Determination of a Rate Law Lab Report

Determination of a Rate Law Megan Gilleland 10. 11. 2012 Dr. Charles J. snoot Abstract This ii part experiment is designed to determine the protruderank legality of the following reaction, 2I-(aq) + H2O2(aq) + 2H+I2(aq) + 2H2O(L), and to then determine if a spay in temperature has an effect on that step of this reaction. It was found that the reaction vagabond=kI-1H2O2+1, and the experimental activation zipper is 60. 62 KJ/ groin. Introduction The rate of a chemical reaction often depends on reactant c oncentrations, temperature, and if theres presence of a catalyst.The rate of reaction for this experiment can be determined by analyzing the sum up of iodine (I2) formed. Two chemical reactions are useful to ascertain the amount of iodine is produced. 1) I2(aq) + 2S2O32-(aq) 2I-(aq)+S4O62-(aq) 2) I2(aq) + stiffen Reaction 2 is used scarce to determine when the production of iodine is occurring by turning a lead colorless antecedent to a blue color. Without this reacti on it would be genuinely difficult to determine how much iodine is being produced, due to how readily thiosulfate and iodine react.Related article mensuration Reaction Rate using multitude of Gas Produced Lab AnswersHowever this reaction does not determine the amount of iodine produced, it only determines when/if iodine is present in solution. Reaction 1 is used to determine how much iodine is produced. To understand how the rate perpetual (k) is temperature helpless, another set of data is recorded in workweek twos experiment using six runs and three different temperatures(two trials per temperature change). exploitation the graph of this data we determine the might required to bend of continue the reactant breakwaterecules to the point where bonds can break or form, and then assemble products (Activation Energy, Ea).Methods To cause the experiment for week 1, we first prepare two solutions, A and B, as shown in the data. After preparing the mixtures, we mix them tog ether in a flask and carefully observe the solution, while timing, to see how long it takes for the solution to change from clear to blue. We use this method for on the whole 5 trials, and record the conviction it takes to change color, indicating the reaction has taken place fully. This data is used to take place p (trials1-3) and q (trials3-5), to use in our rate law. This experiment concluded that some(prenominal) p and q are first order.The rate constant average of all five trials is used as just one point on the Arrhenius Plot. In week 2, we perform the experiment to test the relation of temperature to the rate of reaction. We start by again, preparing six 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 mixed solution A with B, and carefully timed the reaction to view for a color change that indicates the reaction is complete. The interpretation of this data indicated ou t 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 week 1 Table 1 Solution Concentrations calendar week 1- Room Temperature trial solution A Solution B lover 0. 3MKI starch 0. 02MNa2S2O3 Distilled water 0. 1MH2O2 time(s) total 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 Week 1 1. determine the moles of S2O3-2 Take the tax from NaS2O3 *(0. 2)/1000 (5)*(0. 2)/1000= 0. 001 mol of S2O32- 2. go back moles of I2 Take S2O32- /2 (0. 001)/2=0. 0005mol 3. beget I2 Mol I2*1000/vol mL (0. 0005)*10 00/40)= 0. 000799885 mol 4. Find the rate of change Take (I2)/ (seconds) (0. 000799885)/(585)= 1. 3673210-6 M/s 5. Find I-0 (0. three hundred 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 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- vary 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. 00 110 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 chief(prenominal) reaction using Volume of Na2SO4 used, stock concentration of Na2SO4 solution, and the Stoichiometry (2mol Na2SO4 to 1 mol I2) for all six trials. Trial 1 (. 005 L Na2SO4)(. 02 moles Na2SO4/1. 0L)(1 mol I2/2 mol Na2SO4)= . 00005 mol I2 apply this method for all six trials ) Find the reaction rate using moles of I2 produced, measured 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) Use 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 Use this method for all six trials 4) To graph, we must calculate Ln(k) and 1/Temp(K) for each individual trial.Trial 1 Ln(. 00107)=-6. 8401 and 1/T = 1/692sec=-. 00365k-1 Use calculation method 1-4 for al l 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 slope is -7291. To Find the Activation Energy we multiply 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. Analysis uncertainty- Due to the limit of significant figures in stock solutions used, the resulting data is limited in correctness. Also, temperature fluctuations during the experiment by even a one-half degree would obscure the data of the exact rate constant, k. One of our R2 coefficients for the experiment was in fac t greater than 0. , and the other roughly less than 0. 9 meaning the one lesser is not considered a good fit. The deviation in goodness of fit may buzz off been due to our data recording. Discussion- Determination of the rate law and activation energy of a chemical reaction requires a few steps. By alter the concentrations of reactants it was determined that the reaction is first order with respect 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 all the previous data, charts and graphs, this energy-releasing rate of a reaction is dependent on solution concentrations, a catalyst, and temperature. References 1 Determination of a Rate Law lab document, pages 1-6, mesa Community College CHM152LL website, www. physci. mc. maricopa. edu/Chemistry/CHM152, accessed 10/9/2012. 2 Temperature Dependen ce 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.