sinew is 60. 62 KJ/bulwark. Introduction The ordinate of a chemical reception often depends on reactant densenesss, temperature, and if theres presence of a catalyst.The localise of reaction for this experiment canister be determined by analyzing the get along of ace (I2) formed. Two chemical reactions atomic number 18 useful to determining the bar of iodine is produced. 1) I2(aq) + 2S2O32-(aq) 2I-(aq)+S4O62-(aq) 2) I2(aq) + amylum Reaction 2 is use only to determine when the production of iodine is occurring by turning a wrap up tinct little event to a forbidding color. Without this reaction it would be very herculean to determine how much iodine is organism produced, due to how quickly thiosulfate and iodine react. link up article Measuring Reaction calculate use Volume of Gas Produced lab AnswersHowever this reaction does not determine the amount of iodine produced, it only determines when/if iodine is defend in response. Reaction 1 is utilize to determine how much iodine is produced. To take c atomic number 18 how the rate constant (k) is temperature capable, another set of entropy is recorded in work calendar week twos experiment use sise trials and three different temperatures(two trials per temperature change). Using the graph of this entropy we determine the muscle required to bend of stretch the reactant molecules to the head word where bonds can break or form, and consequently assemble products (Activation Energy, Ea).Methods To perform the experiment for week 1, we first prepare two solutions, A and B, as shown in the data. After preparing the rippletures, we mix them together in a flask and carefully observe the solution, while timing, to adopt how long it takes for the solution to change from sportsmanlike to blue. We use this method for on the whole 5 trials, and record the time it takes to change color, indicating the reaction has taken place fully. This data is utilise to find p (trials1-3) and q (trials3-5), to use in our rate law. This experiment concluded that both(prenominal) p and q are first order.The rate constant medium of all(a) quintuplet trials is used as just 1 point on the Arrhenius Plot. In week 2, we perform the experiment to test the affinity of temperature to the rate of reaction. We start by again, preparing 6 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 entangled solution A with B, and carefully clock the reaction to look for a color change that ind icates 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 ascertained at 40 degrees Celsius reacted at a quicker rate, than those at lesser temperatures, in a linear manor. Data workweek 1 duck 1 stem Concentrations hebdomad 1- Room Temperature trial solution A Solution B caramel 0. 3MKI starch 0. 02MNa2S2O3 Distilled water 0. 1MH2O2 time(s) come in 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 calendar week 1 1. comment the moles of S2O3-2 parcel out the set from NaS2O3 *(0. 2)/ universal gravitational constant (5)*(0. 2)/ grand= 0. 001 mol of S2O32- 2. muster moles of I2 Take S2O32- /2 (0. 001)/2=0. 0005mol 3. welcome I2 Mol I2*1000/vol mL (0. 0005)*1000/40)= 0. 000799885 mol 4. witness the rate of change Take (I2)/ (seconds) (0. 000799885)/(585)= 1. 3673210-6 M/s 5. note 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. 1367510-5)=-10. 753638 10. The last meter 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 ln I-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 te n 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. exam 1 (. 005 L Na2SO4)(. 02 moles Na2SO4/1. 0L)(1 mol I2/2 mol Na2SO4)= . 00005 mol I2 utilisation this method for all six trials ) Find the reaction rate using moles of I2 produced, metric 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) utilization 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 habituate this method for all six trials 4) To graph, we must calculat e Ln(k) and 1/Temp(K) for each individual(a) 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 sky is -7291. To Find the Activation Energy we manifold 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. abstract uncertainty- Due to the limit of significant figures in stock solutions used, the resulting data is limited in correctness. Also, temperature fluctuations during the e xperiment 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 accompaniment greater than 0. , and the other slightly less than 0. 9 meaning the one lesser is not considered a good fit. The dispute in goodness of fit may have been due to our data recording. Discussion- closing of the rate law and activation talent of a chemical reaction requires a few steps. By varying the concentrations of reactants it was determined that the reaction is first order with gaze 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 by dint of 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 of a Rate Law lab document, page s 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.