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Chapter 11

Chapter 11. Intermolecular Forces. The forces holding solids and liquids together are called intermolecular forces . The covalent bond holding a molecule together is an intramolecular forces. The attraction between molecules is an intermolecular force.

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Chapter 11

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  1. Chapter 11

  2. Intermolecular Forces • The forces holding solids and liquids together are called intermolecular forces. • The covalent bond holding a molecule together is an intramolecular forces. • The attraction between molecules is an intermolecular force. • Intermolecular forces are much weaker than intramolecular forces • When a substance melts or boils the intermolecular forces are broken (not the covalent bonds). Chapter 11

  3. Intermolecular Forces Chapter 11

  4. Intermolecular Forces Ion-Dipole Forces • Interaction between an ion and a dipole (e.g. water). • Strongest of all intermolecular forces. Chapter 11

  5. Intermolecular Forces Dipole-Dipole Forces • Exist between neutral polar molecules. • Polar molecules need to be close together. • Weaker than ion-dipole forces. • There is a mix of attractive and repulsive dipole-dipole forces as the molecules tumble. • If two molecules have about the same mass and size, then dipole-dipole forces increase with increasing polarity. Chapter 11

  6. Intermolecular Forces Dipole-Dipole Forces Chapter 11

  7. Intermolecular Forces Dipole-Dipole Forces Chapter 11

  8. Intermolecular Forces London Dispersion Forces • Weakest of all intermolecular forces. • It is possible for two adjacent neutral molecules to affect each other. • The nucleus of one molecule (or atom) attracts the electrons of the adjacent molecule (or atom). • For an instant, the electron clouds become distorted. • In that instant a dipole is formed (called an instantaneous dipole). Chapter 11

  9. Intermolecular Forces London Dispersion Forces • Polarizability is the ease with which an electron cloud can be deformed. • The larger the molecule (the greater the number of electrons) the more polarizable. • London dispersion forces increase as molecular weight increases. • London dispersion forces exist between all molecules. • London dispersion forces depend on the shape of the molecule. Chapter 11

  10. Intermolecular Forces London Dispersion Forces • The greater the surface area available for contact, the greater the dispersion forces. • London dispersion forces between spherical molecules are lower than between sausage-like molecules. Chapter 11

  11. Intermolecular Forces London Dispersion Forces Chapter 11

  12. Things to Remember • The more intermolecular forces present, the stronger the bonding between molecules; therefore, boiling pt. is higher and freezing pt. is lower • The stronger the intermolecular forces, the less volatile (does not want to evaporate). • The less volatile a compound is, the lower the vapor pressure. The more volatile a compound is, the higher the vapor pressure. • If # of intermolecular forces are the same, the higher the molar mass, the higher the boiling pt and the lower the freezing pt.

  13. True or False • A. CBr4 is more volatile than CCl4. • B. CBr4 has a higher boiling point than CCl4. • C. CBr4 has weaker intermolecular forces than CCl4. • D. CBr4 has a higher vapor pressure at the same temperature than CCl4.

  14. Phase Changes • Sublimation: solid  gas. • Vaporization: liquid  gas. • Melting or fusion: solid  liquid. • Deposition: gas  solid. • Condensation: gas  liquid. • Freezing: liquid  solid.

  15. Phase Changes

  16. ENERGY ASSOCIATED WITH HEATING CURVES

  17. Topics • Vapor Pressure • Normal Boiling Point • Normal Freezing • Specific Heat • Enthalpy (Heat) of Vaporization • Enthalpy (Heat) of Fusion

  18. Vapor Pressure • THE PRESSURE OF A VAPOR IN EQUILIBRIUM WITH ITS LIQUID (OR ITS SOLID)

  19. NORMAL BOILING POINT & FREEZING POINTS • NORMAL BOILING PT. - THE TEMPERATURE @WHICH VAPOR PRESSURE = 1 atm • NORMAL FREEZING PT. – THE TEMPERATURE @ WHICH THE VAPOR PRESSURE OF THE SOLID AND THE LIQUID ARE THE SAME

  20. Heat Capacity aka Specific Heat (C) • Specific Heat (C) = the amount of energy required to raise the temperature of 1 gram of substance 1 degree celcius

  21. Specific Heat (C) aka Heat Capacity • Units for:specific heat (C) = J/g-oC where J = joules oC = temperature in oC g = mass in grams

  22. Specific Heat (C) Values(aka Heat Capacity) • Example: Water • LIQUID: CLiq = 4.18 J/ (oC . g) • SOLID: Csol = 2.09 J/ (oC . g) • GAS: Cgas = 1.84 J/ (oC . g)

  23. Use of Specific Heat • q = mCDT • q = gm substance x specific heat x DT • where: • M = mass of substance in grams • q = amount of heat (energy) • C = specific heat • And DT = change in temperature

  24. Enthalpy of Vaporizationaka heat of vaporization (DHvap) • Is the amount of heat needed to convert a liquid to a vapor at its normal boiling point

  25. Enthalpy of Fusion aka heat of fusion (DHfus) • Is the amount of heat needed to convert a solid to a liquid at its normal melting (freezing) point

  26. Units for DHvap, DHfus and heat(q) Heat of fusion DHvap = kJ/mol Heat of vaporization DHfus = kJ/mol Heat (q) = Joules

  27. Therefore: • To come up with Joules which is the unit of heat, if: (1) DH is given, then: qvap = DHvap x moles and qfus = DHfus x moles (2) Specific heat (C) is given, then: q = mCDT

  28. For Water Heat of fusion: DHfus = 6.02 kJ/mol Heat of vaporization: DHvap = 40.7 kJ/mol

  29. Sample Problem • Calculate the enthalpy change upon converting 1 mole of ice at -25 oC to steam at 125 oC under a constant pressure of 1 atm? The specific heats are of ice, water and steam 2.09 J/g-K for ice, 4.18 J/g-K for water and 1.84 J/g-K for steam. For water, DHfus= 6.01 kJ/mol, and DHvap = 40.67kJ/mol. • Note: The total enthalpy change is the sum of the changes of the individual steps.

  30. HEATING CURVES • ENERGY ASSOCIATED WITH HEATING CURVES • During a phase change, adding heat causes no temperature change.

  31. Phase Changes Critical Temperature and Pressure • Gases liquefied by increasing pressure at some temperature. • Critical temperature: the minimum temperature for liquefaction of a gas using pressure. • Critical pressure: pressure required for liquefaction.

  32. Phase Diagrams • Phase diagram: plot of pressure vs. Temperature summarizing all equilibria between phases. • Given a temperature and pressure, phase diagrams tell us which phase will exist. • Any temperature and pressure combination not on a curve represents a single phase.

  33. Phase Diagrams

  34. Phase Diagrams The Phase Diagrams of H2O and CO2

  35. Problem • C2Cl2F3 has a normal boiling point of 47.6 oC. The specific heats of C2Cl2F3 (l) and C2Cl2F3 (g) are 0.91 J/g-K and 0.67 J/g-K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol. (a) Calculate the heat required to convert 50.0 g of C2Cl2F3 from a liquid at 10.0 oC to a gas at 85.00 oC. • (b) Calculate the amount of energy needed to convert 50.0 grams of C2Cl2F3 from its normal condensation pt. to its normal boiling point.

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