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Chapter 7.3 Intermolecular Forces nd ChJulia Burdge & Jason Overby The McGraw-Hill Companies, Inc. 7.3 Intermolecular Forces Dipole-Dipole Interactions Hydrogen Bonding Dispersive Forces Ion-Dipole Interactions 2 Intermolecular Forces Intermolecular forces • London dispersion forces • Dipole-dipole interactions van der Waals -Hydrogen bonding • Ion-dipole interactions 3 London dispersion forces London dispersion forces: coulombic attractions between instantaneous dipoles of molecules. Dispersion forces are present in all molecules. Figure 7.6 4 London dispersion forces The larger the molecule, the more polarizable, the stronger the force. 5 Dipole-Dipole Interactions Dipole-dipole: attraction between the partially positive end of one molecule and partially negative end of a neighboring molecule. Figure 7.3 6 How Do We Explain This? Figure 7.5 7 Hydrogen Bonding Hydrogen bonding: attraction between a H-atom attached to a highly electronegative atom (F, O, N) and a nonbonding electron pair on a F, O, N-atom. 8 Ion-Dipole Interactions Ion-Dipole Interactions: coulombic attractions between ions (either positive or negative) and polar molecules. 9 Problem 1. Arrange BaCl , 2 , C2, HF, and Ne in order of increasing boiling point. Answer: H < Ne < CO < HF < BaCl 2 2 10 Chapter 12 Liquids and Solids ChemiSecond Editionrst Julia Burdge & Jason Overby 1 12.1 The Condensed Phases 12.2 Properties of Liquids Surface Tension Viscosity Vapor Pressure of Liquids Boiling Point 12.5 Phase Changes Liquid-Vapor Solid-Liquid Solid-Vapor 12.7 Phase Diagrams 2 Fig 12.2 Surface Tension Surface tension: the amount of energy required to stretch or increase the surface of a liquid by a unit area. Capillary action Fig 12.4 Adhesive forces are Cohesive forces are greater than cohesive greater than adhesive forces forces 3 Viscosity • Viscosity: a measure of a fluid’s resistance to flow. • The stronger the intermolecular forces, the lower / higher the viscosity. • The higher temperature, the lower / higher the viscosity. 4 Vapor Pressure Fig 12.6 • Dynamic equilibrium: liquid molecules evaporate and vapor molecules condense at the same rate. • Vapor pressure: the pressure exerted when liquid and vapor states are in dynamic equilibrium. Problem 1. Which has a higher vapor pressure at 25 C: CCl or 4Br ? 4 Answer: CCl Clausius-Clapeyron Equation The relationship between the vapor pressure and temperature is given by: ln P = -( ∆vap)/(RT) + C where P = vapor pressure T = the absolute temperature R = the gas constant (8.314 J/K•mol) ∆H vap molar heat of vaporization C = a constant molar heat of vaporization (ΔH ): the amount of heat vap required to vaporize a mole of substance at its boiling point. Phase Changes Fig 12.27 molar heat of vaporization (ΔH vap: the amount of heat required to vaporize a mole of substance at its boiling point. molar heat of fusion (ΔH )fushe amount of energy required to melt a mole of solid. ΔH + ΔH = ΔH 7 Liquid-Vapor 8 Heating Curve • Start-A: q = sm∆T E no phase transition • A-B: q = n∆Hfus phase transition Boiling point vapor • B-C: q = sm∆T no phase transition C D Temperatureoint • C-D: q = n∆H vap liquid phase transition A B solid • D-E: q = sm∆T equilibrium no phase transition Time Problem 2. Calculate the amount of heat required to convert 1.00 mol of ice o o at –11 C to steam at 115 C under constant pressure. s(ice) 2.03 J/g C , s (water) 4.18 J/g C, s (steam)1.99 J/g C,o ΔH = 6.01 kJ/mol and ΔH = 40.79 kJ/mol. 9 fus vap Liquid-Vapor Boiling point of a substance: the temperature at which its vapor pressure equals the atmospheric pressure. Normal boiling point of a substance: the boiling point at 1 atm. Critical temperature of a substance: the temperature above which its gas cannot be liquified Critical pressure of a substance: the minimum pressure needed to liquefy a substance at its critical temperature 10 Phase Diagram of Carbon Dioxide • CO c2nnot exist in the liquid state at pressures a b c d Critical point below 5.2 atm; CO 2 sublimes at normal pressures. • Triple point: triple point -57 C, 5.2 atm triple point: the combination of pressure and temperature where three phases of a substance exist in equilibrium. critical point: substance at its critical T and critical P 11 Phase Diagram of Water • Triple point: o 0.01 C, 0.006 atm • Normal b.p.: o 100 C • Normal m.p.: o 0 C 12 Chapter 13 Physical Properties of Solutions Chemistry:Atoms Fedition The McGraw-Hill Companies, Inc. 13.1 Types of Solutions 13.2 A Molecular View of the Solution Process The Importance of Intermolecular Forces Energy and Entropy in Solution Formation 13.3 Concentration Units Molality Percent by Mass Comparison of Concentration Units 13.4 Factors that Affect Solubility Temperature Pressure 2 13.5 Colligative Properties Vapor-Pressure Lowering Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure Electrolyte Solutions 13.6 Calculations Using Colligative Properties 13.7 Colloids 3 Solutions Solutions: homogeneous mixtures of two or more pure substances. the solute is uniformly dispersed throughout the solvent. Unsaturated soln: contains less solute than the solvent has the capacity to dissolve at a specific temperature. Saturated soln: contains the maximum amount of solute that will dissolve in a solvent at a specific temperature. Solubility: amount of solute dissolved in a given volume of a saturated solution at a specified temperature. (e.g. NaCl at 20 C is 36g/100ml water) Supersaturated soln: contains more dissolved solute than is present in a saturated solution and are generally unstable. 4 Solution Formation Process Solvation: solute molecules are separated from one another and surrounded by solvent molecules. Solvation depends on interactions: Solute-Solute, Solvent-Solvent, Solute-Solvent interactions 5 Energy Changes in Solution Formation ∆H 10 : solute-solute interactions ∆H >0 : solvent-solvent interactions 2 ∆H <0 (usually): solute-solvent interactions 3 ∆H = endothermic/exothermic soln Enthalpy change for the solution process: ∆H soln ∆H 1 ∆H +2∆H 3 6 Enthalpy is One Part of the Picture Dissolution of MgSO : 4H soln< 0; exothermic Dissolution of NH NO : ∆H > 0; endothermic 4 3 soln Solution Process depends on: Enthalpy decrease Entropy increase measure of the energy dispersal in a system (disorder) 7 Solubility of a substance in a solvent Solute-solvent interactions: “like dissolves like" Polar substances tend to dissolve in polar solvents. Nonpolar substances tend to dissolve in nonpolar solvents. Miscible: mixing in all proportions (immiscible, if not) 8 “like dissolves like" Vitamin A: fat-soluble Vitamin C: water-soluble 9 “like dissolves like" Problem 1. a. Methanol miscible with water and hexanol is almost insoluble in water. Why? Answer: Methanol is polar but hexanol is almost nonpolar b. Which of the following compound(s) do you expect to be more soluble in benzene than in water? CO , K SO ,C H , Br . 2 2 4 2 6 2 Answer: CO , 2 H ,2B6 . 2 10 Concentration Units Molarity = M = moles of solute liters of solution Molarity is temperature dependent moles of solute Molality = m = mass of solvent (kg) Molality is temperature independent moles of A Mole Fraction A = X A= sum of moles of all components 11 Concentration Units mass of solute Percent by mass = x 100% (Percent by mass is temperature independent)solvent Parts per million = ppm = mass of solute x 106 total mass of solution mass of solute 9 Parts per billion = ppb = total mass of solutionx 10 12 Problem 2: A . Calculate the percentage by mass of NaCl in a soln 1 containing 3.50 g of NaCl in 75.0 g of water. Answer: 4.46% A 2 What is the molarity of the NaCl solution in1A ? Assume the density is 1.00 g/mL. Answer: 0.763 M A 3 What is the molality of the NaCl solution in1A ? Answer: 0.799 m B. A commercial bleaching solution contains 3.62 % by mass sodium hypochlorite, NaOCl. What is the mass (g) of NaOCl in a bottle containing 2.50 kg of bleaching solution? Answer: 90.5 g of NaOCl 13 Factors Affecting Solubility Structure: “like dissolves like" Temperature Pressure 14 Temperature Effects on Solubility Gases in water: solubility↓ as temperature↑ Solid solutes in water: generally solubility↑ as temperature↑ Figure 13.4 Problem 3: 60 g of solute is added to 100 mL of water at 60 C. o Using the graph, determine whether the solution will be saturated or unsaturated in case the solute is NaCl, KBr or CuSO .4(Answer: NaCl and CuSO satura4ed ) 15 Pressure Effects on Solubility Pressure Solubility of solids/liquids is hardly affected. Solubility of gases increases with pressure. Henry’s law: the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution. c = kP c = the molar concentration of the dissolved gas (mol/L) k = Henry’s Law constant (mol/L•atm) P = the pressure of the gas over solution (atm) 16 Pressure Effects Problem 4: What pressure of CO is 2equired to keep the CO concentration in a bottle of clubsoda at 2 0.12 M at 25 C. k = 3.4 x 10 mol/L•atm at 25 C. o CO2 Answer = 3.5 atm 17 Colligative Properties Colligative properties: properties that depend on the number of solute particles in solution. Colligative properties: Vapor-Pressure lowering Boiling-Point elevation Freezing-Point depression Osmotic Pressure 18 Vapor Pressure Lowering (nonelectrolyte solutions) Raoult’s Law: Solution containing a non-volatile solute: P = X P or P = X P 1 1 1 solvent solventsolvent – P 1 partial pressure of solvent above the solution (P solvent – X = the mole fraction of solvent in the solution (X ) 1 solvent – P 1 vapor pressure of the pure solvent (P solvent Or: P-P 1 ∆P = X P2 1 ∆P = decrease in vapor pressure X = mole fraction solute in the solution 2 Solution containing a volatile solvent (A) and volatile solute (B): P A X PA anA P = X B , sB PB= X P t X PA A B B *Ideal solution obeys Raoult’s Law 19 Problem 5: The vapor pressure of pure water at 108 C is 1004 torr. An aqueous solution of ethylene glycoloand water has a vapor pressure of 1.10 atm at 108 C. Assuming Raoult’s Law is obeyed, what is the mole fraction of ethylene glycol in the solution? 1 atm = 760. torr Answer: 0.167 20 Boiling-Point Elevation Freezing-Point Depression (nonelectrolyte solutions) 21 Boiling-Point Elevation T b T - b° = Kbm b K b the molal boiling-point elevation constant of the o solvent ( C/m) m = the molality (mol/kg) Freezing-Point Depression T = T° - T = K m f f f f K f the molal freezing-point depression constant of the o solvent ( C/m) m = the molality (mol/kg) 22 Osmotic pressure (nonelectrolyte solutions) Osmosis: the selective passage of solvent molecules through a semipermeable membrane from a more dilute solution to a more concentrated one. Fig. 13.10 Osmotic pressure of a solution (): the pressure required to stop osmosis. = MRT M = molarity R = ideal gas constant (0.08206 L•atm/mol•K) T = absolute temperature (Kelvin) 23 Osmotic pressure (nonelectrolyte solutions) cell in hypotonic cell in isotoniccell in hypertonic solution solution solution 24 Van ‘t Hoff factor(i) actual number of particles in solution a fter dissociation i number of formulas units initially disso lved in solution Nonelectrolytes: i = 1 Electrolytes: i depends on the electrolyte Fig. 13.11 25 Van ‘t Hoff factor(i) actual number of particles in solution a fter dissociation i number of formulas units initially disso lved in solution Nonelectrolytes: i = 1 Electrolytes: i depends on the electrolyte 26 Problem 6: A. At what temperatures will a 2.30M aqueous CaCl 2 solution boil and freeze? B. At what temperature will a 2.30M aqueous ethylene glycol (C2H 6 )2solution freeze? o o K f H2O 1.86 C/m K b H2O= 0.512 C/m The density of the solutions =1.000 g/ml. Answer: o a. CaCl :2T =b104.74 C o Tf= - 17.2 C o b. ethylene glycol: T =f- 4.99 C 27 Problem 7: The osmotic pressure of a solution containing 5.00 g of an unknown protein per 500. mL of solution was 0.745 torr at 27 C. Find the molar mass. 1 atm = 760. torr R = 0.08206 atm-L/mol-K 5 Answer: 2.51 x 10 g/mol 28 Colloids Colloids: a dispersion of particles of one substance throughout another substance. Colloid particles are much larger than the normal solute molecules, but too small to be settled out by gravity. Hydrophylic: water- loving Tyndall effect Hydrophobic: water-hating 29 Chapter 14 Entropy and Free Energy nd ChJulia Burdge & Jason Overby The McGraw-Hill Companies, Inc. 14.1 Spontaneous Processes 14.2 Entropy AQualitative Description of Entropy AQuantitative Description of Entropy 14.3 Entropy Changes in a System CalculatingΔS sys Standard S° Qualitatively PredictingΔS° sys 14.4 Entropy Changes in the Universe CalculatingΔS surr The Second Law of Thermodynamics The Third Law of Thermodynamics 2 14.5 Predicting Spontaneity Gibbs Free-Energy Change,ΔG Standard Free-Energy Changes,ΔG° UsingΔG andΔG° to Solve Problems 14.6 Thermodynamics in Living Systems 3 First Law of Thermodynamics First Law of Thermodynamics: Energy cannot be created nor destroyed 4 Spontaneous Processes Spontaneous process: process that does occur under a specific set of conditions without ongoing outside intervention. CH 4g) + 2O (2) CO (2) + 2H O2l) ΔH° = -890.4 kJ/mol Exothermic processes: often spontaneous H 2(s) H 2(l) ΔH° = 6.01 kJ/mol Endothermic process 5 Entropy: Qualitative Figure 14.1 Entropy (S): a measure of how dispersed the system’s energy is. 6 Entropy: Quantitative Boltzman: S = k lnW k = the Boltzmann constant (1.38 10 23J/K) W = number of possible arrangements 7 Entropy on a microscopic scale Molecules exhibit several types of motion: – Translational: movement of the entire molecule from one place to another. – Rotational: rotation of the molecule about an axis or bonds. – Vibrational: periodic motion of atoms within a molecule. 8 Entropy Changes in a System: S sys Entropy change: S = S S sys final initial Isothermal processes: V S sys= k lnW final k lnW initialr S sys = nRln final Vinitial S : Absolute entropy of a substance at 1 atm. 9 Calculating Reaction Standard Entropy ΔS° rxnnS°(products) – ΣmS°(reactants) n, m = coefficients in the balanced chemical equation Problem 1: Practice Problem 14.2A Using the standard entropies in Appendix 2, calculate the standard entropy change, ΔS°(J/K•mol), for the following reaction at 298 K: 2CO (g) → 2CO(g) + O (g) 2 2 S° CO2(g) 213.6 J/K•mol, S° CO(g)= 197.9 J/K•mol, S° O2(g) 205.0 J/K•mol Answer: ΔS° rxn= 173.6 J/K•mol 10 Calculating Reaction Standard Entropy Problem 2: Practice Problem 14.2A Using the standard entropies in Appendix 2, calculate the standard entropy change, ΔS° (J/K•mol), for the following reaction at 298 K: 3O 2g) → 2O (g3 S° O2(g) 205.0 J/K•mol, S° O3(g)= 237.6 J/K•mol Answer: ΔS° rxn= -139.8 J/K•mol 11 Qualitative ∆S predictions 1.Temperature changes: as T↑, S↑ 2. Volume changes: as V↑, S↑ 3. Phase changes: increasing entropy solid < liquid < gas • Melting, vaporization, sublimation 4. Molar mass/complexity: MM / complexity↑, S ↑ 5. Chemical reaction: reaction resulting in a greater number of gas molecules, S↑. 2NO(g) + O (g2 2NO (g)2 S 6. Dissolution of a solute: Molecular solutes (i.e. sugar): S↑ Ionic compounds: entropy could decrease or increase 12 Problem 3: Determine the sign for ∆S for the following systems: a. CO (s) → CO (g) + 2 2 b. CaO(s) + CO (g)2→ CaCO (s) 3 - c. Heating bromine vapor from 45 °C to 75 °C + d. NH (g) + HCl(g) → NH Cl(s) - 3 4 e. Condensation of water vapor on a cold surface - 13 Entropy Changes in Surroundings An ice cube spontaneously melts in a room at 25°C. Components ΔS System ice positive ↑ Surroundings everything else negative A cup of hot water spontaneously cools to room temperature. Components ΔS System hot water negative Surroundings everything else positive ↑ 14 Second Law of Thermodynamics Process at constant pressure: Hsys H° Ssurr T = rxn T Second law of thermodynamics: ΔS universe 0 for a spontaneous process as written (in the forward direction). S universeSsys+ S surr ΔS > 0 for a spontaneous process universe ΔS universe0 for a nonspontaneous process ΔS = 0 for an equilibrium process universe 15 Second Law of Thermodynamics Problem 4: Practice Problem 14.4A Calculate S and determine if the process is universe spontaneous, nonspontaneous or at equilibrium at the specified temperature. Use the data in the table below. PCl 3l) → PCl 3g) at 61.2 °C H° fkJ/mol S° (J/K•mol) PCl (l) -319.7 217.1 3 PCl 3g) -288.07 311.7 Answer: at equilibrium 16 Second Law of Thermodynamics Problem 5: Practice Problem 14.4B The reaction NH (g) + HCl(g) → NH Cl(s) is spontaneous in 3 4 the forward direction at room temperature but, because it is exothermic, becomes less spontaneous with increasing temperature. a. Show that the reaction is exothermic. b. Determine the temperature (°C) at which it is no longer spontaneous in the forward direction. H° fkJ/mol S° (J/K•mol) NH 3g) -46.3 193.0 HCl(g) -92.3 187.0 NH Cl(s) -315.39 94.56 4 Answer: a. H° rxn< 0 b. T = 346 °C. 17 Third Law of Thermodynamics Fig. 14.5 • The entropy of a perfect crystalline substance at absolute zero (0 K) is zero: S = 0. for a perfect crystal, S = k ln W = k ln 1 = 0. • The standard or absolute entropy, S ( o J/K•mol ): measured at 1 atm (appendix 2) 18 Gibbs Free Energy S = S + S spontaneous: S > 0 univ sys surr univ H S univ S sys+ sys > 0 -TS univ -TS sys+ H sys< 0 T ΔG = ΔH – TΔS , where G = Gibbs free energy • G < 0; the forward reaction is spontaneous (S > 0) • G > 0; the forward reaction is nonspontaneous (S < 0) • G = 0; the system is at equilibrium. 19 Problem 5: Practice Problem 14.4B The reaction NH (g) + HCl(g) → NH Cl(s) is spontaneous in 3 4 the forward direction at room temperature but, because it is exothermic, becomes less spontaneous with increasing temperature. a. Determine the temperature (°C) at which it is no longer spontaneous in the forward direction. H° fkJ/mol S° (J/K•mol) NH (g) -46.3 193.0 3 HCl(g) -92.3 187.0 NH C4(s) -315.39 94.56 Answer: T = 346 °C. 20 Standard Free Energy Change: ΔG° The standard free energy of reaction (ΔG° ) is free- rxn energy change for a reaction when it occurs under standard-state conditions (1atm, 1M, pure solid/liquid, G = 0 for elements in their most stable f form) G = SnG (products) SmG (reactants) f f n,m : the stoichiometric coefficients G f the standard free energy of formation (free-energy change that occurs when 1 mole of compound is synthesized from its constituent elements, each in its standard state) 21 Thermodynamics in Living Systems alanine + glycine → alanylglycine ΔG° = 29 kJ/mol ATP + H 2 → ADP + H PO 3 4 ΔG° = –31 kJ/mol ΔG° = 29 kJ/mol + –31 kJ/mol = –2 kJ/mol 22
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