Chapter Three – Chemical Equations - the Alfred State College ：三章–化学方程式-艾尔弗雷德州立大学

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CHAPTER 3 CHEMICAL BONDING 化学键

The arrangement of ions in a crystal lattice depends on the relative sizes of the ions
STRUCTURE AND BONDING
The physical properties of a substance depend on its structure and type of bonding present. Bonding determines the type of structure.
• Co-ordinate (dative covalent) bonding
• Molecular solids • Covalent networks • Metallic bonding
STRUCTURE AND BONDING
The physical properties of a substance depend on its structure and type of bonding present. Bonding determines the type of structure.
THE IONIC BOND
Ionic bonds tend to be formed between elements whose atoms need to “lose” electrons to gain the nearest noble gas electronic configuration (n.g.e.c.) and those which need to gain electrons. The electrons are transferred from one atom to the other. Sodium Chloride

chemicalreactionengineering3ededition作者octavelevenspiel课后习题答案

Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING .......................................... 错误!未定义书签。

CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS ........................................................ 错误!未定义书签。

CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA ..................................................... 错误!未定义书签。

CHAPTER 4 INTRODUCTION TO REACTOR DESIGN ............................................................... 错误!未定义书签。

CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR........................................................... 错误!未定义书签。

CHAPTER 6 DESIGN FOR SINGLE REACTIONS ....................................................................... 错误!未定义书签。

CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR ....................................................... 错误!未定义书签。

化学基础英文3_chemical_reactivity化学反应性

Chemical reactions are commonly written as the following equations:
3-2 Reaction Classification
If you scan any organic textbook you will encounter a very large, often intimidating, number of reactions. These are the "tools" of a chemist, and to use these tools effectively, we must organize them in a sensible manner and look for patterns of reactivity that permit us make plausible predictions. Most of these reactions occur at special sites of reactivity known as functional groups, and these constitute an organizational scheme that helps us catalog and remember reactions.
Ultimately, the best way to achieve proficiency in organic chemistry is to understand how reactions take place, and to recognize the various factors that influence their course.
A rearrangement reaction generates an isomer, and again the number of bonds normally does not change.

英国化学教材 IGCSE第3章-Chemical Caculation

The mole:the amount of substance with the Avogadro Constant of the particles is calledthe mole.
Why we use the Avogadro number?
(1)The mass of one atom, or even 1000 atoms, is too small to be weighed on a balance;
So the substances taking part in the reaction are always in the same proportion.
And note that the mass on each side of the arrow is the same mass before and after the reaction. That is to say,the mass does not change in a chemical reaction.
limiting
excess
calculate the mass of product
6 g
Easyfor calculation of mass of product
Quick test:
onepiece of marble, 0.3 g, was added to a solution of hydrochloric acid containing 0.05molofHCl.
7. Knowing simple method for getting empirical formula and do related calculation;
8.Knowing the principle, process and calculating the concentration related to titrations.

315化学复习指南2025

315化学复习指南2025English Answers:Chapter 1: Stoichiometry.Define stoichiometry and explain its importance in chemical reactions.Calculate the mole ratio between reactants and products using balanced chemical equations.Determine the limiting reactant and calculate the theoretical yield of a reaction.Convert between mass, moles, and volume using stoichiometry.Solve problems involving percent yield and excess reactants.Chapter 2: Gases.Describe the properties of gases and the kinetic molecular theory.Apply the ideal gas law (PV = nRT) to solveproblems involving pressure, volume, temperature, and moles.Explain the concepts of partial pressure andDalton's law.Calculate the molar mass of a gas using its density or effusion rate.Understand the behavior of real gases anddeviations from the ideal gas law.Chapter 3: Solutions.Define solutions and their components.Express solution concentrations in units ofmolarity, molality, and percent composition.Calculate the mass, volume, and molarity ofsolutions using dilution and mixing formulas.Explain colligative properties, including vapor pressure lowering, boiling point elevation, and freezing point depression.Perform calculations involving titration and neutralization reactions.Chapter 4: Chemical Reactions.Classify chemical reactions and predict their products.Balance chemical equations using half-reactions or the oxidation number method.Determine the reaction type (e.g., redox, acid-base, precipitation) and identify the reactants and products.Predict the products of reactions involving ionic compounds, acids, and bases.Calculate the enthalpy change (ΔH) and entropy change (ΔS) of reactions using Hess's law and the third law of thermodynamics.Chapter 5: Equilibrium.Define chemical equilibrium and explain the concept of Le Chatelier's principle.Write equilibrium constant expressions andcalculate equilibrium constants from reaction data.Use equilibrium constants to predict the direction of reactions and the extent of reactions.Explain the effects of concentration, temperature, and pressure on equilibrium.Solve equilibrium problems involving gas reactions, acid-base reactions, and solubility equilibria.Chapter 6: Acids and Bases.Define acids and bases according to the Arrhenius, Brønsted-Lowry, and Lewis theories.Calculate pH and pOH using the pH scale and the autoionization of water.Perform acid-base titrations and calculate the equivalence point and molarity of solutions.Explain the buffer system and its role in maintaining pH.Solve problems involving acid-base reactions, including neutralization, hydrolysis, and salt formation.Chapter 7: Thermodynamics.Define entropy (S) and enthalpy (H) and explain the second law of thermodynamics.Calculate ΔS and ΔH for reactions using standard enthalpies of formation and standard entropies.Predict the spontaneity of reactions using ΔG =ΔH TΔS.Explain the concepts of chemical potential and electrochemical cells.Solve problems involving heat transfer, entropy changes, and electrochemical reactions.Chapter 8: Nuclear Chemistry.Describe the structure and properties of atomic nuclei.Explain radioactive decay and calculate half-lives and decay rates.Classify nuclear reactions, including alpha decay, beta decay, and nuclear fission.Discuss nuclear energy, radiation safety, and applications of radioactive isotopes.Solve problems involving nuclear reactions and radioactive decay.Chapter 9: Organic Chemistry.Name and draw structures of organic molecules, including hydrocarbons, alcohols, aldehydes, ketones, and carboxylic acids.Explain the principles of organic reactivity, including nucleophilic substitution and electrophilic addition.Classify organic reactions as addition, elimination, substitution, or rearrangement reactions.Describe the mechanisms and stereochemistry of organic reactions.Solve problems involving organic synthesis and reaction mechanisms.Chapter 10: Biochemistry.Describe the structure and function of biological molecules, including carbohydrates, lipids, proteins, and nucleic acids.Explain the principles of metabolism, including glycolysis, the Krebs cycle, and oxidative phosphorylation.Discuss the role of enzymes in biological reactions and their regulation.Describe the processes of DNA replication, transcription, and translation.Solve problems involving biochemical pathways and enzyme kinetics.中文回答：第一章，化学计量。

CHEMISTRY test

CHEMISTRYUNITS 3 and 4PRETEST DATA PAGEUse the data tables below and the periodic table over the page to help you answer the pre-test questions.Table 1: Formulae and charge of selected ions:Cations Na+ sodium NH4+ ammonium Ag+ silver Mg2+ magnesium Zn2+ zinc Cu2+ copper Pb2+ leadAnions OH hydroxide Cl chloride NO3 nitrate O2 oxide SO42 sulfate CO32 carbonate PO43 phosphateTable 2: Solubility of ions in water:Ions that are soluble in waterCompounds with the following ions are always soluble in waterCompounds with the following ions are mostly soluble in waterAmmonium NH4+ Sodium Na+ Potassium K+ Nitrate NO3Chloride Cl except with silver, Ag+, or lead, Pb2+ Sulfate SO42 except with silver, Ag+, or lead, Pb2+Ions that are not soluble in waterCompounds with the following ions are mostly not soluble in waterHydroxide OH Sulfide S2 Carbonate CO32 Phosphate PO43except with Ammonium NH4+ Sodium Na+ Potassium K+Table 3: Molar mass of selected elements:ElementHCONeNaMolar mass (g per mol)1.012.016.020.123.0ClPb35.5207.2Table 4: Molar mass of gasesSTP standard temperature and pressure temperature 0°C, pressure 101.3 kPaSLC standard laboratory conditions temperature 25°C, pressure 101.3 kPaOne mol of any gas occupies 22.4 litres at STPOne mol of any gas occupies 24.5 litres at SLCBefore you startPage 2CHEMISTRYUNITS 3 and 4PRETEST QUESTIONS1. Use the Periodic Table given on the previous page to help you answer the following questions.(a) What is the atomic number of the following elements(i) silver(ii) radon(iii) radium½ + ½ + ½ = 1½ marks(b) What is the chemical symbol of the following elements(i) gold(ii) silicon(iii) sodium(iv) antimony½ + ½ + ½ + ½ = 2 marks(c) Identify the element with 13 protons and 13 neutrons.1 mark2. Butane, C4H10 is commonly used in LPG and lighter fluid. It’s melting and boiling points are shown below.melting point/ freezing (solidification) point boiling (evaporation) point/ condensation point138 ºC 0.5 ºCCircle the physical state of butane at room temperature, 25 C at fridge temperature, 4 C at 5 C solid solid solid liquid gas liquid gas liquid gas½ + ½ + ½ = 1½ marksUse the table of ions (Table 1) on the data page to help you answer questions 3 and 4.3. Identify the ions present in the following ionic compounds and name the compound.(a) MgCl2(b) Na3PO4(c) Ag2S3 marks4. Work out the chemical formula of the following compounds.(a) Silver hydroxide(b) Magnesium carbonate (c) Zinc nitrate3 marks5. Balance the following chemical equations.(a) Mg(s) + N2(g) Mg3N2(s)(b) Ca(S) + O2(g) CaO(s)(c) H2(g) + O2(g) H2O(l)(d) Al (s) + I2 (l) Al2I6(s)4 marksBefore you startPage 3Pre-test questions continued…6. Use the solubility table (Table 2) on the data page to determine which of the following ionic compound are soluble.(a) lead sulfate(b) silver chloride(c) ammonium chloride(d) zinc carbonate½ + ½ + ½ + ½ = 2 marks7. The compound Cu(NO3)2 dissolves in water by dissociation of ions (a) Identify the ions in the compound. (b) Write the ionic equation of the dissociation reaction. Show the state of each compound or ion. 1 + 2 = 3 marksRefer to the molar mass of elements (Table 3) on the data page to help you answer questions 8, 9 and 10.8. Find the molar mass of the following compounds(a) H2O(b) NaCl2 marks9. Find the mass of 1.3 mol of CH41 mark10. Calculate the number of mol contained in the following samples, rounded to three decimal places.(a) 50 g of lead(b) 62 g of NaCl2 marksRefer to the molar volume of gases (Table 4) on the data page to help you answer the following question. 11. (a) Work out the volume of 1.4 mol of chlorine (Cl2) at STP(b) Work out the mass of 2.8 L of neon (Ne) gas at SLC 1 + 1 = 2 marksThe table below summarises the different types of acid reactions:Reactions of acids 1. acid + metal salt + hydrogenThis reaction does not occur with Cu, Hg, or Ag. 2. acid + metal carbonate salt + water + carbon dioxide3. acid + metal oxide salt + water4. acid + metal hydroxide salt + water Carbonates contain the CO32 ionThe chemical formulae of selected compounds.HCl (aq) Na2CO3 (aq) NaCl (aq) H2O (l) CO2 (g) CuO (s)CuCl2 (aq) H2O (l) H2SO4 (aq) NaOH (aq) Na2SO4 (aq)Before you startPage 4Pre-test questions continued…12. Use the summary of acid reactions shown above to predict the products of the following reactions: (a) HCl (aq) + Mg (s) (b) HCl (aq) + Na2CO3 (aq) (c) HCl (aq) + CuO (s) (d) H2SO4 (aq) + NaOH (aq) 1 + 1 + 1 + 1 = 4 marks13. Are the following reactions oxidation or reduction reactions? (a) Cl2(g) + 2e 2Cl (aq) (b) Pb (s) Pb2+ (aq) + 2e½ + ½ = 1 mark14. The reaction Zn(s) + S(s) ZnS (aq) can be written as two half reactions: Zn(s) Zn2+(aq) + 2e and S(s) + 2e S2 (aq)(a) Identify the oxidant in this redox reaction (d) Identify the reductant in this redox reaction½ + ½ = 1 mark15. Iron reacts with hydrochloric acid according to the ionic equation Fe(s) + 2H+(aq) Fe2+(aq) + H2(g)(a) What has been oxidised in this reaction? (b) Write a half equation for the oxidation reaction. (d) What has been reduced in this reaction? (e) Write a half equation for the reduction reaction.½ + 1 + ½ + 1 = 3 marks16. Hydrogen gas and oxygen gas react to form water according to the reaction 2H2 (g) + O2 (g) 2H2O (l)10 g of oxygen is reacted in excess hydrogen.(a) How many mol of oxygen was reacted? (b) How many mole of hydrogen is required? (c) Work out the mass of hydrogen reacted.1 + 1 + 1 = 3 marksEND OF PRE-TESTBefore you startPage 5PRE TEST ANSWERS1. (a) The atomic number of (i) silver 47(ii) radon 86(iii) radium 88(b) The chemical symbol of (i) gold Au (ii)silicon Si(iii) sodium Na (iv) antimony Sb(c) The element with 13 protons and 13 neutrons is Aluminium2.Circle the physical state of butaneat room temperature, 25 C solid liquid gasat fridge temperature, 4 C solid liquid gasat 5 C solid liquid gas3. Identify the ions present in the following ionic compounds and name the compound.(a) MgCl2 Mg2+ Cl(b) Na3PO4 Na+ PO43(c) Ag2S Ag+ S2Magnesium chlorideSodium phosphateSilver sulfide4. Work out the chemical formula of the following compounds.(a) Silver hydroxide(b) Magnesium carbonateAgOHMgCO3(c) Zinc nitrate Zn(NO3)25. Balance the following chemical equations.(a) 3Mg(s) + N2(g) Mg3N2(s)(b) 2Ca (s) + O2(g) 2CaO(s)(c) 2H2(g) + O2(g) 2H2O(l)(d) 2Al(s) + 3I2(l) Al2I6(s)6. Use the solubility table (Table 2) on the data page to determine which of the following ionic compound are soluble.(a) lead sulfate is insoluble(b) silver chloride is insoluble(c) ammonium chloride is soluble(d) zinc carbonate is insoluble7. The compound Cu(NO3)2 dissolves in water by dissociation of ions (a) The ions in the compound. Cu2+ NO3 (b) The ionic equation of the dissociation reaction, showing the state of each compound or ion.H 2OCu(NO3)2 (s) Cu2+ (aq) + 2NO3(aq)Before you startPage 68. (a) Mr(H2O) = 18.0 g/mol(b) Mr(NaCl) = 58.5 g/molPre-test answers continued…9. The mass of 1.3 mol of CH4 = 1.3 16.0 = 20.8 grams10. The number of mol, to three decimal places(a) 50 g of lead = 50 mol = 0.241 mol207.2(b) 62 g of NaCl = 62 mol = 1.059 mol58.511. (a) (b)The volume of 1.4 mol of chlorine (Cl2) at STP = 1.4 22.4 L = 31.36 Litres Work out the mass of 2.8 L of neon (Ne) gas at SLC = 2.8 20.18 = 2.306 grams24.512. Use the summary of acid reactions shown above to predict the products of the following reactions: (a) 2HCl (aq) + Mg (s) MgCl2 (aq) + H2 (g)(b) 2HCl (aq) + Na2CO3 (aq) 2NaCl (aq) + H2O(l) + CO2(g)(c) 2HCl (aq) + CuO (s) CuCl2 (aq) + H2O(l)(d) H2SO4 (aq) + 2NaOH (aq) Na2SO4 (aq) + 2H2O(l)13. Are the following reactions oxidation or reduction reactions? (a) Cl2(g) + 2e 2Cl (aq) Reduction reaction (b) Pb (s) Pb2+ (aq) + 2e Oxidation reaction14. (a) The oxidant in this redox reaction is S(s) (d) The reductant in this redox reaction Zn(s)15. (a) Fe(s) been oxidised in this reaction? (b) The oxidation reaction is Fe(s) Fe2+(aq) + 2e (d) H+(aq) has been reduced in this reaction? (e) The reduction reaction is 2H+(aq) + 2e H2(g)16. (a)(b) (c)The mol of oxygen was reacted = 10g = 0.3125 mol32.0The mole of hydrogen required = 2 0.3125 = 0.625 mol The mass of hydrogen reacted = 0.625 2.0 = 1.25 gramsBefore you startPage 7Your pre-test result… What we recommend …If your score was less than 20 40If you got less than half the pre-test right you will most likely need a lot of time and support to make a success of Year 12 Chemistry. Past experience has shown that students scoring less than 20 out of 40 are not able to continue with this subject because they find it too difficult. We strongly recommend that you contact the Chemistry teachers to discuss your options.If you scored between 20 and 30 you will most likely need revision support throughout the year to develop the skills expected. This means making extra time available for your studies, and perhaps finding a tutor to help you. We suggest that you consider realistically whether you will be able to make the extra study time available. Contact the Chemistry teachers to discuss your options.If your score was between 20 and 30 40 40If you scored between 30 and 40 you should be able to cope with most of the skills expected. However, you will also need to make regular study time a part of your weekly schedule. If you have any concerns, contact the DECV, and speak to the Chemistry teachers.。

Designing Microreactors in Chemical Synthesis – Residence-time Distribution

Designing Microreactors in ChemicalSynthesis –Residence-time Distributionof Microchannel DevicesKLAUS GOLBIGANSGAR KURSAWEMICHAEL HOHMANNSHAHRIYAR TAGHAVI-MOGHADAMTHOMAS SCHWALBECPC-Cellular Process Chemistry Systems GmbH,Mainz,GermanyThis contribution deals with the application of numerical methods in CPC-Systems’microreaction development process and demonstrates the feasibility and significantbenefits for the suggested design method.It is focused on the design of capillary resi-dence tubes,which is necessary if the residence time provided by the original micro-reactor is not sufficient to complete the reaction.The residence time distribution iscalculated from a straightforward numerical model based on the common assump-tion that axial gradients can be neglected.The results can be adapted easily to othercapillary diameters or reaction conditions.As an example,the method is applied tothe case of sequential synthesis.Keywords :Microreactor;Residence time distribution;Sequential synthesis;Dispersion;Fluid dynamicsIntroductionSynthesis via MicroreactorsMicroreaction devices are beneficial innovative tools in the improvement and opti-mization of reactions as well as in fast supply of sufficient quantities of target com-pounds.This exciting new technology improves control of reaction parameters such as temperature and concentration equipartition and thus allows in many cases higher yields,fewer by-products,and higher selectivities.Therefore it has become increas-ingly interesting for small-scale production and mobile reaction systems,e.g.,for automotive applications (Wegeng and Drost,1998).In order to realize these benefits a proper design of the microreaction system is crucial:heat transfer area,mixing channel dimensions,and flow capillaries must be sized very carefully to achieve,for example,reasonable mixing,pressure drops,andReceived 2March 2001;in final form 19June 2003.Address correspondence to Klaus Golbig,CPC-Systems GmbH,Hanauer Landstrasse 526,G58III,Frankfurt,Mainz 60343,Germany.E-mail:wille@620m.,192:620–629,2005Copyright #Taylor &Francis Inc.ISSN:0098-6445print =1563-5201onlineDOI:10.1080=00986440590495197Designing Microreactors in Chemical Synthesis621 flow equipartition(Ehrfeld et al.,1997).To open the field of general application of microreactors in chemical synthesis,a comprehensive analysis of the requirements for such a microreactor has to be performed.Such an analysis identifies the design parameters for multipurpose microreactors,based on a feed rate corresponding to a typical bench-scale synthesis.The resulting solution should be suitable for a variety of applications with reactions or transformations of miscible liquids.Microreactor Approach to a Better ChemistryCPC-Systems,Cellular Process Chemistry Systems GmbH,located in Mainz (Germany),is engaged in the development and optimization of standardized modu-lar microreaction systems as well as in their application to organic synthesis.This approach is based on the idea of shortening development periods of new active substances by consequent numbering-up.This means that the same microreactor units are used in laboratory and in pilot plant,thus avoiding the scale-up risk.Numerous microreactor applications are stated in the current literature that can be considered as proof of the principle of microreaction technology in general. Microreactors have been used in commodity synthesis,i.e.,ethylene oxide(Richter et al.,1998),the preparation of hydrocyanic acid(Hessel et al.,1999)and for poly-merizations.CPC-Systems’modular microreaction system,based on the CYTOS1 microreactor,can be applied to promising fields in drug discovery and development processes.For instance,chemical syntheses of different targets like quinoline acid derivates(Ciprofloxacin1)and the Paal-Knorrpyrrole synthesis have been investi-gated(Taghavi-Moghadam et al.,2001).Further examples from our day-to-day microreactor lab experience could be found in Schwalbe et al.(2002)and Autze et al.(2000);they cover the complete range from common nitrations over rearrange-ments to Suzuki couplings and Wittig-Horner reactions.Furthermore,the CYTOS1 modular microreaction system can be used in a setup for sequential synthesis to generate compound libraries.In comparison to conventional parallel synthesizers,sequential synthesis offers higher flexibility because the reaction conditions can be controlled independently and individually for each reaction pared to the limited reaction vol-ume in a reaction block of a parallel synthesizer,access to variable amounts of target compounds is ensured by a system running continuously,even if the reaction sequence is demanding or divergent.The automation of the system,which can be established by using an auto-feed-sampler and a fraction collector,leads to perform-ing a maximum of chemistry with minimum effort.The automated microreaction system also opens up an easy way to automatic reaction optimization.For this purpose a mathematical software generates a statisti-cal design of experiments and evaluates the results automatically via inline analytical sensors.Therefore CPC-Systems’microreaction systems are suited for the pro-duction of specimens as well as for the process development.Design Model for Capillary FlowsThe ProblemCPC-Systems’CYTOS1microreaction system,shown in Figure1,is composed of a pumping module,an exchangeable microreaction unit,and an optionalresidence-time providing module,all connected by a convenient bayonet coupling.The residence-time providing unit,for example,a stainless steel capillary,is required for realization of larger reaction times.For sequential operations the residence-time distributions (RTD)of the micro-reactor and the residence module have to be considered carefully because they may suffer from the laminar flow regime in the capillary tubing,especially if dead volumes are present.On the other hand,diffusive mixing can be very fast on a small scale,so that laminar flow is not necessarily considered a drawback.If more than one capillary is operated in parallel,flow equipartition also becomes an important issue.However,laminar flow regime is known to be modeled very precisely with minor effort.In our microreaction system corrosion-resistant ceramic piston pumps are used,in contrast to some micro-analytic devices,e.g.,electrophoresis,which are based on electroosmotically driven flows.The velocity profiles differ in both cases.In a pressure-driven capillary,the laminar velocity profile in Figure 2shows a pro-nounced parabolic shape,where the liquid at the wall is nearly stagnant.Influenced by such a velocity profile a concentration peak injected at the inlet will soonbeFigure 1.CYTOS 1microreaction system,the first commercial turn-key microreaction system.622K.Golbig et al.dispersed,yielding a broad residence-time distribution at the outlet.Nevertheless,the molecular diffusion in radial direction limits this peak broadening.In this article design rules consider both effects.ModelingThere are only a few sources available on this particular problem.In the engineering literature usually only an axial dispersion process is considered (Levenspiel,1958,1999),which was shown first by Taylor (1953)to be sufficient.He developed an analytical solution of the interaction between radial diffusion and axial convection in tubes.On the assumption that axial concentration gradients can be neglected in comparison to radial gradients and the introduction of a relative coordinate system moving with the mean velocity of the fluid,he was successful in reformulating the original problem into a simpler unidimensional axial dispersion process.The role of this process in the case of heterogeneous gas-phase reactions was investigated by Matlosz and coworkers (Commenge et al.,2001).In order to uncover the details of this peak-broadening phenomenon,a numerical approach was developed that is more flexible regarding geometry and boundary con-ditions.The model uses finite volume balances as well as the following assumptions:.Fully developed parabolic velocity profile .Restriction to radial diffusion .Conservation of mass (no reaction).Discretization in axial and radial direction .Limitation of maximum concentration change to 10%by appropriate choice of D t .Length of one axial discretization element D lD l ¼D t Áu maxThe program computes the diffusivec i ;j ;k þ1¼c i ;j ;k þD n i À1;k ÀD n i ;j ;k p D l ðr i Àr i À1Þwith D n i ;k ¼2p D D l r i c i ;j ;k Àc i þ1;j ;k r i þ1Àr iand the convectional mass transport from cell i,j À1to cell i,jc i ;j ;k þ1¼ð1Àg i Þc i ;j ;k þg i c i ;j À1;kin an alternating mode.In the equations n denotes amount of moles,c the cell concen-tration,D the diffusion coefficient,r i the outer radial position of the cell i ,and g itheFigure 2.Velocity profile and peak broadening inside a capillary tube due to combination of axial convection and radial diffusion.Designing Microreactors in Chemical Synthesis 623velocity weighting factor of cell i.The velocity weighting factor g i represents the ratio of fluid leaving cell i,j À1into the next cell i,j downstream during the time period D t .g i ¼1Àr 2i þr 2i À12R 2where R denotes to the inner radius of the capillary.The subscripts contain infor-mation about the radial and axial position of the cell,and the last subscript indicates the numeric operation step.Simulation ResultsAs presented in Figure 3,the resulting residence-time distribution of a short capillary for each individual radial position clearly shows that material is short cutting through the center whereas the RTD curves near the wall are delayed.Generalized RTD Diagram.To avoid time-consuming calculations,the fact is used that,simply put,the calculation procedure is just computing the same figures of dimensionless concentrations in each run.Only the scaling of the time steps differs depending on the diffusion time constants d 2=D .Thus,the resulting dimensionless RTD curve is dependent on only one parameter,the dimensionless diffusion time or Fourier number Fo d :Fo d ¼t mean Dd where t mean is mean residence time,D is diffusivity,and d 2is tube diameter is1.Figure 3.Residence-time distribution for capillary with 5min mean residence time and1.75mm inner diameter (D ¼6.9Á10À10m 2=s).624K.Golbig et al.Thus,previously generated output data can be reused for other conditions.Examples for dimensionless RTD curves are illustrated in Figure 4.Each curve represents a set of many different cases that are analogous from a physical point of view.For the required case,only the Fourier number Fo d has to be determined,and the corresponding calculation run can be taken from the output database,which should be arranged in terms of Fo d number.Dimensionless Dispersion Curve.This straightforward numerical model was vali-dated by experiments with capillaries and toluene as tracer,which was analyzed by gas chromatography (GC).The double logarithmic plot in Figure 5reflects the width of the RTD curve peaks at half of their height showing a minor deviation between the experimental results (plotted as dots)and the numerical calculated dis-persion curve (dashed line),but we found excellent agreement between the theory of Taylor (Levenspiel,1999;Taylor,1953)(gray line)and our calculated dispersion curve.During the evaluation it became apparent that there is a need for faster solvers that give results for high Fourier numbers Fo d greater than three.One possibility to reach this aim is to exploit the fact that the dispersion process in the laminar flow regime can be regarded as a linear signal transmission process.Thus,the trans-fer functions of two capillaries,which have to be determined first,can be convo-luted.Then the convolution is done from each inlet radial position of the first capillary to each outlet position of the second one by averaging over all possible combinations.The advantage of such a procedure is that it proceeds exponentially in time or capillary length.Although convolutions are very time-consuming calculations,they offer a speed advantage for high Fo dnumbers.Figure 4.Generalized RTD curves.Designing Microreactors in Chemical Synthesis 625Application to Sequential SynthesisOne major application of microreaction systems is the generation of a variety of substances by automatically exchanging the starting materials (and if necessary also modifying the reaction conditions).A photo of such a sequential synthesis setup using the SEQUOS 2Lab System is given in Figure 6.An auto-sampler and a fraction collector complete the setup of the CYTOS 1microreaction system.The individual substances are separated by short purge-flushes of solvents.The intensity of these purge-flushes can be modified in order to optimize the process.While too long flushes waste operating time,too short flushes may result in cross-contamination of the different substances.Figure 7illustrates an example with short pumping and purging periods.Reac-tant A is pumped over a time period of t ¼4min 30s,the spacer over 2min 20s.Because of the RTD,the resulting response at the system’s exit after 9min 40s is slightly broadened.The product is collected directly after the elapse of the mean residence time for the adjusted pumping time.While collecting the first reactant A,the next reactant B is already fed into the reactor separated by the spacer.It can be clearly seen that both reactant peaks are nevertheless separated.An inte-gration of the RTD signal yields that 85%of the fed reactants are collected,slightly diluted by spacer solvent.This example clearly demonstrates that the residence-time distribution of the system –microreactor and residence tubing –has to be taken into account and is a crucial part of the system’s functionality.In the example of Figure 8a longer pumping time was applied in order to obtain a sample of higher quality (i.e.,lessFigure 5.Dimensionless dispersion curve.626K.Golbig et al.dilution with spacing solvent),slightly increasing the spacing time from 140s up to 270s and improving the separation of both sequences.This setup allows the collect-ion of 77%of the whole pulse with only 1.3%dilution by spacing solvent.Theore-tically,it is possible to gather all spent material if the purging time is as long as the total peak width from the Diracpulse.Figure 7.Short cycle times for high throughput (SEQUOS 2microstructured residence time modules having 9min 40s mean residence time,pumping time 270s,spacing time 140s).Figure 6.Sequential synthesis with SEQUOS 2microreaction system.Designing Microreactors in Chemical Synthesis 627Conclusion and OutlookThus it can be stated that.CPC-Systems has successfully implemented a realistic numerical model for residence-time distribution of capillaries.Residence capillaries have been designed with adjusted trade-off between peak-broadening and pressure drop demands.Calculated data can be converted easily to actual conditions.Experiments can be planned accurately inadvance.Figure 8.Sequential synthesis with longer cycle times for overall higher product quality and better sequence separation (SEQUOS 2microstructured residence time modules having 9min 40s mean residence time,9min 40s pumping time,and 4min 30s spacingtime).Figure 9.Temperature distribution in a flat microchannel setup (70m m wide capillary gap)with neutralization reaction (55kJ =mol,concentration 4800mol =m 3).628K.Golbig et al.Designing Microreactors in Chemical Synthesis629 This method will be used as a pre-optimization tool for future microreactor projects at CPC-Systems.Future work aims at the extension to more complex cases and flow structures.An example for this is presented in Figure9,where the tempera-ture distribution in a flat rectangular channel is shown.The channel is fed with strong acid and strong base from the left.Instantaneous release of reaction heat(55kJ=mol)is presumed as well as isothermal wall condition.Due to micro dimensions of the channel,the hot spot is limited to4.2K in this case.LiteratureAutze,V.,Kleemann,A.,Golbig,K.,and Oberbeck,S.(2000).Nachr.Chem.,48,p.683–685. Commenge,J.M.,Falk,L.,Corriou,J.P.,and Matlosz,M.(2001).In5th International Conference on Microreaction Technology:Book of Abstracts,143.Ehrfeld,W.,Golbig,K.,Hessel,V.,Lo¨we,H.,and Richter,Th.(1997).In VDI-GVC Jahrbuch,102–116,VCH,Weinheim.Hessel,V.,Ehrfeld,W.,Golbig,K.,and Wo¨rz,O.(1999).GIT Lab.-Fachz.,10,1100–1103. Levenspiel,O.(1958).Ind.Eng.Chem.,50,343.Levenspiel,O.(1999).Chemical Reaction Engineering,3d ed.,ch.13,296–301,John Wiley, New York.Richter,Th.,Ehrfeld,W.,Gebauer,K.,Golbig,K.,Hessel,V.,Lo¨we,H.,and Wolf,A.(1998).In2nd International Conference on Microreaction Technology:Process Miniaturization, Topical Conference Preprints,146–151,American Institute of Chemical Engineers, New York.Taghavi-Moghadam,S.,Kleemann,A.,and Golbig,K.(2001).Microreaction technology as a novel approach to drug design,process development and reliability,Org.Process Res.&Dev.,5,652–658.Taylor,G.I.(1953).Proc.Roy.Soc.,219A,186.Schwalbe,T.,Autze,V.,and Wille,G.(2002).Chimia,56,636–646.Wegeng,R.S.and Drost,M.K.(1998).In2nd International.Conference on Microreaction Technology:Process Miniaturization,3–11.。

chemical-reaction-engineering-3ed-edition作者-octave-Levenspiel-课后习题答案

chemical-reaction-engineeri ng-3ed-edition作者-octave-Levenspiel-课后习题答案Corresponding Solutions for Chemical Reaction EngineeringCHAPTER 1 OVERVIEW OF CHEMICAL REACTION ENGINEERING (1)CHAPTER 2 KINETICS OF HOMOGENEOUS REACTIONS (3)CHAPTER 3 INTERPRETATION OF BATCH REACTOR DATA (7)CHAPTER 4 INTRODUCTION TO REACTOR DESIGN (20)CHAPTER 5 IDEAL REACTOR FOR A SINGLE REACTOR (23)CHAPTER 6 DESIGN FOR SINGLE REACTIONS (27)CHAPTER 10 CHOOSING THE RIGHT KIND OF REACTOR (34)CHAPTER 11 BASICS OF NON-IDEAL FLOW (36)CHAPTER 18 SOLID CATALYZED REACTIONS (45)Chapter 1 Overview of Chemical Reaction Engineering1.1 Municipal waste water treatment plant. Consider a municipal water treatment plant for a small community (Fig.P1.1). Waste water, 32000 m 3/day, flows through the treatment plant with a mean residence time of 8 hr, air is bubbled through the tanks, and microbes in the tank attack and break down the organic material (organic waste) +O 2 −−−→−microbes CO 2 + H 2OA typical entering feed has a BOD (biological oxygen demand) of 200 mg O 2/liter, while the effluent has a megligible BOD. Find the rate of reaction, or decrease in BOD in the treatment tanks.Figure P1.1Solution:)/(1017.2)/(75.183132/100010001)0200()(313200031320001343333s m mol day m mol day molgm L mg g L mg day day m dayday m VdtdN r A A ⋅⨯=⋅=-⨯⨯⨯-⨯-=-=--1.2 Coal burning electrical power station. Large central power stations (about 1000 MW electrical) using fluiding bed combustors may be built some day (see Fig.P1.2). These giants would be fed 240 tons of coal/hr (90% C, 10%H 2), 50% of which would burn within the battery of primary fluidized beds, the other 50% elsewhere in the system. One suggested design would use a battery of 10 fluidized beds, each 20 m long, 4 m wide, and containing solids to a depth of 1 m. Find the rate of reaction within theWaste Waste Clean200 mgMean residenZerobeds, based on the oxygen used.Solution:380010)1420(m V =⨯⨯⨯=)/(9000101089.05.01024033hr bed molc hrkgckgcoal kgc hr coal t N c ⋅-=⨯-=⨯⨯⨯-=∆∆ )/(25.111900080011322hr m kmolO t N V r r c c O ⋅=-⨯-=∆∆-=-=)/(12000412000190002hr bed mol dt dO ⋅=+⨯= )/(17.4800)/(105.113422s m mol hr bed mol dt dO V r O ⋅=⋅⨯==-Chapter 2 Kinetics of Homogeneous Reactions2.1 A reaction has the stoichiometric equation A + B =2R . What is the order of reaction?Solution: Because we don’t know whether it is an elementary reaction or not, we can’t tell the index of the reaction.2.2 Given the reaction 2NO 2 + 1/2 O 2 = N 2O 5 , what is the relation between the ratesof formation and disappearance of the three reaction components? Solution: 522224O N O NO r r r =-=-2.3 A reaction with stoichiometric equation 0.5 A + B = R +0.5 S has the following rateexpression-r A = 2 C 0.5 A C BWhat is the rate expression for this reaction if the stoichiometric equation is written asA + 2B = 2R + SSolution: No change. The stoichiometric equation can’t effect the rate equation, so it doesn’t change.2.4 For the enzyme-substrate reaction of Example 2, the rate of disappearance ofsubstrate is given by-r A =A06]][[1760C E A + , mol/m 3·sWhat are the units of the two constants? Solution: ][]6[]][][[][03A A C E A k s m mol r +=⋅=- 3/][]6[m mol C A ==∴sm mol m mol m mol s m mol k 1)/)(/(/][3333=⋅⋅=2.5 For the complex reaction with stoichiometry A + 3B → 2R + S and withsecond-order rate expression-r A = k 1[A][B]are the reaction rates related as follows: r A = r B = r R ? If the rates are not so related, then how are they related? Please account for the sings , + or - .Solution: R B A r r r 2131=-=-2.6 A certain reaction has a rate given by-r A = 0.005 C 2 A , mol/cm 3·min If the concentration is to be expressed in mol/liter and time in hours, what wouldbe the value and units of the rate constant?Solution:min)()(3'⋅⨯-=⋅⨯-cm molr hr L mol r A A 22443'300005.0106610)(minAA A A A C C r r cm mol mol hr L r =⨯⨯=⋅⨯=-⋅⋅⋅=-∴ AA A A A C C cmmol mol L C cmmolC L mol C 33'3'10)()(=⋅⋅=∴⨯=⨯2'42'32'103)10(300300)(AA A A C C C r --⨯=⨯==-∴ 4'103-⨯=∴k2.7 For a gas reaction at 400 K the rate is reported as -dtdp A= 3.66 p 2 A , atm/hr (a) What are the units of the rate constant?(b) What is the value of the rate constant for this reaction if the rate equation isexpressed as-r A = - dtdN V A1 = k C2 A , mol/m 3·s Solution:(a) The unit of the rate constant is ]/1[hr atm ⋅ (b) dtdN V r AA 1-=-Because it’s a gas reaction occuring at the fined terperatuse, so V=constant, and T=constant, so the equation can be reduced to22)(66.366.3)(1RT C RTP RT dt dP RT dt dP VRT V r A A A A A ==-=-=-22)66.3(A A kC C RT ==So we can get that the value of1.12040008205.066.366.3=⨯⨯==RT k2.9 The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol.How much faster the decomposition at 650℃ than at 500℃?Solution:586.7)92311731()10/(314.8/300)11(3211212=-⋅⋅=-==KK K mol kJ mol kJ T T R E k k Ln r r Ln7.197012=∴r r2.11 In the mid-nineteenth century the entomologist Henri Fabre noted that French ants (garden variety) busily bustled about their business on hot days but were rather sluggish on cool days. Checking his results with Oregon ants, I findRunning speed, m/hr150160230295370Temperatu re, ℃13 16 22 24 28 What activation energy represents this change in bustliness? Solution:RTE RTE RTE ek eak t cons ion concentrat f let ion concentrat f ek r ---=⋅⋅=⋅='00tan )()(RET Lnk Lnr A 1'-=∴ Suppose Tx Lnr y A 1,==, so ,REslope -= intercept 'Lnk =)/(1-⋅h m r A 150 160 230 295 370 A Lnr-3.1780 -3.1135 -2.7506 -2.5017 -2.2752CT o / 13 16 22 24 28 3101-⨯T3.4947 3.4584 3.3881 3.3653 3.3206-y = 5417.9x - 15.686R2 = 0.9712340.00330.003350.00340.003450.00351/T-L n r-y = -5147.9 x + 15.686Also K REslope 9.5147-=-=, intercept 'Lnk == 15.686 , mol kJ K mol J K E /80.42)/(3145.89.5147=⋅⨯-=Chapter 3 Interpretation of Batch Reactor Data3.1 If -r A = - (dC A /dt) =0.2 mol/liter·sec when C A = 1 mol/liter, what is the rate ofreaction when C A = 10 mol/liter? Note: the order of reaction is not known.Solution: Information is not enough, so we can’t answer this kind of question.3.2 Liquid a sedomposes by first-order kinetics, and in a batch reactor 50% of A isconverted in a 5-minute run. How much longer would it take to reach 75% conversion?Solution: Because the decomposition of A is a 1st -order reaction, so we can express the rate equation as:A A kC r =-We know that for 1st -order reaction, kt C C LnAAo=, 11kt C C LnA Ao =, 22kt C CLn A Ao = Ao A C C 5.01=, Ao A C C 25.02=So 21)24(1)(11212Ln kLn Ln k C C Ln C C Ln k t t A Ao A Ao =-=-=- equ(1) min 521)(111===Ln kC C Ln k t A Ao equ(2) So m in 5112==-t t t3.3 Repeat the previous problem for second-order kinetics. Solution: We know that for 2nd -order reaction, kt C C A A =-011, So we have two equations as follow:min 511211101k kt C C C C C AoAo Ao A A ===-=-, equ(1)2123)1(31411kt kt C C C C C AoAo Ao Ao A ===-=-, equ(2) So m in 15312==t t , m in 1012=-t t3.4 A 10-minute experimental run shows that 75% of liquid reactant is converted to product by a 21-order rate. What would be the fraction converted in a half-hour run?Solution: In a-21order reaction: 5.0AA A kC dt dC r =-=-, After integration, we can get:5.015.02A Ao C C kt -=, So we have two equations as follow:min)10(5.0)41(15.05.05.05.015.0k kt C C C C C Ao Ao AoA Ao ===-=-, equ(1) min)30(25.025.0k kt C C A Ao ==-, equ(2)Combining these two equations, we can get:25.05.1kt C Ao =, but this means 05.02<A C , whichis impossible, so we can conclude that less than half hours, all the reactant is consumed up. So the fraction converted 1=A X .3.5 In a hmogeneous isothermal liquid polymerization, 20% of the monomer disappears in 34 minutes for initial monomer concentration of 0.04 and also for 0.8 mol/liter. What rate equation represents the disappearance of the monomer?Solution: The rate of reactant is independent of the initial concentration of monomers, so we know the order of reaction is first-order,monomer monomer kC r =-And k C C Lnoomin)34(8.0= 1min 00657.0-=kmonomer monomer C r )min 00657.0(1-=-3.6 After 8 minutes in a batch reactor, reactant (C A0 = 1 mol/liter) is 80% converted; after 18 minutes, conversion is 90%. Find a rate equation to represent this reaction. Solution:In 1st order reaction, 43.1511111111212==--=Ln Ln X Lnk X Ln k t t A A , dissatisfied.In 2nd order reaction, 49/4/912.0111.01)11(1)11(11212==--=--=Ao Ao Ao Ao Ao Ao Ao A Ao A C C C C C C C C k C C k t t, satisfied.According to the information, the reaction is a 2nd -order reaction.3.7 nake-Eyes Magoo is a man of habit. For instance, his Friday evenings are all alike —into the joint with his week’s salary of ＄180, steady gambling at “2-up” for two hours, then home to his family leaving ＄45 behind. Snake Eyes’s betting pattern is predictable. He always bets in amounts proportional to his cash at hand, and his losses are also predictable —at a rate proportional to his cash at hand. This week Snake-Eyes received a raise in salary, so he played for three hours, but as usual went home with ＄135. How much was his raise? Solution:180=Ao n , 13=A n , h t 2=,135'=A n , h t 3;=, A A kn r α-So we obtain kt n n LnAAo=, ''')()(tn n Ln t n n Ln AAo A Ao= 3135213180'Ao n Ln Ln =, 28'=An3.9 The first-order reversible liquid reactionA ↔ R , C A0 = 0.5 mol/liter, C R0=0takes place in a batch reactor. After 8 minutes, conversion of A is 33.3% while equilibrium conversion is 66.7%. Find the equation for the this reaction. Solution: Liquid reaction, which belongs to constant volume system,1st order reversible reaction, according to page56 eq. 53b, we obtain121112102110)(1)(-+-+=+-==⎰⎰AX A A tX k k k k Lnk k X k k k dX dt t Amin 8sec 480==t , 33.0=A X , so we obtain eq(1)33.0)(1min8sec 480211121k k k k Ln k k +-+= eq(1) Ae AeAe c X X M C C k k K -+===1Re 21, 0==AoRo C C M , so we obtain eq(2) 232132121=-=-==AeAe c X X k k K ,212k k =∴ eq(2)Combining eq(1) and eq(2), we obtain1412sec 108.4m in 02888.0---⨯==k 14121sec 1063.9m in 05776.02---⨯===k kSo the rate equation is )(21A Ao A AA C C k C k dtdC r --=-=- )(sec 1063.9sec 108.401414A A A C C C -⨯-⨯=----3.10 Aqueous A reacts to form R (A→R) and in the first minute in a batch reactor itsconcentration drops from C A0 = 2.03 mol/liter to C Af = 1.97 mol/liter. Find the rate equation from the reaction if the kinetics are second-order with respect to A.Solution: It’s a irreversible second -order reaction system, according to page44 eq 12, we obtainmin 103.2197.111⋅=-k , so min015.01⋅=mol Lkso the rate equation is 21)min 015.0(A A C r -=-3.15 At room temperature sucrose is hydrolyzed by the catalytic action of the enzymesucrase as follows:Aucrose −−→−sucraseproductsStarting with a sucrose concentration C A0 = 1.0 millimol/liter and an enzyme concentrationC E0= 0.01 millimol/liter, the following kinetic data are obtained in a batch reactor (concentrations calculated from optical rotation measurements):Determine whether these data can be reasonably fitted by a knietic equation of the Michaelis-Menten type, or-r A =MA E A C C C C k +03 where C M = Michaelis constantIf the fit is reasonable, evaluate the constants k 3 and C M . Solve by the integral method.Solution: Solve the question by the integral method:AA M A A Eo A A C k Ck C C C C k dt dC r 5431+=+=-=-, M Eo C C k k 34=, MC k 15= AAo A Ao A Ao C C C C Lnk k k C C t -⋅+=-4451hrt ,AC ,mmol /L A Ao AAo C C C C Ln-AAo C C t -1 0.84 1.0897 6.25 20.681.20526.25C A , millimol /liter0.84 0.68 0.53 0.38 0.27 0.16 0.09 0.04 0.018 0.006 0.0025t,hr 1 2 3 4 5 6 7 8 9 10 113 0.53 1.3508 6.38304 0.38 1.5606 6.45165 0.27 1.7936 6.8493 6 0.16 2.1816 7.14287 0.09 2.6461 7.69238 0.04 3.3530 8.33339 0.018 4.0910 9.1650 10 0.006 5.1469 10.0604 110.00256.006511.0276Suppose y=A Ao C C t-, x=AAo A Ao C C C C Ln-, thus we obtain such straight line graphy = 0.9879x + 5.0497R 2 = 0.99802468101201234567Ln(Cao/Ca)/(Cao-Ca)t /(C a o -C a )9879.0134===Eo M C k C k Slope , intercept=0497.545=k k So )/(1956.00497.59879.015L mmol k C M ===, 14380.1901.09879.01956.0-=⨯==hr C C k k Eo M3.18 Enzyme E catalyzes the transformation of reactant A to product R as follows: A −−→−enzyme R, -r A =min22000⋅+liter molC C C A E AIf we introduce enzyme (C E0 = 0.001 mol/liter) and reactant (C A0 = 10mol/liter) into a batch rector and let the reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/liter. Note that the concentration of enzyme remains unchanged during the reaction.. Solution:510001.020021+=⨯+=-=-AA A A A C C C dC dt r Rearranging and integrating, we obtain:10025.0025.0100)(510)510(⎥⎦⎤⎢⎣⎡-+=+-==⎰⎰A Ao A Ao A A tC C C C Ln dC C dt t min 79.109)(5025.01010=-+=A Ao C C Ln3.20 M.Hellin and J.C. Jungers, Bull. soc. chim. France, 386(1957), present the data in Table P3.20 on thereaction of sulfuric acid with diethylsulfate in a aqueous solution at22.9℃:H 2SO 4 + (C 2H 5)2SO 4 → 2C 2H 5SO 4HInitial concentrations of H 2SO 4 and (C 2H 5)2SO 4 are each 5.5 mol/liter. Find a rate equation for this reaction.Table P3.20 t, minC 2H 5SO 4H , mol/li ter t, minC 2H 5SO 4H , mol/li ter1804.1141 1.18 194 4.31 48 1.38 212 4.45 55 1.63 267 4.86 75 2.24 318 5.15 96 2.75 368 5.32 127 3.31 379 5.35 146 3.76 410 5.42 1623.81∞(5.80)Solution: It’s a constant -volume system, so we can use X A solving the problem: i) We postulate it is a 2nd order reversible reaction system R B A 2⇔+ The rate equation is: 221R B A A A C k C C k dtdC r -=-=- L mol C C Bo Ao /5.5==, )1(A Ao A X C C -=, A A Ao Bo B C X C C C =-=, A Ao R X C C 2=When ∞=t , L mol X C C Ae Ao /8.52Re == So 5273.05.528.5=⨯=Ae X , L mol X C C C Ae Ao Be Ae /6.2)5273.01(5.5)1(=-⨯=-== After integrating, we obtaint C X k X X X X X LnAo AeA Ae A Ae Ae )11(2)12(1-=--- eq (1)The calculating result is presented in following Table.t,mi nLmol C R /,Lmol C A /,AXAAe AAe Ae X X X X X Ln---)12()1(AeAX X Ln -0 0 5.5 0 0 041 1.18 4.91 0.10730.2163 -0.227548 1.38 4.81 0.12540.2587 -0.271755 1.63 4.685 0.14820.3145 -0.329975 2.24 4.38 0.20360.4668 -0.488196 2.75 4.125 0.25 0.6165 -0.642712 7 3.31 3.8450.30090.8140 -0.845614 6 3.76 3.620.34181.0089 -1.044916 2 3.81 3.5950.34641.0332 -1.069718 0 4.11 3.4450.37361.1937 -1.233119 4 4.31 3.3450.39181.3177 -1.359121 2 4.45 3.2750.40451.4150 -1.4578267 4.86 3.07 0.4418 1.7730 -1.8197 318 5.15 2.925 0.4682 2.1390 -2.1886 368 5.32 2.84 0.4836 2.4405 -2.4918 379 5.35 2.825 0.4864 2.5047 -2.5564 4105.42 2.79 0.4927 2.6731 -2.7254 ∞5.82.60.5273——Draw AAe AAe Ae X X X X X Ln---)12(~ t plot, we obtain a straight line:y = 0.0067x - 0.0276R 2= 0.998800.511.522.530100200300400500tL n0067.0)11(21=-=Ao AeC X k Slope ,min)/(10794.65.5)15273.01(20067.041⋅⨯=⨯-=∴-mol L kWhen approach to equilibrium, BeAe c C C C k k K 2Re 21==, so min)/(10364.18.56.210794.642242Re 12⋅⨯=⨯⨯==--mol L C C C k k Be Ae So the rate equation ism in)/()10364.110794.6(244⋅⨯-⨯=---L mol C C C r R B A Aii) We postulate it is a 1st order reversible reaction system, so the rate equation isR A AA C k C k dtdC r 21-=-=- After rearranging and integrating, we obtaint k X X X Ln AeAe A '11)1(=-eq (2) Draw )1(AeAX X Ln -~ t plot, we obtain another straight line: -y = 0.0068x - 0.0156R 2 = 0.998600.511.522.530100200300400500x-L n0068.0'1-==AeX k Slope ,So 13'1m in 10586.35273.00068.0--⨯-=⨯-=k133Re '1'2min 10607.18.56.210586.3---⨯-=⨯⨯-==C C k k AeSo the rate equation ism in)/()10607.110586.3(33⋅⨯+⨯-=---L mol C C r R A AWe find that this reaction corresponds to both a 1st and 2nd order reversible reaction system, by comparing eq.(1) and eq.(2), especially when X Ae =0.5 , the two equations are identical. This means these two equations would have almost the same fitness of data when the experiment data of the reaction show that X Ae =0.5.(The data that we use just have X Ae =0.5273 approached to 0.5, so it causes to this.)3.24 In the presence of a homogeneous catalyst of given concentration, aqueous reactant A is converted to product at the following rates, and C A alone determines this rate:C A ,mol/liter1 2 4 6 7 9 12-r A , mol/liter·hr0.06 0.1 0.25 1.0 2.0 1.0 0.5We plan to run this reaction in a batch reactor at the same catelyst concentration as used in getting the above data. Find the time needed to lower the concentration of A from C A0 = 10 mol/liter to C Af = 2 mol/liter.Solution: By using graphical integration method, we obtain that the shaped area is 50 hr.04812162002 4 68 10 12 14Ca-1/Ra3.31 The thermal decomposition of hydrogen iodide 2HI → H 2 + I 2is reported by M.Bodenstein [Z.phys.chem.,29,295(1899)] as follows:T,℃ 508427 393 356 283k,cm 3/mol·s0.10590.003100.00058880.9×10-60.942×10-6Find the complete rate equation for this reaction. Use units of joules, moles, cm 3,and seconds.According to Arrhenius’ Law,k = k 0e -E/R Ttransform it,- In(k) = E/R·(1/T) －In(k 0)Drawing the figure of the relationship between k and T as follows:y = 7319.1x - 11.567R 2= 0.987904812160.0010.0020.0030.0041/T-L n (k )From the figure, we getslope = E/R = 7319.1 intercept = - In(k 0) = -11.567E = 60851 J/mol k 0 = 105556 cm 3/mol·sFrom the unit [k] we obtain the thermal decomposition is second-order reaction, so the rate expression is- r A = 105556e -60851/R T ·C A 2Chapter 4 Introduction to Reactor Design4.1 Given a gaseous feed, C A0 = 100, C B0 = 200, A +B→ R + S, X A = 0.8. Find X B ,C A ,C B . Solution: Given a gaseous feed, 100=Ao C , 200=Bo C , S R B A +→+0=A X , find B X , A C , B C0==B A εε, 202.0100)1(=⨯=-=A Ao A X C C4.02008.01001=⨯⨯==Bo A Ao B C X bC X 1206.0200)1(=⨯=-=B Bo B X C C4.2 Given a dilute aqueous feed, C A0 = C B0 =100, A +2B→ R + S, C A = 20. Find X A , X B , C B .Solution: Given a dilute aqueous feed, 100==Bo Ao C C ,S R B A +→+2, 20=A C , find A X , B X , B CAqueous reaction system, so 0==B A εε When 0=A X , 200=V When 1=A X , 100=VSo 21-=A ε, 41-==Ao Bo A B bC C εε8.01002011=-=-=Ao A A C C X , 16.11008.010012>=⨯⨯=⋅=Bo A Ao B C X C a b X , which is impossible. So 1=B X , 100==Bo B C C4.3 Given a gaseous feed, C A0 =200, C B0 =100, A +B→ R, C A = 50. Find X A , X B , C B . Solution: Given a gaseous feed, 200=Ao C , 100=Bo C ,R B A →+, 50=A C .find A X , B X , B C75.02005011=-=-=Ao A A C C X , 15.1>==BoAAo B C X bC X , which is impossible. So 100==Bo B C C4.4 Given a gaseous feed, C A0 = C B0 =100, A +2B→ R, C B = 20. Find X A , X B , C A . Solution: Given a gaseous feed, 100=+Bo Ao C C ，R B A →+2, 20=Bo C , Find A X , B X , A C0=B X , 200100100=+=B A V ,1=B X 15010050=+=R A V25.0200200150-=-=B ε, 5.01002110025.0-=⨯⨯-=-A ε842.02025.010020100=⨯--=B X , 421.0100842.010021=⨯⨯=A X34.73421.05.01421.0110011=⨯--⨯=+-=A A A AoA X X C C ε4.6 Given a gaseous feed, T 0 =1000 K, π0＝5atm, C A0=100, C B0=200, A +B→5R,T =400K, π=4atm, C A =20. Find X A , X B , C B .Solution: Given a gaseous feed, K T o 1000=, atm 50=π, 100=Ao C , 200=Bo CR B A 5→+, K T 400=, atm 4=π, 20=A C , find A X , B X , B C .1300300600=-=A ε, 2==Ao Bo AB bC C a εε,5.0410********=⨯⨯=ππT T According to eq page 87,818.05.010020115.0100201110000=⨯⨯+⨯-=+-=ππεππT T C C T T C C X Ao A AAo A A409.0200818.0100=⨯==Bo A Ao B aC X bC X130818.011200)818.0100200(1)(0=⨯+⨯-=+-=A A Ao A Ao Bo B X C T T X a b C C C εππ4.7 A Commercial Popcorn Popping Popcorn Popper. We are constructing a 1-literpopcorn to be operatedin steady flow. First tests in this unit show that 1 liter/min of raw corn feed stream produces 28 liter/minof mixed exit stream. Independent tests show that when raw corn pops its volumegoes from 1 to 31.With this information determine what fraction of raw corn is popped in the unit.Solution: 301131=-=A ε, ..1u a C Ao =, ..281281u a C C Ao A ==%5.462813012811=⨯+-=+-=∴AA Ao A Ao A C C C C X εChapter 5 Ideal Reactor for a single Reactor5.1 Consider a gas-phase reaction 2A → R + 2S with unknown kinetics. If a spacevelocity of 1/min is needed for 90% conversion of A in a plug flow reactor, find the corresponding space-time and mean residence time or holding time of fluid in the plug flow reactor.Solution: min 11==sτ,Varying volume system, so t can’t be found.5.2 In an isothermal batch reactor 70% of a liquid reactant is converted in 13 min.What space-time and space-velocity are needed to effect this conversion in a plug flow reactor and in a mixed flow reactor? Solution: Liquid reaction system, so 0=A ε According to eq.4 on page 92, min 130=-=⎰AX AAAo r dC C t Eq.13, AAAo A A Ao R F M r X C r C C -=--=..τ, R F M ..τ can’t be cert ain. Eq.17, ⎰-=AX AAAo R F P r dX C 0..τ, so m in 13...==R B R F P t τ5.4 We plan to replace our present mixed flow reactor with one having double thebolume. For the same aqueous feed (10 mol A/liter) and the same feed rate find the new conversion. The reaction are represented byA → R, -r A = kC 1.5 ASolution: Liquid reaction system, so 0=A εA A Ao Ao r X C F V -==τ, 5.1)]1([)(A Ao A A Ao A Ao X C k X r C C C -=-- Now we know: V V 2=', Ao Ao F F =', Ao Ao C C =', 7.0=A X So we obtain5.15.15.15.1)1()2)1(2A Ao A A Ao A Ao Ao X kC X X kC X F VF V -='-'==''52.8)7.01(7.02)1(5.15.1=-⨯='-'∴A AX X794.0='A X5.5 An aqueous feed of A and B (400liter/min, 100 mmol A/liter, 200 mmol B/liter) isto be converted to product in a plug flow reactor. The kinetics of the reaction is represented byA +B→ R, -r A = 200C A C Bmin⋅liter molFind the volume of reactor needed for 99.9% conversion of A to product.Solution: Aqueous reaction system, so 0=A εAccording to page 102 eq.19,⎰⎰-=-==Af AfX AA X A A AoAo Ao r dX r dC C C t F V 001⎰-==AfX AAAo or dX C Vντ, m in /400liter o =ν, L r dX r dX C V AAX A A o Ao Af3.1244001.0999.000=-⨯=-=∴⎰⎰ν5.9 A specific enzyme acts as catalyst in the fermentation of reactant A. At a givenenzyme concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (C A0 =2 mol/liter ). The kinetics of the fermentation at this enzyme concentration is given byA −−→−enzymeR , -r A = litermolC C A A ⋅+min 5.011.0Solution: P.F.R, according to page 102 eq.18, aqueous reaction, 0=ε⎰-=A X AA Ao r dX F V 0 )11(21251.05.010A AX A A A Ao X X Ln dX C C F V A+-⨯=+=∴⎰\L Ln4.986)95.005.01(125=+=5.11 Enzyme E catalyses the fermentation of substrate A (the reactant) to product R.Find the size of mixed flow reactor needed for 95% conversion of reactant in a feed stream (25 liter/min ) of reactant (2 mol/liter) and enzyme. The kinetics of the fermentation at this enzyme concentration are given byA −−→−enzyme R , -r A =litermolC C A A ⋅+min 5.011.0Solution: min /25L o =ν, L mol C Ao /2=, m in /50mol F Ao =, 95.0=A X Constant volume system, M.F.R., so we obtainmin 5.199205.05.01205.01.095.02=⨯⨯+⨯⨯⨯=-==AAAo or X C Vντ,39875.4min /25min 5.199m L V o =⨯==τν5.14 A stream of pure gaseous reactant A (C A0 = 660 mmol/liter) enters a plug flowreactor at a flow rate of F A0 = 540 mmol/min and polymerizes the as follows3A → R, -r A = 54min⋅liter mmolHow large a reactor is needed to lower the concentration of A in the exitstream to C Af = 330 mmol/liter?Solution: 321131-=-=A ε, 75.0660330321660330111=⨯--=+-=Ao A A Ao A A C C C C X ε 0-order homogeneous reaction, according to page 103 eq.20A Ao AoAooX C F VC kVkk ===ντ So we obtainL X k C C F V A Ao Ao Ao 5.75475.05401=⨯==5.16 Gaseous reactant A decomposes as follows:A → 3 R, -r A = (0.6min -1)C AFind the conversion of A in a 50% A – 50% inert feed (υ0 = 180 liter/min, C A0 =300 mmol/liter) to a 1 m 3 mixed flow reactor.Solution: 31m V =, M.F.R. 1224=-=A εAccording to page 91 eq.11, AAAoAAo AAAo oX X C X C r X C V+-=-==116.0ντmin/1801000)1(6.0)1(L LX X X A A A =-+=So we obtain 667.0=A XChapter 6 Design for Single Reactions6.1 A liquid reactant stream (1 mol/liter) passes through two mixed flow reactors in aseries. The concentration of A in the exit of the first reactor is 0.5 mol/liter. Find the concentration in the exit stream of the second reactor. The reaction is second-order with respect to A and V 2/V 1 =2.Solution:V 2/V 1 = 2, τ1 =011υV =A A A r C C --10 , 2τ = 022υV = 221A A A r C C --C A0=1mol/l , C A1=0.5mol/l , 0201υυ=, -r A1=kC 2 A1 ,-r A2=kC 2 A2 (2nd-order) , 2×2110A A A kC C C -=2221A A A kC C C - So we obtain 2×(1-0.5)/(k0.52)=(0.5-C A2)/(kC A22)C A2= 0.25 mol/l6.2 Water containing a short-lived radioactive species flows continuously through awell-mixed holdup tank. This gives time for the radioactive material to decay into harmless waste. As it now operates, the activity of the exit stream is 1/7 of the feed stream. This is not bad, but we’d like to lower it still more.One of our office secretaries suggests that we insert a baffle down the middle ofthe tank so that the holdup tank acts as two well-mixed tanks in series. Do you think this would help? If not, tell why; if so calculate the expected activity of the exit stream compared to the entering stream.Solution: 1st-order reaction, constant volume system. From the information offeredabout the first reaction,we obtain1τ=01100117171A A A A A A C k C C kC C C V ⋅-=-=υ If a baffle is added,022220212122212υυτττV V +=+==011υV =2222221210A A A A A A kC C C kC C C -+-=007176A A kC C =6/k …… ①。

H2_CO_CH4反应机理

3.2.OXIDATION OF HYDROGEN AND CARBON MONOXIDEA basic understanding of the oxidation mechanisms of hydrogen and carbon monox-ide is important for two reasons.First,hydrogen and carbon monoxide are major sources of fuel themselves.Second,their oxidation mechanisms are subsets of those of hydrocarbons,which are mainly made up of hydrogen and carbon.We shall there-fore separately discuss their oxidation mechanisms.The complete reaction mecha-nism for hydrogen and CO oxidation is shown in Table 3.1,which also lists the reaction numbers as well as the rate constants deﬁned in Eq.(2.2.3).A reverse reaction will be indicated by a negative sign in the reaction number.3.2.1.Explosion Limits of Hydrogen–Oxygen MixturesIn Chapter 2,we studied the role of the chain mechanism in the chemical runaway of a homogeneous mixture when it is also subjected to wall deactivation.We shall now consider the hydrogen–oxygen system,which exhibits a Z -shaped,pressure–temperature explosion-limit boundary,shown in Figure 3.2.1for the response of such a mixture in a heated and pressurized chamber.It is seen that whereas explo-sion is always possible at high temperatures,the response is highly nonmonotonic at moderate temperatures in that,by steadily increasing the pressure from a low value,a nonexplosive mixture would ﬁrst become explosive,then nonexplosive,and ﬁnally explosive again.The phenomenon of interest here,involving the Z -shaped response,is of a general nature in that similar responses have also been observed,both experimentally and computationally,for hydrogen–oxygen mixtures in other systems,for example the ignition temperature of a uniform ﬂow for a given pressure and residence time.Since initiation reactions involve only the reactant species,there are three possible initiation reactions for the present H 2–O 2system,namely the dissociation of H 2,the dissociation of O 2,and the reaction between H 2and O 2,as inH 2+M →H +H +M(H5)O 2+M →O +O +M(−H6)H 2+O 2→HO 2+H .(−H10)Table3.1.Oxidation of H2----CO mixturesNo.Reaction B[cm,mol,s]αE a(kcal/mol) H2----O2Chain Reactions(1)H+O2 O+OH1.9×1014016.44(2)O+H2 H+OH5.1×1004 2.67 6.29(3)OH+H2 H+H2O2.1×1008 1.51 3.43(4)O+H2O OH+OH3.0×10062.0213.40H2----O2Dissociation/Recombination(5)H2+M H+H+M4.6×1019−1.40104.38(6)O+O+M O2+M6.2×1015−0.500(7)O+H+M OH+M4.7×1018−1.00(8)H+OH+M H2O+M2.2×1022−2.00Formation and Consumption of HO2(9)H+O2+M HO2+M6.2×1019−1.420(10)HO2+H H2+O26.6×101302.13(11)HO2+H OH+OH1.7×101400.87(12)HO2+O OH+O21.7×10130−0.40(13)HO2+OH H2O+O21.9×1016−1.000Formation and Consumption of H2O2(14)HO2+HO2 H2O2+O24.2×1014011.981.3×10110−1.629(15)H2O2+M OH+OH+M1.2×1017045.50(16)H2O2+H H2O+OH1.0×101303.59(17)H2O2+H H2+HO24.8×101307.95(18)H2O2+O OH+HO29.5×10062.03.97(19)H2O2+OH H2O+HO21.0×1012005.8×101409.56Oxidation of CO(1)CO+O+M CO2+M2.5×10130−4.54(2)CO+O2 CO2+O2.5×1012047.69(3)CO+OH CO2+H1.5×10071.3−0.765(4)CO+HO2 CO2+OH6.0×1013022.95 Source:Kim,T.J.,Yetter,R.A.&Dryer,F.1994.New results on moist CO oxidation:high-pressure, high-temperature experiments,and comprehensive kinetic bust.Inst.25,759–766. The endothermicities of these three reactions are respectively104,118,and 55kcal/mole.Since the activation energy of a dissociation reaction is roughly equal to its endothermicity,reaction(−H10)is the most important initiation reaction under almost all conditions.Reaction(H5)may also contribute to initiation,but only at high temperatures.Reaction(−H6)is usually not preferred because oxygen has a larger dissociation energy than hydrogen.With the production of H from either(H5)or(−H10),the following chain reaction is initiated:H+O2→O+OH(H1)O+H2→H+OH(H2)OH+H2→H+H2O.(H3)3.2.Oxidation of Hydrogen and Carbon Monoxide 9110-310-210-1100101650700750800850900P r e s s u r e (a t m )Temperature (K)Figure 3.2.1.Explosion limits of a stoichiometric H 2----O 2mixture.The reverse reactions for (H1)and (H2)are not important at this stage because they involve the collisions between two radical species whose concentrations are very low.The reverse reaction of (H3)is also not important because of the low concentration of the product species,H 2O,during the initiation stage of explosion.For sufﬁciently low temperatures and pressures,explosion is not possible even with the addition of some H or OH.This is because the key chain-branching step (H1)is endothermic by 17kcal/mole and is therefore not favored at low temperatures.Fur-thermore,at low pressures these active species either rapidly diffuse to the chamber wall where they are destroyed,H ,O ,OH →wall destruction ,or react too slowly relative to an imposed ﬁnite residence time.As pressure increases,collision becomes more frequent and reactions are facil-itated.As the ﬁrst explosion limit is crossed,the rate of branching becomes over-whelmingly fast relative to either the rate of removal at the wall or the ﬁnite residence time,and explosion occurs.By further increasing pressure,the three-body reaction,H +O 2+M →HO 2+M,(H9)becomes more frequent and will eventually replace(H1)as the dominant reaction between H and O2.Being relatively inactive,the HO2radicals can survive many collisions and eventually will either diffuse to the wall where they are destroyed,or exhausted in aﬂow ofﬁnite residence time.Reaction(H9)is therefore an effective termination step of the radical chain process.As a result,the chain sequence(H1)–(H3)is broken.The second explosion limit is therefore determined by the competition between the growth(H1)–(H3)and destruction(H9)of the H atom.To determine the p-T dependence of this limit,we write the rate equations for the H,O,and OH radicals asd[H]dt=−k1[H][O2]+k2[O][H2]+k3[OH][H2]−k9[H][O2][M](3.2.1) d[O]dt=k1[H][O2]−k2[O][H2](3.2.2) d[OH]dt=k1[H][O2]+k2[O][H2]−k3[OH][H2].(3.2.3) Assuming steady state for the O and OH radicals,that is,d[O]/dt=0,and d[OH]/dt=0,we obtaink2[O][H2]=k1[H][O2]k3[OH][H2]=k1[H][O2]+k2[O][H2]=2k1[H][O2].Putting the above expressions into Eq.(3.2.1),we haved[H]dt=2k1[H][O2]−k9[H][O2][M](3.2.4a)=(2k1−k9[M])[H][O2],(3.2.4b) which is simplyd[H]dt=2ω1−ω9.(3.2.5) Following the same discussion as that for branched-chain explosions in Sec-tion2.4.2,Eq.(3.2.4b)shows that[H]varies exponentially with time,growing for (2k1−k9[M])>0and decaying otherwise.This then implies that the second limit is given by the neutral condition,2k1=k9[M].(3.2.6) Since p=[M]R◦T,and if the fall-off parameters of all third body species in(H9) are the same,then the pressure and temperature relationship of the second limit is explicitly given byp=2k1k9R◦T.(3.2.7)3.2.Oxidation of Hydrogen and Carbon Monoxide93Based on the rate constant expressions listed in Table3.1,we obtain the p–T re-lationship shown in Figure3.2.1.It is seen that the second explosion limit is well described by this relation,which is called the crossover temperature.It is to be noted that although Eq.(3.2.6)corresponds to the condition d[H]dt≡0, it is not a consequence of the steady-state assumption for[H].Indeed,while the steady-state assumption for[O]and[H]holds over extensive regimes in p and T,the transition boundary d[H]/dt≡0holds only along the crossover temperature line. As the third explosion limit is crossed,with further increase in pressure,the con-centration of HO2becomes higher.The reactionsHO2+H2→H2O2+H(−H17)H2O2+M→OH+OH+M(H15)are more frequent and overtake the stability of HO2.Therefore,instead of having HO2lost either to the wall or through theﬂow,it reacts with H2and generates the active species,H,O,and OH,to again induce explosion.At higher temperatures,say about900K,more radicals are produced and reactions between them become important.The HO2radical can either react with itself,HO2+HO2→H2O2+O2,(H14)followed by(H15),or with the H and O radicals,HO2+H→OH+OH(H11)HO2+O→OH+O2.(H12)In this situation,reaction(H9)is a part of the chain propagation process,and thus explosion will always occur.In Table3.1the H2----O2chain reaction sequence is given in Nos.1–4,the H2----O2 recombination and dissociation reactions in Nos.5–8,the formation and consump-tion of the hydroperoxyl radical,HO2,in Nos.9–13,and the chemistry of hydrogen peroxide,H2O2,in Nos.14–19.We also note from Table3.1that some reactions have negative activation ener-gies.This often occurs for reactions without intrinsic energy barriers such as those involving two radicals.The negative activation energy then implies that it is easier for a cold radical to combine with another cold radical than is for two hot radicals to combine.In the latter case,the combined kinetic energy of the two hot radicals are so large that the colliding radicals may justﬂy by without forming a bond.Table3.1further shows that reaction(H14)has two sets of reaction constants. This reﬂects the fact that depending on the initial point of contact,the reaction may proceed by two different routes.One of the routes dominates at low temperatures while the other dominates at high temperatures.For this reason,the overall rate constant is the sum of these two elementary reactions.3.2.2.Carbon Monoxide OxidationIt is difﬁcult to ignite and sustain a dry CO----O2ﬂame because the direct reaction between CO and O2,CO+O2→CO2+O,(CO2)has a high activation energy(48kcal/mol)and therefore is a very slow process even at high temperatures.Furthermore,the O atom produced does not lead to any rapid chain-branching reactions(Lewis&von Elbe1987).However,in the presence of even a small quantity of hydrogen,say20ppm,OH radicals are formed through reactions(H1−H3,H5,and−H10).Then the formation of CO2via the reactionCO+OH→CO2+H(CO3)becomes the dominant path for CO oxidation.The H atoms produced in(CO3)feed the chain-branching reactions(H1)–(H3),and thereby accelerate the CO oxidation rate.In moist air,water can also catalyze CO oxidation throughO2+M→O+O+M,(−H6) followed byO+H2O→OH+OH,(H4)which provides the OH radicals needed by(CO3).At high pressures,the reactionCO+HO2→CO2+OH(CO4)provides another route of CO to CO2conversion.Thus CO oxidation can be modeled by adding mainly(CO3)and(CO4)to the hydrogen–oxygen mechanism,as shown in Table3.1.Since trace amounts(ppm)of moisture are invariably present in practical systems and laboratory experiments,the relevant oxidation mechanism to use in understand-ing the response of these systems is that of wet instead of dry CO.Indeed,sometimes it maybe useful in experiments to deliberately add a small amount of hydrogen or water in the mixture,exceeding the background moisture contamination,so as to accurately deﬁne the composition of the mixture.3.2.3.Initiation Reactions in FlamesThe mechanisms of fuel oxidation inﬂames,including those of hydrogen and car-bon monoxide,are quite different from those of ignition of homogeneous reactive mixtures discussed above because the radical pool concentration in aﬂame is al-ways much larger than that generated during the induction period of homogeneous ignition.In addition,the initiation chemistry responsible for homogeneous ignition is often less important forﬂames because there are always abundant radicals that can back diffuse from the high-temperatureﬂame region to the colder,unburned3.3.Oxidation of Methane95regions of fuel and oxidizer,resulting in a different initiation chemistry.For exam-ple,for hydrogenﬂames,the dominant initiation reaction is the H+O2branching reaction(H1),because of the abundance and mobility of the highly reactive H atom, instead of reactions(H5)and(−H10)for homogeneous mixtures.The global activa-tion energy is,of course,also affected.3.3.OXIDATION OF METHANE3.3.1.General Considerations of Hydrocarbon OxidationRecently,considerable progress has been made in our understanding of the detailed reaction mechanisms of hydrocarbon oxidation.The general consensus is that the two most important reactions in a combustion process areH+O2→O+OH(H1)CO+OH→CO2+H,(CO3)which do not involve the speciﬁc hydrocarbon fuel at all.Reaction(CO3)is respon-sible almost exclusively for converting CO to CO2.It also generates some of the H atoms needed by(H1).In addition to the similarity of the most important reactions for hydrocarbon fuels, it is recognized that although the initial fuel breakdown is fuel speciﬁc,its rate is in general too fast to limit the overall rate of combustion.Furthermore,the initial fuel breakdown always leads to C1,C2,and C3fragments.A comprehensive understand-ing of hydrocarbon oxidation must then be strongly hierarchical in that mechanisms for the oxidation of complex fuels contain within them the submechanisms of simpler molecules.The key to the understanding of the oxidation mechanism of hydrocar-bon fuels,therefore,must start from an understanding of the simpler hydrocarbon oxidation mechanisms.Methane,being the lightest hydrocarbon fuel,is naturally the starting point.Before proceeding to the discussion of methane oxidation,it is important to rec-ognize that there exist diverse ranges of conditions in which a combustion process can take place.The dominant reaction mechanisms may thus vary substantially,de-pending on the local thermodynamic states of the combustion process.In particular, as we have just shown in the previous section,(H1)has a sufﬁciently large activation energy whereas(H9)has none.Therefore(H1)is the dominant chain-branching step at high temperatures.However,in the intermediate-to low-temperature regimes,the reaction chemistry is often dominated by that of the HO2radicals produced through (H9)as well as the initiation reactions such as(−H10)for hydrogen oxidation and (M2)for methane oxidation,which will be discussed in the following section.It is therefore necessary to distinguish the reaction mechanisms in the high-temperature regime(usually above1,100K at atmospheric pressure)from those in the low-to intermediate-temperature regimes.CHCH 3COCO 2HCOC 2H 6C H 5C 2H 4C 2H 3C 2H 22+CH 322Higher molecular weight species+O CH 3, CH 2O CH 3, HCO+O 2Products +O CH 2, CO, HCCO (CH 2*)Figure 3.3.1.Reaction pathways in methane ﬂames (adapted from Warnatz 1981).In this section we study the ignition of methane and the methane ﬂame chemistry,including a detailed discussion of the formaldehyde (CH 2O)oxidation mechanism,which is an integral part of the methane oxidation process.The reaction pathways for the oxidation involving methane ﬂames are depicted in Figure 3.3.1,with the thickness of the arrows indicating the relative importance of individual pathways.Much of the following discussion on methane autoignition is also included in this diagram.3.3.2.Methane AutoignitionFor methane,the ignition delay time τ,also called the induction time,has been empirically correlated byτ=2.5×10−15exp(26700/T )[CH 4]0.32[O 2]−1.02,where τis in seconds (s)and [CH 4]and [O 2]are respectively the molar concen-trations (mole/cm 3)of methane and oxygen in the reactant mixture.This correla-tion represents shock tube measurements (Tsuboi &Wagner 1975)in argon over the temperature range of 1,200to 2,100K,and shows that τdecreases with in-creasing temperature and pressure.The inhibitive effect of methane in the global3.3.Oxidation of Methane97consideration,as indicated by the negative reaction order for methane,is particu-larly worth noting.The preignition chemistry of methane is dictated by the rate of radical accumula-tion.The chain reaction is initiated primarily by the following two reactions:CH4+M CH3+H+M(M1)CH4+O2 CH3+HO2.(M2)Since(M1)is a large activation-energy,unimolecular reaction,it may be favored over(M2)only at high temperatures.Thus if(M1)is the dominant initiation step, the reactions that follow areH+O2 OH+O(H1) CH4+(H,O,OH) CH3+(H2,OH,H2O).(M3,M4,M5)Reaction(M3)plays an inhibiting role in the ignition process because it competes with the chain-branching step(H1)for H and converts the active H atoms to the less active CH3radicals.This is the primary reason for the positive dependence of the induction timeτon the methane concentration.The initiation reaction(M1)can also play the role of inhibiting ignition in that when the radical pool begins to build up, its backward reaction becomes increasingly fast,leading to increased radical chain termination.When(M2)is the dominant initiation step,the following reactions may participate in further radical growth:CH4+HO2 CH3+H2O2(M6)H2O2+M OH+OH+M.(H15)The similarity between(M6)and(−H17)is to be noted.The methyl radicals produced in the above reactions subsequently react with molecular oxygen through two reaction channels,CH3+O2 CH3O+O(M7a)CH2O+OH.(M7b)The branching ratio,that is,the relative contributions of the two channels to the overall rate constant,appears to be crucial to methane ignition because channel b is a chain-carrying step,producing formaldehyde,CH2O,whereas channel a leads to chain branching.Close to ignition,the reactionCH3+HO2 CH3O+OH(M8)becomes the dominant step for methyl oxidation.The CH3O radical,produced from (M7a)and(M8),is highly active.It is converted to formaldehyde rapidly viaCH3O+M CH2O+H+M(M9)CH3O+O2 CH2O+HO2.(M10)The formaldehyde produced in(M9)and(M10)subsequently reacts with OH and O2,CH2O+OH HCO+H2O(M11)CH2O+O2 HCO+HO2,(M12)producing the highly active formyl radical,HCO,which is consumed by reactions similar to the destruction of the CH3O radical,HCO+M H+CO+M(M13)HCO+O2 CO+HO2.(M14)The further conversion of CO to CO2via reaction(CO3)occurs after ignition.Al-though theﬁnal temperature of the burned mixture depends on the conversion of CO to CO2,the induction time is usually insensitive to(CO3).It is therefore clear that through the above sequence of reactions we have pro-gressively reduced the hydrogen content of the original fuel molecule,CH4,to CH3 and CH3O,then to CH2O,HCO,andﬁnally to CO.The hydrogen atoms abstracted at different stages are eventually oxidized to form H2O,while the CO that is formed through(M14)at the end of the sequence is oxidized to form CO2through the CO oxidation mechanism described earlier.The hierarchical nature of the oxidation pro-cess in terms of the progressive reduction of the parent hydrocarbon fuel species to the constituents of the H2and CO oxidation systems,through which theﬁnal products H2O and CO2are formed,is particularly noteworthy.The above mechanism describes the route through which the size of the original fuel molecule is reduced.There is,however,another route through which the size of the fuel molecule is increased.Speciﬁcally,upon the formation of the methyl radical CH3,it can self react throughCH3+CH3+M C2H6+M(M15)CH3+CH3 C2H5+H.(M16)Reaction(M15)is a radical termination step,which inhibits ignition,while reaction (M16)is chain carrying,leading to the formation of the ethyl radical C2H5and the highly active H atoms.Since(M15)tends to be more important at higher pressures because it is a third-order radical–radical recombination reaction,it has also shown strong sensitivity for ignition near atmospheric pressure.It is important to note that through reactions such as(M15)and(M16),species containing two carbon atoms are generated even though the original fuel molecule, methane,consists of only one carbon atom.These larger molecules will subsequently undergo oxidation themselves.Thus the oxidation mechanism of methane must in-clude not only those of the smaller species such as CH2O and CO,but also those of the larger,C2species.The inﬂuence of C2chemistry is particularly important for rich mixtures because of the abundance of CH3radicals.3.3.Oxidation of Methane99By extending this concept further,we can anticipate that,in principle,the C2H5 radicals produced through,say,(M16)can self react to generate the even larger molecule C4H10,while the recombination between the CH3and C2H5radicals pro-duces C3H8.Consequently the oxidation mechanism of CH4,and indeed that of any hydrocarbon,will necessarily involve the oxidation of all hydrocarbons of larger molecules.In reality,the concentrations of these larger molecules,except those pro-duced through theﬁrst generation of recombination,are so small that their effects on the overall reaction response are negligible.3.3.3.Methane FlamesSimilar to the discussion on the H2and COﬂames,the main difference between the autoignition andﬂame chemistry is the abundance of the radicals H,O,and OH that are produced at theﬂame and back diffuse upstream.Consequently,the destruction of methane is achieved mainly through H abstraction by H,O,and OH, that is,reactions(M3)–(M5),producing the methyl radical.The methyl radical is then consumed mainly by the O atom throughCH3+O CH2O+H,(M17) and by HO2through(M8).Upon further hydrogen abstraction of CH2O by H,CH2O+H→HCO+H2,(M18)and by OH through(M11),the formyl radical is produced.It then decomposes rapidly to yield CO and H,or undergoes the H-abstraction by H and O2.Some CH3radicals will also react with the OH radical to yield the highly active, singlet methylene radical(CH∗2),CH3+OH CH∗2+H2O.(M19)The CH∗2radical has no unpaired electrons,but it has an empty orbital which makes it a highly energetic and active species.Most of the CH∗2are de-energized by colli-sion with other molecules to the more stable triplet CH2radical,with two unpaired electrons,CH∗2+M→CH2+M.(M20)The CH∗2radical can also react with O2to provide secondary chain branching,pro-ducing the highly active H and OH radicals,CH∗2+O2→CO+H+OH(M21)to speed up the overall oxidation.Even if reaction(M20)is the dominant CH∗2removal channel,the resulting triplet CH2radicals are still highly active.They react readily with O2to provide secondary chain branching,CH2+O2→HCO+OH.(M22)《燃烧学基础》　第三章：化学反应动力力学阅读材料100Oxidation Mechanisms of Fuels Some of the CH 2will form the CH radical through the H-abstraction reaction,CH 2+H CH +H 2.(M23)The CH radicals are quickly consumed by H 2O or O 2,CH +H 2O →CH 2O +H ,(M24)CH +O 2→HCO +O .(M25)In rich mixtures,the relative abundance of CH 3enhances their recombination toyield C 2species via reactions (M15)and (M16)and via the fast reaction of CH 3and CH 2,CH 3+CH 2 C 2H 4+H .(M26)These C 2species either undergo further oxidation or may survive the attack by oxygenated species,eventually leading to acetylene according to the reaction steps shown in Figure 3.3.1.The production of acetylene is crucial to the formation of soot,as will be discussed later.3.4.OXIDATION OF C 2HYDROCARBONS We shall now discuss the oxidation mechanisms of the C 2hydrocarbons,namelyethane (C 2H 6),ethylene (C 2H 4),and acetylene (C 2H 2).We ﬁrst note that not only is ethane a major intermediate during the rich combustion of methane,as shown above,it is also the second most important constituent of natural gas.Similar to ethane,ethylene and acetylene are fuels by themselves.Furthermore,they are the major intermediates of ethane and higher hydrocarbon oxidation.We shall discuss the current understanding of the oxidation mechanisms of the three C 2hydrocarbons in high-temperature ﬂames.The oxidation of ethane in ﬂames starts from the H-abstraction of C 2H 6by H,O,and OH,producing the ethyl radical,C 2H 6+(H ,O ,OH) C 2H 5+(H 2,OH ,H 2O).(C 21)The ethyl radical is not very stable.It reacts rapidly with H and O 2,or decomposes to ethylene and an H atom,C 2H 5+(H ,O 2) C 2H 4+(H 2,HO 2)(C 22)C 2H 5+M C 2H 4+H +M ,(C 23)or it reacts with O 2to produce acetaldehyde (CH 3CHO),C 2H 5+O 2→CH 3CHO +OH .(C 24)Acetaldehyde subsequently reacts with H,O,and OH,producing the CH 3CO radical,followed by its unimolecular decomposition,leading to CH 3and CO,CH 3CHO +(H ,O ,OH) CH 3CO +(H 2,OH ,H 2O)(C 25)CH 3CO +M CH 3+CO +M .(C 26)《燃烧学基础》　第三章：化学反应动力力学阅读材料该材料来自于 Chung K. Law 教授的专著"Combustion Physics"中的第三章。

Lehninger Principles of Biochemistry 习题答案chapter 3

12
11.30 9.60 (IV)
P O
C
NH2
(V)
10
8 5.97
pH
6
(III)
4 2.34 2 (I) 0.5 1.0 1.5 2.0 OH (equivalents) (II)
0
Note: before considering statements (a) through (o), refer to Figure 3–10. The three species involved in the titration of glycine can be considered in terms of a useful physical analogy. Each ionic species can be viewed as a different floor of a building, each with a different net charge:
S-26
2608T_ch03sm_S26-S43
2/1/08
11:45AM
Page 27 ntt 102:WHQY028:Solutions Manual:Ch-03:
Chapter 3 Amino Acids, Peptides, and Proteins
S-27
H3NOCH2OCOOH H3NOCH2OCOO H2NOCH2OCOO
CH3 H3N C H C O O H2N CH3 C H C OH O
Zwitterionic
Uncharged
(a) Why is alanine predominantly zwitterionic rather than completely uncharged at its pI? (b) What fraction of alanine is in the completely uncharged form at its pI? Justify your assumptions. Answer (a) The pI of alanine is well above the pKa of the a-carboxyl group and well below the pKa of the a-amino group. Hence, at pH pI, both groups are present predominantly in their charged (ionized) forms. (b) From Table 3–1, the pI of alanine is 6.01, midway between the two pKa values 2.34 and 9.69. From the Henderson-Hasselbalch equation, pH pKa log ([A ]/[HA]). For the carboxyl group: log [A ] [HA] [HA] [A ] 6.01 10

Chapter 3 Stoichiometry Calculations with Chemical Formulas and Equations

--Antoine Lavoisier, 1789
© 2012 Pearson Education, Inc. Stoichiometry
Chemical Equations
Chemical equations are concise representations of chemical reactions.
Stoichiometry © 2012 Pearson Education, Inc.
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C: H: N: O: 1 mol 12.01 g 1 mol 5.14 g x 1.01 g 1 mol 10.21 g x 14.01 g 1 mol 23.33 g x 16.00 g 61.31 g x = 5.105 mol C = 5.09 mol H = 0.7288 mol N = 1.456 mol O
© 2012 Pearson Education, Inc.
Stoichiometry
Combustion Reactions
• Combustion reactions are generally rapid reactions that produce a flame. • Combustion reactions most often involve hydrocarbons reacting with oxygen in the air.
• Examples:
– CH4(g) + 2O2(g) CO2(g) + 2H2O(g) – C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

ChemicalReactionNSCCNetIDPersonalWebSpace化学反应和个人网络空间的用户名

The number of atoms in the reactants is equal to the number of atoms in the products.
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
3
Symbols Used in Equations
Symbols are used in chemical equations to show
TABLE 5.2
The states of the reactants.
The states of the products.
The reaction conditions.
Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings
4
Chemical Equations are Balanced
Chemical equations must be balanced!
Atoms are not gained or lost.
5
A Balanced Chemical Equation
Ine must be the same number of each type of atom on the reactant side and on the product side.
1
Chemical Reaction
In a chemical reaction,
A chemical change produces one or more new substances.

Chemistry Unit 3 - Heterogeneous

5
The Carbonate System: Solids, Gases, Weak Acids & Bases Major Players Carbon dioxide (gas phase and aqueous), CO2 Carbonic acid, H2CO3 Bicarbonate, HCO3Carbonate, CO3-2 Calcium Carbonate, CaCO3(s)
[A b+ ]a [பைடு நூலகம்a - ]b K= [A a Bb (s)]
Solubility Product = Ks = [Ab+]a[Ba-]b Some engineering applications: 1. Hardness removal 2. Heavy metal removal 3. Phosphate (P) removal 4. SOx control (make CaSO4(s))
12
Example: Formaldehyde in a Natural System
100 kg of formaldehyde
13
Organic Chemistry (p 61-64)
Number of carbons 1 (CH4) 2 (C2H6) 3 (C3H8) 4 (C4H10) 5 (C5H12) 6 (C6H14) 7 (C7H16) 8 (C8H18) 9 (C9H20) 10 (C10H22) prefix Meth (methane) Eth (ethane) Prop (propane) But (butane) Pent (pentane) Hex (hexane) Hept (heptane) Oct (octane) Nona (nonane) Deca (decane)

化学与化工专业英语2-化学基础知识

3）smell
odourless
pungent
penetrating
choking
offensive
sour sweet bitter
4）solubility
soluble
insoluble
slightly soluble
very soluble
5）observations
brisk effervescence
75 45 37 44 62 21
镭氡
铼铑铷钌钐钪
Selenium* Silicon* Silver* Sodium* Strontium* Sulfur* Tantalum* Technetium* Tellurium* Terbium* Thallium* Thorium Thulium Tin* Titanium* Tungsten* Uranium Vanadium* Xenon* Ytterbium Yttrium* Zinc* Zirconium*
Exercise
试用英语描述氧气、氮气和金属铁的物理性质。
Part 2 Chemical Equations
化学方程式 1.反应名称：

disproportionation neutralization; hydrolysis exothermic reaction endothermic reaction reversible reaction forward reaction reverse reaction spontaneous reaction nonspontaneous reaction
over quotient

化工最重要的三本专业课英语作文

化工最重要的三本专业课英语作文The Three Most Important Lessons in Chemical EngineeringHi there! My name is Timmy, and I'm going to tell you about the three most important lessons in chemical engineering. Even though I'm just a kid, I've learned a lot about this cool subject from my older brother who studies it at university.The first really important lesson is all about chemistry. You see, chemical engineers need to understand how different substances react with each other. They learn about atoms, molecules, and how to mix and separate different chemicals. This is super important because they use this knowledge to create new materials, medicines, and all sorts of useful products.The second big lesson is about physics. Chemical engineers need to know how stuff moves and behaves under different conditions, like heat, pressure, and gravity. They learn about things like fluid dynamics, thermodynamics, and heat transfer. This helps them design and operate machines and equipment that can handle chemical processes safely and efficiently.The third crucial lesson is all about math. Chemical engineers use a lot of complicated equations and calculations to solve problems and predict how things will work. They study subjectslike calculus, statistics, and numerical analysis. This allows them to model and optimize chemical processes, making sure everything runs smoothly and efficiently.I know these lessons might sound a bit boring to some kids, but they're actually really cool and important. Chemical engineers use what they learn to make our lives better in so many ways. They help create life-saving medicines, develop clean energy sources, and even make our favorite snacks and drinks!So, even though I'm still just a little kid, I can see how awesome and important chemical engineering is. And now you know the three most important lessons that chemical engineers need to learn: chemistry, physics, and math. Who knows, maybe one day you'll become a chemical engineer too and use these lessons to make the world a better place!。

温州2024年01版小学六年级上册F卷英语第三单元综合卷[有答案]

温州2024年01版小学六年级上册英语第三单元综合卷[有答案]考试时间：80分钟（总分:100）A卷考试人：_________题号一二三四五总分得分一、综合题(共计100题共100分)1. 听力题：Can you help me ________ my homework?2. 选择题：What do you call a baby crocodile?A. HatchlingB. PupC. KitD. Calf答案: A3. 听力题：A chemical equation uses symbols to represent a _____.4. 选择题：Which fruit is often mistaken for a vegetable?A. TomatoB. OnionC. CarrotD. Broccoli答案:A5. 听力题：A Newton is the force needed to accelerate a mass of one ______ (kilogram).6. 填空题：I enjoy making _________ (手工艺品) using my old _________ (玩具).7. 填空题：The ________ was a significant era in American history for social change.She is wearing a ________ dress.9. 听力题：A ________ is a large area of land with few trees.10. 选择题：What do we call the process of taking in oxygen and releasing carbon dioxide?A. InhalationB. RespirationC. PhotosynthesisD. Ventilation答案:B11. 选择题：What is the name of the ocean located on the east coast of the United States?A. Atlantic OceanB. Pacific OceanC. Indian OceanD. Arctic Ocean答案: A12. 选择题：What do we call a person who studies the impact of culture on society?A. AnthropologistB. SociologistC. HistorianD. Psychologist答案: A13. 听力题：The chemical formula for carbon monoxide is _______.14. 选择题：How many continents are there in the world?a. Fiveb. Sixc. Sevend. Eight答案:c15. 听力题：A __________ is a large area of flat land near the coast.16. 填空题：I enjoy playing ________ (棋类游戏) with my family.The ______ (温室效应) can impact plant habitats.18. 选择题：What is the capital of Kenya?A. NairobiB. MombasaC. KisumuD. Nakuru答案:A. Nairobi19. 听力题：The __________ can indicate areas at risk of environmental degradation.20. 填空题：The country known for its maple syrup is ________ (以枫糖浆闻名的国家是________).21. 听力题：A hydrate is a compound that contains ______ molecules.22. 听力题：The ____ is known for its impressive migration patterns.23. 选择题：What is the capital of China?A. BeijingB. ShanghaiC. Hong KongD. Guangzhou24. 选择题：Which color is a banana?A. BlueB. YellowC. RedD. Green答案: B25. 听力题：The flowers bloom in _____ (spring/fall).26. 填空题：I feel _______ when I help others.27. 选择题：What do we call the first human-made object to orbit the Earth?A. Apollo 11B. Voyager 1C. Sputnik 1D. Hubble Space Telescope28. 选择题：What is the name of the famous statue in Rio de Janeiro?A. Christ the RedeemerB. The ThinkerC. Statue of LibertyD. David答案: A29. 填空题：My favorite game is _______ (棋盘游戏).30. 填空题：We have a ______ (精彩的) event planned for next week.31. 填空题：The hawk uses its keen eyesight to spot ________________ (猎物) from high above.32. 填空题：My ________ (玩具名称) creates lasting joy.33. 填空题：The _____ (小狗) barks happily when it sees its owner.小狗看到它的主人时高兴地叫。

BalancingChemicalEquations：化学方程式的配平

Balancing Chemical EquationsChemical equations do not come already balanced. This must be done before the equation can be used in a chemically meaningful way.All chemical calculations to come must be done with a balanced equation.A balanced equation has equal numbers of each type of atom on each side of the equation.The Law of Conservation of Mass is the rationale for balancing a chemical equation. The law was discovered by Antoine Laurent Lavoisier (1743-94) and this is his formulation of it, translated into English in 1790 from the Traité élémentaire de Chimie (which was published in 1789):"We may lay it down as an incontestible axiom, that, in all the operations of art andnature, nothing is created; an equal quantity of matter exists both before and after theexperiment; the quality and quantity of the elements remain precisely the same; andnothing takes place beyond changes and modifications in the combination of theseelements."A less wordy way to say it might be:"Matter is neither created nor destroyed."Therefore, we must finish our chemical reaction with as many atoms of each element as when we started.Here is the example equation for this lesson:H + O ---> H O222It is an unbalanced equation (sometimes also called a skeleton equation). This means that there are UNEQUAL numbers at least one atom on each side of the arrow.In the example equation, there are two atoms of hydrogen on each side, BUT there are two atoms of oxygen on the left side and only one on the right side.Remember this: A balanced equation MUST have EQUAL numbers of EACH type of atom on BOTH sides of the arrow.An equation is balanced by changing coefficients in a somewhat trial-and-error fashion. It is important to note that only the coefficients can be changed, NEVER a subscript.The coefficient times the subscript gives the total number of atoms.Three quick examples before balancing the equation.(a) 2 H - there are 2 x 2 atoms of hydrogen (a total of 4).22(b) 2 H O - there are 2 x 2 atoms of hydrogen (a total of 4) and 2 x 1 atoms of oxygen (a total of2).42(c) 2 (NH )S - there are 2 x 1 x 2 atoms of nitrogen (a total of 4), there are 2 x 4 x 2 atoms ofhydrogen (a total of 16), and 2 x 1 atoms of sulfur (a total of 2).So, now to balancing the example equation:222H + O ---> H OThe hydrogen are balanced, but the oxygens are not. We have to get both balanced. We put a two in front of the water and this balances the oxygen.222H + O ---> 2 H OHowever, this causes the hydrogen to become unbalanced. To fix this, we place a two in front of the hydrogen on the left side.2222 H + O ---> 2 H OThis balances the equation.Two things you CANNOT do when balancing an equation.1) You cannot change a subscript.You cannot change the oxygen's subscript in water from one to two, as in:2222H + O ---> H O 22True, this balances the equation, but you have changed the substances in it. H O is a completely 2different substance from H O.2) You cannot place a coefficient in the middle of a formula.The coefficient goes at the beginning of a formula, not in the middle, as in:222H + O ---> H 2O2Water only comes as H O and you can only use whole formula units of it.There is another thing you should avoid. Make sure that your final set of coefficients are all whole numbers with no common factors other than one. For example, this equation is balanced:2224 H + 2 O ---> 4 H OHowever, all the coefficients have the common factor of two. Divide through to eliminate common factors like this.The equation just above is correctly balanced, but it is not the BEST answer. The best answer has all common factors greater than one removed.Balance this equation: H + Cl ---> HCl22Remember that the rule is: A balanced equation MUST have EQUAL numbers of EACH type of atom on BOTH sides of the arrow.The correctly balanced equation is:H + Cl ---> 2 HCl22Placement of a two in front of the HCl balances the hydrogen and chlorine at the same time.Balance this equation: O ---> O23Hint: think about what the least common multiple is between 2 and 3. That's right - six.The LCM tells you how many of each atom will be needed. Your job is to pick coefficients that get you to the LCM.The correctly balanced equation is:3 O ---> 2 O23Practice ProblemsHow many oxygens are indicated: 3 Ca(NO)32Balance these equations:Zn + HCl ---> ZnCl + H22KClO ---> KCl + O32S + F ---> SF826Fe + O ---> Fe O223C H + O ---> CO + H O26222The last problem above involved the use of fractional coefficients. Balance these three equations using ONLY fractional coefficients:S + F ---> SF826C H + O ---> CO + H O410222S + O ---> SO832Be careful on using fractions. For example, 1/2 H O is not a correct use of fractions. Why not?221/2 H = one H atom, but 1/2 O = one half of one atom. You cannot split atoms in chemical reactions.2Generally speaking, fractions are mostly used with diatomics (with O is the most common). However, know that you do not have to use fractions in balancing equations. The “official”method for balancing says that you should go back to the beginning and change the coefficient of the first compound to the next number (a “1" becomes a “2,” etc.) and then go through the entire equation again with this new starting point. So, fractions or starting over – both methods will give the same answer. I usually use the fraction approach when it occurs because it is easier for me. Find what works best for you.Want more balancing practice? Try Balancing Worksheet #1 with 50 problems and answers.Want still more? Try Balancing Worksheet #2 with 60 more problems and answers.Have fun!ANSWERS TO THE PRACTICE PROBLEMSHow many oxygens are indicated: 18Balance these equations:22Equation 1) Zn + HCl ---> ZnCl + H Upon examining this equation, you see that there is already one Zn on each side of the equation.We will attempt to leave it alone, if at all possible, since it is already balanced.One the right side, we see two chlorines and two hydrogens, with only one of each on the left.Putting a two in front of the HCl doubles the number of chlorine and hydrogen on the left side.This now leaves us with two chlorines and two hydrogens on each side of the arrow, making them both balanced.Since the zinc was already balanced, the entire equation is now balanced.32Equation 2) KClO ---> KCl + O Start by noticing the the K and the Cl are ALREADY balanced in the skeleton equation.However, the oxygen is out of balance with three on the left and two on the right.It is important to emphasize that the oxygen on the left will increase only in steps of three, while the oxygen on the right will increase only in steps of two. The question to ask yourself is "What is the least common multiple between 2 and 3?" The answer of course is six. We need sixoxygens on each side of the equation. We use a two on the left side since 2 x 3 = 6 and we use a three on the right side since 3 x 2 = 6.This causes the K and the Cl to become unbalanced, but putting a two in front of the KCl on the right side fixes that.This problem is interesting because you focused on the oxygens first. Normally, oxygen is the last (or next-to-last) element to be balanced.826Equation 3) S + F ---> SF 6An eight in front of the SF will balance the sulfurs.2This now gives us 48 fluorines on the right-hand side, since 8 x 6 = 48. Use a 24 in front of F since 24 x 2 also equals 48.223Equation 4) Fe + O ---> Fe O In the unbalanced equation, there was only one Fe on the left and two on the right. Putting a two in front of the Fe on the left brings the irons into balance.The situation balancing the oxygen is quite common. You saw it in a previous example. This time, I'll try to lay it out in steps.1. The oxygen on the left ONLY comes in twos, while the right-hand side oxygen comes onlyin threes.2. We have to get an equal number of oxygens. (Remember, we can only adjust the value ofthe coefficient. We cannot change the subscript.)3. The least common multiple between two and three is six. This means we will need sixoxygens on each side of the equation.4. To get this, we put a three in front of the O since 3 x 2 = 6 and we put a two in front of the2Fe O since 2 x 3 = 6.23The Fe was balanced, but has become unbalanced as a consequence of our work with the oxygen. Putting a four in front of the Fe on the left solves this.Equation 5) C H + O ---> CO + H O26222First, balance the carbons with a two in front of the CO. Then balance the hydrogens with a2three in front of the H O. This leaves the following equation:2C H + O ---> 2 CO + 3 H O26222Only the oxygens remain to be balanced, but there is a problem. On the right side of the equation, there are seven oxygen atoms, BUT oxygen only comes in a group of two atoms on the left side.Another way to say it - with O it is impossible to generate an ODD number of oxygen atoms.2However, that is true only if you were using whole number coefficients. It is allowable to use FRACTIONAL coefficients in the balancing process. That means I can use seven-halves as a coefficient to balance this equation, like this:C H + (7/2) O ---> 2 CO + 3 H O26222Generally, the fractional coefficient is not retained in the final answer. Multiplying the coefficients through by two gets rid of the fraction. here is the final answer:2 C H + 7 O ---> 4 CO + 6 H O26222Also, improper fractions like 7/2 should be used rather than a mixed number like 3 1/2.FOR THE FRACTIONAL COEFFICIENT PROBLEMS:Equation 1) (1/8) S + 3 F ---> SF ===> S + 24 F ---> 8 SF826826Equations 2) C H + (13/2) O ---> 4 CO + 5 H O ===> 2 C H + 13 O ---> 8 CO + 10 H O 410222410222Equation 3) (1/8) S + (2/3) O ---> SO===> [see below for the final balanced equation]832Equation 3 alternate) (3/8) S + 2 O ---> 3 SO ===> 3 S + 16 O ---> 24 SO832832。

化学的英文翻译

化学的英文翻译Chemistry is a branch of science that deals with the properties, structure, composition, reactions, and transformations of matter. It involves the study of various substances, their changes, and their interactions with each other. The word "chemistry" is derived from the Arabic word "kīmiyā," which means the study of transmutation or change.Chemistry has a wide range of applications and is considered a central science, as it connects and overlaps with other scientific disciplines such as physics, biology, and geology. It is crucial in industries such as pharmaceuticals, agriculture, materials science, and environmental sciences.One of the fundamental concepts in chemistry is the atom, which is the basic unit of matter. Atoms are made up of protons, neutrons, and electrons. Protons carry a positive charge, neutrons carry no charge, and electrons carry a negative charge. The arrangement and behavior of these subatomic particles determine the properties of the elements.Elements are substances that cannot be broken down into simpler substances by chemical means. There are over 100 known elements, each with its unique atomic number, symbol, and properties. Elements can combine to form compounds through chemical reactions. Compounds are substances composed of two or more different elements in fixed ratios.Chemical reactions involve the rearrangement of atoms to form new substances. Reactants are the starting substances, and productsare the resulting substances. Chemical equations are used to represent these reactions, with the reactants on the left side and the products on the right side of the equation. The reaction is balanced by ensuring that the number of atoms of each element is the same on both sides of the equation.Chemical bonding occurs when atoms share or transfer electrons to achieve a more stable electron configuration. There are three main types of chemical bonding: ionic bonding, covalent bonding, and metallic bonding. Ionic bonding involves the transfer of electrons between atoms, resulting in the formation of ions. Covalent bonding involves the sharing of electrons between atoms. Metallic bonding occurs in metals, where electrons are free to move throughout the lattice structure.Chemical reactions can be classified into various categories, such as synthesis reactions, decomposition reactions, combustion reactions, acid-base reactions, and redox reactions. Synthesis reactions combine two or more substances to form a new compound. Decomposition reactions break down a compound into its individual elements or simpler compounds. Combustion reactions involve the rapid combination of a fuel source with oxygen, releasing heat and light. Acid-base reactions involve the transfer of protons between substances. Redox reactions involve the transfer of electrons, resulting in the oxidation and reduction of substances.Chemical analysis is an essential aspect of chemistry, involving techniques to determine the composition, structure, and properties of substances. Analytical methods include spectroscopy,chromatography, spectrometry, and various other techniques. Overall, chemistry is a vast field of study that explores the nature of matter and its interactions. It plays a crucial role in understanding the world around us, developing new materials, and solving pressing global challenges.。

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EXAMPLE
Determine whether or not the following compounds are soluble or insoluble in water. 1) NaBr
2) Ba(OH)2
3) Calcium carbonate
Chemical Equations for Aqueous Reactions
Solubility Rules
Solubility Soluble Examples:
Insoluble -
Examples:
Solubility Table
Solubility guidelines - soluble salts
Solubility Table
Solubility guidelines - insoluble salts
Example:
Balancing Chemical Equations
The Law of Conservation of Matter (Conservation of Mass):
Balancing Chemical Equations
Chemical equations balanced via inspection. Numbers used to balance chemical
Balancing Chemical Equations
Write balanced equations for: 1. Sodium hydroxide reacting with carbon
dioxide to form water and sodium carbonate 2. Sodium phosphate reacting with magnesium chloride to form magnesium phosphate and sodium chloride
solution as freely moving ions What happens when you
apply an electrical source (battery) to the solution? What about pure water?
Electrolytes and Nonelectrolytes
CHEM 5013
Applied Chemical Principles
Chapter Three Professor Bensley Alfred State College
Chapter Objectives
Explain balancing a chemical reaction as an application of the law of conservation of mass.
Reactants form products
Writing Chemical Equationsቤተ መጻሕፍቲ ባይዱ
Reactants and products are represented by chemical formulas.
The phase for the compounds must be given.
Electrolyte:
Examples:
Electrolytes and Nonelectrolytes
Nonelectrolyte:
Examples:
Strong and Weak Electrolytes
Strong electrolyte: Example: Weak electrolyte: Example:
KCl (aq) + AgNO3 (aq) KNO3 (aq) + AgCl (s) Potassium Chloride + Calcium Hydroxide
Balancing Chemical Equations
DO NOT change the subscripts!
BUT, if you have a polyatomic ion like SO42- (sulfate) this stays as one unit!!!
So to balance, place a new subscript outside the parenthesis as in:
Remember chemical equations? When ions are involved, we can write
them in three forms:
1) 2) 3)
Example: Called metathesis or precipitate or double
equations are called coefficients. Example: Burning of natural gas (methane)
(Combustion Reaction)
Example: Write the complete balanced combustion reaction of propane.
Write balanced chemical equations for simple reactions, given either an unbalanced equation or a verbal description.
Chapter Objectives
Distinguish between electrolytes and nonelectrolytes and explain how their solutions differ.
Describe the species expected to be present (ions, molecules, etc.) in various simple solutions.
Writing Chemical Equations
What does a Chemical equation represent?
substitution reactions.
Molecular and Ionic Equations
Molecular Equation: Complete Ionic Equation: Net Ionic Equation: Spectator Ions:
Example
Write a net ionic equation for the molecular equations below:
Solutions, Solvents, and Solutes
Aqueous solutions: Solutions: (review)
Solvent: Solute:
Ions in Aqueous Solution
Dissolve NaCl in water. Na+ ions and Cl- ions go into
When an ionic solid dissolves in water, what happens to the compound?
This type of reaction is called a dissociation reaction.
N + N a + ( -C
Molecular and Ionic Equations
Balancing Chemical Equations
We normally use the smallest whole numbers possible as coefficients.
Trial and Error
HINT – Balance the atoms first that occur in only one substance on each side
When a covalently bonded material dissolves in water, what happens to the molecules?
C 6 H 1 O 6 2 ( 6 H 1 O s C 6 2 (
NO IONS ARE FORMED!
Chemical Equations for Aqueous Reactions