Transcript Slide 1
June 12, 2009 – Class 41 and 42 Overview • 12.3 Some Properties of Solids – Melting, melting (freezing) point, heat of fusion, sublimation, deposition. • 12.4 Phase Diagrams – • Critical Point 12.9 Energy Changes in the Formation of Ionic Crystals – Determination of lattice energy using a Born-Fajans-Haber cycle June 12, 2009 – Class 41 and 42 Overview • 13.1 Types of Solutions: Some Terminology – • Solvent, solute, solid solutions (alloys), gaseous and liquid solutions 13.2 Solution Concentration – Calculations involving mass percent, volume percent, mass/volume percent, parts per million, billion and trillion, mole fraction and mole percent, molarity, molality Some Properties of Solids • Melting: the transition of a solid to a liquid that occurs at the melting point. The melting point and freezing point of a substance are identical. – Heat of fusion (DHfus) Some Properties of Solids • Sublimation: the passage of molecules from the solid to the gaseous state. • Deposition: the passage of molecules from the gaseous to the solid state. • Enthalpy of sublimation: DH sub DH fus DH vap Sublimation of solid iodine, and the deposition of the vapor to solid on the cooler walls of the flask. Critical Point • Critical Point: refers to the temperature and pressure at which a liquid and its vapor become identical. It is the highest temperature point on the vapor pressure curve. Critical Point A gas can be liquefied only at temperatures below its critical temperature. If room temperature is above Tc, then a gas can cannot be liquefied. Temperature must be lowered to below Tc. Phase Diagrams • Phase Diagrams: a graphical representation of the conditions of temperature and pressure at which solids, liquids and gases exist, either as single phases or states of matter or as two or more phases in equilibrium. Triple point: a condition of temperature and pressure at which three phases of a substance coexist at equilibrium. Phase Diagrams – Supercritical Fluids Above the critical point the state of matter has the high density of a liquid, but the low viscosity of a gas. It is called a supercritical fluid. Phase Diagrams - Iodine Note that the melting-point and triple-point temperatures for iodine are essentially the same. Phase Diagrams – Carbon Dioxide (CO2) Note that a line intersects the sublimation curve at P = 1atm (consistent with behaviour observed for “dry ice”). Phase Diagrams – Water (H2O) Ice exists in different phases at different pressures. Negative slope of the OD line means that the melting point of ice decreases with increasing pressure (think: ice skating) Energy Changes in the Formation of Ionic Crystals • • Enthalpy of formation: Cs(s) + Cl2(g) CsCl(s) DH of 442.8kJ / mol Lattice energy: the quantity of energy released in the formation of one mole of crystalline ionic solid from its separated gaseous ions. Example: Cs+(g) + Cl–(g) CsCl(s) DH 669 .2kJ / mol Energy Changes in the Formation of Ionic Crystals • Lattice energy can be determined indirectly using a Born-Fajans-Haber cycle. Energy Changes in the Formation of Ionic Crystals Enthalpy diagram for the formation of an ionic crystal. Shown here is a fivestep sequence for the formation of one mole of NaCl(s) from its elements in their standard states. The sum of the five enthalpy changes gives DHof[NaCl(s)]. The equivalent onestep reaction for the formation of NaCl(s) directly from Na(s) and Cl2(g) is shown in color. Energy Changes in the Formation of Ionic Crystals • Lattice energy can be determined indirectly using a Born-Fajans-Haber cycle. • This method consists of five steps: 1. 2. 3. 4. 5. Sublime one mole of solid Na. Dissociate 0.5 mole of Cl2(g) into one mole of Cl(g) Ionize one mole of Na(g) to Na+(g). Convert one mole of Cl(g) to Cl-(g). Allow the Na+(g) and Cl-(g) to form one mole of NaCl(s). Energy Changes in the Formation of Ionic Crystals Problem: Determine the lattice energy for MgF2(s) from the following data: DH°f MgF2(s) = -1123 kJ mol-1, enthalpy of sublimation of Mg(s) = 148 kJ mol-1, bond dissociation energy of F2(g) = 159 kJ mol-1, I1 for Mg(g) = 738 kJ mol-1, I2 for Mg(g) = 1450kJ mol-1 EA for F(g) = -328 kJ mol-1. Problem: Determine DH°f CaCl2(s) given the lattice energy of CaCl2 = -2223 kJ mol-1 enthalpy of sublimation of Ca(s) = 178.2 kJ mol-1, bond dissociation energy of F2(g) = 122 kJ mol-1, I1 for Ca(g) = 590 kJ mol-1, I2 for Ca(g) = 1145 kJ mol-1 EA for Cl(g) = -349 kJ mol-1. Types of Solutions: Some Terminology • Solution: homogeneous mixture made up of a solvent and a solute • Solvent: solution component in which one or more solutes is dissolved. – • Usually the solvent is present in greater amounts than the solutes and determines the state of matter in which the solution exists. Solute: the solution component that is dissolved in the solvent. – A solution may have several solutes, with the solutes generally present in lesser amounts than the solvent. Types of Solutions: Some Terminology • Solutions may be one of three phases: solid, liquid or gas • Alloys: a mixture of two or more metals; a solid solution. – Amalgams: metal alloys containing mercury; most are in the liquid phase. Solution Concentration • Standard Solution: A solution that contains a precise mass of solute in a precise volume of solution • Concentration: refers to the composition of a solution. – • One method for expressing concentration is molarity Molarity (M): the number of moles of solute dissolved in one litre of solution; expressed in units of (mol/L) – – Note: Often, the unit (mol/L) is abbreviated (M). Molarity can be calculated from M = nsolute Vsolution Solution Concentration • Other methods can be used to express the concentration of a solution: Percent by Mass (Mass/Mass) • If you dissolve 5.00 g of NaCl in 95.0 g of H2O you get 100.0 g of solution that is 5.00 % NaCl, by mass. Mass Percent (mass/mass) = Mass solute (g) x 100% Mass solution (g) Percent by Mass (Mass/Volume) • If you dissolve 0.9 g of NaCl in 100.0 mL of solution, your solution is 0.9 % NaCl (mass/volume). Mass Percent (mass/volume) = Mass solute (g) x 100 % Volume Solution (mL) Solution Concentration Percent by Mass (Volume/Volume) • If you prepare 100.0 mL of aqueous solution containing 25.0 mL of CH3OH(l) the solution is 25.0 % CH3OH, by volume. Mass Percent (volume/volume) = Volume solute (mL) x 100% Volume solution (mL) Solution Concentration Parts Per Million (Mass/Mass) • If you dissolve 0.5 g of NaCl in 99.5 g of H2O you get 100.0 g of solution that is 5000 ppm NaCl (mass/mass). Parts Per Million (ppm) = Mass solute (g) x 106 Mass solution (g) Parts Per Million (Mass/Volume) • If you dissolve 0.9 g of NaCl in 100.0 mL of solution, your solution is 9000 ppm NaCl (mass/volume). Parts Per Million (ppm) = Mass solute (g) x 106 Volume Solution (mL) Solution Concentration Mole Fraction and Mole Percent • Mole fraction can be determined by: ci = _____amount of component i (in moles)______ total amount of all solution components (in moles) • • Note: ci + cj + ck + …. = 1 Mole percent is mole fraction multiplied by 100%. Molality (m) • The number of moles of solute per kg of solvent (NOT solution) Molality (m) = amount of solute (moles) mass of solvent (kg) Solution Concentration Problem: An ethanol-water solution is prepared by dissolving 10.00 mL of ethanol (C2H5OH; d = 0.789 g/mL) in sufficient water to produce 100.0 mL of a solution with a density of 0.982 g/mL. What is the concentration of ethanol in this solution expressed as: (a) Volume percent (b) Mass percent (c) Mass/volume percent (d) Mole fraction (e) Mole percent (f) Molarity (g) Molality Problem: Laboratory ammonia is 14.8 M NH3(aq) with a density of 0.8980 g/mL. What is cNH3 in this solution?