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Food Chemistry

by: Andreanne Collier

Food Chemistry FD SC 400

Andreanne Collier
Penn State
GPA 3.58


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This 0 page Class Notes was uploaded by Andreanne Collier on Sunday November 1, 2015. The Class Notes belongs to FD SC 400 at Pennsylvania State University taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/233076/fd-sc-400-pennsylvania-state-university in Food Science & Technology at Pennsylvania State University.

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Date Created: 11/01/15
6 Lipids All living matter contains some fat as their membranes are made up from selfassociated lipid molecules In practical terms though the fat content of foods varies from effectively zero many fresh vegetables certain sh to 100 cooking oils Low fat foods are seen in this culture as being almost virtuous but also not as tasty as their high fat alternatives This is probably because of the nutritional and functional roles of fats in foods Nutrition Fats are the densest source of Functionality Fats provide important calories available in the diet 7 kcal per texture to foods gram which is seen as a bad thing in the 0 Liquid fat on food is greasy eg west A high fat diet is associated with fried chicken many health problems heart disease and o Emulsi ed fats give a creamy rich some cancers as well as obesity mouthfeel to foods eg heavy However not all fats are equally quotbad for cream is smoother and richer than youquot and mOdern lipid nutrition light cream which is quotcreamierquot than recommendations are increasingly milk complex AS W611 as the negative 0 Semicrystalline fats are responsible nmlee aSPeCIS 0f fats many llplds are for the plastic spreadability of butter essential components of the diet Oils and margarine act as a solvent for other important Largely crystalline fats are hard molecules such as certain vitamins and solids eg chocolate avors 1 Lipid Structure In this section we will identify the important molecules that occur in the class of compounds known as lipids Fatty Acids The basic building blocks of lipids are fatty acids which are carboxylic acids with long hydrocarbon chains attached In natural fats the hydrocarbons are usually 0 Linear nonbranched 0 Even numbers of carbon atoms 0 Between 4 and 20 carbons long the preponderance being of middle length 12 carbon Fatty acids can be classified according to their length short chain fatty acids 4 10 carbons medium chain fatty acids 1214 carbons long chain fatty acids more than 16 carbons 0 Contain zero unsaturated one monounsaturated or more polyunsaturated double bonds Triple bonds are not normally seen Any double bonds are usually cis but the trans form is thermodynamically more stable so heating can allow the activation energy for transformation to be exceeded cis trans isomerization to occur About 40 of the fat in a French fry and overall approximately 10 of western dietary fat may be trans Trans fatty acids are thought to be harmful to the health as they cannot be correctly metabolized by the body or incorporated into membranes oi fatty acid force a turn in the molecular backbone so polyunsaturated fatty acids can be bent dou le l l Linoleic acid 16 carbons lArachadonic acid 20 ELauric acid 14 carbons double bonds at carbon 9 carbons double bonds no double bonds and 12 note count carbon at carbons 5 8 11 and l 1 as the carbonyl group 14 The important fatty acid structures are listed in Table 1 in the text Common fatty acids have common names eg lauric acid stearic acid but there are also more formal naming conventions for example quotA vrquot oquot u Hexanoic acid 7 Hex six carbons v anoic an unsaturated structure acid a carboxylic aci x Hexenoic acid p39 Hex six carbons VJ enoic an saturated structure acid a carboxylic acid an Glycerol ester 5 Fatt acids do occur in foods but the vast majority N99 of them are on 39 a d as esters of glycerol Glycerol is a three carb polyol It IS S h e W The three alcohol groups can esterify With carboxylic acids a dehydration reaction to form mono di and triglycerides More correctly these are own as mono or di or tri acyl glycerols Mono and diglycerides are amphiphilic and triglycerides are hydrophobic and Water insoluble H C ioH H2070 new l W tho L39I lycero 39 39 39 L substituents Like sugars the Fisher projection is used to differentiate between the chiral c c I I I u I below u with the R 39 quot 39 39 tnan suggest there are multiple Rgroups at different bonding sites CH20F 1 OP CH20F391 1 Rzowu gt R10 H H 0mm TL 39 l L le ie up from page ru u i I I lU MIC two ends carbon one at the top three at the bottom So snSLPO would have stearic acid in position 1 at the top and palmitic and oleic in positions 2 and 3 respectively The triglyceride n t M h 39 39 L 39 39 39 39n39 L A of stereoisorners or ifthe cornposition is unknown or unimportant Phospholipids Glycerol can esterify with molecules other than fatty acids for example 39 39 l l T A 39L at t diesterofphosphoric acid 39 39 39 39 39 39 l l 39 unn nai L39 LU nhtn quot normalfatty acids 39 39 p man u i form membranes 2 Lipid Oxidation In this section we are mainly concerned with lipid autoxidation but we will initially deal brie y with lipolysis Lipolysis is the enzymatic lipase or thermal hydrolytic cleavage of ester links holding fatty acids to glycerol Free fatty acids particularly the small ones are quite volatile and have some aroma so their formation can lead to off avors There are very few free fatty acids in quotnormalquot food oils Avoiding enzymatic lipolysis requires the oil must be separated from the protein enzyme fraction of the food as soon as possible in the extraction process or the enzyme should be rapidly thermally inactivated Once formed free fatty acids can be converted to soaps at high pH ie sodium salts and extracted to remove the avor Autoxidation Lipid autoxidation is the catalyzed radical reaction between unsaturated fatty acids and triglycerides and molecular oxygen Lipid oxidation leads to food spoilage as there is extensive molecular fragmentation leading to the formation of volatile and extremely smelly products The formation of avor from oxidized lipids is known as rancidity In certain cases some fried foods and cheeses some rancid avor is desirable but usually rancid avors are extremely unpleasant Once radical oxidation has started in a food or biological system it may spread to nonlipid components in food this may be as simple as vitamin loss but can also cause protein damage There are also some health concerns as any oxidation induced in vivo can lead to genetic damage and may be carcinogenic Extensive lipid oxidation can lead to some lipid polymerization with consequent increases in viscosity and browning The Radical Reaction All radical reactions proceed via initiation propagation and terminations steps 0 Initiation The homolytic cleavage of a covalent bond to form two free radicals There are no radical groups on the left hand side of the reaction and two on the right Propagation The transfer of the radical character from one molecule to another Propagation always has one radical species on the left hand side of the reaction and one on the right Termination Two radicals come together to form a covalent bond There are two radical groups on the left hand side of the reaction and none on the right In lipids the typical reaction scheme is Initiation A hydrogen atom is abstracted from a methylene group adjacent to a CC bond usually one between two ie CHCHCH2 H 139 CHCH The structure is a l4pentadiene 1139 Sometimes the radical center formed can migrate along an unsaturation to localize on another carbon before further reacting 70HCH70H2CHCH7 a CHCHCHCHCH CH CH CH Q 7 z i i r i CHCH CH CH w p L J icHiCH2CHiCl 12CH7 CHCHCHCHCH i 02 J 13hydroperoxide diene 9hydroperoxide diene A consequence of this is there are often several starting materials for the next phase of the reaction and a mixture of breakdown products The products of resonance stabilization of free radicals in unsaturated fatty acids are conjugated dienes to adjacent CC Propagation The lipid radical rapidly reacts with molecular oxygen to yield a R peroxy radical The peroxy radical reacts with fresh lipid often the same type of 14 pentadiene structure taking a hydrogen atom to form a lipid hydroperoxide and leaving a R00 RH ROOH R new lipid radical which can react with molecular oxygen and continue the cycle The net effect ofthis reaction is to consume oxygen and produce lipid hydroperoxides 02 ROO39 Termination Any two radicals can combine to form a nonradical product TR The rate of these reactions is initially RI TROO39 39 ROOR limited by the relatively small numbers of R00 39 ROO39 39 ROOOOR 7 radicals present in the system So a single radical generation event can lead to the production of large quantities of lipid hydroperoxides Lipid hydroperoxides are relatively innocuous having no bad smell or color but they can and do spontaneously breakdown via another radical mechanism Hydroperoxide breakdown The first step of hydroperoxide breakdown is the homolytic cleavage of the oxygenoxygen bond to produce an alkoxy and hydroxy radical Hydroperoxide 39 Alkoxyl radical Hy dmxy r adic a1 RiCHiR RiCHiR OH I OOH o The next step is for the radical electron to quotfoldquot into the C0 bond along with a single electron from either the CR or CR39 bond to form CO The other electron from CR or R39 moves to the R or R39 and cleaves the lipid molecule The bond that breaks can either be left or right methyl or carbonyl end of the central carbon forming one of two possible pairs of producm Methyl side breakdown OR Acid side breakdovm p R HR39 RCHR39 L 0 O I R Hydrocarbon R Add ester R i Oxoacid R 070 oxoester CZO Aldehyde convert to aldehyde The most important producm are the aldehydes which are volatile and have characteristic smells Each unsaturated fatty acid can produce several types of aldehydes and as there are many fatty acids so rancidity tends to have a quotbroadquot range of perceived flavor and aroma Many of the reaction producm acids hydrocarbons and aldehydes in the figure above are unsaturated and can oxidize further The large number of highly reactive radicals present means the producm of lipid oxidation are very highly complex Question 1 Predict the likely autoxidation producm of this fatty acid 0 WWW note the double bonds are drawn as transfor convenience in a realfat they are Factors affecting autoxidation rate Fatty acid composition In general the rate of lipid oxidation increases very rapidly with the number of unsaturations present in a fatty acid This is particularly important for polyunsaturated fatty acids essential in the diet esp the 06 and 03 fatty acids in sh oils The rapid oxidation of fish oils is believed to contribute to the short shelf life of sh Free fatty acids are more prone to oxidation than those attached to glycerol so lipolysis can lead to autoxidation Temperature The rate of lipid oxidation increases rapidly with temperature Water activity As with most food chemical reactions the rate of lipid oxidation decreases as the water activity is lowered from one towards the monolayer value 02 However many lipid oxidation reactions increase in rate under very low water activities aWltmonolayer and rancidity can be a major problem in dehydrated foods The rate of lipid oxidation is believed to increase because the lipid hydroperoxides are dehydrated and break down more readily and because nonhydrated metal ions are more effective catalysts Metal ions Although unsaturated lipids can spontaneously react with oxygen the rate of the uncatalyzed reaction is many times slower than the catalyzed reaction If the amount of free metal is restricted the rate will be slower Particularly effective metal catalysts are soluble salts of iron and copper and the use of these metals in food processing equipment may cause problems particularly due to corrosion or poor quality welding Control of metal ions is often achieved by the use of a class of antioxidants known as metal chelators which tightly bind free catalytic ions and render them unavailable A good example is EDTA which binds iron Oxygen If oxygen could be completely eliminated from a product it could not oxidize In practice this is very hard to achieve absolutely as only a very small amount of oxidation and therefore oxygen is required to make a food inedible Oxygen is very soluble in both water and oil and there is usually enough present to allow oxygen to proceed to an unacceptable level Even if this oxygen was eliminated most natural fats contain a considerable amount of preexisting hydroperoxides that can break down and lead to oxidation products in the absence of additional oxygen Light Light is a source of energy that can lead to the formation of radical initiators Ultraviolet light is particularly harmful as it is higher energy than visible light It would be better to store most fatty foods in the dark but this is not practical in retail Antioxidants Fat naturally present in animal and plant tissues is relatively resistant to oxidation due to the presence of natural antioxidants e g tocopherols certain pigments some enzymes but when the fat is extracted the puri cation process remove a lot of the antioxidants and the oil is less stable Natural and synthetic antioxidants frequently added back to fatty foods particularly those prone to oxidation There are two main classes of antioxidant molecules which prevent the action of catalysts e g metal chelators and molecules which are oxidized preferentially to the lipid It is the second class we will consider here Sacrificial antioxidants are oxidized more easily than the lipids they are designed to protect The idea is that when an oxidationcausing event occurs e g the generation of the radical the antioxidant will rapidly react and itself take on the radical character The antioxidant radical formed is more stable than a lipid radical so tends not to react further and cause damage to the valuable components of the food In forming a radical the antioxidant is quotdestroyedquot and has no further antioxidant activity so will only delay not completely prevent lipid oxidation In the diagram the process is shown as open1ng up another Annoxidant Antioxidam secondary pathway to get r1d of the Radical antioxidant radical avoiding the unstable Lipid A radical li id radical that would lead to I gvor I I WWWij Sacri cial antioxidants are ammmdam often divided into primary and Lipid radical secondary Primary antioxidants are those which remove the radical Secondary antioxidants remove the radical character from the primary antioxidant radical returning it to a quotworkingquot form Secondary antioxidants extend the lag before oxidation is seen Good examples of primary and secondary antioxidants are vitamins E and C in living cells In order to work a primary antioxidant must 1 React rapidly with radicals 2 Produce a stable radical on reaction ie not react further Examples of natural and synthetic antioxidants permitted as food additives include Antioxidant radical Rancidily BHA BHT 39 BHABHT Butylated hydroxyanisole Butylated hydroxytoluene Butylated hydroxyanisole and butylated hydroxytoluene are similar OH OH synthetic molecules used as primary antioxidants in many foods The phenol group in both BHA and BHT is vulnerable to having a hydrogen atom removed leaving a radical The mm H3C3C CCH33 OCH3 CH3 radical formed is relatively stable because it can ef ciently delocalize around the benzene ring shown below for BHA Butylated hydruxyamsule 0 039 In general a structure is stabilized by cm 11 mean cm extensive resonance delocalization of 4 electrons but BHA is even more stable because W The single highly reactive electron spends a lot of its quottimequot on the benzene carbon adjacent to the tertiary butyl group The 00 Several resunance stabilized stxuemres i CW3 0 CW3 tertiary butyl group is 39 electropositive and draws the electron a little away from the OCH OCH benzene carbon further stabilizing that position The tertiary butyl is extremely bulky and prevents the physical approach of other molecules that might react with the radical if they could get close enough BHT reacts very similarly to BHA o Excm 00H 0 Tocopherols Tocopherols are natural A Ph TmPh m primary antioxidants There 0 are a whole group of very Ha similar molecules all with O CH similar structures Many of M 0 them have Vitamin E CH CH CH CH activity for example 0L tocopherol Tocopherols are lipid soluble molecules but the concentration of hydroxyl groups on one end makes them slightly amphiphilic The hydrogen atoms can be abstracted from the hydroxyl groups and the resulting radical delocalizes extensively around the heterocyclic structure gure 39 in the Fennema text 3 Lipid Crystallization Lipid crystallization is an example of a phase transition and some parallels can be drawn between this and the formations of gas bubbles in a can of soda or avoiding gritiness during lactose crystallization in particular the supersaturation of a solution before the phase transition occurs Triglyceride molecules take on a quottuning forkquot conformation Note unsaturation in any of the fatty acids can make the shape very twisted In a liquid oil the triglyceride molecules are orientated randomly to one another and constantly in motion In a crystalline fat they are tightly packed in a regular repeating pattern To undergo a liquid n to solid transition the oil must first be cooled to reduce the average thermal motion of each molecule enough that they will start to interact with each other Once the thermal energy is low enough a 2 solid can form The interactions between oil molecules in a crystal are largely Van der Waals Crystallization is a liquid to solid phase transition divided into two distinct processes nucleation and growth Nucleation Nucleation is the formation of a solid in a liquid matrix The very smallest crystals have a very small radius of curvature and so the lipid molecules are more soluble The consequence is the smallest crystal embryos tend to redissolve before they grow unless the temperature is considerably below the melting point of the fat when the thermodynamic pressure to crystallize becomes to great and the nuclei begin to grow Therefore a liquid oil may remain liquid below its melting point for a considerable time supercooling Crystals melt at their thermodynamic melting point there is no superheating so there is often hysteresis between solids content on cooling and x lIujl gtl THEF Equot JIn39ll supcrmmling Growth The second phase of crystallization is growth During the growth phase liquid triglycerides align themselves at the surface of the crystal and are incorporated The nuclei will grow until they overlap and ll the container with solid fat Oils contain a mixture of fatty acids with very many different triglyceride molecules present Any pure compound has a characteristic melting point so each triglyceride present will crystallize at a unique temperature E slightly different from each different 39 g 33 triglyceride present As a consequence 0 real food oils have a melting range 00 rather than a melting point The solid 0 fat content of a food can be measured as a function of temperature by nuclear Temperature magnetic resonance N MR or differential scanning calorimetry DSC and will reduce from 100 to 0 liquid oil over a range of as few as 2 or 3 and as many as 20 or more degrees Kelvin H 0 Co Sol1d fat The solids content of common food fats changes from 0100 over the range of temperatures found in the manufacture storage and use of foods The solids content is an important determinant of food texture particular in high fat foods Liquid cooking oil is completely liquid Butter is soft at room temperature but will harden in the refrigerator Chocolate is hard at room temperature but will melt in the mouth Partly crystalline fats can be seen as a dispersed system with solid crystals in a liquid oil continuous phase The individual crystals as 23 um long and needleshaped but they can aggregate to form a viscoelastic network e g margarine Texture changes caused by temperature are because more fat crystallizes and reinforces the network causing a shift from liquidlike to solidlike behavior Polymorphism quotPolymorphic forms are crystalline phases of the same chemical composition that di er amongst themselves in structure but yield the same liquid phase on melting Triglyceride molecules are capable of packing into a number of distinct crystal lattice forms known as polymorphic forms polymorphs Their crystal packing can vary in angle andor spacing of the tuning forks Note in the right hand figure the structure Iquot lullquot 1quot at the top is double chain packing and at if I39 the bottom triple chain packing 3 quot Iquot l and 4 In many fats there is only one reasonable stable polymorph so there is no need to distinguish between solid forms you practically only ever see one but in cocoa butter the basic fat in chocolate there are several variably stable polymorphs and good quality chocolate requires finding the right one Melting Chain point Packing I 5 2 17 double II or 21 double III mixed 255 double IV 5 1 28 double V 52 345 triple VI 51 36 triple Selected properties of the different polymorphic forms of cocoa butter are listed in the table and can be summarized as follows Desirable forms have a high melting point The higher melting point the more stable More stable fats are denser More stable forms are slow to form Like crystals will grow from like Cocoa butter can convert from a less to more stable form These points are exploited in the tempering process Tempering The last step before liquid chocolate is poured into a mold to solidify into a bar is tempering Tempering is required to ensure the fat crystallizes in the right polymorphic form In tempering liquid chocolate is poured into a gently mixed container and the temperature altered according to a fixed program 1 The fat is heated to about 50 C At this temperature all the solids melt so there are no residual nuclei snuc 2 The fat is cooled at about 32 C crystallization of the stable 5 crystals will start but cooling is 320C J 32 quot continues to create a large mixed crop of 1 2 law different crystal 4 polymorphs Reheating the mixed H39ml crystals melts out the less stable polymorphs leaVing only the stable ones When the fat is cooled again the stable crystals NI2 of the fat will act as nuclei for the rest of the lipid and the whole bulk will crystallize into the right form After solidification well tempered chocolate is hard has good quotsnapquot it has a glossy surface and it is readily detached from molds It is also resistant to bloom 1 Mcli 2 foal A m rlyralulliannxJI 3 l mntlniml39u 4 Huh um uuamb E IL NI FATLJIIN39U 4 Bhum Fatblnamxsthz fammnah um mmdd m surface unsung ansahd chncalme bars Elnam can farm sla yvvex Irs days ax mnmhs shing um m mm mmmscap blanm laaks m Jagged bmkzn cryslals sockmg up mm m Mm Elnam 15 m mxw harmful ax mmlld but u laaks unpleasant and is an m 5mg Wm caalmg aws n m Hymn n m causes afblnam are mm m manufacturers camml mm mm and Vexya z cansnmzx ange but as wnh allde was m manufacturer w be ma xespunsfblz mm M eduumng madam n 15 pusgblz a mnd 39ythe mu m mud blanm rammm amnbyaddmg mm m ax an emulsi er e g 5mm mnmsmmte FDSC400 Part 1 Kinetics Supplemental reading Chapter 173 of the Fennema text offers several ways to understand the changes in foods with time and temperature In this course we will focus only on the one closest to classical chemical kinetics The basic theories of kinetics are covered in adequate detail in most general chemistry tests 1 Introduction The application of kinetics provides a scienti c and quantitative understanding of common sense No one expects foods to last forever 7 bread is fresh for a few days then becomes progressively worse a steak cooked for a minute will be rare but after ten minutes will be well done Out experience teaches us that time and temperature are crucial factors affecting the quality of foods If we can understand a change mathematically we can predict how changes in our process will affect the quality of a product This section will be concerned with how the properties of food change with time and how we can use a quantitative understanding of this process to predict shelf life and optimal storage conditions A key concept here is we can be very vague about the reaction we are talking about yet be very precise about how it proceeds We will often define a parameter as food quality which can be any number of factors or combination of factors This could be a real chemical reaction such as the reaction of fat with oxygen but it might also be physical like crunchiness in a cookie or formation of visible defect or most probably consumer perception sensory scores or a combination of factors Later on in the course we will encounter some of the important reactions that determine food quality and the approaches studied here will help govern how they contribute to the success or failure of a product On completing this section you should be able to Identify and provide examples of kinetic changes in foods Understand and be able to properly use the terms reaction rate and reaction order Use rate equations to predict the changes in food as a function of time Use the Arhennius equation to predict changes in rate with respect to temperature Use rate equations and the Arhennius equation together to model the effects of time and temperature on a process Be able to design and understand the limitations of an accelerated shelflife test Elk WP 0 2 Rate Equations A chemist would start a description of chemical kinetics by designing an imaginary reaction where A is converted to B A9B You could then calculate how the concentration of A decreases and B increases as the reaction progresses In chemistry it is relatively easy to provide examples of what A and B are eg H202 reacts to form water and half a molecule of oxygen If we are concerned with the organic chemistry going on in foods we could do something similar For example during pasteurization vitamin C is thermally destroyed and studying the kinetics of the reaction would teach us something about how the nutritional quality of the drink would change In another example we might be interested in how spinach looses its green color during steaming If we know the green color of spinach is due to the pigment chlorophyll then we can use measurements of chlorophyll concentration as a chemical proxy for color Here chlorophyll is not interesting in itself but only as it tells us the color of the plant food However in other cases we might have no idea what chemical reaction is responsible for the changes in food quality we are interested in For example com akes sometimes become soft if they are stored too long coffee stored hot might become bitter and loose its characteristic avor meat roasted for long periods of time might become dry In all of these cases we cannot point to a chemical reaction as responsible for the changes we are interested in We can however say that some sort of reaction must have taken place and so write a general expression Reagents 9 Products We can define reagents and products in whatever way is convenient or us in the earlier chemical examples this would have been simply vitamin C and chlorophyll but in the later food examples it could be instrumental measurement of com ake texture sensory score of coffee quality of change in mass of the meat roast as it dries out in the oven The later examples are not defined chemical reactions but can be measured and will change as the reaction proceeds We will use mathematical models to predict how food will respond to different conditions and use that to estimate a maximum shelflife or optimum processing time If reagents are being converted to products then the rate of loss of one is equal to the rate of gain of the other note Products are always shown as Positive dRdP dt dt The rate of the reaction depends on the mechanism To illustrate this imagine we were modeling the process of bacterial growth Simply put to make a new bacterium the parent splits into two Bananum 9 Bananam Bananum a wuuldlead m At an mm 2 banana an mm 4 banana AL H mm a banana At lZUmm m banana At lSUmm 32 bananana Thenumberufbactenaas m anaaaasangwa nmebutsuxs Wm ai 1 a ma rate ufmcrease an the 7 N am numbaa afbananaas 1 Sn apnas mu man an the sa End in mm ans mm mm than mm mm aaa 5 a a n Essenuallythe same rea1mnquot as aaaumng bn dFdtxspmgessxvely annaasang Wa Rate k Raagans reacuun baaaaasa Reagents as rased tn the puwer 1 nn shuwn m naa equanun The depend un huw many are uaaaa maak panans settu farm a gel Tba panans wa are cuncemed aban are aasaans wbaab maaallas Tu makayagun the malk as fermented waua lanaa aaadbanana that aanvamba Expressed Lhas way the aaanaaan wa are cuncemed waua as am Casaan maaaua aasaan maaaua aasaan damaa 75m The rate quhe reacuun as a dcasem maaaua ma Ur dcasem damerdt an Dada Nabamamaa s 5cm ma aauaaa the rate ufabmzry an culhsmn as paapamanaa Lu W cuncentranun squared Tba obvaaaasay mas asaan a capamfax dmngthz mamas sub placesspxwpzdy casan mxeelles available 1 e Rate uf a gven mxcelle nnnng anuLher une r number avalable tunes the Mal number ufmmelles available x e tutal rate nfenlhsmns x number avalablex numba39 avalable m Rate rrdm1ellesdnmerrkmmellesz 1 reachun absulute rate 15 always ltcuncentrahungt ltnnnegt buttherate cunstzntxs m umts ltcuncenkanungt m ltt1megtl Cunsxda athn39d mure mm example the changem vulume ufsudam a dnnks dispenser m acunvemence stars We cuLle wntethe change as a reacuunquot Sudam cuntama39 Sada mspensea L WEIde expect atypical pluttu be sxmxlzrtu that snnwn m the gure Nutethatthe rate 15 funennn nur are the cups puuredatare lar huweverwecansee atthe 2w 2 vulume 15 appmxxmately 5 n n n e lmearwx tune The en gs g W m Sean vulume m 2 ngen E nnne depends unthe uw uf w v 5 5 n and huw therLy they are nut 391 un huwmuch sudaxsm me 15 cuntamer Sn the rate Rate rdsuda m cuntamer d unne k we the pussxbxhty ufsellmg a gene duesn39t depmd un huw nnuen 15 avalable Building off the logic of the previous paragraph we could hypothesize that k could be expressed as a function of the rate of customers entering the store times their average thirstiness but not on the volume of soda ie kk x ltrate of customgt x lt thirstinessgt So we have identi ed three separate formulae of the rates of reaction with different order Combining them we can see that in general Rate kreagentsquotkproductsn Where k is the rate constant of the reaction and n is the order 02 Finally consider a truly complex reaction for example the formation of brown color on the crust of a loaf as it bakes The color is due to pigments known as melanoidins forms from the reaction between amines from proteins and reducing sugars ie those with a free aldehyde group A simplified mechanism of the reaction is given in the Hodge scheme see later in the notes for a more complete treatment of this famously complex reaction Reducing sugar Amine lntermediates Brown pigment But that would clearly be a gross oversimplification ignoring the potential for cross reactions catalytic intermediates nonpigmented products etc In fact there are many pathways and competing reactions could lead to the colored product we are interested in and in all cases multiple reaction steps are involved each with different rates and orders In steady state we can still say Rate kbrown colorquot where k is the rate of a composite reactipn but because it is a composite reaction we need nolonger assume n has an integer value The equation Rate kreagentsn is a start but not areally that helpful It tells us that if we know two parameters k and n we can calculate the rate of change of something However rate of change is not what we need to know but rather the absolute amount as a function of time eg how much vitamin C is there after 10 min cooking or how much soda will be left by the end of the day We need to integrate the rate equation so if Rate 61 13 kPquot ch ch Then P 1413quot dt 2 Starting with a trivial example zeroth order reactions RatedPdtk therefore 2 In the context of this class we will not be concerned with noninteger reaction orders but you should be aware under what circumstances they may exist PJkdtkt t or RR0kt where R0 is the starting concentration of reagents So in the soda example if we know k is 10 l hr39l we can say that after 2 hours 20 liters of soda will have been dispensed Life gets slightly more complicated for the higher order reactions For the rst order case RatedRdtkR so 1 j dR jkdt R lnR C kt where C is an integration constant Knowing that at time zero R is at a maximum starting value Ro we can say lnR0C kgtlt0 C 1nR0 ln 3 kt R0 R R0 exp kt substituting This equation implies that the logarithm of the relative extent of the reaction EURO should be proportional to time This is easy to verify graphically There are also standard integrals for second order and noninteger reactions see Table 1 on page 1018 of the Fennema text Once we know order and rate constant we can use these equations to calculate concentration as a function of time From the form of the equations we can see that a plot of R with time should give a straight line when plotted against time only for a zeroth order reaction In RRo should give a straight line for a rst order reaction By plotting our data in this manner we can identify the order of the reaction which plot gives the best straight line and the rate constant what is the slope of the straight line plot Question 1 Rate Constants 1 The quality of a bagged salad product was judged by a consumer sensory panel and ranked from 10excellent to 1revolting The average sensory score of the products is shown in the data set provided What is the apparent reaction order of the salad deterioration process What is the rate constant of the process If salads became inedible at a sensory score less than 3 how long could you keep this product 2 Aspartame is an artificial sweetener used in drinks It breaks down during the storage of the product leading to a loss of sweetness Assuming the reaction is first order use the data provided to calculate the rate constant of the process and calculate the time for the concentration of aspartame to reach 50 of its initial value The rate constant for bacterial growth is 01 day39l If the likely number of a given bacteria needed to cause an infection is 1000000 how long would you store a food with an initial load of a l bacteria b 1000 bacteria E Question 2 Fm39ng Burgers 1 Beef burgers loose mass as they are fried Why Assume the reaction is first order Use the data provided to calculate the rate constant Question 3 Toxin Destruction 1 The rate constant for the destruction of an endogenous toxin during the cooking of shell sh is 005 unit depends on order 7 assume time unit to be minutes Assuming the initial toxin content is 15 mg kg39l calculate the toxin content after 1 and 50 min cooking assuming the reaction is 0 l or 2quotd Based on your previous answer and making whatever additional calculations and plots you find helpful how long would you have to run your experiment to confidently calculate the order of the reaction Why N 3 Temperature Dependency The rate of chemical reaction depends on the conditions eg temperature pH presence of catalysts or inhibitors It would be helpful to incorporate these terms into our rate equations so we could extend them to conditions other than those where measurements were made The most successful and general approach to this problem is the use of Arhennius kinetics which relate the rate constant to temperature E kk ex a 0 p RT where k is the rate at absolute temperature T Ea is the activation energy of the reaction a measure of how much faster the reaction goes when heated R is the gas constant 83 J K39l39 mol391and k0 a constant the rate at a reference temperature This equation tells us that as T increases the term in brackets decreases so the left hand side of the equation increases so the right hand side of the equation increases Meaning as temperature increases so does the reaction rate 7 the equation makes qualitative sense It is easier to deal with an equation containing an exponential by taking logs and turning it into a linear expression In this case taking the natural log of both sides 1r k 1r kg rm lT Cumpare ths equanun tn the en equanun ufa straghtlme lt5 any temperature Knuwmg temperature are pmgressmn quhe reacuun at a arffererrl temperature can then be calculated usmg une quherate equanuns develuped earner Ouemnn4 nnm a ml 39 39 39 Flmr ml MnH onlerepnyll x lne Important green plgnenl m wgemblex rnelmngpear anee are m zct two alnerl rdenlreal compound enlerepnyll a and b and bath break dawn dmngfeedpreeerrrngle yaw an nnpleemnl alxw green eeler ll 1 rnperlanl le eplrnrze lne eeelang sandman le nannrze enlerepnyll and nenee eeler relenlren However we do nel needm mow lne neenanrrn efenlerepnyll d2 radanan m erder le 012ng a nedelfer lne eeler lee rnpear Ifyau are rnlererledlne enenrrlry efenlerepnyll I clearly Explmned m Ecmm l2 2 2 m Fennema Read lne paperpmuded m elarr and anrwer lnefellewrng qnemenr 1 What pre smn m temperature and pH 15th reactur develuped by Ryerr Stunehzm capable7 Huw dues are readur autumaucallymcreasethe pH ufa sulunum 2 In Fxgures 375 why arethe authurs cun dent a rst urderreamunxs pruceedmg7 Huw dues Lharate ufnhlumphyll luss change mhpm Temperature 7 In Frgureu huw can yuu be cenam me eeuveuer energny maepenaenl upr 7 z pumts m5 Cnnkm meat l The rate cunstant furbmwn eeler furmanun cm are surface ufgnlled meatwas calculated as afuncnun uftemperature Flutthe data as an Axhenmus plat and calculate the amvatmn energy furthe bruwmng reacnun Temp C mu lZEI 14m mu 18m znu Rate constant h391 0001591 0001875 0002174 0002487 0002811 0003146 2 The rate constant for the loss ofvitamin B on cooking meat is 05 h391 at 100 C The activation energy for the reaction is 18 kJmol391 What would be the rate constant of degradation at 450 F Plot the ratetemperature function for vitamin loss on the same axis as used for the vitamin loss data above Assume you wanted to make a brown product with a high vitamin content 7 use this Figure to explain what cooking temperature would you choose and why LA 4 Accelerated Shelf Life Testing In de ning equations to describe the way foods change over time we have created a model for the process As we saw in the soda example the model need not be perfect but should hopefully give a good approximation of how the process will proceed over time One value to this is we can now predict how a system will change without doing experiments Within the time range we have measured we know the model fits the data well so we have great confidence in the model but to make a prediction outside this range we will run into problems of extrapolation For a simple and well characterized chemical reaction this is often OK but may be less reliable in a food Furthermore food systems are such complex materials full of uncharacterized catalysts and inhibitors of reactions it is hard to use a literature value of rate constant to make a prediction We must often rely on our own measurements of the rate constant to estimate the time course of the reaction Measuring a rate constant is relatively easy provided you have good experimental data For that we need a good measurable change in the parameter of interest ie many times larger than the precision of the experimental error so we can i recognize the shape of the function and decide an appropriate order for the reaction and ii f1r a rate equation to the data and measure the rate constant If the reaction is very slow e g the rate of lipid oxidation in dried beef we could calculate the rate constant only after making measurements over several years This approach clearly cannot work in product development Instead we must find a way to accelerate the reaction usually by heating to get a measurable rate constant and order in a reasonable time then calculate how much more slowly the reaction proceeds at a lower temperature Thus the experimental protocol is i Calculate the rate of the reaction at several elevated temperatures by measuring property vs time and fitting an appropriate rate equation ii Plot the rate data In k vs reciprocal absolute temperature lT in an Arhennius plot Fit a straight line to the data and extrapolate to the temperature the product is actually going to be stored Read off the rate of the reaction at that temperature iii Plug the rate into a rate equation to predict how property changes with time at the storage temperature We have predicted the shelf life of a product in less time than it would take for the shelf life to degrade naturally ie this is an accelerated shelf life test There are substantial problems with this approach and it must be treated with caution First amongst these is to question whether the reaction you are studying at the elevated temperatures is the same reaction taking place at the storage temperature For example 0 Bacterial growth is a substantial issue in vegetables stored at warmer temperatures while enzymatic degradation is important at low temperatures 0 Ice coarsening limits the shelf life of ice cream at 710 C but clearly not at elevated temperatures A more systematic explanation of this problem is to suggest there are two reactions with different activation energies and reaction rates at a given high temperature During the accelerated tests we would measure the quotredquot reaction because that is what occurs faster extrapolating this would give us no idea of the situation at lower temperatures when the quotbluequot reaction would dominate If you labeled blue as quality and red T as microbial safety this diagram would give a 1 lot quotColdquot good explanation of high temperature short time processing An example of this is the storage of dried mashed potato at lipid oxidation dominates at low temperatures and Maillard browning dominates at high temperatures Accelerated amp Storage temp Log Rate A second major problem is accounting for experimental error during the multiple curve tting steps in this procedure Particularly in the Arhennius step the natural log of the rate constant is calculated by tting a straight line to the available data There is clearly some error in this procedure and even a tiny error in lnrate constant can lead to a huge difference in rate constant for example miscalculating lnrate constant as 41 rather than 40 N15 error would lead to an almost 10 error in rate constant Rather than calculate the whole degradation curve it is sometimes preferable to assign a critical value for the measured parameter that is the minimum acceptable the shelf life is the time taken for that parameter to deteriorate to the critical value under storage conditions Ted Labuza proposes a systematic series of steps to conduct accelerated shelf life test and see section 1736 in the Fennema text ASLT 1 Decide What are the acceptable limits for your product Consider the nutritional and safety implications of spoilage rst then the loss of quality Most product qualities will degrade over time what are the limits your customers are willing to accept and which reaction is limiting ASLT 2 What kinds of reactions are likely to be important Finding out a chemical mechanism for the degradation is important as it will give you insights into how to control it how to measure it and possible harmful outcomes you have not considered ASLT 3 How will you pack the food You should package your samples for shelf life testing in a similar way to the real food is stored especially consider light exposure and humidity bearing in mind that there may be limited space to store experimental samples ASLT 4 What storage temperatures should you use This is a key question Labuza makes some suggestions for different types of food but in each example consider that you will need enough heating to produce a change in reaction rate and not enough to fundamentally change the product or the reactions spoiling it ASLT 5 How long do you need to store at each temperature You want your sample to change appreciably but you don t want to make unnecessary measurements ASLT 6 How will you test and how often The analytical method chosen can be as specific as a chromatographic determination of an important compound or as general as a sensory score for the product Remember to select an analytical method that corresponds as closely as possible to the limits set out in 1 Test as often as you can afford Try to space out your testing so the amount of change in parameter between tests is constant rather than the amount of time before tests ASLT 7 Plot kinetic curves and calculate k and 11 Find the best rate equation to fit your data ASLT 8 Plot 10 a ainst 1T ie Arrhenius and calculate k at storage temperature ASLT 9 Plot the kinetic curve at the storage temperature and calculate shelf life Use the same equation used in 7 and the rate constant calculated in 8 The limit of parameter is that set out in l Reading exercise Executive summary of the WHO report on the development of nutrition bars for refugee populations What chemical issues must be overcome to develop a useful nutrition bar Do you think it is advantageous to supplement the bar How would you decide how much of a given vitamin to supplement the bar with Question 6 ASLT Testing Use this data from an ASLT test to calculate the level of folic acid supplementation needed for a refugee bar to be a useful source of the vitamin after 30 months storage at 30 C The vitamin content was measured as a function of time at four elevated temperatures Assume vitamin degradation follows rst order kinetics calculate 1 The rate constant at each elevated temperature 2 Plot an Arhennius curve and calculate the rate constant at 30 C 3 If one serving of the bar must contain the RDA of this vitamin after 30 months at 30 C at what level must the bar be supplemented Vitamin content normalized to initial value Time at 80 C at 100 C at 120 C at 140 C at 160 C months 0 1020 1000 1000 1000 1000 05 0985 0933 0821 0637 0361 1 0974 0874 0652 0375 0136 15 0949 0812 0557 0258 0028 2 0914 0749 0447 0167 0006 25 0876 0707 0389 0072 0009 3 0844 0636 0298 0053 0001 35 0826 0607 0270 0059 0023 0835 0569 0225 0019 0024 45 0781 0517 0190 0001 0002 5 0801 0482 0122 0021 0021 55 0772 0461 0098 0008 0008 6 0729 0428 0072 0012 0006 Do you recommend supplementation


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