1 Thermodynamics and Kinetics Lecture 14 Properties of Mixtures Raoult s Law Henry s Law Activity NC State University
2 Measures of concentration There are three measures of concentration: molar concentration per unit volume (molarity) c = n/(liter of solution) molar concentration per unit mass (molality) m = n/(kg of solution) mole fraction x j = n j n i i
3 Raoult's law states P j = x j P j Ideal solutions where P j is the vapor pressure of pure component j. The vapor pressure of component j in an ideal solution is given by the product of its mole fraction and P j. The chemical potential can be expressed as: m j = m j + RTln(P j /P j ) where P j is vapor the pressure of the pure component j in the standard state.
4 Ideal solutions The significance of this expression is that we can now consider the equilibrium between vapor and solution to write: m j soln = m j vap = m j 0 + RTln(P j /P j0 ) but for the vapor P j 0 = 1 bar. In the limit that the vapor becomes the pure vapor we have: m j = m j vap = m j 0 + RTln(P j /P j0 ) Thus m j soln = m j + RTln(P j /P j ) keeping in mind the notation means the pure component.
5 Ideal solutions The equation below is central equation of binary solution mixtures. m j soln = m j + RTln(P j /P j ) Using Raoult's law x j = P j /P j we see that the chemical potential can be expressed as: m j soln = m j + RTln(x j ) This equation defines an ideal solution.
6 The free energy of mixing The free energy for formation of a solution from individual components is given by G mix = G soln G 1 G 2 Since G soln = n 1 m 1 + n 2 m 2, G 1 = n 1 m 1 and G 2 = n 2 m 2. we have that G mix = n 1 m 1 + n 2 m 2 n 1 m 1 n 2 m 2 = nrt x 1 ln x 1 + x 2 ln x 2 for an ideal solution there is no enthalpy of mixing. The volume of the mixture also does not change for an ideal solution. The entropy change can be obtained from in agreement with a previous derivation.
7 Two component phase diagrams The total vapor pressure over an ideal solution is given by P total = P 1 + P 2 = x 1 P 1 + x 2 P 2 = (1 - x 2 )P 1 + x 2 P 2 = P 1 + x 2 (P 2 - P 1 ) A plot of the total pressure has the form of a straight line.
8 Liquid composition: specific example Consider the example in the book of 1-propanol and 2-propanol, which have P 1 = 20.9 torr and P 2 = 45.2 torr, respectively. So in this example, the phase diagram has the appearance:
9 The vapor mole fraction The value of the mole fraction in the vapor is not necessarily equal to that of the liquid. In the vapor phase the relative numbers of moles is given by Dalton's law. Applying Dalton's law we find: y 1 = P 1 /P total = x 1 P 1 /P total or y 2 = P 2 /P total = x 2 P 2 /P total The vapor curve is not the same as the liquid line.
10 Deriving the vapor curve Using the equation of the liquid line we find: When substituted into y 2 = x 2 P 2 /P total we have Solving for P total we find: x 2 = P total P 1 P 2 P 1 y 2 = P 2 P total P 1 P total P 2 P 1 y 2 P total = P 2 P total P 1 P total P 1 = P 2 P 2 P 1 P 2 P P 2 1 P 2 P 1 P 2 y 2 P 2 P 1 P total = P 2 P 1 P total = P 2 P 1 P 2 y 2 P 2 P 1
11 The vapor curve The is shown in the Figure below. The purple line was calculated using the Dalton's law expression. What lies between the blue and purple lines? This is the two phase region. Two Phase Region
12 The two phase region If we pick a composition and pressure that is inside this region then we can use a tie line to indicate the composition of each phase.
13 Boiling in a two component system If we reduce the pressure above a two component mixture of 1-propanol and 2-propanol with a mole fraction of propanol what is the composition of the vapor?
14 What is the composition of the vapor? a. y 2 = 0.60 b. y 2 = 0.48 c. y 2 = 0.78 d. y 2 = 0.98
15 What is the composition of the vapor? a. y 2 = 0.60 b. y 2 = 0.48 c. y 2 = 0.78 d. y 2 = 0.98
16 Boiling in a two component system If we continue to reduce the pressure the system moves into the two-phase region. In the two-phase region the composition of the liquid and the vapor is not the same.
17 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70
18 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70
19 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70
20 Boiling in a two component system If we continue to reduce the pressure the system reaches the boundary with pure vapor.
21 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70
22 What is the composition? a. x 2 = 0.40, y 2 = 0.60 b. x 2 = 0.50, y 2 = 0.70 c. x 2 = 0.50, y 2 = 0.50 d. x 2 = 0.40, y 2 = 0.70
23 The tie line The tie line shown in Figure (red line) is at a total pressure of 30 torr. You can read the x 2 and y 2 values from the plot (or calculate them using the equations above used to generate the blue and purple curves in the composition-pressure plot.
24 The lever rule The tie line can be used to define the quantity of each phase present in the two phase region. The total composition x a can be used together with x 2 and y 2. z a y 2 - z a z a
25 The lever rule The lever rule states that the amount of each phase present is inversely proportional to the length of distance along the tie line from the phase boundary to the total composition, z a. z a y 2 - z a z a
26 The lever rule states: The lever rule n liquid n vapor = y 2 z a z a x 2 The total composition is z a. z a y 2 - z a z a
27 Non-ideal solutions Many solutions are not ideal. For ideal solutions the role of intermolecular interactions can be ignored. This may be because they are small or because two components have the same interaction with each other that they have with themselves. In other words similar solvents will form ideal solutions. However, in many cases, intermolecular interactions cause deviations from Raoult's law. We can consider the "like" interactions between molecules of same species and "unlike" interactions between molecules of different species. If the unlike-molecule interactions are more attractive than the like molecule interactions, the vapor pressure above a solution will be smaller than we would calculate using Raoult's law. If the unlike-molecule interactions are more repulsive, then the vapor pressure is greater than for the ideal solution.
28 Henry s law The statement of Henry s law is: P 1 = x 1 k H,1 It looks like Raoult s law except that you have this funny constant k H,1 instead of P 1 (the vapor pressure of component 1). This law is only valid for dilute solutions, i.e. when component 1 is the solute. Under these conditions the vapor pressure of component 1 really does not matter, because component 1 is mostly surrounded by component 2 and so its properties really quite different from the properties of pure 1. Henry s law can be applied to mixtures of solvents And also to solutions of gases in liquids. For example, see the problems on the concentration of O2 and N2 in water at the end of the lecture.
29 Henry s law Attractive interactions between unlike molecules leads to negative deviations from Raoult's law (lower vapor pressure than ideal) and repulsive interactions lead to positive deviations (higher vapor pressure than ideal). As any solution approaches a mole fraction of one (i.e. approaches a pure solution of one component) it becomes an ideal solution. In other words, P 1 x 1 P 1 as x 1 1. However, as x 1 0 the component is surrounded by unlike molecules and the solution has the maximum deviation from ideal behavior. For this case we define Henry's law, P 1 x 1 k H,1 as x 1 0. In this expression k H,1 is the Henry's law constant. Although we have focused on component 1 the same holds true for component 2.
30 Non-ideal solutions For the example shown in the plot above we have assumed that P 1 = 100 torr. Note that as x 1 1 the slope approaches the ideal slope obtained from Raoult's law. However, as x 1 0 (and therefore x 2 1) the slope is quite different from ideal behavior. Note that the slope has the value of the Henry's law constant k H,1. This is depicted in the Figure below (purple line). The Henry's law value can be quite different from the ideal value.
31 Activity The activity in non-ideal solutions corresponds to mole fraction in ideal solutions. a j = P j P j The activity replaces mole fraction in the expression for the chemical potential. When considering a non-ideal solution the above expressions hold and thus the mole fraction x j is no longer equal to P j /P j. However, as the mole fraction approaches unity (a pure substance) the solution becomes ideal. Thus, as x j 1, a j x j. m j soln = m j + RT ln a j
32 The activity coefficient We can define an activity coefficient g 1 such that g 1 = a 1 /x 1. The property of the activity coefficient is that it approaches a value of 1 (ideal behavior) as the composition approaches the pure solvent: a 1 x 1 and g 1 1 as x 1 1. Thus, the solution This definition is based on a Raoult's law standard state, which is also known as a solvent standard state. The activities or chemical potentials are meaningless unless we know the standard state
33 Henry s law constant and the solubility of gases Henry s law constants in H 2 O (atm x 10 3 ) He 131 N 2 86 CO 57 O 2 43 Ar 40 CO Problem: Divers get the bends if bubbles of N 2 form in their blood because they rise too rapidly. Calculate the molarity of N 2 in water (i.e. blood) at sea level and 100 m below sea level. At sea level: a N2 = P N2 /k H,N2 = 0.8 atm/86 x 10 3 atm = 9.3 x 10-6 c N2 = 55.6 a N2 = 5 x 10-4 mol/l At 100 m: a N2 = P N2 /k H,N2 = 9.8 atm/86 x 10 3 atm = 1.1 x 10-4 c N2 = 55.6 a N2 = 6 x 10-3 mol/l
34 A note on conversion from mole fraction to molarity The conversion from mole fraction to molarity can be solved analytically. n x 1 = 1 c = 1 n 1 + n 2 c 1 + c 2 since n 1 = c 1 V and V cancels x 1 (c 1 + c 2 ) = c 1 x 1 c 2 = c 1 (1 x 1 ) x 1 c 2 = c 1 x 2 c 2 = c 1 x 2 x 1 Note that for water as solvent (component 1) x 1 ~ 1 and the Concentration of water is c 1 ~ 55.6 so that the conversion For a dilute solute such as a gas is c 2 ~ 55.6 x 2
35 Question Henry s law constants in H 2 O (atm x 10 3 ) He 131 N 2 86 CO 57 O 2 43 Ar 40 CO A species of fish requires a concentration of O 2 > 100 mm. A marine biologist is trying to determine the depth profile for O 2 in sea water. The first step is to calculate the concentration of O 2 in sea water at sea level. As an assistant you perform the calculation and find that the O 2 concentration is: A. 250 mm B. 430 mm C. 760 mm D mm
36 Question Henry s law constants in H 2 O (atm x 10 3 ) He 131 N 2 86 CO 57 O 2 43 Ar 40 CO Most fish require a concentration of O 2 that is greater than 100 mm. A marine biologist is trying to determine the depth profile for O 2 in sea water. The first step is to calculate the concentration of O 2 in sea water at sea level. As an assistant you perform the calculation and find that the O 2 concentration is: A. 250 mm B. 430 mm C. 760 mm D mm a O2 = P O2 /k H,O2 = 0.2 atm/43 x 10 3 atm = 4.7 x 10-6 c O2 = 55.6 a O2 = 2.5 x 10-4 mol/l
37 The Henry s law constant is an equilibrium constant Gases dissolve in liquids to form solutions. This dissolution is an equilibrium process for which an equilibrium constant can be written. For example, the equilibrium between oxygen gas and dissolved oxygen in water is O 2 (aq) <--> O 2 (g). The equilibrium constant for this equilibrium is: K H = p(o 2 )/c(o 2 ). The form of the equilibrium constant shows that the concentration of a solute gas in a solution is directly proportional to the partial pressure of that gas above the solution. The form of the equation can be Rearranged to give: p(o 2 ) = c(o 2 )K H = x(o 2 ) c(h 2 O)K H
38 The Henry s law constant is tabulated for c and x The Henry s law constant can be tabulated for mole fraction (as seen in the previous problem) or for molarity. If the units of the Henry s law constant are atm then it is valid for mole fraction. If the units are atm/(mol/l) then K H is tabulated for molarity. The relationship between the two is: p(o 2 ) = c(o 2 )K H = x(o 2 ) c(h 2 O)K H So that K H = 55.6 (mol/l)k H since the concentration of water is c(h 2 O) = molar. For example, K H is 757 atm/(mol/l). K H = K H 55.6 = = 13.6
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
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)
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:
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
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
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)
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
Lecture 1: Physical Equilibria The Temperature Dependence of Vapor Pressure Our first foray into equilibria is to examine phenomena associated with two phases of matter achieving equilibrium in which the
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
CHEMISTRY The Molecular Nature of Matter and Change Third Edition Chapter 13 The Properties of Mixtures: Solutions and Colloids Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction
Chapter 6 Gases Kinetic Theory of Gases 6.1 Properties of Gases 6.2 Gas Pressure A gas consists of small particles that move rapidly in straight lines. have essentially no attractive (or repulsive) forces.
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
31 Introduction to Chemical Engineering Calculations Lecture 7. Vapor-Liquid Equilibria Vapor and Gas Vapor A substance that is below its critical temperature. Gas A substance that is above its critical
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
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
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,
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
Name: Class: Date: Unit 13 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The dissolution of water in octane (C 8 H 18 ) is prevented by.
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
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
NAME: CHEMISTRY 443, Fall, 2011 (11F) Circle Section Number: 10 11 80 81 Final Exam, December 14, 2011 Answer each question in the space provided; use back of page if extra space is needed. Answer questions
ChemActivity T15 The Ideal Solution Focus Question: An equi-molar mixture of benzene and toluene is prepared. What will be the composition of the vapor in equilibrium with this solution? Model 1: Benzene
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;
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
CHAPTER 12 Gases and the Kinetic-Molecular Theory 1 Gases vs. Liquids & Solids Gases Weak interactions between molecules Molecules move rapidly Fast diffusion rates Low densities Easy to compress Liquids
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
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
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.
Chapter 14. CHEMICAL EQUILIBRIUM 14.1 THE CONCEPT OF EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT Many chemical reactions do not go to completion but instead attain a state of chemical equilibrium. Chemical
Chapter 16 Thermodynamics GCC CHM152 Thermodynamics You are responsible for Thermo concepts from CHM 151. You may want to review Chapter 8, specifically sections 2, 5, 6, 7, 9, and 10 (except work ). Thermodynamics:
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
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
Common Equations Used in Chemistry Equation for density: d= m v Converting F to C: C = ( F - 32) x 5 9 Converting C to F: F = C x 9 5 + 32 Converting C to K: K = ( C + 273.15) n x molar mass of element
Sample Exercise 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for K c for the following reactions: Solution Analyze: We are given three equations and are asked to write
Chapter Thirteen Physical Properties Of Solutions 1 Solvent: Solute: Solution: Solubility: Types of Solutions Larger portion of a solution Smaller portion of a solution A homogeneous mixture of 2 or more
CHEM 10123/10125, Exam 1 February 8, 2012 (50 minutes) Name (please print) 1. SHOW ALL WORK. A saturated, aqueous NaCl solution is 5.40 M NaCl (58.44 g/mol) and is 26.0% NaCl by weight. a.) (10 points)
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
1 Phase Equilibria  Vapor-liquid equilibrium VLE) is the state of coexistence between liquid and vapor phases. For a one component system, this is easily visualized, as shown in Figure 1. As the number
Energy Balances on Closed Systems A system is closed if mass does not cross the system boundary during the period of time covered by energy balance. Energy balance for a closed system written between two
Colligative properties CH102 General Chemistry, Spring 2014, Boston University here are four colligative properties. vapor-pressure lowering boiling-point elevation freezing-point depression osmotic pressure
Chapter 5 The Gaseous State I) Pressure Pressure is the force exerted per unit area. A) Devices used to measure pressure 1) barometer used to measure the atmospheric pressure at seal level and 0 o C, P
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
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,
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 cellular-level molecular phenomena.
Cautions Butane is toxic and flammable. No OPEN Flames should be used in this experiment. Purpose The purpose of this experiment is to determine the molar mass of butane using Dalton s Law of Partial Pressures
Colligative Properties Vapor pressures have been defined as the pressure over a liquid in dynamic equilibrium between the liquid and gas phase in a closed system. The vapor pressure of a solution is different
The Gas, Liquid, and Solid Phase When are interparticle forces important? Ron Robertson Kinetic Theory A. Principles Matter is composed of particles in constant, random, motion Particles collide elastically
T-41 Tutorial 6 GASES Before working with gases some definitions are needed: PRESSURE: atmospheres or mm Hg; 1 atm = 760 mm Hg TEMPERATURE: Kelvin, K, which is o C + 273 STP: Standard Temperature and Pressure:
Gases States of Matter States of Matter Kinetic E (motion) Potential E(interaction) Distance Between (size) Molecular Arrangement Solid Small Small Ordered Liquid Unity Unity Local Order Gas High Large
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
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
Chapter 4 The Properties of Gases Significant Figure Convention At least one extra significant figure is displayed in all intermediate calculations. The final answer is expressed with the correct number
Name Period Gas Laws Kinetic energy is the energy of motion of molecules. Gas state of matter made up of tiny particles (atoms or molecules). Each atom or molecule is very far from other atoms or molecules.
inary Solutions Composition as a thermodynamic variable ibbs free energy of binary solutions Entropy of formation and ibbs free energy of an ideal solution Regular solutions: Heat of formation of a solution
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
EQUATIONS OF STATE FOR GASES Questions A gas enters a reactor at a rate of 255 SCMH. What does that mean? An orifice meter mounted in a process gas line indicates a flow rate of 24 ft 3 /min. The gas temperature
CHEM110 Week 9 Notes (Gas Laws) Page 1 of 7 Lecture Notes: Gas Laws and Kinetic Molecular Theory (KMT). Gases Are mostly empty space Occupy containers uniformly and completely Expand infinitely Diffuse
Thermodynamics Chapter 13 Phase Diagrams NC State University Pressure (atm) Definition of a phase diagram A phase diagram is a representation of the states of matter, solid, liquid, or gas as a function
Thermodynamics: An Engineering Approach Seventh Edition in SI Units Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011 Chapter 3 PROPERTIES OF PURE SUBSTANCES Mehmet Kanoglu University of Gaziantep Copyright
PROPERTIES OF GASES or GAS LAWS 1 General Properties of Gases There is a lot of empty space in a gas. Gases can be expanded infinitely. Gases fill containers uniformly and completely. Gases diffuse and
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
Exam 4 Practice Problems 1 1. Which of the following statements is false? a. Condensed states have much higher densities than gases. b. Molecules are very far apart in gases and closer together in liquids
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
roblem Set 4 Answers rofessor: C. E. Loader. At 5.0 ºC, the vapor pressure of ice is 3.03 mm and that of liquid water is 3.63 mm. Calculate G for the change of one mole of liquid water at 5.0 ºC to solid
CHEMISTRY The Central Science Properties of Solutions The Solution Process Solutions: Air; brass; body fluids; sea water When a solution forms some questions we can ask are: What happens on a molecular
Colligative properties 1 1. What does the phrase like dissolves like mean. 2. Why does the solubility of alcohols decrease with increased carbon chain length? Alcohol in water (mol/100g water) Methanol
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 kinetic-molecular
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
Gas Laws v101613_10am Objective: In this lab you will become familiar with the Ideal Gas Law and Dalton s Law of Partial Pressures. You will be able to use the information collected along with stoichiometry
Phase diagram of water Note: for H 2 O melting point decreases with increasing pressure, for CO 2 melting point increases with increasing pressure. WATER Covers ~ 70% of the earth s surface Life on earth
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
CHM1045 Practice Test 3 v.1 - Answers Name Fall 013 & 011 (Ch. 5, 6, 7, & part 11) Revised April 10, 014 Given: Speed of light in a vacuum = 3.00 x 10 8 m/s Planck s constant = 6.66 x 10 34 J s E (-.18x10
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
CHATER 3. The atmosphere is a homogeneous mixture (a solution) of gases.. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. have volumes that depend on their conditions,
AP CHEMISTRY 2009 SCORING GUIDELINES (Form B) Question 3 (10 points) 2 H 2 O 2 (aq) 2 H 2 O(l) + O 2 (g) The mass of an aqueous solution of H 2 O 2 is 6.951 g. The H 2 O 2 in the solution decomposes completely
Gases Section 13.1 The Gas Laws In your textbook, read about the basic concepts of the three gas laws. Use each of the terms below to complete the passage. Each term may be used more than once. pressure
Chapter 6: Thermochemistry (Chemical Energy) (Ch6 in Chang, Ch6 in Jespersen) Energy is defined as the capacity to do work, or transfer heat. Work (w) - force (F) applied through a distance. Force - any
Name Unit 11 Review: Gas Laws and Intermolecular Forces Date Block 1. If temperature is constant, the relationship between pressure and volume is a. direct b. inverse 2. If pressure is constant, the relationship