Περιβαλλοντική Βιοτεχνολογία- Environmental Biotechnology

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1 Περιβαλλοντική Βιοτεχνολογία- Environmental Biotechnology Ενότητα 2: Stoichiometry and Bacterial Energetics Κορνάρος Μιχαήλ Πολυτεχνική Σχολή Τμήμα Χημικών Μηχανικών

2 Introduction Design of systems for biological treatment Mass balances For a given quantity of waste chemicals & end-products Energy (e.g. Ο 2 ) Nutrient (e.g. Ν, P) Environmental needs Excess microorganisms (costly disposal problem) CH 4 (source of energy) (Ca(OH) 2 or H 2 SO 4 for ph control) 2

3 Introduction Stoichiometry Energetics Relationships among reactants & products Balancing for elements, electrons & charge The microbial reactions complicate the stoichiometry. Microbial reactions often involve oxidation and reduction of more than one species. catalysts for the reaction The microorganisms have two roles products of the reaction The microorganisms carry out most chemical reactions in order to capture some of the energy released for cell synthesis and for maintaining cellular activity 3

4 An example stoichiometric equation Porges et al (956) for a casein-containing wastewater carbon source electron donor C 8 H 2 O 3 N 2 3O 2 casein C 5 H 7 O 2 N NH 3 3 CO 2 H 2 O microorganisms Μ. Β (kg/d) Could we predict the stoichiometry of such a reaction? Empirical formula for cells How the electron-donor substrate is partitioned between energy generation and synthesis The proportion of the electron-donor substrate that is used to synthesize new biomass to the energy gained from catabolism and the energy needed for anabolism. 4

5 Empirical formula for microbial cells C 5 H 7 O 2 N empirical formula Depends on : The characteristics of the microorganisms involved The substrates being used for energy The availability of other nutrients required for microbial growth (e.g. in a nitrogen-deficient environment more fatty material or carbohydrates are produced) C n H a O b N c (2n0.5a-.5c-b)/2 O 2 n CO 2 c NH 3 (a-3c)/2 H 2 O COD / weight = (2n 0.5a-.5c-b) 6 2n a 6b 4c n = %C/2T, a = %H/T, b = %O/6T and c = %N/4T T = %C/2 %H %O/6 %N/4 5

6 Substrate partitioning and cellular yield Electron donor f e 0 f s 0 Energy production Cell synthesis Active bacterial cells Reaction end products f e0 f s0 = Decay Cell residual True yield : Y = f s0 (M c g cells/mol cells) / [ (n e e - eq/mol cells)(8 g COD/e - eq donor) ] f s0 = g cell produced per g COD Y = g cell produced per g COD Μ c = empirical formula weight of cells n e = number of electron eq. in a mole of cells 6

7 Substrate partitioning and cellular yield The growth rate of microbial cells : dx dt a = Y ds dt bx a The net yield : Y n = dx a / dt ds / dt = Y X a b ds / dt f e f s = f e > f e 0 f s < f s 0 Y n because X a or ( ds/dt) Maintentance energy: ds / dt b Yn = 0, = = X Y a m 7

8 Energy reactions Microorganisms obtain their energy for growth and maintenance from oxidation reduction reactions. Oxidation-reduction reactions always involve an electron donor and an electron acceptor. Aerobic conditions: electron donor organic matter or e.g. ΝΗ 3 electron acceptor oxygen Anaerobic conditions: electron donor organic matter electron acceptor NO 3-, SO 4 2-, CO 2 Fermentation: electron donor organic matter electron acceptor organic matter 8

9 Energy reactions Example: Glucose is the electron donor Aerobic oxidation C H O 6O 6CO 6H O Denitrification 5C H O 24NO 24H 30CO 42H O 2N Sulfate reduction C H O 6SO 9H 2CO 2H O 3H S 3HS Methanogenesis C H O 3CO 3CH Ethanol Fermentation C H O 2CO 3CH CH OH Free Energy (kj/mol glucose) -2,880-2,

10 Energy reactions Energy reactions using half-reactions. The oxidation half-reaction for glucose: C6H2O6 H 2O CO2 H e The reduction half-reaction for nitrate: 5 The overall balanced reaction: 6 NO H e N2 H 2O C 6 H 2 O 6 5 NO H 4 CO H O 2 0 N 2 5C 6 H 2 O 6 24ΝΟ 3-24Η 30 CO 2 42H 2 O 2N 2 0

11 Energy reactions Steps of half-reactions Step : Write the oxidized form of the element of interest on the left and the reduced form on the right. CO 2 CH3CHNH2COOH Step 2: Add other species that are formed or consumed in the reaction. In oxidationreduction reactions, water is almost always a reactant or a product; here it will be included as a reactant in order to balance the oxygen present in the organic compound. As a reduction half-reaction, electrons must also appear on the left side of the equation. Since the nitrogen is present in the reduced form as an amino group in alanine, N must appear on the left side of the equation in the reduced form, either as NH 3 or NH 4. In this case, we arbitrarily select NH 3 for illustration. CO Organic Compounds H O NH e CH CHNH COOH

12 Energy reactions Steps of half-reactions Step 3: Balance the reaction for the element that is reduced and for all elements except oxygen and hydrogen. In this case, carbon and nitrogen must be balanced. 3CO H O NH e CH CHNH COOH Step 4: Balance the oxygen through addition or subtraction of water. Elemental oxygen is not to be used here, as oxygen must not have its oxidation state changed. 3CO NH e CH CHNH COOH 4H O Step 5: Balance hydrogen by introducing H. 3CO NH 2H e CH CHNH COOH 4H O

13 Energy reactions Steps of half-reactions Step 6: Balance the charge on the reaction by adding sufficient e- to the left side of the equation. 3CO NH 2H 2e CH CHNH COOH 4H O The coefficient on electrons should equal the number of electron equivalents in the reduced compound, or 2e - eq in alanine, which it does. Step 7: Convert the equation to the electron-equivalent form by dividing by the coefficient on e -. CO2 NH3 H e CH3CHNH2COOH H2O

14 Energy reactions Steps of half-reactions Inorganic Compounds Step : Step 2: CrO Cr CrO H e Cr H O Step 3: 2 3 CrO4 H e Cr H2 O ( Cr balanced in Step 2; thus no change here) Step 4: Step 5: CrO H e Cr 4H O CrO 8H e Cr 4H O Step 6: Step 7: CrO 8H 3e Cr 4H O CrO4 H e Cr H2O

15 Energy reactions Steps of half-reactions The conditions in which a reaction takes place determine the form of the products (e.g. ΝΗ 4 instead of ΝΗ 3 at ph neutral). Anaerobic methanogenic fermentation of alanine: CO2 H e CH4 H2O CH3CHNH2COOH H2O CO2 NH4 HCO3 H e Sum : CH CHNH COOH H O CH CO NH HCO

16 Energy reactions Steps of half-reactions For each half-reaction that involves organic nitrogen, HCO 3- enters into the reaction. 2 CO2 HCO3 H e CH3COCOO H2O Different half-reaction forms sometimes used are illustrated: CO2 NH3 H e CH3CHNH2COOH H2O CO2 NH4 HCO3 H e CH3CHNH2COOH H2O CO2 NH4 HCO3 H e CH3CHNH2COO H2O CO2 NH4 HCO3 H2 CH3CHNH2COOH H2O NH4 HCO3 H e CH3CHNH2COO H2O

17 Overall reactions for biological growth Bacterial growth involves two basic reactions : One for energy production and One for cellular synthesis The electron donor provides electrons to the electron acceptor for energy production. Combination of half-reaction for synthesis (R c ) and the acceptor half-reaction (R a ). Nitrogen source (Ν) : ΝΗ 4, ΝΟ 2-, ΝΟ 3-, Ν 2 Cell mass: C 5 H 7 O 2 N Electron acceptor : Ο 2, ΝΟ 3-, Fe 3, SO 4 2-, CO 2 7

18 Overall reactions for biological growth Overall energy reaction : R e = R a - R d Overall synthesis reaction : R s = R c - R d Where R d the donor half-reaction ( - because the donor is oxidized). Overall reaction : R = f e R e f s R s = f e (R a R d ) f s (R c R d ) But also applies : f s f e = R d ( f s f e ) = R d R = f e R a f s R c R d (the net consumption of reactants and production of products when the microorganisms consume one electron equivalent of electron donor) 8

19 Overall reactions for biological growth Synthesis reaction 9 Rc : CO2 NH4 HCO3 H e C5H 7O2N H2O Rd : C6H5COO H2O CO2 HCO3 H e Rs : C6H5COO NH4 HCO3 C5H 7O2N H2O Overall reaction R = f e R e f s R s f R : 0.02C H COO 0.2NO 0.2H 0.2CO 0.06N 0.02HCO 0.H O e e f R : 0.033C H COO 0.02NH HCO 0.02C H O N H O s s R :0.0333C H COO 0.2NO 0.02NH 0.2H C5H 7O2N 0.06N2 0.2CO HCO H 2O 9

20 Overall reactions for biological growth Example: Electron donor : benzoate (C 6 H 5 COO - ) Electron acceptor : ΝΟ 3 - Nitrogen source : ΝΟ 3 - f s = 0.55, f e = 0.45 Overall reaction f R : 0.09NO 0.54H 0.45e 0.045N 0.27H O e a f R : 0.096NO CO H 0.55e 0.096C H O N 0.26H O s c R : C H COO H O 0.2CO HCO H e d R :0.0333C H COO 0.096NO 0.096H C H O N 0.045N 0.08CO HCO H O We see that 0.09 mole nitrate is converted to nitrogen gas, and mole is converted into the organic nitrogen of the cells. 20

21 Bacterial growth of chemolithotrophic microorganisms Example (Nitrification stoichiometry): Electron donor: ΝΗ 4 Electron acceptor : Ο 2 Nitrogen source : ΝH 4 Carbon source : CO 2 f s = 0.0 f e = 0.90 Waste amount (Q) : 000 m 3 /d Concentration ΝΗ 4 -Ν : 22 mg/l For each 0.3(4)=.82 g ammonium-nitrogen, 0.225(32)=7.2 g oxygen is consumed. Also, 0.005(3)=0.565 g cells and 0.25(4)=.75 g NO 3- -N are produced. The amount of ammonium-nitrogen treated= (22 mg/l)(000 m 3 )(03 liters/m 3 )(kg/0 6 mg)=22 kg. Thus, Oxygen consumption=22 kg(7.2 g/.82 g)=87 kg Cell dry weight produced=22 kg(0.565 g/.82 g)=6.83kg Effluent NO 3- -N conc.=22 mg/l(.75 g/.82 g)=2 mg/l fr : 0.225O 0.9H 0.9e 0.45HO e a 2 2 f R : 0.02CO 0.005NH 0.005HCO 0.H 0.e 0.005C H O N 0.045H O s c R : 0.25NH 0.375H O 0.25NO.25H e d R :0.3NH 0.225O 0.02CO 0.005HCO C H O N 0.25NO 0.25H 0.2HO

22 Bacterial growth of methanogenic Example (methanogenesis stoichiometry): Electron donor : C 8 Η 7 Ο 3 Ν Electron acceptor : CO 2 Nitrogen source : ΝH 4 Carbon source : C 8 Η 7 Ο 3 Ν f s = 0.08 f e = 0.92 Waste amount (Q) : 50 m 3 /d Concentration C 8 Η 7 Ο 3 Ν : 23,000 mg/l Process efficient : 95% Conditions: Τ = 35 o C, P = atm microorganisms The formula weight of C 8 H N is 75, and the equivalent weight is 0.025(75), or g. Methane fermentation of one equivalent of organic matter produces 0.5 mol methane and mol carbon dioxide. Thus, Methane produced=[(273 35)/273][ m 3 gas/mol] [3,280,000g/d][0.5 mol/4.375 g]=2,80 m 3 /d Percent methane=00[0.5/( )]=72 percent 7 7 Rd : NH4 HCO3 CO2 H e C8H7O3 N H2O f R : 0.5CO 0.92H 0.92e 0.5CH 0.23H O e a f R : 0.06CO 0.004NH 0.004HCO 0.08H 0.08e 0.004C H O N 0.036H O s c R : 0.025C H O N 0.35H O 0.025NH 0.025HCO 0.75CO H e d R:0.025CH 8 7ON HO C5H 7O2N 0.5CH CO2 0.02NH4 0.02HCO3 22

23 Simple fermentation reactions Electron donor: organic compound Electron acceptor: organic compound Ra : CO2 H e CH3CH2OH H2O Rd : C6H2O6 H2O CO2 H e C 6 H 2 O 6 2 CO 2 2 CH 3 CH 2 OH R : C H O CH CH OH CO e R : 0.3CO 0.78H 0.78e 0.065CH CH OH 0.95H O a R : 0.044CO 0.0NH 0.0HCO 0.22H 0.22e 0.0C H O N 0.099H O c R : 0.047C H O 0.25H O 0.25CO H e d R :0.047C6H2O6 0.0NH4 0.0HCO3 0.0C 5H7O2N 0.065CH3CH2OH 0.076CO H2O f s = 0.22 f e =

24 Mixed fermentation reactions Electron donor: one or more organic compounds (e.g. municipal wastewaters) Electron acceptor: one or more organic compounds (e.g. E.coli ) where R = a ai ai i= n equivai eai = and ai n e = i= equiv j= n e R aj Here, e ai is the fraction of the n reduced end products that is represented by product ai. Equiv ai represents the equivalents of ai produced. where R = d di di i= n equivdi edi = and di n e = i= equiv j= n e R dj The sum of the fractions of all reduced end products equals. 24

25 Mixed fermentation reactions CITRATE FERMENTATION TO TWO REDUCED PRODUCT Bacteroides sp. converts mol citrate into mol formate, 2 mol acetate, and mol bicarbonate. Write the overall balanced energy reaction (R e ) for this fermentation. The reduced end products are formate and acetate. Bicarbonate, like carbon dioxide, is an oxidized end product and not considered in constructing the electron balance. The first step is to determine the number of equivalents (equiv ai ) formed for each reduced product. e formate =2/(2 6) or 0., and e acetate = 6/(26) or The sum of e formate plus e acetate equals.0. 8 CO 2 8 HCO 3 H e - = 8 CH COO H O 2 2 HCO 3 H e - = 2 HCOO 2 H O 2 25

26 Mixed fermentation reactions 0. R : HCO 0.H 0.e HCOO H O formate R : 0.CO 0.HCO 0.889H 0.889e 0.CH COO 0.333H O acetate R :0.CO 0.66HCO H e HCOO 0.CH COO 0.388H O a The overall energy reaction (R e ) is then found using R a - R d, where R d represents the half-reaction for citrate. Combining these produces the following for R e : ( COO ) CH COH ( COO ) CH COO 0.056H O HCOO 0.CH COO 0.056CO If this equation is normalized by dividing through by , the moles of citrate in one equivalent, the following mole-normalized equation is obtained: ( COO ) CH COH ( COO ) CH COO H O HCOO 2CH COO CO

27 Energetics and bacterial growth Microorganisms carry out oxidation-reduction reactions in order to obtain energy for growth and cell maintenance. The amount of energy released per electron equivalent of an electron donor oxidized varies considerably from reaction to reaction. It is not surprising then that the amount of growth that results from an equivalent of donor oxidized varies considerably as well. Cell maintenance has energy requirements for activities such as cell movement and repair of cellular proteins (that decay because of normal resource recycling or through interactions with toxic compounds). When cells grow rapidly in the presence of non limiting concentrations of all factors required for growth, cells make the maximum investment of energy for synthesis. However, when an essential factor, such as the electron-donor substrate, is limited in concentration, then a larger portion of the energy obtained from substrate oxidation must be used for cell maintenance. Y n dx a / dt X a = = Y b ds / dt ds / dt 27

28 Energetics and bacterial growth The electron equivalents are easily related to measurements of widespread utility in enviromental engineering practice, such as COD : electron equivalent = 8 g Ο 2 - O2 H e = H 2O 4 2 COMPUTING COD' A wastewater contains 2.6 g/l of ethanol. Estimate the e - eq/l and the COD' (g/l) for this wastewater. There are 2 e - eq/mol ethanol. Since mol of ethanol weighs 46 g, the equivalent weight is 46/2 or 3.83 g/e - eq. The ethanol concentration in the wastewater is thus 2.6/3.83 or 3.29 e - eq/l. The COD' thus becomes (8 g OD/e - eq)(3.29 e eq/l) = 26.3 g/l. 6 CO 2 H e - = 2 CH CH OH H O 2 28

29 Free energy of the energy reaction Balanced half-reaction for 2-chlorobenzoate formation: HCO CO Cl H e C H ClCOO H O aq g aq ( aq.0 ) l 3( ) 2( ) ( ) ( ) The free energies of formation for each species are (in kj/e - eq): ( ), ( ), (-3.35), (-39.87), 0, (-237.9), (-237.8) The half-reaction free energy is calculated as the sum of the product free energies minus the sum of the reactant free energies, or ΔG O = kj/e - eq. 29

30 Free energy of the energy reaction One can adjust the reaction free energy for nonstandard concentrations of reactants and products: υ A υ A... υ A υ A... υ A 2 2 m m m m n n which can be written in an even more general form: 0 n = υir Ai i= The value of υ ir is negative if constituent A i appears on the left side of the above equation and positive if it appears on the right side. The nonstandard free energy change for this reaction can be determined from: n 0 r r υir ln i i= G = G RT a Here, υ ir represents the stoichiometric coefficient for constituent A i in reaction r, and a i represents the activity of constituent A i. T is absolute temperature (K). 30

31 Free energy of the energy reaction CH3CH2OH O2 = CO2 H2O For this energy reaction, ΔG r0 equals ΔG 0 r because H is not a component of the equation; n = 4; and υ ik equals -/2, -/4, /6, and /4 for ethanol, oxygen, carbon dioxide, and water, respectively. G = G RT r 0 r ln /6 /4 [ CO2] [ H2O] [ CH CH OH ] [ O ] /2 / We assume that the concentration of ethanol in an aqueous solution is M, the oxygen partial pressure is that in the normal atmosphere at sea level (0.2 atm), the carbon dioxide concentration is that in normal air ( atm), and the temperature is 20 C. We also assume that the activities of the three constituents are equal to their molar concentration or partial pressure. (These values could be corrected if one knew the respective activity coefficients for each constituent. In this case the activity coefficient is likely to be close to.0.) With aqueous solutions in which water is the major solvent, the activity of H 2 0 is sufficiently close to. Then, [ ] [ ] [ ] [ ] /6 /4 G r = 09, (273 20)ln =, 000 J / e eq or kj / e eq /2 / 4 3

32 Yield coefficient and reaction energetics The energy carriers are "spent" to drive cell synthesis or cell maintenance. As with all reactions, a certain amount of thermodynamic free energy is lost with each transfer. 32

33 Yield coefficient and reaction energetics We must determine the energy change resulting from the conversion of the carbon source to pyruvate (as a representative intermediate) (ΔG p ). ΔG p = ΔG c 0 Where ΔG 0 c is the half-reaction free energy of carbon source, which for heterotrophic bacteria is the electron donor. e.g. For pyruvate: ΔG 0 c = 27.4 kj/e - eq. In photosynthesis we can determined the energy involved ΔG 0 c = kj/e - eq (for the half-reaction Η 2 Ο-Ο 2 ), thus ΔG p = (-78.72) = 3.8 kj/e - eq. When pyruvate is converted to cellular carbon, the energy required is : ΔG pc = 3.33 kj/g cells or 8.8 kj/e - eq 33

34 Yield coefficient and reaction energetics Energy is always lost in the electron transfers. The energy requirement for cell synthesis becomes: ΔG s = ΔG p /ε n ΔG pc /ε where ε : term of energy-transfer efficiency ( ), typical value = 0.6 For some electron donors, such as glucose n = - (energy is obtained by its conversion to pyruvate or ΔG p <0 ). In other cases, such as with acetate n = (energy is required in its conversion to pyruvate or ΔG p >0 ). Energy required in A equivalent electron donor for the synthesis of equivalent cells. Energy balance: Α ε ΔG r ΔG S = 0 A = (ΔG p /ε n ΔG pc /ε) / ε ΔG r where ΔG r is the free energy released per equivalent of donor oxidized for energy generation. 34

35 Yield coefficient and reaction energetics Since part of the donor consumed is used for energy (A equivalents in this case) and the other part for synthesis (.0 equivalents in this case), the total donor used is A. f 0 s = 0 0 fe = - fs A = A A 35

36 Yield coefficient and reaction energetics EFFECTS OF ε ON HETEROTROPHIC YIELD Compare estimates for f s0 and Y for aerobic oxidation of acetate, assuming ε=0.4, 0.6, and 0.7, that ph=7, and that all other reactants and products are at unit activity. Ammonium is available for synthesis. G = = 7.69 kj / e eq. p Since this is an aerobic reaction G = kj / e eq. 0' a G = G G = = 06.2 kj / e eq. 0' 0' r a d Since ΔG p is positive, n =. Also, since ammonium is available for cell synthesis, ΔG pc equals 8.8 kj/e - eq. Hence, A ε ε = 06.2ε 36

37 References The images where their origin is not mentioned are derived from the book: Environmental Biotechnology : Principles and Applications, Bruce E. Rittmann and Perry L. McCarty, McGraw-Hill Series in Water Resources and Environmental Engineering 37

38 Τέλος Ενότητας

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