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Chapter 11 Lecture 3: Phase Changes

Chapter 11 Lecture 3: Phase Changes. Molecular Solids Molecules are at the lattice points of the crystalline solid Ice = H 2 O at each lattice point It takes 6 kJ/mol of energy to melt ice (470 kJ/mol to break O—H bond) Weak, intermolecular H-bonds hold ice together

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Chapter 11 Lecture 3: Phase Changes

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  1. Chapter 11 Lecture 3: Phase Changes • Molecular Solids • Molecules are at the lattice points of the crystalline solid • Ice = H2O at each lattice point • It takes 6 kJ/mol of energy to melt ice (470 kJ/mol to break O—H bond) • Weak, intermolecular H-bonds hold ice together • CO2 = no dipole moment; London forces hold solid together (gas at 25 oC) • P4, S8 = strong covalent bonds within the unit, but weak intermolecular forces between molecules = larger molecules, more London force, solids at higher T • Ionic Solids • Ions at lattice points of the crystalline solid • Strong (Coulombic) inter-ionic forces: stable, high melting point solids • Packing • Larger anions are usually in one of the closest packing forms • Smaller cations fit into the holes among the anions to maximize the attraction and minimized repulsions

  2. Types of holes for the smaller ion • Trigonal holes = between 3 spheres in the same layer • Tetrahedral holes = between 3 spheres in one layer and 1 in another • Octahedral holes = between 3 spheres in one layer and 3 in another • Hole size: trigonal < tetrahedral < octahedral Trigonal too small to be occupied in most ionic solids • There are more holes than cations needed, usually • ZnS Example: twice the number of tetrahedral holes as cations needed • Ccp structure: 4 S2- in the unit cell, 8 tetrahedral holes, 4 Zn2+ needed • NaCl Example: ccp Cl-: each octahedral hole surrounded by 6 Cl- • One hole per each Cl-, one Na+ needed for each Cl-

  3. C. Vapor Pressure • Vapor Pressure = pressure of the vapor over a liquid at equilibrium • Equilibrium = molecules evaporating at the same rate as condensing • Dynamic = millions of individual molecules are changing state • Net change in gaseous molecules is Zero 3. Vapor Pressure depends on Kinetic Energy = Temperature

  4. butane 4. Vapor pressure depends on Intermolecular Forces • Volatile Liquid = high vapor pressure i. Small molecular weight ii. Small intermolecular forces: no dipole or H-bonding • Non-volatile Liquid = low vapor pressure i. Large molecular weight (high London Forces): oils ii. Large intermolecular forces (H-bonding): water

  5. Quantitative description of Vapor Pressure • Heat of Vaporization = DHvap = heat added to free molecules to gas • Relationship of DHvap, Pvap, and T (in Kelvin) • Plots as a straight line: y = mx + b R = gas constant C = constant of the compound = “lnb”

  6. The Clausius-Clapeyron Equation: We can calculate Pvap at a new T • Example: Pvap(H2O) = 23.8 torr and DHvap = 43.9 kJ/mol at T = 25 oC. What is Pvap at T = 50 oC?

  7. Changes of State • States of matter can be reversibly changed without changing the substance • Changing states • Melting (Fusion) = Solid to Liquid (opposite is Freezing) [DHfus] • Vaporization = Liquid to Gas (opposite is Condensation) [DHvap] • Sublimation = Solid to Gas (opposite is Vapor Deposition) 2.09 J/goC 4.18 J/goC 2.01 J/goC q = mCiceDT q = nDHfus q = mCliqDT q = nDHvap q = mCgasDT Boiling Sublimation

  8. The Heating Curve for Water 2.09 J/goC 4.18 J/goC 2.01 J/goC q = mCiceDT q = nDHfus q = mCliqDT q = nDHvap q = mCgasDT

  9. How much heat is required to heat 500g H2O at -20oC to steam at 250oC? • (500g)(2.09J/goC)(20oC) = 20,900 J • (500g)(1mol/18g)(6,020J/mol) = 16,700 J • (500g)(4.18J/goC)(100oC) = 209,000 J • (500g)(1mol/18g)(40,700J/mol) = 1,130,000 J • (500g)(2.01J/goC)(150oC) = 151,000 J ______________________________________________________________ 6. Total = 20,900 + 16,700 + 209,000 + 1,130,000 + 151,000 = 1,527,600 J

  10. E. Phase Diagrams • Phase diagrams show us the temperatures and pressures of state changes • Critical Temperature = temp. above which substance is always a gas • Critical Pressure = pressure needed to make liquid at Critical Temp • Critical Point = intersection of critical temp. and critical pressure • Triple Point = single point where all three phases can coexist

  11. 2. Navigating in Phase Diagrams • Phase diagrams for other compounds: Iodine Water is unusual

  12. 2. Applications of the Water Phase Diagram • Solid/Liquid Boundary has a negative slope: density of ice is less than liquid water (this is why ice floats on water) • Ice Skating: narrow skate blade applies large pressure + friction = ice melts to liquid. This lubricates the blade and allows smooth skating. • Elevation Effects: Lower pressure = lower boiling point

  13. The CO2 Phase Diagram • Solid/Liquid boundary has positive slope (normal) • Solid is more dense than liquid • Triple point = 5.1 atm, -56.6 oC • Critical point = 72.8 atm, 31 oC • Solid sublimes directly to gas at 1 atm and 25 oC = Dry Ice • Fire Extinguisher = liquid CO2; high pressure makes it liquid As it is opened to atmosphere, gaseous CO2 puts out fire (fog = water)

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