Presentation on theme: "Chapter 11 Particle Forces. States of Matter Solid- Particles moving about a fixed point Liquid-Particles moving about a moving point Gas-Particles filling."— Presentation transcript:
Chapter 11 Particle Forces
States of Matter Solid- Particles moving about a fixed point Liquid-Particles moving about a moving point Gas-Particles filling the volume of the container with complete random motions.
Ion-Ion Interactions Coulomb’s law states that the energy (E) of the interaction between two ions is directly proportional to the product of the charges of the two ions (Q 1 and Q 2 ) and inversely proportional to the distance (d) between them.
Predicting Forces of Attraction Coulombs Law indicates the increases in the charges of ions will cause an increase in the force of attraction between a cation and an anion. Increases in the distance between ions will decrease the force of attraction between them.
Size of Ions
Lattice Energy The lattice energy (U) of an ionic compound is the energy released when one mole of the ionic compound forms from its free ions in the gas phase. M + (g) + X - (g) ---> MX(s)
Comparing Lattice Energies Lattice Energies of Common Ionic Compounds CompoundU(kJ/mol) LiF-1047 LiCl-864 NaCl-790 KCl-720 KBr-691 MgCl MgO-3791
Practice Determine which salt has the greater lattice energy. A. MgO and NaF B. MgO and MgS
Lattice Energy Using Hess’s Law
Electron Affinity Electron affinity is the energy change occurring when one mole of electrons combines with one mole of atoms or ion in the gas phase. Step 4 in diagram on the last slide. Cl(g) + e - (g) ---> Cl - (g) ΔH Ea = -349 kj/mole
Calculating U Na + (g) + e - (g) ---> Na(g) - H IE1 Na(g) ---> Na(s)- H sub Cl - (g) ---> Cl(g) + e - (g)- H EA Cl(g) ---> 1/2Cl 2 (g) -1/2 H BE Na(s) + 1/2Cl 2 (g) ---> NaCl(s) H f Na + (g) + Cl - (g) ---> NaCl(s) ΔU U = H f - 1/2 H BE - H EA - H sub - H IE1
Lattice energy for NaCl. ΔUΔU
Interactions Involving Polar Molecules An ion-dipole interaction occurs between an ion and the partial charge of a molecule with a permanent dipole. The cluster of water molecules that surround an ion in aqueous medium is a sphere of hydration.
Illustrates of Ion-Dipole Interaction
The Solution Process Bond Breaking Processes Break solute particle forces (expanding the solute), endothermic Break solvent particle forces (expanding the solvent), endothermic
The Solution Process Attractive Forces Energy released when solute solvent are attracted, exothermic Energy is released due to new attractions Ion dipole if the solute is ionic and the solvent polar. London-Dipole for nonpolar solute and polar solvent Dipole-dipole for polar solute and polar solvent
The Solution Process Theromodynamics Enthalpy Entropy (Perfect crystal, assumed to be zero) Gibbs free energy
The Solution Process Oil dissolving in water London forces holding the oil molecules together are large do to the large surface area of the oil The hydrogen bonds holding water molecules together are large The forces of attraction of between nonpolar oil and polar water are weak at best Thus the overall process is highly endothermic and not allowed thermo chemically
The Solution Process Oil dissolving in water Entropy should be greater than zero Free energy should be greater than zero, since the process is highly endothermic Thus the overall process is nonspontaneous
The Solution Process Sodium chloride dissolving in water Large amount of energy is required to break the ionic lattice of the sodium chloride (expand solute) Large amount of energy is required to separate the water molecules to expand the solvent breaking hydrogen bonds Formation of the ion dipole forces releases a large amount of energy, strong forces (why?) The sum of the enthalpies is about +6 kJ (slightly endothermic), which is easily overcome by the entropy of the solution formation.
Water as a Solvent Water most important solvent, important to understand its solvent properties Most of the unusual solvent properties of water stem from it hydrogen bonding nature Consider the following ∆S of solution KCl →75j/K-mole LiF→-36j/K-mole CaS→-138 j/K-mole
Water as a Solvent We would expect ∆S>0 for all solutions, right? But two are negative, why? Obviously, something must be happening for the increased order. Ion-dipole forces are ordering the water molecules around the ions, thus causing more order in water i.e. less positions for water than in the pure liquid state
Water as a Solvent Smaller ions, have stronger ion dipole forces, thus pulling water closer, therefore less positions Also, ions with a charge greater than one will attract to water stronger than a one plus charge, thus more order due to less space between particles
Dipole-Dipole Interactions Dipole-dipole interactions are attractive forces between polar molecules. An example is the interaction between water molecules. The hydrogen bond is a special class of dipole-dipole interactions due to its strength.
Dipole-Dipole Forces Dipole-dipole (Polar molecules) Alignment of polar molecules to two electrodes charged + and δ– Forces compared to ionic/covalent are about 1 in strength compared to a scale of 100, thus 1% H Cl H Cl H Cl δ–δ– δ–δ– δ–δ– δ+δ+δ+δ+ δ+δ+
Slide 28 of 35 Dipole Dipole Interactions
Hydrogen Bonding Hydrogen bonding a stronger intermolecular force involving hydrogen and usually N, O, F, and sometimes Cl –Stronger that dipole-dipole, about 10 out of 100, or 10 –Hydrogen needs to be directly bonded to the heteroatom –Since hydrogen is small it can get close to the heteroatom –Also, the second factor is the great polarity of the bond.
Slide 30 Hydrogen Bonding in HF(g)
Slide 31 Hydrogen Bonding in Water around a molecule in the solid in the liquid
Boiling Points of Binary Hydrides
Interacting Nonpolar Molecules Dispersion forces (London forces) are intermolecular forces caused by the presence of temporary dipoles in molecules. A temporary dipole (or induced dipole) is a separation of charge produced in an atom or molecule by a momentary uneven distribution of electrons.
Strength of Dispersion Forces The strength of dispersion forces depends on the polarizability of the atoms or molecules involved. Poarizability is a term that describes the relative ease with which an electron cloud is distorted by an external charge. Larger atoms or molecules are generally more polarizable than small atoms or molecules.
London Forces (Dispersion) Induced dipoles (Instantaneous ) Strength is surface area dependent More significant in larger molecules All molecules show dispersion forces Larger molecules are more polarizable
Slide 37 Instantaneous and Induced Dipoles
Molar Mass and Boiling Points of Common Species. HalogenM(g/mol)Bp(K)Noble GasM(g/mol)Bp(K) He24 F2F2 3885Ne2027 Cl Ar4087 Br Kr84120 I2I Xe Rn211
HydrocarbonAlcohol Molecular Formula Molar Mass Bp ( o C) Molecular Formula Molar Mass Bp ( o C) CH CH CH 3 OH CH 3 CH 2 CH CH 3 CH 2 OH CH 3 CH(CH)CH CH 3 CH(OH)CH CH 3 CH 2 CH 2 CH CH 3 CH 2 CH 2 OH
The Effect of Shape on Forces
Practice Rank the following compound in order of increasing boiling point. CH 3 OH, CH 3 CH 2 CH 2 CH 3, and CH 3 CH 2 OCH 3
Practice Rank the following compound in order of increasing boiling point. CH 3 OH, CH 3 CH 2 CH 2 CH 3, and CH 3 CH 2 OCH 3 CH 3 OH CH 3 CH 2 CH 2 CH 3 CH 3 CH 2 OCH 3 MM IM Forces London and H-bonding London, only London and Dipole-dipole
Practice Rank the following compound in order of increasing boiling point. CH 3 OH, CH 3 CH 2 CH 2 CH 3, and CH 3 CH 2 OCH 3 CH 3 OH CH 3 CH 2 CH 2 CH 3 CH 3 CH 2 OCH 3 MM IM Forces London and H-bonding London, only London and Dipole-dipole The order is: CH 3 CH 2 CH 2 CH 3
Polarity and Solubility If two or more liquids are miscible, they form a homogeneous solution when mixed in any proportion. Ionic materials are more soluble in polar solvents then in nonpolar solvents. Nonpolar materials are soluble in nonpolar solvents. Like dissolves like
Solubility of Gases in Water Henry’s Law states that the solubility of a sparingly soluble chemically unreactive gas in a liquid is proportional to the partial pressure of the gas. C gas = k H P gas where C is the concentration of the gas, k H is Henry’s Law constant for the gas.
Henry’s Law Constants for Gas Henry’s Law Constants Gask H [mol/(Latm)]k H [mol/(kgmmHg)] He3.5 x x O2O2 1.3 x x N2N2 6.7 x x CO x x 10 -5
Terms A hydrophobic (“water-fearing) interaction repels water and diminishes water solubility. Polar vs. nonpolar A hydrophilic (“water-loving”) interaction attracts water and promotes water solubility. Polar vs. polar, best with hydrogen bonding involved.
Types of Forces Cohesive Forces Intermolecular forces between the same particles. Adhesive Forces Intermolecular forces between the different particles.
Cohesive Forces Example Surface tension (resistance to increasing the surface area) Def: To increase surface area molecules must move from the middle. This requires energy j/m 2 The stronger the IMF the stronger the surface tension Needle or paper clip on top of water Beading or wetting on a surface Rounded surface of liquid mercury in a tube
Adhesive Forces Examples. Capillary rise water forms a meniscus since the forces between the glass and water are stronger than between water and water. Both are hydrogen bonds
Cohesive and Adhesive Forces The left test tube shows adhesive forces due to the attraction of water solvent to the polar glass (SiO 2 ) Hydrogen bonding, right? The right tube shows cohesive forces, since mercury is nonpolar and attracts more strongly to itself, rather than to the glass (SiO 2 )
Terms Capillary rise is the rise of a liquid up a narrow tube as a result of adhesive forces between the liquid and the tube and cohesive forces within the liquid. Viscosity is a measure of the resistance to flow of a fluid.
The Liquid State Adhesive Forces Intermolecular forces between unlike molecules Example Capillary rise Blood up a capillary Meniscus Capillary rise is when the adhesive forces are stronger than the cohesive forces Capillary rise when polar bonds are present in the container walls like glass, SiO 2 Mercury is an example where the cohesive forces are stronger than the adhesive forces
Slide 55 Intermolecular Forces
The Liquid State Viscosity (resistance to flow) How fast liquids flow Due in part to intermolecular forces, but also entanglement Newton’s/m 2 called poise
Change of State (Water) H 2 O(s) H 2 O(l) H 2 O(g) Water Thermodynamic Properties ∆H fus = 6.02 kj/mole ∆H vap = 40.7 kj/mole melting evaporation Condensation freezing sublimation deposition
Change of State (Water) Heat capacity of water = j/g-°C Water has a very large heat capacity since it hydrogen bonds and a lot of energy is required to break these bonds Why water is used in radiators Used to cool animals
Slide 60 of 35 Some Properties of Solids Freezing Point Melting Point ΔH fus (H 2 O) = kJ/mol Super cooling
Super Cooling and Heating Super cooled, when a liquid exists below its freezing point. Super cooling occurs when the rate of cooling is faster than it takes for the molecules to rotate for correct alignment to form crystals. When the crystals rotate and form inter-particle forces, heat is released, thus raising the temperature up to the correct m.p. Super heated Called bumping Use boiling stones, cannot reuse the stones Hot vapor at bottom expands rapidly and bursts
Vapor Pressure Vaporization or evaporation is the transformation of molecules in the liquid phase to the gas phase. Vapor pressure is the force exerted at a given temperature by a vapor in equilibrium with its liquid phase.
Vapor Pressure What evaporates faster pure distilled water in the beaker on the left, or seawater in the beaker on the right?? Both beakers are the same size and at the same temperature.
Slide 64 Intermolecular Forces
Vapor Pressure Yes, pure distilled water evaporates faster, since there are more water molecules on the surface to evaporate?
Vapor Pressure Physical properties that depend on the number of particles, and not on the particle nature are called colligative properties
An Aqueous Solution and Pure Water in a Closed Environment
Slide 68 Vapor Pressure (e) (d) (c) (b) (a) ( ) 1 T Which one is water? -ΔH vap + B R Ln P = Linear B = y-intercept = ∆S R (Entropy of vaporization) No Units! Clausius-Clapeyron Equation
Vapor Pressure Clasius Clapeyron Equation Assume data for two different temperatures and pressures to generate two separate equations By subtracting the equations the y-intercept component is eliminated. ln P 1 = - ΔH/R(1/T 1 ) + C - (ln P 2 = - ΔH/R(1/T 2 ) + C) ln((P 1 /P 2 ) = - ΔH/R(1/T 2 – 1/T 1 ) Another useful version of the two point equation ln((P 1 /P 2 ) = - ΔH/R(T 2 -T 1 )/T 1 T 2
Vapor Pressure As a liquid evaporates in a closed container the concentration of vapor increases, thus the rate of condensation increases As the rate of condensation is increasing eventually it will equal the constant rate of evaporation, then we have vapor in equilibrium with the liquid The pressure of the vapor at equilibrium is called the equilibrium vapor pressure
Raoult’s Law P solution = X solvent (P solvent ) P - vapor pressure X - mole fraction X solvent + X solute = 1
For a Solution that Obeys Raoult's Law, a Plot fo P soln Versus X solvent, Give a Straight Line
Vapor Pressure of Solvent and Solution
July 2009General Chemistry: Chapter 11 Slide 74 of 46 Liquid-Vapor Equilibrium
Two Volatile Liquids Positive deviation Ideal Solution Negative deviation Positive deviation exists when experimental value is larger than calculated value, weaker solute solvent attraction; more evaporation. Negative deviation exists when experimental value is smaller than calculated value; stronger solvent solute attraction; less evaporation
July 2009General Chemistry: Chapter 11 Slide 76 of 46 Fractional Distillation
July 2009General Chemistry: Chapter 11 Slide 77 of 46 Fractional Distillation
Practice A solution contains mL of water and mol of ethanol. What is the mole fraction of water and the vapor pressure of the solution at 25 o C, if the vapor of pressure of pure water is 23.8 torr?
Practice A solution contains mL of water and mol of ethanol. What is the mole fraction of water and the vapor pressure of the solution at 25 o C, if the vapor of pressure of pure water is 23.8 torr? 100.0mL
Practice A solution contains mL of water and mol of ethanol. What is the mole fraction of water and the vapor pressure of the solution at 25 o C, if the vapor of pressure of pure water is 23.8 torr? 100.0mL mL 1.00 g 18.0 g mole
Practice A solution contains mL of water and mol of ethanol. What is the mole fraction of water and the vapor pressure of the solution at 25 o C, if the vapor of pressure of pure water is 23.8 torr? 100.0mL mL 1.00 g 18.0 g mole mole C 2 H 6 O = 5.56 mole 6.06 mole X HOH = = 0.917
Practice A solution contains mL of water and mol of ethanol. What is the mole fraction of water and the vapor pressure of the solution at 25 o C, if the vapor of pressure of pure water is 23.8 torr? 100.0mL mL 1.00 g 18.0 g mole mole C 2 H 6 O = 5.56 mole 6.06 mole X HOH = = P HOH = 0.917(23.8 torr) P HOH = 21.8 torr
Boiling Point Vs. Pressure
Phase Diagrams A phase diagram is a graphic representation of the dependence of the stabilities of the physical states of a substance on temperature and pressure.
Phase Diagram for Water Triple Point Critical Point Critical Temperature Critical Pressure Supercritical Fluid
The Critical Point
Phase Diagram Terms The triple point defines the temperature and pressure where all three phases of a substance coexist. The critical point is that specific temperature and pressure at which the liquid and gas phases of a substance have the same density and are indistinguishable for each other. A supercritical fluid is a substance at conditions above its critical temperature and pressure.
Phase Diagram for CO 2
Colligative Properties of Solutions Colligative properties of solutions depend on the concentration and not the identity of particles dissolved in the solvent. Sea water boils at a higher temperature than pure water.
Calculating Changes in Boiling Point T b = K b m T b is the increase in Bp K b is the boiling-point elevation constant m is a new concentration unit called molality Molality (m) = moles solute Kg solvent
Practice Calculate the molality of a solution containing mol of glucose (C 6 H 12 O 6 ) in 1.5 kg of water.
Practice Calculate the molality of a solution containing mol of glucose (C 6 H 12 O 6 ) in 1.5 kg of water mole 1.5 kg
Practice Calculate the molality of a solution containing mol of glucose (C 6 H 12 O 6 ) in 1.5 kg of water mole 1.5 kg = 0.58 m
Practice Seawater contains M Cl - at the surface at 25 o C. If the density of sea water is g/mL, what is the molality of Cl - in sea water?
Practice Seawater contains M Cl - at the surface at 25 o C. If the density of sea water is g/mL, what is the molality of Cl - in sea water? 10 3 mL solution mL g= 1022 g solution
Practice Seawater contains M Cl - at the surface at 25 o C. If the density of sea water is g/mL, what is the molality of Cl - in sea water? 10 3 mL solution mL g= 1022 g solution mole Cl g Cl - mole Cl - = g Cl -
Practice Seawater contains M Cl - at the surface at 25 o C. If the density of sea water is g/mL, what is the molality of Cl - in sea water? 10 3 mL solution mL g= 1022 g solution mole Cl g Cl - mole Cl - = g Cl g solution – g Cl - = g H 2 O
Practice Seawater contains M Cl - at the surface at 25 o C. If the density of sea water is g/mL, what is the molality of Cl - in sea water? 10 3 mL solution mL g= 1022 g solution mole Cl g Cl - mole Cl - = g Cl g solution – g Cl - = g H 2 O mole Cl g H 2 O 10 3 g Kg = m
Practice Cinnamon owes its flavor and odor to cinnamaldehyde (C 9 H 8 O). Determine the boiling-point elevation of a solution of 100 mg of cinnamaldehyde dissolved in 1.00 g of carbon tetrachloride (K b = 2.34 o C/m).
Practice Cinnamon owes its flavor and odor to cinnamaldehyde (C 9 H 8 O). Determine the boiling-point elevation of a solution of 100 mg of cinnamaldehyde dissolved in 1.00 g of carbon tetrachloride (K b = 2.34 o C/m). 100 mg C 9 H 8 O mg mmole 1.00 g CCl 4 mmole mole 10 3 g Kg = m 2.34 °C m m = 1.77°C
Freezing-point Depression T f = K f m K f is the freezing-point depression constant and m is the molality.
Practice The freezing point of a solution prepared by dissolving 1.50 X 10 2 mg of caffeine in 10.0 g of camphor is 3.07 Celsius degree lower than that of pure camphor (K f = 39.7 o C/m). What is the molar mass of caffeine?
The van’t Hoff Factor T b = iK b m & T f = iK f m van’t Hoff factor, i is the number of ions in one formula unit
The van’t Hoff Factor Used for ionic compounds, why not osmolarity? The value of i assumes that all of the salt dissolves and dissociates in to its component ions This is not always true, for example 0.10m NaCl I is 1.87 Ion pairing often occurs in solutions Ion pairing most important in concentrated solutions Ion pairing important in highly charged solutions
Values of van’t Hoff Factors
Practice CaCl 2 is widely used to melt frozen precipitation on sidewalks after a winter storm. Could CaCl 2 melt ice at -20 o C? Assume that the solubility of CaCl 2 at this temperature is 70.0 g/100.0 g of H 2 O and that the van’t Hoff factor for a saturated solution of CaCl 2 is 2.5 (K f for water is C/m).
Osmotic Pressure Osmotic pressure ( ) is the pressure that has to be applied across a semipermeable membrane to stop the flow of solvent form the the compartment containing pure solvent or a less concentrated solution towards a more concentrated solution. = iMRT where i is the van’t Hoff factor, M is molarity of solute, R is the idea gas constant ( latm/(molK)), and T is in Kelvin
Osmosis at the Molecular Level
Osmotic pressure Equation from the ideal gas law (pv = nRT) = MRT Semi permeable membrane Isotonic same concentration Cells placed in lower concentration hypotonic, cell will swell called hemolosis If concentration on the outside of the cells is greater then the solution is called hypertonic and the cells shrink called crenation
Osmosis Figure In osmosis, solvent passes through a semipermeable membrane to balance the concentration of solutes in solution on both sides of the membrane.
ChemTour: Lattice Energy Click to launch animation PCPC | MacMac Students learn to apply Coulomb’s law to calculate the exact lattice energies of ionic solids. Includes Practice Exercises.
ChemTour: Intermolecular Forces Click to launch animation PCPC | MacMac This ChemTour explores the different types of intermolecular forces and explains how these affect the boiling point, melting point, solubility, and miscibility of a substance. Includes Practice Exercises.
ChemTour: Henry’s Law Click to launch animation PCPC | MacMac Students learn to apply Henry’s law and calculate the concentration of a gas in solution under varying conditions of temperature and pressure. Includes interactive practice exercises.
ChemTour: Molecular Motion Click to launch animation PCPC | MacMac Students use an interactive graph to explore the relationship between kinetic energy and temperature. Includes Practice Exercises.
ChemTour: Raoult’s Law Click to launch animation PCPC | MacMac Students explore the connection between the vapor pressure of a solution and its concentration as a gas above the solution. Includes Practice Exercises.
ChemTour: Phase Diagrams Click to launch animation PCPC | MacMac Students use an interactive phase diagram and animated heating curve to explore how changes in temperature and pressure affect the physical state of a substance.
ChemTour: Capillary Action Click to launch animation PCPC | MacMac In this ChemTour, students learn that certain liquids will be drawn up a surface if the adhesive forces between the liquid on the surface of the tube exceed the cohesive forces between the liquid molecules.
ChemTour: Boiling and Freezing Click to launch animation PCPC | MacMac Students learn about colligative properties by exploring the relationship between solute concentration and the temperature at which a solution will undergo phase changes. Interactive exercises invite students to practice calculating the boiling and freezing points of different solutions.
ChemTour: Osmotic Pressure Click to launch animation PCPC | MacMac Students discover how a solute can build up pressure behind a semipermeable membrane. This tutorial also discusses the osmotic pressure equation and the van’t Hoff factor.
Solubility of CH4, CH2Cl2, and CCl4 Which of the following three compounds is most soluble in water? A) CH 4 (g) B) CH 2 Cl 2 (λ) C) CCl 4 (λ)
Solubility of CH4, CH2Cl2, and CCl4 Consider the following arguments for each answer and vote again: A.A gas is inherently easier to dissolve in a liquid than is another liquid, since its density is much lower. B.The polar molecule CH 2 Cl 2 can form stabilizing dipole-dipole interactions with the water molecules, corresponding to a decrease in ΔH° soln. C.The nonpolar molecule CCl 4 has the largest molecular mass, and so is most likely to partially disperse into the water, corresponding to an increase in ΔS° soln.