Thermodynamics of Mixing


 Amy Stokes
 2 years ago
 Views:
Transcription
1 Thermodynamics of Mixing Dependence of Gibbs energy on mixture composition is G = n A µ A + n B µ B and at constant T and p, systems tend towards a lower Gibbs energy The simplest example of mixing: What is the Gibbs free energy if we take two pure ideal gases and mix them together? Consider gas A and gas B, both in separate containers at pressure p at temperature T. The chemical potentials are at their pure values at this point. Gibbs energy is G = n A µ A + n B µ B = n A µ o A + RT ln p p o + n µ o + RT ln p B B p o We can simplify things by letting p denote the pressure relative to p o, writing G = n A ( µ o A + RT ln p)+ n B ( µ o B + RT ln p) Prof. Mueller Chemistry Fall 2003 Lecture 161
2 Gibbs Energy of Mixing After mixing, the partial pressures of the gases are p A and p B, where the total pressure is p = p A + p B. The total Gibbs energy is then G = n A ( µ o A + RT ln p A )+ n B µ o B + RT ln p B ( ) The difference in Gibbs energies, G f  G i, is the Gibbs energy of mixing G = n A RT ln p A + n B RT ln p B p p We use mole fractions, replacing n J with x J n : G = nrt( x A ln x A + x B ln x B ) Since the mole fractions are never greater than 1, the ln terms are negative, and G < 0 This allows is to conclude that mixing processes are spontaneous, and gases mix spontaneously in all proportions Prof. Mueller Chemistry Fall 2003 Lecture 162
3 Calculating Gibbs Energy of Mixing Two containers of equal volume are partitioned from one another, with one containing 3.0 mol H 2 and the other 1.0 mol N 2 at 25 o C. Calculate the Gibbs energy of mixing when the partition is removed We assume ideal gas behavior, with pressure of N 2 being p, pressure of H 2 being 3p. Volume of each gas will double, and therefore partial pressures fall by a factor of 2 Prof. Mueller Chemistry Fall 2003 Lecture 163
4 Calculating Gibbs Energy of Mixing Two containers of equal volume are partitioned from one another, with one containing 3.0 mol H 2 and the other 1.0 mol N 2 at 25 o C. Calculate the Gibbs energy of mixing when the partition is removed o o G i = ( 3.0 mol ( )µ H 2 + RT ln 3p)+ ( 1.0 mol) ( µ N2 + RT ln p) o G f = ( 3.0 mol) µ H 2 + RT ln 3p o + ( 1.0 mol) µ 2 N2 + RT ln p 2 Gibbs energy of mixing is the difference between the above quantities 3p G = ( 3.0 mol) RT ln mol 3p = ( 3.0 mol)rt ln2 ( 1.0 mol)rt ln2 = ( 4.0 mol)rt ln2 = 6.9 kj ( ) RT ln p 2 p Prof. Mueller Chemistry Fall 2003 Lecture 164
5 Since Other Thermodynamic Mixing Functions G T p,n = S, and G = nrt( x A ln x A + x B ln x B ) it follows that for a mixture of perfect gases, the entropy of mixing is S = G T p,n A,n B Since ln x < 0, then S > 0 for mixtures of all compositions We expect this increase IF WE SPEAK ABOUT DISORDER, since dispersal of one gas into another implies greater disorder in the system. What is the preferred interpretation? Example: = nr( x A ln x A + x B ln x B ) G = ( 4.0 mol)rt ln2 = 6.9 kj S = ( 4.0 mol)rln2 =+23 J K 1 The isothermal isobaric enthalpy of mixing of two gases is found from G = H T S, and we find H = 0 This is expected for a system in which there are no interactions between the molecules in the mixture. Prof. Mueller Chemistry Fall 2003 Lecture 165
6 Chemical Potentials of Liquids Some nomenclature: Quantities relating to pure substances will be denoted with asterisks, so the chemical potential of pure liquid A is µ A* (l). Since the vapor pressure of pure liquid A is p A*, the chemical potential of A in the vapor is µ o + RT ln p A * (p A * as relative pressure p A* /p o ). Chemical potentials are at equilibrium: µ A = µ A o + RT ln p A If another substance is present (e.g., a solute in the liquid), chemical potential of A in the liquid is µ A and vapor pressure is p A µ A = µ A o + RT ln p A Combine to eliminate the standard potential: µ A = µ A + RT ln p A p A The chemical potential of A depends on its partial vapor pressure. It follows that chemical potential of liquid A is related to its partial pressure Prof. Mueller Chemistry Fall 2003 Lecture 166
7 Raoult s Law After experimenting with mixtures of similar liquids, the French chemist Francois Raoult found that the ratio of the partial vapor pressure of each component to its vapor pressure as a pure liquid, p A /p A*, is approximately equal to the mole fraction of A in the liquid mixture p A = x A p A Some mixtures obey Raoult s law very closely, especially when they are structurally similar these are called ideal solutions, and they obey µ A = µ A + RT ln x A Prof. Mueller Chemistry Fall 2003 Lecture 167
8 Molecular Interpretation of Raoult s Law We have to consider the rate at which molecules leave and enter solution to understand Raoult s law The presence of a second component actually hinders molecules from leaving the solution, but does not inhibit the rate at which they return The rate at which A molecules leave the surface is proportional to the number of them at the surface, which is proportional to their mole fraction rate of vaporization = k x A where k is a proportionality constant. The rate at which molecules return is proportional to their gas phase concentration, which is proportional to their partial pressure rate of condensation = k p A At equilibrium, rates of vaporization and condensation are equal, so For a pure liquid, x A = 1, so kx A = k p A p A = kx A k p A = k k Prof. Mueller Chemistry Fall 2003 Lecture 168
9 Departure from Raoult s Law, Dilute Solutions Some solution mixtures behave very differently from Raoult s law, notably with mixtures of structurally and chemically dissimilar liquids Mixture of CS 2 and acetone: When the solute is dilute, and solvent obeys Raoult s Law very closely When the solvent is nearly pure, it has a vapor pressure proportional to the mole fraction with slope p B * When it is the minor component (the solute) the vapor pressure has a different constant of proportionality Prof. Mueller Chemistry Fall 2003 Lecture 169
10 Henry s Law The English chemist William Henry found that for real solutions at low concentrations, although the vapor pressure of the solute is proportional to the mole fraction, the constant of proportionality is not the vapor pressure of the pure substance p B = x B K B Here x B is the mole fraction of the solute B, and K B is an empirically determined constant with pressure dimensions. K B is chosen so that a plot of vapor pressure of B against mole fraction is tangent to the experimental curve at x B = 0 Such solutions are called idealdilute solutions Solvent behaves like a slightly modified pure liquid Solute behaves entirely differently from its pure state (since its molecules are surrounded by solvent molecules), unless the molecules of the two components are very similar Prof. Mueller Chemistry Fall 2003 Lecture 1610
11 The Validity of Raoult s & Henry s Laws Consider an acetone (A) and chloroform (C) mixture at 35 o C x C p C (Torr) p A (Torr) K = 175 Torr for (A) and K = 165 Torr for (C) When mole fractions of (A) or (C) are near 1, Raoult s Law predicts the partial pressures When mole fractions of (A) or (C) are near 0, Henry s Law predicts the partial pressures In both cases, there are some deviations from this predicted behavior Prof. Mueller Chemistry Fall 2003 Lecture 1611
12 Using Henry s Law Estimate the molar solubility of oxygen in water at 25 o C and partial pressure of 120 Torr, its partial pressure at sea level The amount of O 2 dissolved is very small, and x J = p J /K J, so x( O 2 )= ( ) ( ) n ( O 2) n( H 2 O) n O 2 n( O 2 )+ n H 2 O n( O 2 )= x( O 2 )n( H 2 O)= p O 2 ( )n H 2 O K ( ) = ( ) Torr ( 160 Torr) 55.5 mol kg 1 = mol kg 1 So the molar solubility of oxygen is 2.69 x 104 mol kg 1 corresponding to concentration 2.7 x 104 mol L 1 Henry s law constants are very well known by biochemists for treatment of gas behavior in fats and lipids  very important in understanding respiration processes: especially when partial pressure of O 2 is abnormal, such as in mountaineering, scuba diving, and in the use of gaseous anaesthetics. Prof. Mueller Chemistry Fall 2003 Lecture 1612
Thermodynamics and Kinetics. Lecture 14 Properties of Mixtures Raoult s Law Henry s Law Activity NC State University
Thermodynamics and Kinetics Lecture 14 Properties of Mixtures Raoult s Law Henry s Law Activity NC State University Measures of concentration There are three measures of concentration: molar concentration
More informationSimple Mixtures. Atkins 7th: Sections ; Atkins 8th: The Properties of Solutions. Liquid Mixtures
The Properties of Solutions Simple Mixtures Atkins 7th: Sections 7.47.5; Atkins 8th: 5.45.5 Liquid Mixtures Colligative Properties Boiling point elevation Freezing point depression Solubility Osmosis
More informationVapor Pressure Diagrams and Boiling Diagrams
Vapor Pressure Diagrams and Boiling Diagrams We are now ready to begin talking about phase diagrams involving two components. Our first few phase diagrams will involve only the liquid and gas (or vapor)
More informationChapter 7 : Simple Mixtures
Chapter 7 : Simple Mixtures Using the concept of chemical potential to describe the physical properties of a mixture. Outline 1)Partial Molar Quantities 2)Thermodynamics of Mixing 3)Chemical Potentials
More informationvap H = RT 1T 2 = 30.850 kj mol 1 100 kpa = 341 K
Thermodynamics: Examples for chapter 6. 1. The boiling point of hexane at 1 atm is 68.7 C. What is the boiling point at 1 bar? The vapor pressure of hexane at 49.6 C is 53.32 kpa. Assume that the vapor
More informationSample Test 1 SAMPLE TEST 1. CHAPTER 12
13 Sample Test 1 SAMPLE TEST 1. CHAPTER 12 1. The molality of a solution is defined as a. moles of solute per liter of solution. b. grams of solute per liter of solution. c. moles of solute per kilogram
More informationEXERCISES. 16. What is the ionic strength in a solution containing NaCl in c=0.14 mol/dm 3 concentration and Na 3 PO 4 in 0.21 mol/dm 3 concentration?
EXERISES 1. The standard enthalpy of reaction is 512 kj/mol and the standard entropy of reaction is 1.60 kj/(k mol) for the denaturalization of a certain protein. Determine the temperature range where
More information48 Practice Problems for Ch. 17  Chem 1C  Joseph
48 Practice Problems for Ch. 17  Chem 1C  Joseph 1. Which of the following concentration measures will change in value as the temperature of a solution changes? A) mass percent B) mole fraction C) molality
More informationMajor chemistry laws. Mole and Avogadro s number. Calculating concentrations.
Major chemistry laws. Mole and Avogadro s number. Calculating concentrations. Major chemistry laws Avogadro's Law Equal volumes of gases under identical temperature and pressure conditions will contain
More informationThermodynamics Answers to Tutorial # 1
Thermodynamics Answers to Tutorial # 1 1. (I) Work done in free expansion is Zero as P ex = 0 (II) Irreversible expansion against constant external pressure w = P ex (V 2 V 1 ) V 2 = nrt P 2 V 1 = nrt
More informationThermodynamics and Equilibrium
Chapter 19 Thermodynamics and Equilibrium Concept Check 19.1 You have a sample of 1.0 mg of solid iodine at room temperature. Later, you notice that the iodine has sublimed (passed into the vapor state).
More informationTypes of Solutions. Chapter 17 Properties of Solutions. Types of Solutions. Types of Solutions. Types of Solutions. Types of Solutions
Big Idea: Liquids will mix together if both liquids are polar or both are nonpolar. The presence of a solute changes the physical properties of the system. For nonvolatile solutes the vapor pressure, boiling
More informationThe first law: transformation of energy into heat and work. Chemical reactions can be used to provide heat and for doing work.
The first law: transformation of energy into heat and work Chemical reactions can be used to provide heat and for doing work. Compare fuel value of different compounds. What drives these reactions to proceed
More informationCHEMISTRY The Molecular Nature of Matter and Change
CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 13 The Properties of Mixtures: Solutions and Colloids Copyright The McGrawHill Companies, Inc. Permission required for reproduction
More informationPHASE CHEMISTRY AND COLLIGATIVE PROPERTIES
PHASE CHEMISTRY AND COLLIGATIVE PROPERTIES Phase Diagrams Solutions Solution Concentrations Colligative Properties Brown et al., Chapter 10, 385 394, Chapter 11, 423437 CHEM120 Lecture Series Two : 2011/01
More informationCOLLIGATIVE PROPERTIES:
COLLIGATIVE PROPERTIES: A colligative property is a property that depends only on the number of solute particles present, not their identity. The properties we will look at are: lowering of vapor pressure;
More informationEntropy Changes & Processes
Entropy Changes & Processes Chapter 4 of Atkins: he Second Law: he Concepts Section 4.3, 7th edition; 3.3, 8th edition Entropy of Phase ransition at the ransition emperature Expansion of the Perfect Gas
More information1.3 Saturation vapor pressure. 1.3.1 Vapor pressure
1.3 Saturation vaor ressure Increasing temerature of liquid (or any substance) enhances its evaoration that results in the increase of vaor ressure over the liquid. y lowering temerature of the vaor we
More informationChemical Process calculation III
Chapter 7 Ideal and Real Gases Gas, Liquid, and Solid Chemical Process calculation III Gas: a substance in a form like air, relatively low in density and viscosity Liquid: a substance that flows freely
More informationPhysical pharmacy. dr basam al zayady
Physical pharmacy Lec 7 dr basam al zayady Ideal Solutions and Raoult's Law In an ideal solution of two volatile liquids, the partial vapor pressure of each volatile constituent is equal to the vapor pressure
More informationChapter 13 Properties of Solutions. Classification of Matter
Chapter 13 Properties of Solutions Learning goals and key skills: Describe how enthalpy and entropy changes affect solution formation Describe the relationship between intermolecular forces and solubility,
More informationApplied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur
Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur Lecture  7 Ideal Gas Laws, Different Processes Let us continue
More informationVapor Pressure Lowering
Colligative Properties A colligative property is a property of a solution that depends on the concentration of solute particles, but not on their chemical identity. We will study 4 colligative properties
More informationChemistry: The Central Science. Chapter 13: Properties of Solutions
Chemistry: The Central Science Chapter 13: Properties of Solutions Homogeneous mixture is called a solution o Can be solid, liquid, or gas Each of the substances in a solution is called a component of
More informationCHAPTER 3 PROPERTIES OF NATURAL GASES
CHAPTER 3 PROPERTIES OF NATURAL GASES The behavior of natural gas, whether pure methane or a mixture of volatile hydrocarbons and the nonhydrocarbons nitrogen, carbon dioxide, and hydrogen sulfide, must
More informationChemistry Notes for class 12 Chapter 2 Solutions
1 P a g e Chemistry Notes for class 12 Chapter 2 Solutions Solution is a homogeneous mixture of two or more substances in same or different physical phases. The substances forming the solution are called
More informationGibbs Free Energy and Chemical Potential. NC State University
Chemistry 433 Lecture 14 Gibbs Free Energy and Chemical Potential NC State University The internal energy expressed in terms of its natural variables We can use the combination of the first and second
More informationChem. 1A Final Exam Review Problems From ch. 11, 12 & 13
Chem. A Final Exam Review Problems From ch., 2 & 3 f Multiple Choice Identify the choice that best completes the statement or answers the question.. Place the following cations in order from lowest to
More informationColligative Properties. Vapour pressure Boiling point Freezing point Osmotic pressure
Colligative Properties Vapour pressure Boiling point Freezing point Osmotic pressure Learning objectives Describe meaning of colligative property Use Raoult s law to determine vapor pressure of solutions
More informationStandard Free Energies of Formation at 298 K. Average Bond Dissociation Energies at 298 K
1 Thermodynamics There always seems to be at least one free response question that involves thermodynamics. These types of question also show up in the multiple choice questions. G, S, and H. Know what
More informationChapter 20. Thermodynamics p. 811 842. Spontaneity. What have we learned about spontaneity during this course?
Chapter 20 p. 811 842 Spontaneous process: Ex. Nonspontaneous process: Ex. Spontaneity What have we learned about spontaneity during this course? 1) Q vs. K? 2) So.. Spontaneous process occurs when a system
More informationIf you like us, please share us on social media. The latest UCD Hyperlibrary newsletter is now complete, check it out.
Sign In Forgot Password Register ashwenchan username password Sign In If you like us, please share us on social media. The latest UCD Hyperlibrary newsletter is now complete, check it out. ChemWiki BioWiki
More informationThe Ideal Solution. ChemActivity T15
ChemActivity T15 The Ideal Solution Focus Question: An equimolar mixture of benzene and toluene is prepared. What will be the composition of the vapor in equilibrium with this solution? Model 1: Benzene
More informationChapter 11 Properties of Solutions
Chapter 11 Properties of Solutions 11.1 Solution Composition A. Molarity moles solute 1. Molarity ( M ) = liters of solution B. Mass Percent mass of solute 1. Mass percent = 1 mass of solution C. Mole
More information1 Exercise 5.5b pg 201
In this solution set, an underline is used to show the last significant digit of numbers. For instance in x 2.51693 the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the
More information= T T V V T = V. By using the relation given in the problem, we can write this as: ( P + T ( P/ T)V ) = T
hermodynamics: Examples for chapter 3. 1. Show that C / = 0 for a an ideal gas, b a van der Waals gas and c a gas following P = nr. Assume that the following result nb holds: U = P P Hint: In b and c,
More informationChapter 19. Chemical Thermodynamics. The reverse reaction (two eggs leaping into your hand with their shells back intact) is not spontaneous.
Chapter 19. Chemical Thermodynamics SOURCE: Chemistry the Central Science: Prentice hall I. Spontaneous Processes Thermodynamics is concerned with the question: will a reaction occur? First Law of Thermodynamics:
More informationLecture 3: Models of Solutions
Materials Science & Metallurgy Master of Philosophy, Materials Modelling, Course MP4, Thermodynamics and Phase Diagrams, H. K. D. H. Bhadeshia Lecture 3: Models of Solutions List of Symbols Symbol G M
More informationChapter 13. Properties of Solutions
13.4 Ways of Expressing Concentration All methods involve quantifying the amount of solute per amount of solvent (or solution). Concentration may be expressed qualitatively or quantitatively. The terms
More informationFinal Exam CHM 3410, Dr. Mebel, Fall 2005
Final Exam CHM 3410, Dr. Mebel, Fall 2005 1. At 31.2 C, pure propane and nbutane have vapor pressures of 1200 and 200 Torr, respectively. (a) Calculate the mole fraction of propane in the liquid mixture
More informationPhase Equilibrium: Fugacity and Equilibrium Calculations. Fugacity
Phase Equilibrium: Fugacity and Equilibrium Calculations (FEC) Phase Equilibrium: Fugacity and Equilibrium Calculations Relate the fugacity and the chemical potential (or the partial molar Gibbs free energy)
More informationa) Use the following equation from the lecture notes: = ( 8.314 J K 1 mol 1) ( ) 10 L
hermodynamics: Examples for chapter 4. 1. One mole of nitrogen gas is allowed to expand from 0.5 to 10 L reversible and isothermal process at 300 K. Calculate the change in molar entropy using a the ideal
More informationProblem Set 3 Solutions
Chemistry 360 Dr Jean M Standard Problem Set 3 Solutions 1 (a) One mole of an ideal gas at 98 K is expanded reversibly and isothermally from 10 L to 10 L Determine the amount of work in Joules We start
More informationCHEMICAL EQUILIBRIUM
Chemistry 10 Chapter 14 CHEMICAL EQUILIBRIUM Reactions that can go in both directions are called reversible reactions. These reactions seem to stop before they go to completion. When the rate of the forward
More information= 1.038 atm. 760 mm Hg. = 0.989 atm. d. 767 torr = 767 mm Hg. = 1.01 atm
Chapter 13 Gases 1. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. Gases have volumes that depend on their conditions, and can be compressed or expanded by
More informationAbsorption of Heat. Internal energy is the appropriate energy variable to use at constant volume
6 Absorption of Heat According to the First Law, E = q + w = q  P V, assuming PV work is the only kind that can occur. Therefore, E = q V. The subscript means that the process occurs at constant volume.
More informationPV (0.775 atm)(0.0854 L) n = = = 0.00264 mol RT 11
catalyst 2 5 g ¾¾¾¾ 2 4 g 2 g DH298 = rxn DS298 C H OH( ) C H ( ) + H O( ) 45.5 kj/mol ; = 126 J/(K mol ) ethanol ethene water rxn 1 atm 760 torr PV (0.775 atm)(0.0854 L) n = = = 0.00264 mol RT 11 (0.08206
More informationIdeal Gas and Real Gases
Ideal Gas and Real Gases Lectures in Physical Chemistry 1 Tamás Turányi Institute of Chemistry, ELTE State roerties state roerty: determines the macroscoic state of a hysical system state roerties of single
More informationChapter 13. Properties of Solutions
Sample Exercise 13.1 (p. 534) By the process illustrated below, water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is Na 2 SO 4(s) + 10 H 2
More informationDetermination of Molecular Weight by Freezing Point Depression
Determination of Molecular Weight by Freezing Point Depression Freezing point depression is a kind of colligative properties as treated in high school chemistry course. Normally used alcohol or mercury
More informationChemical Thermodynamics
Chemical Thermodynamics David A. Katz Department of Chemistry Pima Community College Tucson, AZ 85709, USA First Law of Thermodynamics The First Law of Thermodynamics was expressed in the study of thermochemistry.
More informationChapter 6 Multiphase Systems
Chapter 6 Multiphase Systems SingleComponent Systems Phase Diagram: a plot that shows conditions under which a pure substance exists in a particular phase e.g. a liquid, a solid, or a gas. Often, the
More informationFUNDAMENTALS OF ENGINEERING THERMODYNAMICS
FUNDAMENTALS OF ENGINEERING THERMODYNAMICS System: Quantity of matter (constant mass) or region in space (constant volume) chosen for study. Closed system: Can exchange energy but not mass; mass is constant
More informationReading. Spontaneity. Monday, January 30 CHEM 102H T. Hughbanks
Thermo Notes #3 Entropy and 2nd Law of Thermodynamics Monday, January 30 CHEM 102H T. Hughbanks Reading You should reading Chapter 7. Some of this material is quite challenging, be sure to read this material
More informationSolutions. Occur in all phases. Ways of Measuring. Ways of Measuring. Energy of Making Solutions. 1. Break apart Solvent. Page 1
s Occur in all phases The solvent does the dissolving. The solute is dissolved. There are examples of all types of solvents dissolving all types of solvent. We will focus on aqueous solutions. Ways of
More information12.3 Colligative Properties
12.3 Colligative Properties Changes in solvent properties due to impurities Colloidal suspensions or dispersions scatter light, a phenomenon known as the Tyndall effect. (a) Dust in the air scatters the
More informationReview of Chemical Equilibrium Introduction
Review of Chemical Equilibrium Introduction Copyright c 2016 by Nob Hill Publishing, LLC This chapter is a review of the equilibrium state of a system that can undergo chemical reaction Operating reactors
More informationCHEM 36 General Chemistry EXAM #1 February 13, 2002
CHEM 36 General Chemistry EXAM #1 February 13, 2002 Name: Serkey, Anne INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show
More informationSolutions. Occur in all phases. Ways of Measuring. Ways of Measuring. Page 1
Solutions Occur in all phases The solvent does the dissolving. The solute is dissolved. There are examples of all types of solvents dissolving all types of solvent. We will focus on aqueous solutions.
More informationSaturation Vapour Pressure above a Solution Droplet
Appendix C Saturation Vapour Pressure above a Solution Droplet In this appendix, the Köhler equation is derived. As in Appendix B, the theory summarised here can be found in Pruppacher & Klett (1980) and
More informationThermodynamics of Adsorption
CTI_CHAPTER_21.qxd 6/7/24 3:31 PM Page 243 CHAPTER 21 Thermodynamics of Adsorption ALAN L. MYERS 1 Introduction The attachment of molecules to the surface of a solid by adsorption is a broad subject. This
More informationSOLUTIONS MODULE  3. Objectives. Chemistry. States of matter. Notes
Chemistry 9 SOLUTIONS Y ou know that when sugar or salt is added to water, it dissolves. The resulting mixture is called a solution. Solutions play an important role in our life. In industry, solutions
More informationVaporLiquid Equilibria
31 Introduction to Chemical Engineering Calculations Lecture 7. VaporLiquid Equilibria Vapor and Gas Vapor A substance that is below its critical temperature. Gas A substance that is above its critical
More informationAP CHEMISTRY 2006 SCORING GUIDELINES
AP CHEMISTRY 2006 SCORING GUIDELINES Question 3 3. Answer the following questions that relate to the analysis of chemical compounds. (a) A compound containing the elements C, H, N, and O is analyzed. When
More information4. Thermodynamics of Polymer Blends
4. Thermodynamics of Polymer Blends Polymeric materials find growing applications in various fields of everyday life because they offer a wide range of application relevant properties. Blending of polymers
More informationColligative properties. Chemistry 433. Freezing point depression. Freezing point depression. Freezing point depression 10/28/2008
Chemistry 433 Lecture 20 Colligative Properties Freezing Point Depression Boiling Point Elevation Osmosis NC State University Colligative properties There are a number of properties of a dilute solution
More informationSolutions. ... the components of a mixture are uniformly intermingled (the mixture is homogeneous). Solution Composition. Mass percentageof solute=
Solutions Properties of Solutions... the components of a mixture are uniformly intermingled (the mixture is homogeneous). Solution Composition 1. Molarity (M) = 4. Molality (m) = moles of solute liters
More informationThe gas laws Equations of state The state of any sample of substance is specified by giving the values of the following properties: V the volume the
The gas laws Equations of state The state of any sample of substance is specified by giving the values of the following properties: V the volume the sample occupies p the pressure of the sample T the temperature
More informationAnswers: Given: No. [COCl 2 ] = K c [CO][Cl 2 ], but there are many possible values for [CO]=[Cl 2 ]
Chemical Equilibrium What are the concentrations of reactants and products at equilibrium? How do changes in pressure, volume, temperature, concentration and the use of catalysts affect the equilibrium
More informationChapter 13 Properties of Solutions
Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 13 Properties of are homogeneous mixtures of two or more pure substances. In a solution,
More informationRange of Competencies
CHEMISTRY Content Domain Range of Competencies l. Nature of Science 0001 0003 18% ll. Matter and Atomic Structure 0004 0006 18% lll. Energy and Chemical Bonding 0007 0010 23% lv. Chemical Reactions 0011
More informationChapter 7. Mixtures Colligative properties DebyeHückel. Chemical Thermodynamics : Georg Duesberg
Chapter 7 Mitures Colligative properties DebyeHückel Chemical Thermodynamics : Georg Duesberg Deviations from Raoult s Law Similar liquids can form an ideal solution obeying Raoult s Law Raoult s law
More informationEntropy and The Second Law of Thermodynamics
The Second Law of Thermodynamics (SL) Entropy and The Second Law of Thermodynamics Explain and manipulate the second law State and illustrate by example the second law of thermodynamics Write both the
More informationFugacity, Activity, and Standard States
Fugacity, Activity, and Standard States Fugacity of gases: Since dg = VdP SdT, for an isothermal rocess, we have,g = 1 Vd. For ideal gas, we can substitute for V and obtain,g = nrt ln 1, or with reference
More informationChapter 18 Homework Answers
Chapter 18 Homework Answers 18.22. 18.24. 18.26. a. Since G RT lnk, as long as the temperature remains constant, the value of G also remains constant. b. In this case, G G + RT lnq. Since the reaction
More informationExploring Creation With Chemistry Table of Contents
Exploring Creation With Chemistry Table of Contents MODULE #1: Measurement and Units...1 Introduction... 1 Experiment 1.1: Air Has Mass... 1 Experiment 1.2: Air Takes Up Space... 2 Units of Measurement...
More informationConsider: N 2 (g) + 3H 2 (g) 2NH 3 (g) G = 32.90 kj/mol. conc. time
5.111 Lecture Summary #19 CHEMICAL EQUILIBRIUM (Chapter 9 Section 9.09.9) Topics Nature of Chemical Equilibrium Meaning of K Relationship between Equilibrium Expressions External Effects on K 19.1 Chemical
More informationChemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 10 Gases
Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 10 Gases A Gas Has neither a definite volume nor shape. Uniformly fills any container.
More informationRate of Reaction and the Collision Theory. Factors that Affect the Rate of a Chemical Reaction
Chemical Kinetics and Thermodynamics Chemical Kinetics concerned with: 1. Rates of Chemical Reactions # of moles of reactant used up or product formed Unit time Or 2. Reaction Mechanisms Rate of Reaction
More informationWed Sep 12, 2007 THE GASEOUS STATE
Chapter 5: Gases Gas Stoichiometry Partial Pressure Kinetic Theory Effusion and Diffusion Wed Sep 12, 2007 Exam #1  Friday, Sep 14 Attendance is mandatory! Practice exam today in recitation Week 3 CHEM
More informationName Class Date. In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question.
Assessment Chapter Test A Chapter: States of Matter In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. 1. The kineticmolecular
More information1. Define the term colligative property and list those physical properties of a solution that can be classified as colligative properties.
Solutions Colligative Properties DCI Name Section 1. Define the term colligative property and list those physical properties of a solution that can be classified as colligative properties. Colligative
More informationChapter 4 Practice Quiz
Chapter 4 Practice Quiz 1. Label each box with the appropriate state of matter. A) I: Gas II: Liquid III: Solid B) I: Liquid II: Solid III: Gas C) I: Solid II: Liquid III: Gas D) I: Gas II: Solid III:
More informationEverest. Leaders in Vacuum Booster Technology
This article has been compiled to understand the process of Solvent Recovery process generally carried out at low temperatures and vacuum. In many chemical processes solute is to be concentrated to high
More informationCHEMISTRY. Matter and Change. Section 13.1 Section 13.2 Section 13.3. The Gas Laws The Ideal Gas Law Gas Stoichiometry
CHEMISTRY Matter and Change 13 Table Of Contents Chapter 13: Gases Section 13.1 Section 13.2 Section 13.3 The Gas Laws The Ideal Gas Law Gas Stoichiometry State the relationships among pressure, temperature,
More informationName Date Class THERMOCHEMISTRY. SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510)
17 THERMOCHEMISTRY SECTION 17.1 THE FLOW OF ENERGY HEAT AND WORK (pages 505 510) This section explains the relationship between energy and heat, and distinguishes between heat capacity and specific heat.
More informationChapter 15 Chemical Equilibrium
Chapter 15 Chemical Equilibrium Chemical reactions can reach a state of dynamic equilibrium. Similar to the equilibrium states reached in evaporation of a liquid in a closed container or the dissolution
More informationOverview of Physical Properties of Gases. Gas Pressure
Overview of Physical Properties of Gases! volume changes with pressure! volume changes with temperature! completely miscible! low density gases: < 2 g/l liquids and solids: 1000 g/l Gas Pressure force
More informationc. Applying the first law of thermodynamics from Equation 15.1, we find that c h c h.
Week 11 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More informationR = J/mol K R = L atm/mol K
version: master Exam 1  VDB/LaB/Spk This MC portion of the exam should have 19 questions. The point values are given with each question. Bubble in your answer choices on the bubblehseet provided. Your
More informationColligative Properties Discussion Chem. 1A
Colligative Properties Discussion Chem. 1A The material covered today is found in sections Chapter 12.5 12.7 This material will not be covered in lecture, you will have homework assigned. Chem. 1A Colligative
More informationColligative Properties of Nonvolatile Solutes 01. Colligative Properties of Nonvolatile Solutes 02. Colligative Properties of Nonvolatile Solutes 04
Colligative Properties of Nonvolatile Solutes 01 Colligative Properties of Nonvolatile Solutes 02 Colligative Properties: Depend on the amount not on the identity There are four main colligative properties:
More informationUnit 4: Chemical Reactions Indicate date when step is accomplished Learning Target Intro Pract Master Verified
Unit 4: Chemical Reactions 1. I can use moles to convert between mass, particles, and volume (of a gas at STP) of any substance. 2. I can calculate percent composition from laboratory data or standard
More informationSolutions & Colloids
Chemistry 100 Bettelheim, Brown, Campbell & Farrell Ninth Edition Introduction to General, Organic and Biochemistry Chapter 6 Solutions & Colloids Solutions Components of a Solution Solvent: The substance
More informationAbsolute and Relative Gas Concentration: Understanding Oxygen in Air. Bruce Bugbee and Mark Blonquist
Absolute and Relative Gas Concentration: Understanding Oxygen in Air Bruce Bugbee and Mark Blonquist Absolute and Relative Gas Concentration Gas concentration is described in two ways, absolute and relative
More informationEntropy in Chemistry
LECURE 5 Can we predict: If a reaction or process can occur? How will it proceed? How fast will it go? Entropy in Chemistry hermodynamics tells us nothing about rates tells everything about the rest So
More informationChemistry 433. The Third Law of Thermodynamics. Residual Entropy. CO: an Imperfect Crystal. Question. Question. Lecture 12 The Third Law
Chemistry 433 Lecture 12 he hird Law he hird Law of hermodynamics he third law of thermodynamics states that every substance has a positive entropy, but at zero Kelvin the entropy is zero for a perfectly
More informationChapter 12 Solutions
Chapter 12 Solutions 12.1 Ideal solutions Solutions are arguably the most important kind of system studied by chemical engineers, whether in large chemical processing or cellularlevel molecular phenomena.
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
A.P. Chemistry Practice Test: Ch. 11, Solutions Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Formation of solutions where the process is
More informationCHEMICAL THERMODYNAMICS
CHEMICAL THERMODYNAMICS Spontaneous reactions Nonspontaneous reactions Enthalpy change (ΔH) Exothermic and endothermic reactions Entropy change (ΔS) Gibbs free energy change, (ΔG) Free energy of formation
More information