'Saponifacation of Ethyl Acetate and Soldium Hydroxide\r'

' ethyl radical group radical radical group Acetate †NaOH reply Kinetics try Martin Novick Group 14, chemical Engineering science laboratory Submitted to Prof. David B. Henthorn September 25, 2012 Summary The goal of this jut was to take root the pre-exponential situationor, k o , the energizing energy, E, and the chemical response commit constants, k, of the saponification process of ethyl acetate using sodium hydroxide (NaOH) at 5 temperature amongst 15 and 25 degrees Celsius. Two trails were performed at temperatures 16, 18, 20, 22, and 24 degrees Celsius. The main equipment of the working class were the jacketed beaker quite a little reactor and the LabPro conduction analyze.The solution’s conduction throughout the chemical chemical chemical reception was put in and plan in a educated diagram against snip to get hold ?????? assess for each trial. The ordinate law was assumed to be ???????????????????????? = ????????????? ???????????? ?????????????????? , where ?????????????????? and ?????????????????? argon the constrictions of sodium hydroxide and ethyl acetate respectively. The ln? k) look ons were patchted against the inverse temperatures to ( unidimensionalize the Arrhenius comparability. The k o re repute and E take account from the li adjoiningized Arrhenius speckle of land were found to be 15 ± 3M ? 1 s ? 1 and ? 6402 ± 8191? j ? mol? 1 respectively. The E time value universe nix suggests the chemical reaction is exothermic. The large exemplificationised computer errors of the ???????????? and ?????? set were in every last(predicate) probability actord by the crushed number of information acids collected or the assumed step law was wrong. Introduction The objective of this project was to determine the pre-exponential factor, k o , the activation energy, E, and the reaction count constants, k, of the saponification process of ethyl acetate using sodium hydroxide (NaOH). Sapon ification is a chemical process heavily utilise in industry, especi onlyy in soap production.Knowing the personal effects of temperature on the reaction stray in al subalterns better control on the whole over the reaction process and find the optimizing point of production. The right temperature maximizes production and minimizes reactants and heating or cooling energy, finding the maximum interchangeablely provista. 2 The saponification of ethyl acetate with sodium hydroxide (NaOH) is an equi-molar reaction given up(p) as NaOH + CH3 COOCH2 CH3 > CH3 COONa + CH3 CH2 OH. [2] The rate was assumed to be fleck hunting lodge overall, simply frontmost order relative to either reactant,[2] with the disappearance rate of sodium hydroxide given as:[4] dCOH = ? COH Cac , dt (1) where COH is the NaOH c one timentration, Cac is the tightness of ethyl acetate, and k is the rate constant. COH is represented with x, and Cac is assumed to be in excess with a starting intentness of a. The concentration of ethyl acetate throughout the reaction was given as Cac = a ? xo + x, where xo is the initial starting concentration of NaOH. substituting the expressions for COH and Cac into compare 1 departs in: dx = ? kx(a ? xo + x). dt comparison 2 is then separated and incorporate shown as the followers: ? t dx = ? k ? dt. xo x(a ? xo + x) 0 x (2) (3)The result of comparison 3 would become: ln ( xo (a ? xo + x) ) ax = kt. a ? xo (4) Since a 20% excess ethyl acetate solution was used, the initial ethyl acetate concentration was 1. 25 measure of the initial starting concentration of NaOH, so a = 1. 25xo . After the substitution of a = 1. 25xo , Equation 4 was simplified to: 3 0. 2x ln ( x o + 0. 8) = kt. 0. 25xo The relationship in the midst of xo x (5) and conduction of the solution was given as:[2] ???????????? ???????????? ? ??????? = , ?????? ?????? ? ??????? (6) where ???????????? is the initial conductivity of NaOH in the archetypical place adding ethyl acetate, ??????? s the conductivity by and by the reaction has reached boundary, and ?????? is the conductivity at each point of time during the reaction process. modify Equation 6 into Equation 5 gives the following: ???????????? ? ?????? ln (0. 2 ?????? ? ?????? ? + 0. 8) ? = kt. 0. 25xo (7) Since ???????????? , ???????????? and ??????? are constants, the solely variable in the left heap look of Equation 7 world ??????. Each ?????? point bottom be substituted into the left hand side of Equation 7 and maculationted against time to retrieve a linear plot with the angle macrocosmness the k value. To find the value of the pre-exponential factor, ???????????? and activation energy, ?????? , would require linearizing the Arrhenius equation given as: k = k o ?????? ????????????? , where R is the gas constant and T is the temperature the given k is at in degrees Kelvin. Equation 8 is linearized by fetching the natural log of some(prenominal) sides: ln(k) = ln(k o ) ? E . R T 1 ?????? ?????? (8) (9) to retrieve a linear plot with Equation 9 shows a linear relationship amidst ln(k) and 4 the y- stop being ln(???????????? ) and the hawk being ??????. Hence: k o = ey? intercept . ?????? = ?????????????????????????????? ? ?????? (10) (11) ??????Equipment, Materials, and Method The equipment used were a jacketed atomic pile reactor beaker, cooling peeing circulation system, computer, LabPro temperature probe and conductivity probe, mixing stand and magnetic stir bar. The materials used for this reaction were a 0. 08M NaOH solution and a 0. 1M ethyl acetate solution. A 20% excess ethyl radical acetate was used to look into NaOH was the control reactant. [1] NaOH was chosen for the limiting reactant because of its racy conductivity relative to Ethyl acetate. The extent of the reaction was monitored by measuring the conductivity throughout the reaction.With NaOH being the limiting reactant, the change in conductivity is more visible, and the terminati on of the reaction merchantman be more considerably observed. The each experimental trail was apparatus as shown in regards A1 and A2 of supplement A. The temperature and conductivity probes were completely submerged under the scratch of the reacting solution. genius info reading was retrieved per second. in the beginning the reaction process begun, the cooling water circulation system was set at want temperature and the reactants were cooled to the target temperature to keep an 5 isothermal reaction environment.In Equation 8, the k value was barely a function of temperature; if temperature varied throughout the reaction, the k value would in addition vary throughout the reaction. The reaction was started once thermal equilibrium between the batch reactor beaker and cooling water was established. For rough the first half a infinitesimal, alone the conductivity of NaOH in the reactor was collected, to ensure the initial conductivity in conductivity was more easily obse rved. The initial conductivity was the first conductivity point by and by the major drop of conductivity collect to the addition of ethyl acetate.The reaction was started by and by just about half a minute afterward the start of the entropy arrangement by adding the ethyl acetate. The solution was well(p) mixed throughout the reaction to ensure a uniform temperature of the solution, to prevent any local anaesthetic k value being unalike from the overall k value. If thither any temperature gradients throughout the system existed during the reaction process, in that respect would be different local k values throughout the reactor. selective information collection was terminated at least(prenominal) half a minute after the reaction had gone to completion, when no conductivity change was observed.The final reaction completion conductivity, ???????????? , was the conductivity of the solution after the reaction has completed, and was used to find k value. Hysteresis effects we re removed by using a random number generator to determine the order of trials. 6 Results and Discussion 1. 2 1. 1 1 0. 9 ln(k) 0. 8 0. 7 0. 6 0. 5 0. 4 0. 3 0. 00335 0. 00337 0. 00339 0. 00341 1/T (T-1) y = -4378. 4x + 15. 713 R? = 0. 7383 0. 00343 1 0. 00345 0. 00347 grade 1 the plot of the linearized Arrhenius equation where ln(?????? ) ???????????? . ?????? The linearized Arrhenius equation plot shown in meet 1 has a R2 value of 0. 383 suggesting the data does non bind a good linear fit. The leave out of data points whitethorn keep up been the cause of the low R2 value. The ln(k) values of the corresponding temperature mostly seem to have small variations, unless overall has the predicted turn out of decreasing in ln(k) value as ?????? ?1 increases. The plot has a negative overall slide with and hence retrieving a negative E value. The negative E value suggests the dismissal of energy as the reaction proceeds. prorogue B1 in app curiosityix B shows the retrieved valu es of k, ko and E. The positive ko value was expected harmonise to Equation 8.Since k values are linearly proportional to ko, for k values to be positive, ko must also be positive. The resulting negative E value was an expected result, since the reaction was said to be very spontaneous. The standard errors of ko and E being about 20% to 25% of the original value, suggests the data as inexact. The imprecision of the data can also be observed from the low R2 value. 7 Looking at the imprecision of the plot in persona 1, there is a nonable difference between the k values at the same temperature for = 0. 003364 (24oC) and ?????? 1 1 ?????? = 0. 03455(16oC). This may be due to the variance in concentration of solutions since solutions were remade each laboratory day. Another affirmable cause maybe that the table of contents of solutions may have reacted with the melody introducing extra contents to the reaction. The k values all have standard errors less than 1%. Figures B1 to B5 o f Appendix B shows the plots of ???????????????????? ln( 1 ???????????? ???????? ( +1)) 1. 25 ?????????????? 0. 25xo used to find the k values at each temperature. The plots all have a positive trend which is reasonable according to Equation 7.The resulting slope is the k value of that trial, and the k values cannot be negative. as well as the R2 values of the plots were all above 0. 95, suggesting a good fit with the linear crush fit line model. Despite the high R2 values, the a clear parabolical trend was observed in all the repose plots shown in Figures B6 to B14 of Appendix B, the linear attribute of the plots are questioned. Figure B15 of Appendix B shows an example plot of the full data from 16oC trail 1. A clear linear trend was observed from the beginning of the plot, but an obvious curve was observed from the end of the plot.This observation was probably due to the fact that as the reaction proceeded, the reaction started behaving like a first order reaction. The first order reaction behavior is because near the end of the reaction of the excess ethyl acetate concentration was much greater than the NaOH concentration and can be treat as a constant. With the concentration of ethyl acetate being constant the only resulting variable would be 8 the NaOH concentration. terminus and Recommendations The retrieved ko value and E value were 15 ± 0. 3 and ? 36402 ± 8191.Since the standard errors of ko and E values were about 20% and 25% of the original value respectively, and the R2 value was only 0. 7383 for the plot in Figure 1, the data is reason to be imprecise and that more data points should be collected to improve the precision. The assumption of the rate being second order overall, but first order relative to either reactant was reason out to be inappropriate. Though the R2 value for all the trials were above 0. 95, because a clear parabolic trend was observed in all the equalizer plots, Figures B6 to B14 of Appendix B, it was concluded th at all the trial data do not have a linear relationship.Since it has been concluded that the assumed rate equation, Equation 1, does not describe the reaction, a new rate equation is needed to describe the reaction. The low R2 value in Figure 1 can be improved by increasing the amount of data collected. more than trails can be done at each temperature and more temperatures can be tested within the given interval. One large bag solution can be made and used for all the trails, but stored and isolated from the atmosphere. The error caused by the variation in concentrations of the reactants at different trails can be solved by using the one large bulk of reactants.The problem of the reactants reacting with the atmosphere can be solved by isolating the reactants from the atmosphere during storage. 9 References [1] Crismyre, Bobby, Do, Bryan, and Kronmiller, Mandy. â€Å"Ethyl Acetate †NaOH Reaction Kinetics Experiment Standard operate Procedures,” (2010) [2] Julio F. Mata -Segreda, Hydroxide as a popular Base in the Saponification of Ethyl Acetate. ledger of the American Chemical Society, (2010), 124 (10), 2259-2262. [3] Keusch Fachdidaktik Chemie. â€Å"Reaction sulphur Order,” Institute of Organic Chemistry, Universitat Regensburg. [4] Schmidt, Larry D. â€Å"The Engineering of Chemical Reactions,” 2 Ed.Press, New York (2005) Oxford University 10 addition A: Experimental Set Up Figure A 1: Shows the conjectural setup and flow of each experiment. Figure A 2: Shows the actual set up used for each experiment. 11 APPENDIX B: Experimental Results and data Table B 1 shows the result of the desired information Set Temperature (oC) 16 ( running game 1) 16 ( outpouring 2) 18 ( exertion 1) 20 ( examination 1) 20 (Trial 2) 22 (Trial 1) 22 (Trial 2) 24 (Trial 1) 24 (Trial 2) ???????????? (??????? 1 ?????? ?1 ) ?????? ( ?????? ) ?????????????????? Actual Temperature (oC) 16. 45 ± 0. 05 16. 44 ± 0. 04 18. 33 ± 0. 03 20. 30 ± 0. 05 20. 30 ± 0. 5 22. 24 ± 0. 04 22. 25 ± 0. 05 16. 45 ± 0. 05 16. 45 ± 0. 05 15 ± 3 ? 36402 ± 8191 k-values (??????? 1 min? 1 ) 1. 674±0. 001 2. 023±0. 001 1. 921±0. 002 2. 241±0. 002 2. 247±0. 003 2. 244±0. 006 2. 169±0. 002 2. 572±0. 003 3. 140±0. 002 12 25 20 y = 2. 0232x R? = 0. 9994 15 Y (L/mol) y = 1. 6744x R? = 0. 9972 10 5 0 0 2 ln( 1 ???????????? ???????? ( +1)) 1. 25 ?????????????? 4 6 prison term (min) 8 ln( 10 1 ???????????? ???????? ( +1)) 1. 25 ?????????????? 12 Figuer B 1 maculation of ???????????????????? 0. 25xo at 16 degrees Celsius, where ?????? = 0. 25xo . 16 14 12 10 8 6 4 2 0 0 -2 ln( 1 ???????????? ???????? +1)) 1. 25 ?????????????? y = 1. 9206x R? = 0. 9974 Series1 linear (Series1) 1 2 3 4 5 6 7 8 Figuer B 2 Plot of ???????????????????? 0. 25xo at 18 degrees Celsius, where ?????? = ln( 1 ???????????? ???????? ( +1)) 1. 25 ?????????????? 0. 25xo . 13 25 y = 2. 247x R? = 0. 9964 y = 2. 2407x R? = 0. 9982 Y (L/mol ) 15 Trial 1 Trial 2 10 Linear (Trial 1) Linear (Trial 2) 5 20 0 0 2 4 6 Time (min) 1 ???????????? ???????? ( +1)) 1. 25 ?????????????? 1 ???????????? ???????? ( +1)) 1. 25 ?????????????? 8 10 12 Figuer B 3 Plot of ???????????????????? ln( 0. 25xo at 20 degrees Celsius, where ?????? = ln( 0. 25xo . 0 18 16 14 Y (L/mol) 12 10 8 6 4 2 0 0 2 4 Time (min) 1 ???????????? ???????? ( +1)) 1. 25 ?????????????? 1 ???????????? ???????? ( +1)) 1. 25 ?????????????? y = 2. 244x R? = 0. 9909 y = 2. 1687x R? = 0. 9985 Trial 1 Trial 2 Linear (Trial 1) Linear (Trial 2) 6 8 10 Figuer B 4 Plot of ???????????????????? ln( 0. 25xo at 22 degrees Celsius, where ?????? = ln( 0. 25xo . 14 25 20 y = 2. 5723x R? = 0. 9982 15 Y (L/mol) y = 3. 1405x R? = 0. 9992 Trial 1 Trial 2 Linear (Trial 1) Linear (Trial 2) 10 5 0 0 1 2 3 4 Time (min) 5 6 7 8 Figuer B 5 Plot of ???????????????????? ln( 1 ???????????? ???????? ( +1)) 1. 25 ?????????????? . 25xo at 24 degrees Celsius, where ?????? = ln( 1 ???????????? ??????? ? ( +1)) 1. 25 ?????????????? 0. 25xo . X covariant 1 equalizer Plot 0. 8 0. 6 residuums 0. 4 0. 2 0 -0. 2 -0. 4 0 2 4 6 X variant 1 8 10 12 Figure B 6 The balance wheel plot for 16 degrees Celsius Trial 1. 15 X versatile 1 symmetricalness Plot 0. 6 0. 4 Residuals 0. 2 0 -0. 2 -0. 4 0 2 4 6 8 10 12 X shifting 1 Figure B 7 The residual plot for 16 degrees Celsius Trial 2. X versatile 1 Residual Plot 0. 8 0. 6 Residuals 0. 4 0. 2 0 -0. 2 -0. 4 0 1 2 3 4 X changeable 1 5 6 7 8 Figure B 8 The residual plot for 18 degrees Celsius Trial 1. X variable quantity 1 Residual Plot 0. 8 0. 6 Residuals 0. 4 0. 2 0 -0. 2 0 -0. 4 -0. 6 X Variable 1 2 4 6 8 10 12 Figure B 9 The residual plot for 20 degrees Celsius Trial 1. 16 X Variable 1 Residual Plot 1. 5 1 Residuals 0. 5 0 0 -0. 5 -1 X Variable 1 2 4 6 8 10 12 Figure B 10 The residual plot for 20 degrees Celsius Trial 2. X Variable 1 Residual Plot 1. 5 1 Residuals 0. 5 0 -0. 5 -1 0 1 2 3 4 5 6 7 8 X Variable 1 Figure B 11 The residual plot for 22 degrees Celsius Trial 1. X Variable 1 Residual Plot 0. 8 0. 6 Residuals 0. 4 0. 2 0 -0. 2 -0. 4 0 1 2 3 4 5 6 7 8 9 X Variable 1 Figure B 12 The residual plot for 22 degrees Celsius Trial 12 7 X Variable 1 Residual Plot 1 Residuals 0. 5 0 0 -0. 5 1 2 3 4 X Variable 1 5 6 7 8 Figure B 13 The residual plot for 24 degrees Celsius Trial 1. X Variable 1 Residual Plot 0. 6 0. 4 Residuals 0. 2 0 -0. 2 -0. 4 0 1 2 3 4 X Variable 1 5 6 7 8 Figure B 14 The residual plot for 24 degrees Celsius Trial 2. 140 120 100 Y (L/mol) 80 60 40 20 0 0 5 10 15 20 25 30 35 40 45 Time (min) Figure B 15 The plot of all the data from 16oC trail 2 18 APPENDIX C: warning Calculations playscript of 1M NaOH solution needed to coordinate 300mL 0. 08M NaOH solution: 0. 3?????? ? 0. 08?????? = 0. 024?????? = 24????????????? ?????? Volume of ethyl acetate needed to uprise a 300mL 0. 1M ethyl acetate solution: 1000???????????? 88. 105?????? ???????????? 3 0. 001?????? 0. 10?????? ? ? ? ? ? 0. 3 = 2. 96? ??????????? ?????? ?????????????????? 0. 897?????? ???????????? 3 finding E value: ?????? ?????????????????? Calculating the k value for time being 1 minute in trial 2 of 16 degrees: ?????? = ?????????????????????????????? ? ?????? = ? 2904. 9 ? 8. 3145 = ? 24153 1 ???????????? ? ??????? ln ( ( + 1)) 1. 25 ?????? ? ??????? 0. 25xo 1 16324. 71 ? 4698. 16 ln ( ( + 1)) 1. 25 8259. 03 ? 4698. 16 0. 25(0. 8) = kt = = 14. 89 19APPENDIX D: fault Analysis If the result (R) is compute by the following equation: ?????? = ? ???????????? ?????? where ???????????? is the exponent of the self-sustaining variable, ???????????? . The standard error would be calculated by the following equation: 2 2 ???????????? ???????????? = ?????? (? ( ?????????????????? ) ) ???????????? 1 ?????? (1) (2) where????????????? is taken from equation 5, ?????????????????? is the standard error of each exclusive independent variable. Since the equation used is the following: ???????????? ? ?????? ln (0. 8 ( ?????? ? ?????? ? + 1)) ? 0. 25???????????? = ???????????? , 3) where ?????? is the rate constant, ???????????? is the initial conductivity, ??????? is the conductivity at the end of the reaction, ?????? is the conductivity at any time, and ???????????? is the initial concentration of the NaOH solution. The error of the right hand side of equation 3 with ???????????? = 36. 662 is show as the follow: 36. 662 (( 0. 0339 ) ) = 15. 536. 0. 08 1 2 2 ???????????? was given the value of 0. 08?????? with an misgiving of ±0. 0339??????. The reaction rate constant, k, were found by the stovepipe fit line of the plot, so the uncertainty of the k were found by taking the linear regression.The rate constant was the slope of the best fit line qualification the calculation of the error with a 95% confidence take aim as: ?????????????????????????????? = 1. 96 ? ???????????? ,?????? v? ???????????? 2 ? ????????????? 2 Where ???????????? ,?????? is the standard error of estimate, ???????????? is the x values of the plot, n being the number of points, and ???????????? = ? ??????? ???????????? ?????? =1 ?????? , (4) . As for the y intercept, or the activation energy, the error with a 95% confidence level would be: 20 ?????????????????????????????? = 1. 96 ? ???????????? ,?????? v??????? 1 (? ??????????? )2 + 2 , ?????? (? ???????????? 2 ? ????????????? 2 ) (5) Where ???????????? ,?????? is the standard error of estimate, ???????????? is the x values of the plot, n being the number of points, and ???????????? = ? ??????? ???????????? ?????? =1 ?????? . The standard error of estimate, ???????????? ,?????? , seen in both equation 4 and equation 5 is given as: ???????????? ,?????? =v ? (???????????? ? ???????????? )2 , ?????? ? 2 (6) where ???????????? is the y value of the points, ???????????? is the corresponding y value on the best fit line, and n is the number of data points.\r\n'