Hydrogen peroxide decomposition by Baker’s yeast, Study Guides, Projects, Research of Chemistry

Download Hydrogen peroxide decomposition by Baker’s yeast and more Study Guides, Projects, Research Chemistry in PDF only on Docsity! TKP4105/TKP4110 Hydrogen peroxide decomposition by Baker’s yeast Report Audun F. Buene audunfor@stud.ntnu.no Elise Landsem elisel@stud.ntnu.no Group B19 Supervisor: Naresh Doni Jayavelu Lab: K4-317 Date: 18th of September 2012 Summary In this experiment, the enzyme catalysed reaction of hydrogen peroxide decompo- sition was investigated. The parameter that was altered in order to investigate the enzymatic acticvity was the concentration of yeast. It was found that the initial rate of decomposition increased with increasing concentration of yeast, following the linear approximation for a volume of 30 mL: rH2O2 (gyeast) = 5 · 10−4 · gyeast − 4 · 10−6. 1 2 Theory 2.1 Decomposition process Yeast is a eukaryote singlecellular microorganism. It has got a rich variety of enzymes, in order to get nutrition as well as to protect itself. Yeast is a lot less complicated than other eukaryotes, but all the more interesting because it has a lot of similar enzymes as its fellow and more complex eukaryotes. For instance the enzyme, catalase, that catalyse the reaction in which hydrogen peroxide is converted to water and oxygen. 2H2O2 → 2H2O+O2 (2.1) The reason why the reaction is catalyzed by catalase is that the structure of the enzyme is made to match with the molecular structure of H2O2. The active sites induce the breaking of chemical bonds, and promote the making of new ones. En- zymes lower the activation energy of chemical reactions by promoting a different reaction mechanism, and this is the reason for the increase in the initial reaction rates. These enzymes can not be consumed or altered in the reaction, which is the definition of a catalyst. [3] The initial reaction rate of this reaction will be the objective of this experiment, with respect to the consentration of yeast. There will be other factors to consider, conducting an experiment with living cells, for instance the effect of change in temperature, the concentraion of H2O2, pH of the mixture or other enzymes and reactions also occuring in the same batch. These effects are not to be investigated during this experiment. The yeast is a living organism, so its activity will vary depending on a number of different factors. Therfore it is necessary to check the catalytic activity of the yeast that will be used. This test is explained in Section 3.2. 4 2.2 Data analysis Because this experiment was run as if in a batch reactor, we’re only interested in the initial reaction rate, because there is no easy way of measuring the H2O2- concentration. The data gathered is then converted and plotted, to find the initial reaction rate as a function of the concentraion of yeast. This is possible, due to the fact that the yeast concentration is constant for each series. There are two different ways to estimate the initial reaction rate. The first is by estimating the tangent to the curve of the first measuring point. Secondly one can estimate the slope from a straight line through the 3-4 first measuring points. The last approach is more widely used, and also what is to be used for the calculations in this experiment. 2.3 Statistical analysis In order to estimate the standard errors in the slope and the intersection of the initial reaction rate curves as a function of yeast consentration, statistical analysis is used. The method used will be the method of least squares. This is usually done with computers directly from the plots of the measurements. It is based on calculating the distance between the actual data point, and the estimated function. This distance is denoted di: di = yi − a · xi − b (2.2) where yi is the actual data point, and a·xi−b is the calculated value from the linear approximation. Furthermore one can calculte the standard error in the y-values, sy, using the following equation: sy = √∑n i=1(d 2 i ) n− 2 (2.3) where d2i is used to ensure all the values are positive. This is the basic thought behind the method of least squares. The error in the slope of the linear estimate, 5 sa, is given by: sa = sy · √ n n · ∑n i=1(x 2 i )− ( ∑n i=1 xi) 2 (2.4) At last the error in the intersection of the linear estimate with the y-axis, sb, can be calculated from the following equation: sb = sy · √ ∑n i=1(x 2 i ) n · ∑n i=1(x 2 i )− ( ∑n i=1 xi) 2 (2.5) 6 4 Results The complete set of measurements are found in Appendix E. 4.1 Preliminary test For the preliminary test it took 28.15 seconds for the reaction to produce 10 mL of O2-gas. This was a bit too fast, so all the volumes of yeast suspension were halved. This was done in order to get more accurate measurements. 4.2 Rate of decomposition The initial rate of decomposition of H2O2 as a function of the weight of yeast was found using a linear approximation from Figure 4.1. The data for the plot is shown in Appendix D. The initial decomposition rate for a 30 mL solution was found to be: rH2O2 (gyeast) = 5 · 10−4 · gyeast − 4 · 10−6 (4.1) 9 Figure 4.1: All the initial reaction rates plotted as a function of the weight of yeast in the reaction mixture. A linear approximation to the curve has been added. 4.3 Statistical analysis A statistical analysis was preformed on the measurements using the formulaes un- der Section 2.3. This gave the following errors for the inital reaction rate, sy, slope, sa and intersection with the y-axis, sb: sy = 9.9211 · 10−6 [mol H2O2/s] sa = 1.5568 · 10−4 [ mol H2O2/s gyeast ] sb = 8.3849 · 10−6 [mol H2O2/s] 10 5 Discussion The preliminary test indicated that the enzymatic activity of the yeast that was used, was higher than expected. A consequence of this was that the concentration of yeast had to be reduced in order to get proper measurements. During the measurements of volume of O2 gas, a slight delay was noticed. The plots of the individual measurement series also suggest a slight delay, because all the trend lines intersect the y-axis at a considerable negative volume. This indi- cates that the the entire plot is shifted due to the reaction delay. It is possible that the delay was caused by either lack of stirring or diffusion in and out of the cells. It was also noticed that for low concentrations of yeast, the plots of volume against time were approximatly linear, whilst for higher concentrations, these plots vere slightly curved. From Figure 4.1 it is clear that the initial reaction rate of H2O2 decomposition increases with incresing amount of yeast. This is due to the increasing amount of active sites available to decompose hydrogen peroxide. This result was as ex- pected. The trend line has a slightly negative intersection with the y-axis. This is clearly wrong, since this indicates that in abscense of catalase, hydrogen peroxide would be formed. The estimated error in the intersection is larger than the nega- tive intersection value, which indicates that the plot is within what is reasonable. 5.1 Sources of error There are several possible sources of error in this experiment. First of all there will allways be a human factor to consider, and this will affect the measurements. However if a certain consistancy is present, all the measurements will be affected in the same way, and relative to each other, the error would be quite small. A few assumptions have been made, for instance that oxygen behaves as an ideal gas and that the syringe is frictionless. Furthermore there could have been leakage 11 A Example calculations A.1 Calculations on volume of gas The following assumptions have been made for this calculation: • Room temperature in reaction flask • No biproducts of the decomposition • Oxygen as an ideal gas • Amount of O2 absorbed in the water of the mixture is negligible. The overall reaction is: 2H2O2 → 2H2O+O2 (A.1) 4 mL of a 3 wt.% solution of hydrogen peroxide is used. This gives a total weight of H2O2 in the reaction mix: mH2O2 = 4 · 10−3kg · 0.03 = 0.12 gram (A.2) This molecular weight of H2O2 is 34.015 g/mol, so this gives the total number of moles of H2O2: nH2O2 = mH2O2 MmH2O2 = 0.12 gram 34.015 g/mol = 3.528 · 10−3 mol (A.3) From the stoichiometry of the reaction, the total number of moles of oxygen gas is obtained: nO2 = 1 2 · nH2O2 = 1.764 · 10−3 mol (A.4) The mass of O2 can be obtained by: mO2 =MmO2 · nO2 = 32 g/mol · 1.764 · 10−3 mol = 0.0564 grams (A.5) 14 The total theoretical volume of O2 is found by using the density of O2, ρO2 = 1.309 · 10−3 g/m3. VO2 = 0.0564 1.309 · 103 g/m3 = 4.309 · 10−5m3 = 43.09 mL (A.6) A.2 Calculation of initial reaction rate As only the first measurements are of importance, the first 4-6 data points for the volume of produced oxygen gas were plotted against time. These plots can be found in Appendix D. From the linear function given by the computational program, the slope was read off. This slope indicates how fast the okxygen gas is produced. The volume of oxygen is then converted to moles of oxygen. Finally the number of moles of H2O2 decomposed per unit of time can be found from the stoichiometry of the reaction. To demonstrate the calculations, test number 6 will be used. For this test, 6 mL of yeast suspension was used, 4 mL 3 wt.% H2O2 and 20 mL of water. The measurements were done with a 20 mL frictionless syringe and a stopwatch, and are shown in Table A.1. Table A.1: Measurements from test no. 6. VO2 is the collected volume of O2-gas. Measurement no. VO2 [mL] Time [s] 1 1 14.16 2 2 17.29 3 3 19.66 4 4 22.6 These data are shown in Figure A.1, as well as the linear approximation made by the computational program. 15 Figure A.1: Values of volume of produced O2-gas plotted as a function of time. From Figure A.1, the linear approximation made by the computational program is: y = 0.3602x− 4.1381 (A.7) This gives a slope of 0.3602 mL/s, which in turn gives the initial reaction rate: dV (O2) dt = 0.3602 [mL/s] (A.8) From the volume of oxygen, one can easily find the number of moles of oxygen produced, using the molar volume of an ideal gas. dn(O2) dt = 0.3602 mL/s 22414 mL/mol O2 = 1.61 · 10−5 [mol O2/s] (A.9) From the stoichiometry of the reaction, it is obvious that for each mole of O2-gas produced, two moles of H2O2 have been decomposed. This gives: dn(H2O2) dt = (−2) · 1.61 · 10−5 [mol O2/s] = −3, 21 · 10−5 [mol H2O2/s] (A.10) which is the initial reaction rate of the reaction. 16 si de 1 a v 2 14 .0 9. 20 12 N TN U R is ik ov ur de rin g N um m er D at o H M S- av d. 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V  H2O  [mL] V  O2  [mL] t  [s] 1(1mL  yeast) 1 25 0.5 61.89 10mL  syringe 2 25 1 77.04 3 25 1.5 91.7 4 25 2 102.45 5 25 2.5 115.11 6 25 3 126.83 dV(O2)/dt 0.0388 mL/s dn(O2)/dt 1.73106E-­‐06 mol/s dn(H2O2)/dt -­‐3.46212E-­‐06 mol/s y  =  0.0388x  -­‐  1.9668   0   0.5   1   1.5   2   2.5   3   3.5   60   70   80   90   100   110   120   130   V( O 2)  [m L]   t  [  s  ]   D Graphs for estamation of initial reaction rates Test  nr. Measure  nr. V  H2O  [mL] V  O2  [mL] t  [s] 4  (4mL  yeast) 1 22 1 21.41 10  mL  syringe 2 22 2 26.04 3 22 3 31.2 4 22 4 36.54 dV(O2)/dt 0.1976 mL/s dn(O2)/dt 8.81592E-­‐06 mol/s dn(H2O2)/dt -­‐1.76318E-­‐05 mol/s y  =  0.1976x  -­‐  3.1909   0   0.5   1   1.5   2   2.5   3   3.5   4   4.5   20   22   24   26   28   30   32   34   36   38   V( O 2)  [m L]   t  [  s  ]   Test  nr. Measure  nr. V  H2O  [mL] V  O2  [mL] t  [s] 5  (5mL  yeast) 1 21 1 15.54 20mL  syringe 2 21 2 19.07 3 21 3 4 21 4 25.63 5 21 5 29.85 dV(O2)/dt 0.2837 mL/s dn(O2)/dt 1.26573E-­‐05 mol/s dn(H2O2)/dt -­‐2.53145E-­‐05 mol/s y  =  0.2837x  -­‐  3.3886   0   1   2   3   4   5   6   15   17   19   21   23   25   27   29   31   V( O 2)  [m L]   t  [  s  ]   Test  nr. Measure  nr. V  H2O  [mL] V  O2  [mL] t  [s] 6  (6mL  yeast) 1 20 1 14.16 20mL  syringe 2 20 2 17.29 3 20 3 19.66 4 20 4 22.6 dV(O2)/dt 0.3602 mL/s dn(O2)/dt 1.60703E-­‐05 mol/s dn(H2O2)/dt -­‐3.21406E-­‐05 mol/s y  =  0.3602x  -­‐  4.1381   0   0.5   1   1.5   2   2.5   3   3.5   4   4.5   14   16   18   20   22   24   V( O 2)  [m L]   t  [  s  ]   Measurements from Experiment 1: Dependence of the rate of the yeast consentration Measurement V02 (g) [mU Time [sj 1 no. 2 ç 3 3 — — 4 S 35 6 8 tO Measurement V02 (g) [mU Time [SIno. 1 i 3 3 2. Z(p0L2 4 5 5• (o6 7 1bO-7 8 cC -i-v-&’ 1 2 3 4 5 6 7 8 -OL 3((j4 S 9- S Measurements from Experiment 1: Dependence of the rate of the yeast consentration Measurement no. V02 (g) [mL] Time [s] I \fo ‘ 322 ci Measurement V02 (g) [mL]no. CD CD C CD Ci ) C C Ci ) CD 0 CD Ci ) CD CD Cl ) 0 CD I CD CD CD CD C 7 F i CD CD CD