Forces Between Ions and Molecules and Colligative Properties Chapter 10 Forces Between Ions and Molecules and Colligative Properties

States of Matter The state of a pure substance (at a given T & P) is governed by intermolecular (non-bonding) forces

At room temperature and pressure, He, Ar, Kr, Xe, and Rn are monatomic gases; H2, N2, O2, F2, and Cl2 are diatomic gases; Hg and Br2 are liquids. All other elements are solids.

Bonding Forces

Intermolecular Forces E(kJ/mol)

Intermolecular Forces Ion-ion Interactions Ion-dipole Interactions Dipole-dipole Interactions Hydrogen bonds Dispersion Forces (London Forces) Dipole-induced Dipole interactions

Induced dipole* – Induced dipole* 0.05 – 2 Intermolecular ENERGY Interaction Type (kJ/mol) Ion – Ion 300 – 600 Ion – Dipole 10 – 100 Dipole – Dipole 1 – 10 H-Bonds 20 – 40 Ion – Induced Dipole 1 – 10 Dipole – Induced Dipole* 0.05 – 10 Induced dipole* – Induced dipole* 0.05 – 2 *Induced dipole forces also called London Dispersion Forces

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 (Q1 and Q2) and inversely proportional to the distance (d) between them.

Substances that possess charges will be attracted or repelled by one-another. The energy of these electrostatic interactions is described by Coulomb’s Law: E = 2.31E -21 J۰nm (Q1Q2/d) Where Q represents the charge on the substance (e.g. an ion) and d is the distance separating the two charges. E is negative when the charges have opposite sign (+/-), and positive otherwise.

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)

Ionic compounds are generally solids at room temperature Ionic compounds are generally solids at room temperature. They are solids because intermolecular attraction between ions is the strongest type of intermolecular interaction. The cations and the anions forming an ionic compound often are arranged in well-defined crystal lattice. Lattice energies are proportional to the Coulombic interaction energies, but also depend on the specific arrangement of the ions.

Comparing Lattice Energies Lattice Energies of Common Ionic Compounds Compound U(kJ/mol) LiF -1047 LiCl -864 NaCl -790 KCl -720 KBr -691 MgCl2 -2540 MgO -3791

Practice Determine which salt has the greater lattice energy. MgO and NaF MgO and MgS

Determining 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)

Successive Electron Affinities (EA) may be defined, e.g. O(g) + e- → O-(g) DHEA1 = - 141 kJ O-(g) + e- → O2-(g) DHEA2= + 744 kJ O(g) + 2e- → O2-(g) Overall DHEA= + 603 kJ

Calculating U Na(g) ---> Na(s) -Hsub Na+(g) + e-(g) ---> Na(g) -HIE1 Na(g) ---> Na(s) -Hsub Cl-(g) ---> Cl(g) + e-(g) -HEA Cl(g) ---> 1/2Cl2(g) -1/2HBE Na(s) + 1/2Cl2(g) ---> NaCl(s) Hf Na+(g) + Cl-(g) ---> NaCl(s) U U = Hf - 1/2HBE - HEA - Hsub - HIE1

Ionization energy of K = 425 kJ/mol Problem Calculate the lattice energy of potassium chloride from the following data: Ionization energy of K = 425 kJ/mol Electron Affinity of Cl for 1 e- = -349 kJ/mol Energy to vaporize K = 89 kJ/mol Cl2 bond energy = 240 kJ/mol Energy change for the reaction: K(s) + ½ Cl2(g) → KCl (s) is -438 kJ/mol .

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

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.

Boiling Points of Binary Hydrides

Interaction Involving 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.

Induced Dipoles dipole-induced dipole interaction induced dipole-induced dipole interaction

Strength of Dispersion Forces The strength of dispersion forces depends on the polarizability of the atoms or molecules involved. Polarizability (units of cm3) 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.

Molar Mass and Boiling Points of Common Species. Halogen M(g/mol) Bp(K) Noble Gas He 2 4 F2 38 85 Ne 20 27 Cl2 71 239 Ar 40 87 Br2 160 332 Kr 84 120 I2 254 457 Xe 131 165 Rn 211

Hydrocarbon Alcohol Molecular Formula Molar Mass Bp (oC) CH4 16.04 -161.5 CH3CH3 30.07 -88 CH3OH 32.04 64.5 CH3CH2CH3 44.09 -42 CH3CH2OH 46.07 78.5 CH3CH(CH)CH3 58.12 -11.7 CH3CH(OH)CH3 60.09 82 CH3CH2CH2CH3 -0.5 CH3CH2CH2OH 97

The Effect of Shape on Forces

Practice CH3OH CH3CH2CH2CH3 CH3CH2OCH3 CH3CH2CH3 Rank the following compounds in order of increasing boiling point: CH3OH CH3CH2CH2CH3 CH3CH2OCH3 CH3CH2CH3

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.

Section 9.5: Polarity and Solubility Solutes tend to dissolve in solvents in which similar intermolecular interaction are formed, i.e. dipole-dipole, ion-dipole, or induced dipole-induced dipole. This phenomena is generally stated as “like dissolves like”.

Section 9.5: Polarity and Solubility Solutes tend to dissolve (or are miscible) in solvents in which similar intermolecular interactions are formed, i.e. dipole-dipole, ion-dipole, or induced dipole-induced dipole. This phenomena is generally stated as “like dissolves like”. Aqueous solubility of Alcohols: CH3(CH2)nOH 1-propanol n = 2 Miscible 1-butanol n = 3 1.1 M 1-pentanol n = 4 0.30 M 1-hexanol n = 5 0.056 M

Terms A hydrophobic (“water-fearing”) interaction repels water and diminishes water solubility. A hydrophilic (“water-loving”) interaction attracts water and promotes water solubility.

Phospholipids are essential for forming cell membranes… The bilayer is a semipermeable barrier which allows transport of only small non-polar molecules

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. Cgas = kHPgas where C is the concentration of the gas, kH is Henry’s Law constant for the gas.

From Henry’s Law an increase in the partial pressure of a gas in the system will result in an increase concentration of dissolved gas.

Henry’s Law Constants for Gas Solubility in Water at 20oC kH[mol/(L•atm)] kH[mol/(kg•mmHg)] He 3.5 x 10-4 5.1 x 10-7 O2 1.3 x 10-3 1.9 x 10-6 N2 6.7 x 10-4 9.7 x 10-7 CO2 3.5 x 10-2 5.1 x 10-5

[N2]aq = KH۰PN2 KH = slope

KH values decrease with increased temperatures. O2(g) ↔ O2(aq) DHsoln < 0

Arterial Blood contains about 0. 25 g of oxygen per liter at 37°C (98 Arterial Blood contains about 0.25 g of oxygen per liter at 37°C (98.6°F) and standard atmospheric pressure. What is the Henry’s Law constant for O2 dissolved in arterial blood? Compare this value to KH for O2 (aq)

Strengths of Intermolecular Forces Affects: Solubility Vapor Pressure viscosity surface tension Freezing Point Boiling Point

Vapor Pressure Vaporization or evaporation is the transformation of molecules in the liquid phase to the gas phase. Vapor pressure is the force exerted by a vapor in equilibrium at a given temperature with its liquid phase.

Vapor Pressure The normal boiling point of a liquid is the temperature at which its vapor pressure equals 1 atmosphere.

Vapor Pressure

Vapor Pressure and Solute Concentration Raoult’s Law Psolution = Xsolvent (Psolvent) P = vapor pressure X = mole fraction An Ideal Solution has vapor pressures consistent with Raoult’s Law

Problem A solution contains 100.0 mL of water and 0.500 mol of ethanol. What is the mole fraction of water and the vapor pressure of the solution at 25oC, if the vapor of pressure of pure water is 23.8 torr?

Physical State and Phase Transformations A phase diagram is a graphic representation of the dependence of the stabilities of the physical states of a substance on temperature and pressure (or even concentration).

Phase Diagram for Water Triple Point Critical Point Critical Temperature Critical Pressure Supercritical Fluid

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 CO2

Surface Tension

Cohesive and Adhesive Forces Produce a Meniscus

Terms Capillary action 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.

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 Tb = Kbm Tb is the increase in Bp Kb is the boiling-point elevation constant m is the molality

Practice Calculate the molality of a solution containing 0.875 mol of glucose (C6H12O6) in 1.5 kg of water. (180.16 g/mol)

Practice Seawater contains 0.558 M Cl- at the surface at 25oC. If the density of sea water is 1.022 g/mL, what is the molality of Cl- in sea water?

Practice Cinnamon owes its flavor and odor to cinnamaldehyde (C9H8O). Determine the boiling-point elevation of a solution of 100 mg of cinnamaldehyde dissolved in 1.00 g of carbon tetrachloride (Kb = 2.34oC/m). 120.15 g/mol

Freezing-point Depression Tf = Kfm Kf is the freezing-point depression constant and m is the molality.

Practice The freezing point of a solution prepared by dissolving 1.50 X 102 mg of caffeine in 10.0 g of camphor is 3.07 Celsius degree lower than that of pure camphor (Kf = 39.7oC/m). What is the molar mass of caffeine?

The van’t Hoff Factor Tb = iKbm & Tf = iKfm van’t Hoff factor, i is the number of ions in one formula unit

Values of van’t Hoff Factors

Practice CaCl2 is widely used to melt frozen precipitation on sidewalks after a winter storm. Could CaCl2 melt ice at -20oC? Assume that the solubility of CaCl2 at this temperature is 70.0 g/100.0 g of H2O and that the van’t Hoff factor for a saturated solution of CaCl2 is 2.5 (Kf for water is 1.86 0C/m).

Osmosis In osmosis, solvent passes through a semipermeable membrane to balance the concentration of solutes in solution on both sides of the membrane. Figure 10.30

Osmosis at the Molecular Level

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 (0.00821 l•atm/(mol•K)), and T is in Kelvin

ChemTour: Lattice Energy Click to launch animation PC | Mac 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 PC | Mac 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.

Click to launch animation PC | Mac ChemTour: Henry’s Law Click to launch animation PC | Mac 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 PC | Mac 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 PC | Mac 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 PC | Mac 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 PC | Mac 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 Points Click to launch animation PC | Mac 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 PC | Mac 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.

Clicker Questions Note Note to instructors: The following review question is also applicable to chapter 17, The Colorful Chemistry of Transition Metals.

Solubility of CH4, CH2Cl2, and CCl4 Which of the following three compounds is most soluble in water?   © 2008 W. W. Norton & Company Inc. All rights reserved. A) CH4(g) B) CH2Cl2(λ) C) CCl4(λ) Solubility of CH4, CH2Cl2, and CCl4

Consider the following arguments for each answer and vote again: A gas is inherently easier to dissolve in a liquid than is another liquid, since its density is much lower. The polar molecule CH2Cl2 can form stabilizing dipole-dipole interactions with the water molecules, corresponding to a decrease in ΔH°soln. The nonpolar molecule CCl4 has the largest molecular mass, and so is most likely to partially disperse into the water, corresponding to an increase in ΔS°soln. Answer: B Solubility of CH4, CH2Cl2, and CCl4