Determination of a Rate Law Megan Gilleland 10. 11. 2012 Dr. Charles J. Horn Abstract: This two part experiment is designed to determine the rate law of the following reaction, 2I-(aq) + H2O2(aq) + 2H+I2(aq) + 2H2O(L), and to then determine if a change in temperature has an effect on that rate of this reaction. It was found that the reaction rate=k[I-]^1[H2O2+]^1, and the experimental activation energy is 60. 62 KJ/mol. Introduction The rate of a chemical reaction often depends on reactant concentrations, temperature, and if there’s presence of a catalyst.

The rate of reaction for this experiment can be determined by analyzing the amount of iodine (I2) formed. Two chemical reactions are useful to determining the amount of iodine is produced. 1) I2(aq) + 2S2O32-(aq) 2I-(aq)+S4O62-(aq) 2) I2(aq) + starch Reaction 2 is used only to determine when the production of iodine is occurring by turning a clear colorless solution to a blue color. Without this reaction it would be very difficult to determine how much iodine is being produced, due to how quickly thiosulfate and iodine react.

Related article: Measuring Reaction Rate Using Volume of Gas Produced Lab Answers

Methods To perform 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 together 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 all 5 trials, and record the time it takes to change color, indicating the reaction has taken place fully. This data is used to find p (trials1-3) and q (trials3-5), to use in our rate law. This experiment concluded that both 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 look 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 Week 1 Table 1: Solution Concentrations Week 1- Room Temperature trial #| solution A| | | | | Solution B| | | | | | buffer| 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. Find the moles of S2O3-2 Take the value from NaS2O3 *(0. 2)/1000 (5)*(0. 2)/1000= 0. 001 mol of S2O32- 2. Find moles of I2 Take S2O32- /2 (0. 001)/2=0. 0005mol 3. Find I2 Mol I2*1000/vol mL (0. 0005)*1000/40)= 0. 000799885 mol 4. Find the rate of change Take (I2)/ (seconds) (0. 000799885)/(585)= 1. 36732×10-6 M/s 5. Find [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 [H2O2]0 Take (0. 10 M H2O2)*(6. 00mL)/ ( final volume)=0. 015 M 8. Ln of [H2O2]0 Ln(0. 015)= -4. 19970508 9. Find the Ln of rate: Ln(2. 13675×10-5)=-10. 753638 10. The last step for week one calculations is to calculate the average value of k. Rate= k [I-]1[H2O2]. (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| [H2O2]0| ln[H2O2]o| 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| 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 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. Trial 1: (. 005 L Na2SO4)(. 02 moles Na2SO4/1. 0L)(1 mol I2/2 mol Na2SO4)= . 00005 mol I2 Use 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 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 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 half degree would obscure the data of the exact rate constant, k. One of our R^2 coefficients for the experiment was in fact greater than 0. , and the other slightly less than 0. 9 meaning the one lesser is not considered a good fit. The deviation in goodness of fit may have been due to our data recording. Discussion- Determination of the rate law and activation energy of a chemical reaction requires a few steps. By varying 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 exothermic 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 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.