Krishna

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Krishna

  1. 1. Law s of therm Submit Search Powered by JRankBiology ReferenceBiology » Re-Se » Secondary Metabolites in Plants - Biology EncyclopediaSecondary Metabolites in PlantsAds by GoogleLeica Super-Resolution - The Greatest Confocal Resolution With the New Leica TCS SP8STED. - www.Leica-Microsystems.com/SP8IGS N2 & O2 Plants - Worldwide supply of Nitrogen and Oxygen Generators - www.igs-italia.comCyclodextrins - 1st time in India DMF available - www.cyclodextrin.com
  2. 2. Photo by: pashaboAds by GoogleMedicinal Chemistry JournalPublish Your Research Article InInternational Journal:IOSR JOURNALSiosrjournals.orgEnzyme ManufacturerFood processingNow in India!www.KenEncoreGroup.comSecondary metabolites are chemicals produced by plants for which no role has yet beenfound in growth, photosynthesis, reproduction, or other "primary" functions. Thesechemicals are extremely diverse; many thousands have been identified in several majorclasses. Each plant family, genus, and species produces a characteristic mix of thesechemicals, and they can sometimes be used as taxonomic characters in classifyingplants. Humans use some of these compounds as medicines, flavorings, or recreationaldrugs.
  3. 3. Secondary metabolites can be classified on the basis of chemical structure (for example,having rings, containing a sugar), composition (containing nitrogen or not), theirsolubility in various solvents, or the pathway by which they are synthesized (e.g.,phenylpropanoid, which produces tannins). A simple classification includes three maingroups: the terpenes (made from mevalonic acid, composed almost entirely of carbonand hydrogen), phenolics (made from simple sugars, containing benzene rings,hydrogen, and oxygen), and nitrogen-containing compounds (extremely diverse, mayalso contain sulfur).The apparent lack of primary function in the plant, combined with the observation thatmany secondary metabolites have specific negative impacts on other organisms such asherbivores and pathogens , leads to the hypothesis that they have evolved because oftheir protective value. Many secondary metabolites are toxic or repellant to herbivoresand microbes and help defend plants producing them. Production increases when aplant is attacked by herbivores or pathogens. Some compounds are released into the airwhen plants are attacked by insects; these compounds attract parasites and predatorsthat kill the herbivores. Recent research is identifying more and more primary roles forthese chemicals in plants as signals, antioxidants , and other functions, so "secondary"may not be an accurate description in the future.Consuming some secondary metabolites can have severe consequences. Alkaloids canblock ion channels, inhibit enzymes , or interfere with neurotransmission, producinghallucinations , loss of coordination, convulsions, vomiting, and death. Somephenolics interfere with digestion, slow growth, block enzyme activity and cell division,or just taste awful.Most herbivores and plant pathogens possess mechanisms that ameliorate the impactsof plant metabolites, leading to evolutionary associations between particular groups ofpests and plants. Some herbivores (for example, the monarch butterfly) can store(sequester) plant toxins and gain protection against their enemies. Secondarymetabolites may also inhibit the growth of competitor plants (allelopathy). Pigments(such as terpenoid carotenes, phenolics, and flavonoids) color flowers and, together withterpene and phenolic odors, attract pollinators.
  4. 4. Secondary chemicals are important in plant use by humans. Most pharmaceuticals arebased on plant chemical structures, and secondary metabolites are widely used forrecreation and stimulation (the alkaloids nicotine and cocaine; the terpene cannabinol).The study of such plant use is called ethnopharmacology. Psychoactive plant chemicalsare central to some religions, and flavors of secondary compounds shape our foodpreferences. The characteristic flavors and aroma of cabbage and relatives are caused byClass Example Example Some Effects and Uses Compounds SourcesNITROGEN-CONTAININGAlkaloids nicotine cocaine tobacco coca plant interfere with theobromine chocolate (cocao) neurotransmission, block
  5. 5. Class Example Example Some Effects and Uses Compounds Sources enzyme actionNITROGEN-ANDSULFUR-CONTAININGGlucosinolates sinigrin cabbage, relativesTERPENOIDSMonoterpenes menthol linalool mint and relatives, interfere with many plants neurotransmission, block ion transport, anestheticSesquiterpenes parthenolid Parthenium and contact dermatitis relatives ( Asteraceae )Diterpenes gossypol cotton block phosphorylation; toxicTriterpenes, cardiac digitogenin Digitalis (foxglove) stimulate heart muscle, alterglycosides ion transportTetraterpenoids carotene many plants antioxidant; orange coloringTerpene polymers rubber Hevea (rubber) gum up insects; airplane tires trees, dandelionSterols spinasterol spinach interfere with animal hormone actionPHENOLICS
  6. 6. Class Example Example Some Effects and Uses Compounds SourcesPhenolic acids caffeic, all plants cause oxidative damage, chlorogenic browning in fruits and wineCoumarins umbelliferone carrots, parsnip cross-link DNA, block cell divisionLignans podophyllin mayapple poison cathartic, vomiting, allergic urushiol ivy dermatitisFlavonoids anthocyanin, almost all plants flower, leaf color; inhibit catechin enzymes, anti- and pro- oxidants, estrogenicTannins gallotannin, oak, hemlock trees, bind to proteins, enzymes, condensed birdsfoot trefoil, block digestion, antioxidants tannin legumesLignin lignin all land plants structure, toughness, fibernitrogen-and sulfur-containing chemicals, glucosinolates, which protect these plantsfrom many enemies. The astringency of wine and chocolate derives from tannins. Theuse of spices and other seasonings developed from their combined uses as preservatives(since they are antibiotic) and flavorings.SEE ALSO Flowers ; Herbivory and Plant Defenses ; Metabolism, Cellular ; PoisonsJack SchultzBibliography
  7. 7. Agosta, William. Bombardier Beetles and Fever Trees: A Close-up Look at ChemicalWarfare and Signals in Animals and Plants. Reading, MA: Addison-Wesley, 1996.Bidlack, Wayne R. Phytochemicals as Bioactive Agents. Lancaster, PA: TechnomicPublishers, 2000.Karban, Richard, and Ian T. Baldwin. Induced Responses to Herbivory. Chicago:University of Chicago Press, 1997.Rosenthal, Gerald A., and May R. Berenbaum. Herbivores, Their Interactions withSecondary Plant Metabolites. San Diego, CA: Academic Press, 1991.Ads by GoogleChemical Crystallography - Inexpensive benchtop XRD system to determine small moleculestructure - www.rigaku.comUser Contributions: 1aarcee varteMar 4, 2008 @ 11:23 pmplease explain how 2 extract secondary metabolites which is not secreted out from plantsand remain inside the plants onlyComment about this article, ask questions, or add new information about thistopic:Name:
  8. 8. E-mail: Show my email publiclyType the code shown:Public Comment: (50-4000 characters) SendSecondary Metabolites in Plants - Biology Encyclopedia forum« Science WriterSeed Germination and Dormancy » Copyright © 2013 Advameg, Inc.Read more: http://www.biologyreference.com/Re-Se/Secondary-Metabolites-in-Plants.html#ixzz2IdgaAPMfCelluloseFrom Wikipedia, the free encyclopedia
  9. 9. Jump to: navigation, search Cellulose[1] Identifiers CAS number 9004-34-6 UNII SMD1X3XO9M EC-number 232-674-9 ChEMBL CHEMBL1201676 Properties Molecular formula (C6H10O5)n Appearance white powder Density 1.5 g/cm3 Melting point decomp. Solubility in water none
  10. 10. Hazards EU Index not listed NFPA 704 1 1 0 Related compounds Related compounds Starch (verify) (what is: / ?) Except where noted otherwise, data are given for materials in their standard state (at 25 °C, 100 kPa) Infobox referencesCellulose is an organic compound with the formula (C6H10O5)n, a polysaccharide consisting of alinear chain of several hundred to over ten thousand β(1→4) linked D-glucose units.[2][3]Cellulose is the structural component of the primary cell wall of green plants, many forms ofalgae and the oomycetes. Some species of bacteria secrete it to form biofilms. Cellulose is themost common organic compound on Earth. About 33% of all plant matter is cellulose (thecellulose content of cotton fiber is 90%, that of wood is 40–50% and that of dried hemp isapproximately 75%).[4][5][6]For industrial use, cellulose today is mainly obtained from wood pulp and cotton. Cellulose ismainly used to produce paperboard and paper; to a smaller extent it is converted into a widevariety of derivative products such as cellophane and rayon. Converting cellulose from energycrops into biofuels such as cellulosic ethanol is under investigation as an alternative fuel source.Some animals, particularly ruminants and termites, can digest cellulose with the help ofsymbiotic micro-organisms that live in their guts. Humans can digest cellulose to some
  11. 11. extent,[7][8] however it mainly acts as a hydrophilic bulking agent for feces and is often referredto as "dietary fiber".Contents 1 History 2 Products o 2.1 Cellulose source and energy crops 3 Structure and properties 4 Assaying a cellulose-containing material 5 Biosynthesis 6 Breakdown (cellulolysis) 7 Hemicellulose 8 Derivatives 9 References 10 External links[edit] HistoryCellulose was discovered in 1838 by the French chemist Anselme Payen, who isolated it fromplant matter and determined its chemical formula.[2][9][10] Cellulose was used to produce the firstsuccessful thermoplastic polymer, celluloid, by Hyatt Manufacturing Company in 1870.Hermann Staudinger determined the polymer structure of cellulose in 1920. The compound wasfirst chemically synthesized (without the use of any biologically derived enzymes) in 1992, byKobayashi and Shoda.[11]Cellulose in a plant cell.[edit] Products
  12. 12. Cotton fibres represent the purest natural form of cellulose, containing more than 90% of thiscarbohydrate.See also: dissolving pulp and pulp (paper)The kraft process is used to separate cellulose from lignin, another major component of plantmatter. Cellulose is the major constituent of paper, paperboard, and card stock and of textilesmade from cotton, linen, and other plant fibers.Cellulose can be converted into cellophane, a thin transparent film, and into rayon, an importantfiber that has been used for textiles since the beginning of the 20th century. Both cellophane andrayon are known as "regenerated cellulose fibers"; they are identical to cellulose in chemicalstructure and are usually made from dissolving pulp via viscose. A more recent andenvironmentally friendly method to produce a form of rayon is the Lyocell process. Cellulose isthe raw material in the manufacture of nitrocellulose (cellulose nitrate) which is used insmokeless gunpowder and as the base material for celluloid used for photographic and moviefilms until the mid 1930s.Cellulose is used to make water-soluble adhesives and binders such as methyl cellulose andcarboxymethyl cellulose which are used in wallpaper paste. Microcrystalline cellulose (E460i)and powdered cellulose (E460ii) are used as inactive fillers in tablets[12] and as thickeners andstabilizers in processed foods. Cellulose powder is for example used in Krafts Parmesan cheeseto prevent caking inside the tube.Cellulose is used in the laboratory as the stationary phase for thin layer chromatography.Cellulose fibers are also used in liquid filtration, sometimes in combination with diatomaceousearth or other filtration media, to create a filter bed of inert material. Cellulose is further used tomake hydrophilic and highly absorbent sponges.Cellulose insulation made from recycled paper is becoming popular as an environmentallypreferable material for building insulation. It can be treated with boric acid as a fire retardant.Cellulose consists of crystalline and amorphous regions. By treating it with strong acid, theamorphous regions can be broken up, thereby producing nanocrystalline cellulose, a novelmaterial with many desirable properties.[13] Recently, nanocrystalline cellulose was used as the
  13. 13. filler phase in bio-based polymer matrices to produce nanocomposites with superior thermal andmechanical properties. [14][edit] Cellulose source and energy cropsMain article: Energy cropThe major combustible component of non-food energy crops is cellulose, with lignin second.Non-food energy crops are more efficient than edible energy crops (which have a large starchcomponent), but still compete with food crops for agricultural land and water resources.[15]Typical non-food energy crops include industrial hemp (though outlawed in some countries),switchgrass, Miscanthus, Salix (willow), and Populus (poplar) species.Some bacteria can convert cellulose into ethanol which can then be used as a fuel; see cellulosicethanol.A strand of cellulose (conformation Iα), showing the hydrogen bonds (dashed) within and betweencellulose molecules.[edit] Structure and propertiesCellulose has no taste, is odorless, is hydrophilic with the contact angle of 20–30,[16] is insolublein water and most organic solvents, is chiral and is biodegradable. It can be broken downchemically into its glucose units by treating it with concentrated acids at high temperature.Cellulose is derived from D-glucose units, which condense through β(1→4)-glycosidic bonds.This linkage motif contrasts with that for α(1→4)-glycosidic bonds present in starch, glycogen,and other carbohydrates. Cellulose is a straight chain polymer: unlike starch, no coiling orbranching occurs, and the molecule adopts an extended and rather stiff rod-like conformation,aided by the equatorial conformation of the glucose residues. The multiple hydroxyl groups onthe glucose from one chain form hydrogen bonds with oxygen atoms on the same or on aneighbor chain, holding the chains firmly together side-by-side and forming microfibrils with
  14. 14. high tensile strength. This strength is important in cell walls, where the microfibrils are meshedinto a carbohydrate matrix, conferring rigidity to plant cells.A triple strand of cellulose showing the hydrogen bonds (cyan lines) between glucose strandsCompared to starch, cellulose is also much more crystalline. Whereas starch undergoes acrystalline to amorphous transition when heated beyond 60–70 °C in water (as in cooking),cellulose requires a temperature of 320 °C and pressure of 25 MPa to become amorphous inwater.[17]Several different crystalline structures of cellulose are known, corresponding to the location ofhydrogen bonds between and within strands. Natural cellulose is cellulose I, with structures Iαand Iβ. Cellulose produced by bacteria and algae is enriched in Iα while cellulose of higher plantsconsists mainly of Iβ. Cellulose in regenerated cellulose fibers is cellulose II. The conversion ofcellulose I to cellulose II is irreversible, suggesting that cellulose I is metastable and cellulose IIis stable. With various chemical treatments it is possible to produce the structures cellulose IIIand cellulose IV.[18]Many properties of cellulose depend on its chain length or degree of polymerization, the numberof glucose units that make up one polymer molecule. Cellulose from wood pulp has typical chainlengths between 300 and 1700 units; cotton and other plant fibers as well as bacterial cellulosehave chain lengths ranging from 800 to 10,000 units.[11] Molecules with very small chain lengthresulting from the breakdown of cellulose are known as cellodextrins; in contrast to long-chaincellulose, cellodextrins are typically soluble in water and organic solvents.Plant-derived cellulose is usually found in a mixture with hemicellulose, lignin, pectin and othersubstances, while microbial cellulose is quite pure, has a much higher water content, and consistsof long chains.Cellulose is soluble in cupriethylenediamine (CED), cadmiumethylenediamine (Cadoxen), N-methylmorpholine N-oxide and lithium chloride / dimethylformamide.[19] This is used in theproduction of regenerated celluloses (such as viscose and cellophane) from dissolving pulp.[edit] Assaying a cellulose-containing materialGiven a cellulose-containing material, the carbohydrate portion that does not dissolve in a 17.5%solution of sodium hydroxide at 20 °C is α cellulose, which is true cellulose. Acidification of theextract precipitates β cellulose. The portion that dissolves in base but does not precipitate withacid is γ cellulose.
  15. 15. Cellulose can be assayed using a method described by Updegraff in 1969, where the fiber isdissolved in acetic and nitric acid to remove lignin, hemicellulose, and xylosans. The resultingcellulose is allowed to react with anthrone in sulfuric acid. The resulting coloured compound isassayed spectrophotometrically at a wavelength of approximately 635 nm.In addition, cellulose is represented by the difference between acid detergent fiber (ADF) andacid detergent lignin (ADL).[edit] BiosynthesisLocation and arrangement of cellulose microfibrils in the plant cell wallIn vascular plants cellulose is synthesized at the plasma membrane by rosette terminal complexes(RTCs). The RTCs are hexameric protein structures, approximately 25 nm in diameter, thatcontain the cellulose synthase enzymes that synthesise the individual cellulose chains.[20] EachRTC floats in the cells plasma membrane and "spins" a microfibril into the cell wall.RTCs contain at least three different cellulose synthases, encoded by CesA genes, in an unknownstoichiometry.[21] Separate sets of CesA genes are involved in primary and secondary cell wallbiosynthesis.Cellulose synthesis requires chain initiation and elongation, and the two processes are separate.CesA glucosyltransferase initiates cellulose polymerization using a steroid primer, sitosterol-beta-glucoside, and UDP-glucose.[22] Cellulose synthase utilizes UDP-D-glucose precursors toelongate the growing cellulose chain. A cellulase may function to cleave the primer from themature chain.Cellulose is also synthesised by animals, particularly in the tests of ascidians (where the cellulosewas historically termed "tunicine") although it is also a minor component of mammalianconnective tissue.[23][edit] Breakdown (cellulolysis)Cellulolysis is the process of breaking down cellulose into smaller polysaccharides calledcellodextrins or completely into glucose units; this is a hydrolysis reaction. Because cellulosemolecules bind strongly to each other, cellulolysis is relatively difficult compared to thebreakdown of other polysaccharides.[24] Processes do exist however for the breakdown of
  16. 16. cellulose such as the Lyocell process [25] which uses a combination of heated water and acetoneto break down the cellulose strands.Most mammals have only very limited ability to digest dietary fibres such as cellulose. Someruminants like cows and sheep contain certain symbiotic anaerobic bacteria (like Cellulomonas)in the flora of the rumen, and these bacteria produce enzymes called cellulases that help themicroorganism to break down cellulose; the breakdown products are then used by the bacteriafor proliferation. The bacterial mass is later digested by the ruminant in its digestive system(stomach and small intestine). Similarly, lower termites contain in their hindguts certainflagellate protozoa which produce such enzymes; higher termites contain bacteria for the job.Some termites may also produce cellulase of their own.[26] Fungi, which in nature are responsiblefor recycling of nutrients, are also able to break down cellulose.The enzymes utilized to cleave the glycosidic linkage in cellulose are glycoside hydrolasesincluding endo-acting cellulases and exo-acting glucosidases. Such enzymes are usually secretedas part of multienzyme complexes that may include dockerins and carbohydrate-bindingmodules.[27][edit] HemicelluloseMain article: HemicelluloseHemicellulose is a polysaccharide related to cellulose that comprises about 20% of the biomassof most plants. In contrast to cellulose, hemicellulose is derived from several sugars in additionto glucose, especially xylose but also including mannose, galactose, rhamnose, and arabinose.Hemicellulose consists of shorter chains – around 200 sugar units. Furthermore, hemicellulose isbranched, whereas cellulose is unbranched.[edit] DerivativesThe hydroxyl groups (-OH) of cellulose can be partially or fully reacted with various reagents toafford derivatives with useful properties like mainly cellulose esters and cellulose ethers (-OR).In principle, though not always in current industrial practice, cellulosic polymers are renewableresources.Ester derivatives include: Cellulose Reagent Example Reagent Group R esterOrganic Organic Acetic acid and acetic Cellulose acetate H or -(C=O)CH3esters acids anhydride Acetic acid and acetic Cellulose triacetate -(C=O)CH3 anhydride
  17. 17. Cellulose propionate Propanoic acid H or -(C=O)CH2CH3 Cellulose acetate Acetic acid and propanoic H or -(C=O)CH3 or - propionate acid (C=O)CH2CH3 Cellulose acetate H or -(C=O)CH3 or - Acetic acid and butyric acid butyrate (C=O)CH2CH2CH3Inorganic Inorganic Nitrocellulose Nitric acid or another H or -NO2esters acids (cellulose nitrate) powerful nitrating agent Sulfuric acid or another Cellulose sulfate H or -SO3H powerful sulfuring agentThe cellulose acetate and cellulose triacetate are film- and fiber-forming materials that find avariety of uses. The nitrocellulose was initially used as an explosive and was an early filmforming material. With camphor, nitrocellulose gives celluloid.Ether derivatives include: Water E Cellulose Group R = H Reagent Example Reagent solubilit Application numbe ethers or y r Cold Halogenoalkan Methylcellulo ChloromethanAlkyl -CH3 water E461 es se e soluble A commercial thermoplasti c used in Water coatings, Ethylcellulose Chloroethane -CH2CH3 insolubl E462 inks, binders, e and controlled- release drug tablets Ethyl methyl Chloromethan -CH3 or - E465 cellulose e and CH2CH3
  18. 18. chloroethane Cold/ho Gelling andHydroxyalk Hydroxyethyl Ethylene Epoxides -CH2CH2OH t water thickeningyl cellulose oxide soluble agent Hydroxypropy - Cold Propylene l cellulose CH2CH(OH)C water E463 oxide (HPC) H3 soluble Chloromethan Hydroxyethyl Cold Production of e and -CH3 or - methyl water cellulose ethylene CH2CH2OH cellulose soluble films oxide Viscosity Hydroxypropy Chloromethan modifier, -CH3 or - Cold l methyl e and gelling, CH2CH(OH)C water E464 cellulose propylene foaming and H3 soluble (HPMC) oxide binding agent Ethyl Chloroethane -CH2CH3 or— hydroxyethyl and ethylene E467 CH2CH2OH cellulose oxide Often used as its sodium Carboxymethy Cold/HoCarboxyalk Halogenated Chloroacetic salt, sodium l cellulose -CH2COOH t water E466yl carboxylic acids acid carboxymeth (CMC) soluble yl cellulose (NaCMC)The sodium carboxymethyl cellulose can be cross-linked to give the croscarmellose sodium(E468) for use as a disintegrant in pharmaceutical formulations.[edit] References 1. ^ Nishiyama, Yoshiharu; Langan, Paul; Chanzy, Henri (2002). "Crystal Structure and Hydrogen-Bonding System in Cellulose Iβ from Synchrotron X-ray and Neutron Fiber Diffraction". J. Am. Chem. Soc 124 (31): 9074–82. doi:10.1021/ja0257319. PMID 12149011.. ab 2. ^ Crawford, R. L. (1981). Lignin biodegradation and transformation. New York: John Wiley and Sons. ISBN 0-471-05743-6.
  19. 19. 3. ^ Updegraff DM (1969). "Semimicro determination of cellulose in biological materials". Analytical Biochemistry 32 (3): 420–424. doi:10.1016/S0003-2697(69)80009-6. PMID 5361396.4. ^ Cellulose. (2008). In Encyclopædia Britannica. Retrieved January 11, 2008, from Encyclopædia Britannica Online.5. ^ Chemical Composition of Wood6. ^ G. Buschle-Diller, C. Fanter, F. Loth (April 1999). "Structural changes in hemp fibers as a result of enzymatic hydrolysis with mixed enzyme systems". Textile Research Journal 69 (4): 244-251. http://www.globalhemp.com/1999/04/structural-changes-in-hemp-fibers-as-a-result-of-enzymatic- hydrolysis-with-mixed-enzyme-systems.html.7. ^ Slavin, JL; Brauer, PM; Marlett, JA (1981). "Neutral detergent fiber, hemicellulose and cellulose digestibility in human subjects.". The Journal of Nutrition 111 (2): 287–97. PMID 6257867.8. ^ Joshi, S; Agte, V (1995). "Digestibility of dietary fiber components in vegetarian men.". Plant foods for human nutrition (Dordrecht, Netherlands) 48 (1): 39–44. doi:10.1007/BF01089198. PMID 8719737.9. ^ A. Payen (1838) "Mémoire sur la composition du tissu propre des plantes et du ligneux" (Memoir on the composition of the tissue of plants and of woody [material]), Comptes rendus, vol. 7, pages 1052-1056. Payen added appendices to this paper on December 24, 1838 (see: Comptes rendus, vol. 8, page 169 (1839)) and on February 4, 1839 (see: Comptes rendus, vol. 9, page 149 (1839)). A committee of the French Academy of Sciences reviewed Payens findings in : Jean-Baptiste Dumas (1839) "Rapport sur un mémoire de M. Payen, relatif à la composition de la matière ligneuse" (Report on a memoir of Mr. Payen, regarding the composition of woody matter), Comptes rendus, vol. 8, pages 51-53. In this report, the word "cellulose" is coined and author points out the similarity between the empirical formula of cellulose and that of "dextrine" (starch). The above articles are reprinted in: Brongniart and Guillemin, eds., Annales des sciences naturelles ..., 2nd series, vol. 11 (Paris, France: Crochard et Cie., 1839), pages 21-31.10. ^ Young, Raymond (1986). Cellulose structure modification and hydrolysis. New York: Wiley. ISBN 0-471- 82761-4. ab11. ^ Klemm, Dieter; Brigitte Heublein, Hans-Peter Fink, Andreas Bohn (2005). "Cellulose: Fascinating Biopolymer and Sustainable Raw Material". ChemInform 36 (36). doi:10.1002/chin.200536238.12. ^ Weiner, Myra L.; Lois A. Kotkoskie (2000). Excipient Toxicity and Safety. New York ; Dekker. p. 210. ISBN 0-8247-8210-0.13. ^ Peng, B. L., Dhar, N., Liu, H. L. and Tam, K. C. (2011). "Chemistry and applications of nanocrystalline cellulose and its derivatives: A nanotechnology perspective.". The Canadian Journal of Chemical Engineering 89 (5): 1191–1206. http://www.arboranano.ca/pdfs/Chemistry%20and%20applications%20of%20nanocrystalline%20cellulos e%20and%20its%20derivatives%20A%20nanotechnology%20perspective-2011.pdf.14. ^ Lawrence Pranger and Rina Tannenbaum "Biobased nanocomposites prepared by in situ polymerization of furfuryl alcohol with cellulose whiskers or montmorillonite clay" Macromolecules 41 (2008) 8682. http://dx.doi.org/10.1021/ma802021315. ^ Holt-Gimenez, Eric 2007. Biofuels: Myths of the Agrofuels Transition. Backgrounder. Institute for Food and Development Policy, Oakland, CA. 13:216. ^ Charles A. Bishop, ed. (2007). Vacuum deposition onto webs, films, and foils, Volume 0, Issue 8155. p. 165. ISBN 0-8155-1535-9. http://books.google.com/books?id=vP9E3z7o6iIC&pg=PA165.17. ^ Deguchi, Shigeru; Tsujii, Kaoru; Horikoshi, Koki (2006). "Cooking cellulose in hot and compressed water". Chemical Communications (31): 3293. doi:10.1039/b605812d.18. ^ Structure and morphology of cellulose by Serge Pérez and William Mackie, CERMAV-CNRS, 2001. Chapter IV.19. ^ Stenius, Per (2000). "1". Forest Products Chemistry. Papermaking Science and Technology. 3. Finland: Fapet OY. p. 35. ISBN 952-5216-03-9.20. ^ Kimura, S; Laosinchai, W; Itoh, T; Cui, X; Linder, CR; Brown Jr, RM (1999). "Immunogold labeling of rosette terminal cellulose-synthesizing complexes in the vascular plant vigna angularis". The Plant cell 11 (11): 2075–86. doi:10.2307/3871010. JSTOR 3871010. PMC 144118. PMID 10559435. //www.ncbi.nlm.nih.gov/pmc/articles/PMC144118/.21. ^ Taylor, N. G. (2003). "Interactions among three distinct CesA proteins essential for cellulose synthesis". Proceedings of the National Academy of Sciences 100 (3): 1450. doi:10.1073/pnas.0337628100.
  20. 20. 22. ^ Peng, L; Kawagoe, Y; Hogan, P; Delmer, D (2002). "Sitosterol-beta-glucoside as primer for cellulose synthesis in plants". Science 295 (5552): 147–50. doi:10.1126/science.1064281. PMID 11778054. 23. ^ Endean, The Test of the Ascidian, Phallusia mammillata, Quarterly Journal of Microscopical Science, Vol. 102, part 1, pp. 107-117, 1961. 24. ^ David G. Barkalow, Roy L. Whistler, "Cellulose", in AccessScience, McGraw-Hill, doi:10.1036/1097- 8542.118200. Retrieved 11 January 2008. 25. ^ H.Lyocell, "Cellulose" Issue 41, pp 419 26. ^ Tokuda, G; Watanabe, H (22 June 2007). "Hidden cellulases in termites: revision of an old hypothesis". Biology Letters 3 (3): 336–339. doi:10.1098/rsbl.2007.0073. PMC 2464699. PMID 17374589. http://rsbl.royalsocietypublishing.org/content/3/3/336.long 27. ^ Brás, Natércia; N. M. F. S. A. Cerqueira, P. A. Fernandes, M. J. Ramos (2008). "Carbohydrate Binding Modules from family 11: Understanding the binding mode of polysaccharides". International Journal of Quantum Chemistry 108 (11): 2030–2040. doi:10.1002/qua.21755.[edit] External links Structure and morphology of cellulose by Serge Pérez and William Mackie, CERMAV-CNRS Cellulose, by Martin Chaplin, London South Bank University Clear description of a cellulose assay method at the Cotton Fiber Biosciences unit of the USDA. Cellulose films could provide flapping wings and cheap artificial muscles for robots - TechnologyReview.com Using cellulase enzymes in the bioethanol process A list of cellulolytic bacteria v t e Types of carbohydrates Aldose Furanose General Ketose Pyranose Anomer Geometry Cyclohexane conformation Mutarotation
  21. 21. Aldodiose Dioses o Glycolaldehyde Aldotriose o Glyceraldehyde Trioses Ketotriose o Dihydroxyacetone Aldotetroses o Erythrose Tetroses o Threose Ketotetrose o ErythruloseMonosaccharides Aldopentose o Arabinose o Lyxose o Ribose o Xylose Pentoses Deoxy sugar o Deoxyribose Ketopentose o Ribulose o Xylulose Aldohexose Hexoses o Allose o Altrose
  22. 22. o Galactose o Glucose o Gulose o Idose o Mannose o Talose Deoxy sugar o Fucose o Fuculose o Rhamnose Ketohexose o Fructose o Psicose o Sorbose o Tagatose Ketoheptose Heptoses o Mannoheptulose o Sedoheptulose Octose >7 Nonose o Neuraminic acid CellobioseMultiple Disaccharides Lactose Maltose Sucrose
  23. 23. Trehalose Turanose Maltotriose Trisaccharides Melezitose RaffinoseTetrasaccharides Stachyose Acarbose Fructooligosaccharide (FOS) Other Galactooligosaccharide (GOS)oligosaccharides Isomaltooligosaccharide (IMO) Maltodextrin Mannan-oligosaccharides (MOS) Beta-glucan o Lentinan o Sizofiran o Zymosan CellulosePolysaccharides Chitin Dextrin / Dextran Fructose / Fructan o Inulin Galactose / Galactan Glucose / Glucan o Glycogen
  24. 24. Levan beta 2→6 Mannan Starch o Amylopectin o Amylose biochemical families: carbohydrates o alcohols o glycoproteins o glycosides lipids o eicosanoids o fatty acids / intermediates o phospholipids o sphingolipids o steroids nucleic acids o constituents / intermediates proteins o amino acids / intermediates tetrapyrroles / intermediates v t e Paper History Wood pulpMaterials Fiber crop Papyrus
  25. 25. Paper chemicals Blotting Bond Construction Copy Cotton Crêpe Glassine India Kraft Laid Manila NewsprintTypes Onionskin Origami Rag Rice Security Seed Tar Tissue Tracing Transfer Wallpaper Waterproof Wax Wood-free
  26. 26. Wove Size Specifications Density Papermaking Paper engineering Paper mill Paper machine Production Sulfite process Kraft process Soda pulping Paper recycling List of paper mills In Europe Industry In the United States In Japan Bleaching of wood pulp Issues Environmental impact of paper Paper pollution Category CommonsRetrieved from "http://en.wikipedia.org/w/index.php?title=Cellulose&oldid=531575776"Categories: Cellulose Excipients Papermaking Polysaccharides
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  30. 30. IntroductionSome useful constants in thermodynamics: 1 eV = 9.6522E4 J/mol k Boltzmanns constant = 1.38E-23 J/K volume: 1 cm3 = 0.1 kJ/kbar = 0.1 J/bar mole: 1 mole of a substance contains Avogadros number (N = 6.02E23) of molecules. Abbreviated as mol. atomic weights are based around the definition that 12C is exactly 12 g/mol R gas constant = Nk = 8.314 J mol-1 K-1Units of Temperature: Degrees Celsius and KelvinThe Celsius scale is based on defining 0 °C as the freezing point of water and 100°C as the boiling point.The Kelvin scale is based on defining 0 K, "absolute zero," as the temperature at zero pressurewhere the volumes of all gases is zero--this turns out to be -273.15 °C. This definition means thatthe freezing temperature of water is 273.15 K. All thermodynamic calculations are done inKelvin!kilo and kelvin: write k for 1000s and K for kelvin. Never write °K.Units of Energy: Joules and CaloriesJoules and calories and kilocalories: A calorie is defined as the amount of energy required to raise thetemperature of 1 g of water from 14.5 to 15.5°C at 1 atm.4.184 J = 1 cal; all food calories are really kcal.
  31. 31. Many times it is easiest to solve equations or problems by conducting "dimensional analysis,"which just means using the same units throughout an equation, seeing that both sides of anequation contain balanced units, and that the answer is cast in terms of units that you want. As anexample, consider the difference between temperature (units of K) and heat (units of J). Twobodies may have the same temperature, but contain different amounts of heat; likewise, twobodies may contain the same heat, but be at different temperatures. The quantity that links thesetwo variables must have units of J/K or K/J. In fact, the heat capacity C describes the amount ofheat dQ involved in changing one mole of a substance by a given temperature increment dT: dQ = CdTThe heat capacity C is then C = dQ/dTand must have units of J K-1 mol-1. (The specific heat is essentially the same number, but is expressed pergram rather than per mole.)Dont forget significant digits. 1*2=2; 1.1*2=2; 1.1*2.0=2.2; 1.0*2.0=2.0Why Thermodynamics?Think about some everyday experiences you have with chemical reactions.Your ability to melt and refreeze ice shows you that H2O has two phases and that the reactiontransforming one to the other is reversible--apparently the crystallization of ice requires removing someheat.Frying an egg is an example of an irreversible reaction.If you dissolve halite in water you can tell that the NaCl is still present in some form by tasting the water.Why does the NaCl dissolve? Does it give off heat? Does it require energy?How is it that diamond, a high-pressure form of C, can coexist with the low pressure form, graphite, atEarths surface? Do diamond and graphite both have the same energy? If you burn graphite anddiamond, which gives you more energy?When dynamite explodes, why does it change into a rapidly expanding gas, which provides the energyrelease, plus a few solids?Chemical thermodynamics provides us with a means of answering these questions and more.A Few DefinitionsA system is any part of the universe we choose to consider.Matter and energy can flow in or out of an open system but only energy can be added to or subtractedfrom a closed system. An isolated system is one in which matter and energy are conserved.A phase is a homogeneous body of matter. The components of a system are defined by a set ofchemical formula used to describe the system.The phase rule:
  32. 32. F + P = C + 2.Extensive parameters are proportional to mass (e.g., V, mass, energy).Intensive parameters are independent of mass (e.g., P, T); these are the "degrees of freedom" Fcontained in the phase rule.Thermodynamics: Power and LimitationsThermodynamics allows you to predict how chemical systems should behave from a supra-atomic"black-box" level--it says nothing about how chemical systems will behave. Thermodynamics alsopertains to the state of a system, and says nothing about the path taken by the system in changing fromone state to another.Chemical Reactions and EquationsHow to write chemical reactions; stoichiometry.Mass and charge balance: e.g., 2Fe3+ + 3H2O = Fe2O3 + 6H+Reaction-Produced Change in Mass, Density, VolumeThe change in volume rV of a reaction is the volume V of the products minus the volume of thereactants: rV = Vproducts - VreactantsThus, if the products are smaller than the reactants, rV < 0.In a generalized reaction such as aA + bB ... = cC + dD... rV = cVC + dVD - aVA - bVBThis sort of additive relationship is true for other state variables and is usually stated as r i iwhere i are the stoichiometric coefficients, positive for products and negative for reactants.What Actually Drives Reactions? Is it Energy? Can We Just Calculate or Measure the EnergyDifference of Reactants and Products and Know Which Way the Reaction Will Go?
  33. 33. For many years people felt that chemical reactions occurred because the reactants had some kind ofenergy to give up (i.e., use to do work)--and that therefore the energy of the products would be lessthan the energy of the reactants. However, we all know that when ice melts it consumes rather thanreleases heat, so there must be more to the story behind why chemical reactions occur.Le Chateliers Principle"If a change is made to a system, the system will respond so as to absorb the force causing the change."EquilibriumA mechanical analogy for chemical change is that of a ball rolling down a slope with multiple valleys; weexplain the balls behavior by saying that mechanical systems have a tendency to reduce their potentialenergy.At equilibrium none of the properties of a system change with time. A system at equilibrium returns toequilibrium if disturbed."Stable" describes a system or phase in its lowest energy state."Metastable" describes a system or phase in any other energy state.The figure above shows the mechanical analogy for H2O at -5°C and + 5°C and 1 atm. Left: at -5°C, solidH2O has the lowest possible energy state. Right: at +5°C, liquid H2O has the lowest possible energy state.When solid H2O is actually present at +5°C, the difference between the free energy of solid H2O andliquid H2O is available to drive the reaction to form the stable solid H2O phase, and the reaction will goto completion if kinetically possible.
  34. 34. Energy: How Do We Calculate and Measure Energy and How Can We Use this Knowledge toPredict Reaction Behavior?Thermodynamics works equally well to describe any kind of work or energy: magnetic, potential, kinetic,etc. For geological systems we typically talk about pressure-volume work, which, because mechanicalwork is F x, you can imagine might be P Vor V PBecause we noted that rV < 0 if the products are smaller than the reactants, we choose to write the P-V work term as -P Vso that a decrease in volume - V is seen as positive work or that an increase in volume + V results ina decrease in crystal energy.The absolute energy of a body can be calculated from Einsteins equation U=mc2, but the presence ofthe c2 term means that the energy of any system is quite large and that measuring this energy isimpractical. It is more practical to measure differences in energy U, and we always discuss or measuredifferences relative to some arbitrary standard state. Analogous to this might be if someone in Namibiaasked you to measure the elevation of the crests of waves at Campus Point--without agreement onsome kind of standard, you wouldnt be able to do much more than measure the heights of individualwaves. If however, you could both agree on an equivalent "sea level" at both localities, you could thencompare the absolute elevations of the wave crests.A typical thermodynamic standard state is normal laboratory conditions: 25°C (298.15 K) and 1atm (often called STP for standard temperature and pressure).The internal energy U of a mineral is the sum of the potential energy stored in the interatomicbonds and the kinetic energy of the atomic vibrations. Thus, you might expect that weaklybonded minerals have relatively low potential energy and thus low internal energy, and when amineral is cold such that its atomic vibrations are slow it will have low kinetic energy and thuslow internal energy. Internal energies are always defined relative to some non-zero standardstate, so we typically talk about changes in internal energy dU. An Aside on Differences and DifferentialsWhats the difference among , d , and ? is used to indicate any kind of difference.
  35. 35. d is used to indicate a differential. is used to indicate a partial differential. For example, the partial differential, with respect to y, off(x,y) = x3y4is= 4x3y3First Law of ThermodynamicsAdding heat Q to a crystal increases its internal energy U: dU dQ( indicates proportional) but if the crystal is allowed to expand, some of the added energy will beconsumed by expansion dV, so the total energy of the crystal is reduced: dU = dQ - PdVThis is effectively the First Law of Thermo: that total energy (heat + P-V work) is conserved.Heat CapacityHeat capacity C describes the amount of heat required to change the temperature of a substance: C=By definition, the heat capacity of water at 15°C is 1 cal K-1 g-1 or 18 cal K-1 mol-1 (i.e., the heat requiredto heat 1 gram of water from 14.5 to 15.5°C is 1 calorie).Heat capacities of solids approach zero as absolute zero is approached: C=0The heat capacity is written with a subscript P or V depending on whether it obtains for constantpressure CP or constant volume CV.As an aside, 2 CP = CV + TV /
  36. 36. where and are the expansivity and compressibility--for solids the difference between CP and CV isminimal and can be ignored as a first approximation. For gases, CP = CV + R, and is quite significant.Heat capacities are measured by calorimetry and often fit by a function of the form: CP = a + bT + cT-2 + dT-0.5but there are other expansions for the heat capacity involving more or fewer terms.Below are some examples of heat capacities of minerals. Note how silicates have a nearly constant heatcapacity of ~1 J K-1 g-1 above 400K.
  37. 37. EnthalpyWe have already talked about the familiar concept of heat as energy.Lets define another measure of energy called enthalpy H--a kind of measure of the thermal energy of acrystal. As we will see below,
  38. 38. dH = dQ + VdPRecall that we interpreted dU = dQ - PdVto mean that the internal energy change is the heat change minus the energy lost to relaxation of thecrystal. Thus, dH = dQ + VdPmeans that the enthalpy change is the heat change plus the energy the crystal gains by virtue of notbeing allowed to expand.Enthalpy includes the vibrational and bonding energy at absolute zero H0°, plus the energy required toincrease temperature: H = H0° + CPdTi.e., we can find the enthalpy change H produced by changing temperature by integrating the heatcapacity CP: H= CPdT Integration ReminderHow to integrate the heat capacity (to determine change in enthalpy H): CP dT = (a + bT + cT-2 + dT-0.5) =aT + bT2/2 - c/T + 2dT0.5and is evaluated as =a(T2 - T1) + b(T22 - T12)/2 - c/(T2-1 - T1-1) + 2d(T20.5 - T10.5)How to integrate the heat capacity divided by T (to determine entropy S): dT = (a/T + b + cT-3 + dT-1.5)
  39. 39. = a ln T + b T - c T-2/2 - 2 d T -0.5and is evaluated as a(ln T2 - ln T1) + b(T2 - T1) - c(T2-2 - T1-2)/2 - 2d(T2-0.5 - T1-0.5)As an example, lets calculate the change in enthalpy H°298-1000 that results from heating quartzfrom 298 K to 1000 K, given the following heat capacity expansion coefficients:a = 104.35, b = 6.07E-3, c = 3.4E+4, d = -1070 (CP dT = (a + bT + cT-2 + dT-0.5) =aT + bT2/2 - c/T + 2dT0.5evaluated from 298 to 1000K =a*(1000-298) + b*(10002-2982)/2 - c*(1000-1-298-1) + 2d*(10000.5-2980.5) = 45.37 kJ/molRelation Among Enthalpy, Heat, and Heat Capacity ( HP= QP)An important relationship between enthalpy change H and heat change Q is revealed bydifferentiating H = U + PV to obtain the total differential dH = dU + PdV + VdPsubstituting dU = dQ - PdV we get dH = dQ + VdPdividing by dT gives = -Vat constant pressure, = 0, leaving =
  40. 40. which is equal to CP: = = CPDetermining EnthalpiesThus, if we want to measure how the internal energy U of a crystal changes U with increasingtemperature at constant pressure, we want to know H, and we can get that by integrating the heatcapacity CP over the temperature range of interest.Theres another way to measure H, though: calorimetry. By dissolving a mineral in acid and measuringthe heat produced by the dissolution, we get a heat of dissolution (usually positive). The enthalpy of"formation" fH° of the mineral is then just the opposite of the heat of dissolution (usually negative).Exceptions to the "usually positive/negative" rule include CN, HCN, Cu2+, Hg2+, NO, Ag+, and S2-.Enthalpies of formation appear in tables of thermodynamic data and are usually referenced to 298 K and1 atm.Enthalpy of ReactionTo get an enthalpy of reaction rH° we can measure the enthalpies of formation of the reactants andproducts fH° and then take the difference between them as rH° = fH°products- fH°reactantsFor example, we can compute the enthalpy of the reaction anhydrite + water = gypsum: CaSO4 + 2H2O = CaSO4 2H2Ofrom Ca + S + 2O2 = CaSO4 fH° = -1434.11 kJ/mol H2 + 0.5O2 = H2O fH° = -285.830 kJ/mol Ca + S + 3O2 + 2H2 = CaSO4 2H2O fH° = -2022.63 kJ/molThus, rH° = fH°gypsum - fH°anhydrite - fH°water = -16.86 kJ/mol.Exothermic vs. Endothermic
  41. 41. If rH° < 0 the reaction produces a reduction in enthalpy and is exothermic (heat is given up by the rockand gained by the surroundings). If rH° > 0 the reaction produces an increase in enthalpy and isendothermic (heat from the surroundings is consumed by the rock). An easy way to remember this isthat spontaneous reactions produce a decrease in internal energy, and because we know thatUP HPa decrease in HP is also a decrease in UP.Calculating fH° at Temperatures Other Than 298 KSo far we know how to calculate the change in enthalpy caused by heating and we know that we can getenthalpies of formation from tables. What if we want to know the enthalpy of formation of a mineral ata temperature other than 298 K?We do this by calculating rCP for the reaction that forms the mineral of interest: rCP = rCPproducts - rCPreactantsand then integrating. Thus, for example if we want to know fH° for quartz at 1000 K, we getcoefficients for the heat capacities of Si, O2 and SiO2:compound a b c dSi 31.778 5.3878E-4 -1.4654E5 -1.7864E2O2 48.318 -6.9132E-4 4.9923E5 -4.2066E2SiO2 104.35 6.07E-3 3.4E-4 -1070for the reaction Si + O2 = SiO2and we calculate a = 24.254 b = 6.2225E-3 c = -3.5E5 d = -470.7and thus, fH°1000 - fH°298 = CP dT = a*(1000-298) + b/2*(10002-2982) - c*(1000-1-298-1) 0.5 0.5 +2 d*(1000 -298 ) = 5.511 kJ/mol
  42. 42. This is the change in the enthalpy of formation that results from heating. We add this to the enthalpy offormation at 298 K to get the enthalpy of formation at 1000 K: fH°1000 =( fH°1000 - fH°298) + fH°298 = 5.511 - 910.700 = -905.2 kJ/molIn other words, forming quartz from the elements at 1000 K yields slightly less heat than at 298 K.Compare this with the change in enthalpy H°298-1000 that results from heating quartz from 298 K to1000 K, which we calculated is 45.37 kJ/mol.EntropyWe have discussed the intuitive statement that reactions probably proceed because the reactants candecrease their internal energy by reacting. We also noted that internal energy scales with enthalpy,suggesting that reactions might go because of a decrease in enthalpy. However, we then noted that notall reactions give off heat--some, such as the melting of ice, proceed in spite of consuming heat.Moreover, there are other processes that proceed in the apparent absence of any heat change: e.g.,mixing of gases or the spreading of dye in water. What is it that causes these reactions to proceedspontaneously even if the heat change is zero or endothermic?The answer is entropy S, which is a measure of the order or disorder. Entropy has three sources:configurational, electronic, and vibrational.Configurational entropy refers to the entropy resulting from imperfect mixing of differentatoms in the same site in a crystal, and is described by the Boltzmann distribution: Sconfigurational = k ln (This is engraved on Boltzmanns tomb in Vienna!)where is the probability that a given number of atoms in a given number of sites will have a particularconfiguration.For N atomic sites that can contain fraction XA A atoms and XB B atoms, =
  43. 43. N is always large where moles of material are concerned, so we can simplify this (using Stirlingsapproximation) to S = - n R (XA ln XA + XB ln XB)where n is the number of sites per mole. For example in cordierite there are 4 Al atoms and 5 Si atomsdistributed over 9 tetrahedral sites. For a random distribution the entropy is S = - 9 R (4/9 ln 4/9 + 5/9 ln 5/9) = 51.39 J mol-1 K-1Note that the form of the configurational entropy equation (and electronic entropy as well) indicatesthat if XA or XB are 0 or 1, Sconfig is zero:Electronic entropy arises when an electron in an unfilled orbital can occupy more than one orbital; e.g.,for Ti3+, the single 3d electron can occupy one of three possible t2g orbitals and Selectronic = 9 J mol-1 K-1.Vibrational (or calorimetric) entropy arises because the energy of lattice vibrations can only increase ordecrease in discrete steps and the energy quanta (phonons) can be distributed within the possibleenergy steps in different ways. Vibrational entropy is very difficult to calculate from statistical mechanicsbut can be calculated easily from heat capacity. Heres why:The entropy of a system always increases during irreversible processes; i.e., for a reversible process, dS =0, whereas for irreversible processes dS >0. This is the Second Law of Thermo--better known as "Youcant feed s**t into the rear of a horse and get hay out the front."If a mineral becomes more ordered during a reaction, reducing its entropy, the heat liberated mustincrease the entropy of the surroundings by an even greater amount. Thus, we write dS >then
  44. 44. >and recalling that CP =then >and S= dTIn other words, the vibrational entropy can be found by integrating the heat capacity divided bytemperature.In a perfectly ordered, pure crystalline material the entropy is zero. This is a simple statement of theThird Law of Thermo, which follows from the fact that heat capacities approach zero at zero K: C=0However, because the rate of atomic diffusion also goes to zero at 0 K, all compounds have some zero-point entropy S°0.Entropy is thus the only thermodynamic potential for which we can calculate an absolute value. Whatwe typically do is determine the heat capacity from near absolute zero to ambient conditions and thenintegrate it to get the (absolute) entropy (in fact this gives us only the vibrational entropy and ignoresconfigurational and electronic contributions to entropy).
  45. 45. Entropy Change of ReactionJust like rH and rV, we can calculate entropies of reactions by using absolute entropies S andcalculating a difference in entropy rS. For example, if we know that S°CaSO4 = 106.7 J mol-1 K-1 S°Ca = 41.42 J mol-1 K-1 S°S = 31.80 J mol-1 K-1 S°O2 = 205.138 J mol-1 K-1then the entropy of the reaction Ca + S + 2O2 = CaSO4is rS° = 106.7 - 41.42 - 31.80 - 2 * 205.138 = -376.8 J mol-1 K-1
  46. 46. Energy Associated With EntropyThe units of entropy suggest that the energy associated with S scales with temperature: dU -TS(The minus sign is there for reasons similar to the -PV we encountered earlier.)The energy associated with configurational entropy in the Al4Si5 cordierite we talked about earlier lookslike this:The energy associated with vibrational entropy in tremolite, quartz, and chalcopyrite looks likethis:(Josiah Willard) Gibbs Free Energy of a PhaseThe Gibbs free energy G is the thermodynamic potential that tells us which way a reaction goes at agiven set of physical conditions--neither the enthalpy change nor the entropy change for a reaction
  47. 47. alone can provide us with this information. The two measures of energy (enthalpy H and entropicenergy TS) are brought together in the Gibbs free energy equation: (the chemical potential is theequivalent for a component) G = U + PV - TSwhich says that the Gibbs free energy G is the internal energy of the crystal U plus the energy the crystalgains by virtue of not being allowed to expand minus the entropic energy TS. Recalling that H = U + PVwe can write this in a more understandable way G = H - TSwhich says that G is the difference between the heat energy and the entropic energy.Relationship Among G, S, and VIf we differentiate G = U + PV - TSto obtain dG = dU + PdV + VdP - TdS - SdTand substitute
  48. 48. TdS = dU + PdV(this comes from dS = dQ/T and dU = dQ - PdV); we are left with dG = VdP - SdTmeaning that changes in Gibbs free energy are produced by changes in pressure and temperature actingon the volume and entropy of a phase.Realize that when we write dG = VdP - SdTwe are implicitly writing dG = dP - dTwhich means that =Vand = -S
  49. 49. These relations indicate that the change in Gibbs free energy with respect to pressure is the molarvolume V and the change in Gibbs free energy with respect to temperature is minus the entropy S.Gibbs Free Energy of FormationThe defining equation for Gibbs free energy G = H - TScan be written as G= H-T Ssuch that the Gibbs free energy of formation fG° is fG° = fH° -T fS°For example, to calculate the Gibbs free energy of formation of anhydrite, we can use fH°CaSO4 = -1434.11 kJ/mole S°CaSO4 = 106.7 J mol-1 K-1 S°Ca = 41.42 J mol-1 K-1
  50. 50. S°S = 31.80 J mol-1 K-1 S°O2 = 205.138 J mol-1 K-1and we calculate the entropy of formation of anhydrite fS° = S°CaSO4 - S°Ca - S°S - 2 * S°O2 = -376.796 J mol-1 K-1and then use fG° = fH° -T fS° = -1434,110 - 298.15 * -376.796 = -1321.77 kJ/molGibbs Free Energy of ReactionWe can write the Gibbs free energy of reaction as the enthalpy change of reaction minus the entropicenergy change of reaction rG = rH -T rSIf the heat energy equals the entropic energy rH =T rSthen rG =0and there is no reaction. Finally we have come to a satisfying point--we can now determine whether agiven reaction will occur if we know H and S, and both of these are measurable or can becalculated.
  51. 51. If rG < 0, the Gibbs free energy of the products is lower than the Gibbs free energy of the reactantsand the reaction moves to produce more products. If rG > 0, the Gibbs free energy of the products isgreater than the Gibbs free energy of the reactants and the reaction moves to produce more reactants.For example, to calculate rG° at STP for the reaction aragonite = calcitewe use rH° = 370 J rS° = 3.7 J mol-1 K-1to calculate rG° = 370 - 298.15 * 3.7 = -733 J/molThe negative value of G tells us that calcite has lower Gibbs free energy and that the reaction runsforward (aragonite calcite).Clapeyron RelationThere is a useful relation between the slope of a reaction in PT space (i.e., dP/dT) and the entropy andvolume changes of the reaction that follows from rG = VrdP - SrdTAt equilibrium G = 0, such that
  52. 52. rVdP = rSdTor =So, the P-T slope of a reaction is equal to the ratio of the entropy change to the volume change.Alternatively, along the equilibrium curve, the changes in pressure times the volume change are equal tochanges in temperature times the entropy change. This is the Clapeyron Equation.So, a phase diagram is a kind of free energy map. = along an equilibrium, < athigh P and low T, and > at high T and low P. Along the equilibrium boundary the GibbsFree energies of the reactants and products are equal and the Gibbs Free energy of reaction rG,is zero.Shortcutting H and S and Finding G DirectlyLike other thermodynamic potentials, we can write the change in Gibbs free energy of reaction as rG° = fG°reactants- fG°productsInstead of using fH° and fS°, it is often possible to obtain fG° values for most compounds fromelectronic data bases. For example, if the following Gibbs free energies of formation are known: fG°CaSO4 = -1707.280 kJ/mol 2H2O fG°CaSO4 = -1321.790 kJ/mol fG°H2O = -237.129 kJ/mol
  53. 53. then for CaSO4 + 2H2O = CaSO4 2H2O rG° = -1.232 kJ/molGibbs Free Energy at Any Pressure and TemperatureWe know many ways to determine rG at STP--but how do we calculate rG for other pressures andtemperatures? Recall that the changes in Gibbs free energy with pressure and temperature are given bytwo of Maxwells relations = rV and =- rSIf we recast these as = rV P and =- rS Tand integrate, we get rGdP = rGP - rGPref = rVdP or rGP = rGPref + rVdPand rGdT = rGT - rGTref =- rSdT or rGT = rGTref - rSdTthus rGPT = rGPrefTref + rVdP - rSdTSolving the Pressure Integral at Constant TemperatureTo a first approximation, we can ignore the expansivity and compressibility of solids and use
  54. 54. rVsdP = rVs(P - 1)as a simplification. Dont forget that this approximation is valid for solids only! An even more commonassumption for P>>1 is rVsdP = rVsPFor example, calculate the change in Gibbs free energy for the reaction2 jadeite = albite + nephelineif pressure increases from 1 bar to 10 kbar, given nepheline = 54.16 cm3 3 albite = 100.43 cm 3 jadeite = 60.40 cmFirst we calculate rV and find r = nepheline + albite -2 jadeite = 33.79 cm3 = 3.379 J/barand thus rGPT - rG1,T = rVsP = 33.79 kJ/molSolving the Temperature Integral at Constant PressureRecall that the effect of temperature on the entropy change of reaction rS depends on the heatcapacity change of reaction rCP: rS = dTThus rGT = rGTref - rSdTexpands to rGT = rGTref - STref + dT dTIf the form of the heat capacity expansion is
  55. 55. CP = a + bT + cT-2 + dT-0.5then the above double integral is a(T - T ln T) - bT2/2 - cT-1/2 + 4 dT0.5 - aTref - bTref2/2 + cTref-1 - 2 dTref0.5 + aTlnTref + bTTref - cTTref-2/2 - 2 dTTref-0.5 - T rSTref + Tref rSTrefNote that this considers only vibrational entropy and ignores configurational entropy. This means ofsolving for rG requires that you know rG at the reference temperature.An alternative path that requires that you know the enthalpy change rH at the reference temperatureis rGT = rHTref + CPdT - T rSTref + dTSolving the Temperature and Pressure Integrals for rGP,TTo calculate the Gibbs free energy change of a reaction at any pressure and temperature, we can useeither of the following equations, depending on whether we know rH or rG rGP,T = rG1,Tref - rSTref + dT dT + rVsP rGP,T = rH1,Tref + CPdT - T rSTref + dT + rVsPIf you dont have heat capacity data for the reaction of interest, these equations can be roughlyapproximated as rGP,T = rG1,Tref - rS1,Tref(T - Tref) + rVsP rGP,T = rH1,Tref - T rS1,Tref + rVsPFor example, calculate rG for jadeite + quartz = albite at 800 K and 20 kbar. The data at 298 K and 1bar are rH° = 15.86 kJ/mol -1 -1 rS° = 51.47 J K mol 3 rVs° = 1.7342 J/bar = 17.342 cm /molUsing rGP,T = rH1,Tref - T rS1,Tref + rVsP = 15,860 - 800 * 51.47 + 1.7342 * 20,000 = 9.37 kJ/molIf we had used the complete equation for solids, integrating the heat capacities, we would haveobtained an answer of 9.86 kJ/mol--not horrifically different.
  56. 56. Calculating the PT Position of a ReactionIf we say that rG = 0 at equilibrium, then we can write our solids-only and constant-heat-capacityapproximations as 0= rG1,T - rS1,Tref(T - Tref) + rVsP 0= rH1,Tref -T rS1,Tref + rVsPand thus we can calculate the pressure of a reaction at different temperatures by P= rG1,Tref - rS1,Tref(T - Tref) /- rVs P= rH1,Tref -T rS1,Tref /- rVsand we can calculate the temperature of a reaction at different pressures by T = Tref + rG1,Tref + rVsP / rS1,Tref T = Tref + rH1,Tref + rVsP / rS1,TrefLets do this for the albite = jadeite + quartz reaction at T = 400 K and T = 1000 K: P = (15,860 - 5147 * 400) / -1.7342 = 2.7 kbar P = (15,860 - 5147 * 1000) / -1.7342 = 20.6 kbarAssuming that dP/dT is constant (a bad assumption, we know), the reaction looks like thisIntroduction to the Equilibrium ConstantA bit farther down the road we will encounter a monster called the equilibrium constant K: K = exp(- rG°/RT)
  57. 57. or ln K = - rG°/RTAt equilibrium, where rG°= 0, ln K = 0 and K = 1. Lets see what K looks like for jadeite + quartz = albiteat 800 K and 20 kbar: ln K = - ( rH1,Tref - T rSTref + rVsP)/ RT = -(15,860 - 800 * 51.47 + 1.7342 * 20,000)/(8.314*800) = -1.4If we do this for all of PT space, we can contour PT space in terms of lnK:Solutions
  58. 58. Almost no phases are pure, but typically are mixtures of components. For example, olivine varies frompure forsterite Mg2SiO4 to pure fayalite Fe2SiO4, and can have any composition in between--it is a solidsolution. We need a way to calculate the thermodynamic properties of such solutions.As a measure of convenience, we use mole fraction to describe the compositions of phases that are solidsolutions. For example, a mix of 1 part forsterite and 3 parts fayalite yields an olivine with 25 mol%forsterite and 75 mol% fayalite, which can be written as (Mg0.25Fe0.75)2SiO4 or fo25fa75, etc. Mole fractionsare denoted as Xi.We need a way of splitting up the Gibbs free energy of a phase among the various components of thephase--how for example do we decide how much of the Gibbs free energy of an olivine is related to theforsterite component and how much derives from the fayalite component? Likewise, how does theGibbs free energy of a phase vary with composition--is the relationship linear between endmembers??We address these issues by defining a partial Gibbs free energy for each component at constantpressure and temperature and constant composition of other components, called the partial molarGibbs free energy or chemical potential i =where n is the amount of substance. For olivine solid solution composed of fayalite and forsteritecomponents or endmembers, we write dG = dnfayalite + dnforsteriteVolume of MixingImagine that mole fractions of phase A and phase B with molar volumes VA and VB, are mixed together.We can describe the volume of the mixture as V = XAVA° + XBVB°and it is a linear mixing of the two endmember volumes. We call this ideal mixing or mechanical mixing.Real solutions, however, do not behave this way, and the mixing is always non ideal, althoughsometimes only weakly so. The figure shows mixing that produces a smaller volume than expected, butit is not possible to predict the shapes and positions of such mixing curves.
  59. 59. Partial Molar VolumeThe partial molar volume is defined as iIf you mix two compounds A and B together and find a volume of mixing that is non-ideal, how can youdetermine the contribution that A and B each make to the volume? That is, what are the partial molarvolumes of A and B, A and B?? Graphically, the partial molar volumes are the A and B axis interceptsof the tangent to the mixing curve, and can be described by the simple relationship: Vmix = XA A + XB B or Vmix = Xi iThe behavior of this function is such that when XA is 1, Vmix = VA and when XA is 0, Vmix = VB. Alternatively,
  60. 60. A = (Vmix - XB B) / XAEntropy of MixingThe entropy of mixing is never zero because mixing increases entropy. As we discussed days ago, theentropy of mixing (i.e., the configurational entropy) is Smix = -R (Xi ln Xi)where i = 1..n is the number of sites over which mixing is occurring.Enthalpy of MixingEnthalpies also do not combined ideally (linearly) in mixtures because the mixture may have strongerbonds than were present in either of the unmixed phases. The excess enthalpy is Hmix = 0.5 * N z XAXB [2 AB - AA - BB]where AB is the interaction energy among A-B atoms, AA is the interaction energy among A-A atoms,and BB is the interaction energy among B-B atoms.Gibbs Free Energy of MixingRecall that all spontaneous processes/reactions occur because of a decrease in Gibbs free energy. Itshould therefore not surprise you that the Gibbs free energy of mixing is always negative--otherwisemixing would not occur. The fact that A < G°A and B < G°B illustrates why compounds combinespontaneously--each compound is able to lower its free energy.
  61. 61. The above figure is hypothetical because we cannot measure or calculate the absolute Gibbs freeenergy of phases. For this reason, is always expressed as a difference from some standard statemeasurement, as , - °, or - G°.The difference between the absolute Gibbs free energy G° per mole ° of a pure compound andthe chemical potential per mole of dissolved compound is A - G°A = A - °A = RT lnXAThis function has the following shape:
  62. 62. implying that when the mineral is pure (X = 1) then = 0, and when the mineral is infinitelydilute (X = 0) then the chemical potential is undefined. For example, in a two-componentmineral if XA = 0.4, at T = 298 K, A- °A = 8.314 * 298 ln 0.4 = -2271 J B- °B = 8.314 * 298 ln 0.6 = -1266 JThe equation of the Gmix line is the sum of the chemical potentials of the endmembers: Gmix = RT (XA ln XA + XB ln XB) or Gmix = RT (Xi ln Xi)which looks like this for two components:Actually, all this discussion has been predicated on the assumption that Hmix = 0. If this is nottrue, Gmix is not a simple function of composition, but has the general form:
  63. 63. Depending on the relative values of Hmix and -T Smix, the free energy of mixing may benegative throughout the whole composition range if the entropic energy contribution outweighsthe enthalpy increase; this is more likely at higher temperature.The two free energy minima in the above figure indicate that minerals of intermediatecompositions can reduce their free energy by unmixing into two phases. This explains theappearance and driving force for exsolution. Note that this can only be true if Hmix > 0, i.e., if2 AB > AA + BB, which makes sense because it means that the A-B bonds have a higher freeenergy than the sum of the free energies of separate AA and BB bonds.ActivitiesIn reality, no phases behave ideally--that is, their chemical potentials are never simple logarithmicfunctions of composition as A - °A = RT lnXAimplies. Instead, we say that the chemical potential is a simple logarithmic function of activity anddefine activity asa = ( X)where a is the activity of a compound, is the "site occupancy coefficient" (e.g., = 2 for Mg inMg2SiO4), and is the activity coefficient that describes the non-ideal behavior. Thus we write A - °A = RT lnaAFor pure compounds a=1 because X=1. For ideal compounds =1. As a specific example, the chemicalpotential of the almandine (Fe3Al2Si3O12) component of a garnet solid solution ((Fe, Mg, Ca,Mn)3Al2Si3O12) is alm = °alm + RT lnXalm
  64. 64. To be clear, ° is the chemical potential of the component in its pure reference state and varies as afunction of pressure and temperature; this we measure with calorimetry. is the chemical potential asit actually occurs and varies as a function of phase composition; this we measure with an electronmicroprobe. The activity forms a bridge between idealized behavior and real behavior.The Equilibrium ConstantAt equilibrium the sum of the Gibbs free energies of the reactants equals the sum of the Gibbs freeenergies of the products. Equally, the sum of the partial molar Gibbs free energies (chemical potentials)of the reactants equals the sum of the partial molar Gibbs free energies (chemical potentials) of theproducts. In other words, for SiO2 + 2H2O = H4SiO4at equilibrium, H4SiO4 = SiO2 + 2 H2OMore generally, for aA + bB = cC + dDthen c C+d D=a A+b Bor, at equilibrium r =0=c C+d D-a A-b Bwhich we can reformat as r = i iwhere i is the stochiometric coefficient of a product or reactant and is positive if for a product andnegative if for a reactant. If we then remember that - ° = RT lnaand rewrite it as = ° + RT lnawe can reformat the earlier equation as
  65. 65. r = 0 = c( C° + RT lnaC) + d( D° + RT lnaD) - a( A° + RT lnaA) - b( B° + RT lnaB)which looks nicer as r = r ° + RT ln (aCc aDd/aAa aBb)To be completely general we write r = r ° + RT ln ai ( means to multiply all i terms)This equation is invariably simplified to r = r ° + RT lnQand Q is the activity product ratio. The activities in the Q term change as the reaction progresses towardequilibrium.To be clear again, r ° is the difference in the Gibbs free energies of the products and reactants wheneach is in its pure reference state and varies as a function of pressure and temperature. r is thedifference in the Gibbs free energies of the products and reactants as they actually occur and varies as afunction of phase composition.At equilibrium, the product and reactant activities have adjusted themselves such that r = 0. Wewrite this (with K instead of Q, to signify equilibrium) as 0= rG° = -RT ln KK is called the equilibrium constant. If K is very large (ln K positive), the combined activities of thereaction products are enormous relative to the combined activities of the reactants and the reaction willlikely progress. On the other hand, if K is small (ln K negative), there is unlikely to be any reaction.The utility of K is that it tells us for any reaction and any pressure and temperature, what the activityratios of the phases will be at equilibrium. For example, for the reaction albite = jadeite + quartzlets say that at a particular P and T, rG° = -20.12 kJ/molUsing rG° = -RT lnKwe calculate
  66. 66. log K = 3.52This means that at equilibrium, (ajadeite aquartz / aNaAlSi3O8) = e3.52Where ajadeite is the activity of NaAlSi2O6 in clinopyroxene and aalbite is the activity of NaAlSi3O8 inplagioclase.Alternative Route to the Equilibrium ConstantWhen we think of mass balance in a reaction, we can explicitly write 0= iMiwhere i are the stoichiometric coefficients and Mi are the masses or the phase components.Analogously, we can explicitly write a similar balance among the chemical potentials: 0= i iFor each chemical potential we can write i= °i + RT lnaiCombining these two equations we find 0= i °i + iRTln ai 0= i °i + RTln (ai) 0= i °i + RTln ai 0= i °i + RTln Kand eventually 0= rG° + RT ln Kor rG° = -RT ln KThe equation K= ai
  67. 67. is the law of mass action (which actually discusses the action of chemical potential rather than mass).We can also write for 298 K and 1 atm rH° -T rS° = - RT ln Kand for any P and T of interest: rH1,Tref + CPdT - T rSTref + dT + rVP = - RT ln KThis has been called "the most important equation in thermodynamics," so youd better like it(!) Theequilibrium constant K is a function of 1/T -ln K = ( rG° / RT) = [( rH / RT) - ( rS / R)]Which looks likeActivity Models (Activity-Composition Relations) for Crystalline SolutionsGarnets are solid solutions ofcomponent abbrev. Formulapyrope prp Mg3Al2Si3O12almandine alm Fe3Al2Si3O12grossular grs Ca3Al2Si3O12spessartine sps Mn3Al2Si3O12andradite and Ca3Fe23+Si3O12Mixing models derive from entropy considerations. In particular the relation
  68. 68. Smix = -R Xi ln Xialthough we will not go through the derivation.Mixing on a Single SiteThe simplest type of useful activity model is the ionic model, wherein we assume that mixing occurs oncrystallographic sites. For a Mg-Fe-Ca-Mn garnet with mixing on one site, which we can idealize as(A,B,C,D) Al2Si3O12, the activities are 3 3 aprp = Mg XMg 3 3 aalm = Fe XFe 3 3 agrs = Ca XCa 3 3 asps = Mn XMnThe pyrope activity is shown in the above figure.In general, for ideal mixing in a mineral with a single crystallographic site that can containions, ai = Xjwhere a, the activity of component i, is the mole fraction of element j raised to the power. For non-ideal mixing, we include an activity coefficient ai = j XjMixing on a Several SitesFor minerals with two distinct sites and the general formula (A,B) (Y,Z)
  69. 69. there are four possible end member components A Y , A Z , B Y , and B Z .The ideal activities of these components are aA Y = XA XY aA Z = XA XZ aB Y = XB XY aB Z = XB XZFor non-ideal garnet activities we write aprp = XMg3 XAl2 or Mg3 XMg3 Al2 XAl2 aalm = XFe3 XAl2 or Fe3 XFe3 Al2 XAl2 agrs = XCa3 XAl2 or Ca3 XCa3 Al2 XAl2 asps = XMn3 XAl2 or Mn3 XMn3 Al2 XAl2 aand = XCa3 XFe3+2 or Ca3 XMn3 Fe3+2 XFe3+2where the X3 term describes mixing on the 8-fold trivalent site and the X2 term describes mixing on theoctahedral divalent site.The pyrope activity is shown in the above figure for Mg from 0 3 and Al from 0 2.It is common to modify such models that are based on completely random mixing of elementswith models that consider local charge balance on certain sites or the Al-avoidance principle.Geothermometry and GeobarometryExchange ReactionsMany thermometers are based on exchange reactions, which are reactions that exchange elements butpreserve reactant and product phases. For example:
  70. 70. Fe3Al2Si3O12 + KMg3AlSi3O10(OH)2 = Mg3Al2Si3O12 + KFe3AlSi3O10(OH)2 almandine + phlogopite = pyrope + anniteWe can reduce this reaction to a simple exchange vector: (FeMg)gar+1 = (FeMg)bio-1Popular thermometers include garnet-biotite (GARB), garnet-clinopyroxene, garnet-hornblende, andclinopyroxene-orthopyroxene; all of these are based on the exchange of Fe and Mg, and are excellentthermometers because rV is small, such that =is large (i.e., the reactions have steep slopes and are little influenced by pressure). Lets write theequilibrium constant for the GARB exchange reaction K = (aprpaann)/(aalmaphl)thus rG = -RT ln (aprpaann)/(aalmaphl)This equation implies that the activities of the Fe and Mg components of biotite and garnet are afunction of Gibbs free energy change and thus are functions of pressure and temperature.If we assume ideal behavior ( = 1) in garnet and biotite and assume that there is mixing on only 1 site aalm = Xalm3 = [Fe/(Fe + Mg + Ca + Mn)]3 aprp = Xprp3 = [Mg/(Fe + Mg)]3 aann = Xann3 = [Fe/(Fe + Mg)]3 aphl = Xphl3 = [Mg/(Fe + Mg)]3Thus the equilibrium constant is K = (XMggar XFebio)/(XFegar XMgbio)When discussing element partitioning it is common to define a distribution coefficient KD, which is justthe equilibrium constant without the exponent (this just describes the partitioning of elements and notthe partitioning of chemical potential): KD = (XMggar XFebio)/(XFegar XMgbio) = (Mg/Fe)gar /(Mg/Fe)bio = K1/3
  71. 71. Long before most of you were playground bullies (1978) a couple of deities named John Ferry and FrankSpear measured experimentally the distribution of Fe and Mg between biotite and garnet at 2 kbar andfound the following relationship:If you compare their empirical equation ln KD = -2109 / T + 0.782this immediately reminds you of ln K = - ( rG° / RT) = -( rH / RT) - (P rV / RT) + ( rS / R)and you realize that for this reaction rS = 3*0.782*R = 19.51 J/mol K(the three comes from the site occupancy coefficient; i.e., K = KD3) and -( rH / R) - (P rV / R) = -2109or rH = 3*2109*R -2070* rVMolar volume measurements show that for this exchange reaction rV = 0.238 J/bar, thus rH = 52.11 kJ/molThe full equation is then
  72. 72. 52,110 - 19.51*T(K) + 0.238*P(bar) + 3RT ln KD = 0To plot the KD lines in PT spaceNet-Transfer ReactionsNet-transfer reactions are those that cause phases to appear or disappear. Geobarometers are oftenbased on net-transfer reactions because rV is large and relatively insensitive to temperature. Apopular one is GASP: 3CaAl2Si2O8 = Ca3Al2Si3O12 + 2Al2SiO5 + SiO2 anorthite = grossular + kyanite + quartzwhich describes the high-pressure breakdown of anorthite.For this reaction rG = -RT ln [(aqtzaky2agrs) / aan3] = -RT ln agrs / aan3(the activities of quartz and kyanite are one because they are pure phases). A best fit through theexperimental data for this reaction by Andrea Koziol and Bob Newton yields P(bar) = 22.80 T(K) - 7317for rV = -6.608 J/bar. Again, if we use

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