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Intermolecular Forces and States of Matter: Understanding the Molecular Level

Explore the concept of intermolecular forces and their impact on the states of matter. Learn about the different types of attractive forces between molecules and how they affect physical properties such as boiling points and melting points. Understand the relative strength of attractions and the role of dispersion forces, dipole-dipole forces, and hydrogen bonding. Discover the influence of molecular weight and polarizability on intermolecular forces.

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Intermolecular Forces and States of Matter: Understanding the Molecular Level

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  1. Chapter 11Liquids and Intermolecular Forces

  2. Intermolecular Forces • The attractions between molecules are not nearly as strong as the intramolecular attractions (bonds) that hold compounds together. • Many physical properties reflect intermolecular forces, like boiling points, melting points, viscosity, surface tension, and capillary action.

  3. The States of Matter

  4. States of Matter • The fundamental difference between states of matter is the strength of the intermolecular forces of attraction. • Stronger forces bring molecules closer together. • Kinetic energy keeps them apart and moving. Remember that average kinetic energy is related to temperature!

  5. Intermolecular Forces – What is happening on the molecular level that causes a solid to be a solid ? • The attractive forces between molecules – What is happening on the molecular level that causes a liquid to be a liquid ? – What is happening on the molecular level that causes a gas to be a gas ?

  6. Relative Strength of Attractions • Intermolecular attractions are weaker than bonds. • Hydrogen bonds are NOT chemical bonds.

  7. Types of Intermolecular Force between Neutral Molecules • Weakest to strongest forces: • Dispersion forces (or London dispersion forces or induced dipole-induced dipole interactions) • Dipole–dipole forces • Hydrogen bonding (a special dipole–dipole force) Note: The first two types are also referred to collectively as van der Waals forces.

  8. Types of Intermolecular forces 1. London dispersion forces– the very short lived attractive forces caused by the instantaneous displacement of electrons Fritz London1900 – 1954

  9. Electron cloud momentarily distorts to give rise to non-symmetric electron cloud producing an instantaneous dipole

  10. Instantaneous dipole– distorted electrons give rise to a temporary dipole before instantaneous dipoles: He nonpolar instantaneous dipoles occurring: He slightly polar Instantaneous dipoles provide some additional intermolecular ordering on the molecular level

  11. London Dispersion Forces Extent of London dispersion is determined by MW London dispersion forces increase as MW increases

  12. Arrange the following in order of increasing boiling points; C3H8 , CH4 , C8H18

  13. Factors That Affect Amount of Dispersion Force in a Molecule • Number of electrons in an atom (more electrons, more dispersion force) • Size of atom or molecule/molecular weight • Shape of molecules with similar masses (more compact, less dispersion force)

  14. Polarizabilityand Boiling Point • Atomic/molecular weight increases down a group in the periodic table. • Higher atomic/molecular weights translate into stronger dispersion forces, which lead to higher boiling points.

  15. Types of Intermolecular forces 2. Dipole - Dipole– the attractive forces between polar molecules - + Br-Cl is a polar molecule with a permanent dipole moment

  16. Dipole-Dipole Forces dipole-dipole IM attractive force Since unlike charges attract, polar molecules are attracted to each other on the molecular level

  17. Dipole-Dipole Forces Dipole-dipole interactions cause molecules to orient to maximize attractive forces between molecules Molecules line up and assume a somewhat “ordered” state

  18. The quantitative measure of the extent of intermolecular forces is reflected in the molecules melting point or boiling point! B A A solid at 25 C liquid at 25 C Which possesses a greater degree of order ? MP of A > 25 C while MP of B < 25 C Substance A exhibits more IM attractive forces vs. B

  19. Dipole-Dipole Forces • The higher the boiling point, the more order the substance exhibits …..(in other words)…. • The higher the boiling point, the greater the extent of the intermolecular forces between molecules

  20. Arrange the following in order of increasing boiling point: F2 , H2S

  21. Which Have a Greater Effect:Dipole–Dipole Interactions or Dispersion Forces? • If two molecules are of comparable size and shape, dipole–dipole interactions will likely be the dominating force. • If one molecule is much larger than another, dispersion forces will likely determine its physical properties.

  22. Types of Intermolecular Forces 3. Hydrogen Bonding– occurs between polar molecules that contain a hydrogen atom that is attached to a F, or O, or N “H–F” or “H–O” or “H–N” H-bonding is an unusually strong dipole-dipole that provides more order on the molecular level than a routine dipole-dipole

  23. What Forms Hydrogen Bonds? • Hydrogen bonding arises in part from the high electronegativity of nitrogen, oxygen, and fluorine. • These atoms interact with a nearly bare hydrogen nucleus (which contains one proton but NO inner electrons).

  24. H-bonding is responsible for holding the two strands of DNA together!

  25. Ice Compared to Liquid Water • Hydrogen bonding makes the molecules farther apart in ice than in liquid water. Thus H2O (s) is less dense than H2O () causing ice to float.

  26. Arrange the following in order of increasing boiling point; Kr, H2S, NaCl, Ne, NH3, F2

  27. Ion–Dipole Interactions • Ion–dipole interactions are found in solutions of ions. • The strength of these forces is what makes it possible for ionic substances to dissolve in polar solvents, like water.

  28. Determining Intermolecular Forces in a Substance • Note that ALL chemicals exhibit dispersion forces. • The strongest force dictates the extent of attractions between molecules.

  29. Liquid Properties Affected by Intermolecular Forces • Boiling point (previously discussed) and melting point • Viscosity • Surface tension • Capillary action

  30. Viscosity • Resistance of a liquid to flow is called viscosity. • The greater the viscosity, the slower a liquid flows. • Viscosity increases with stronger intermolecular forces and decreases with higher temperature.

  31. Which species has the greater viscosity ? C6H14 () or CH3CH2CH2CH2CH2OH ()

  32. Which species is more viscous (the greater viscosity) ?CH3OH () at 15 C or CH3OH () at 45 C

  33. Surface Tension – the characteristic “skin” a liquid’s surface develops Surface tension increases with increasing intermolecular attractive forces Surface tension decreases with increasing temperature

  34. Which species has the greater surface tension ? H2O () at 15 C or H2O () at 45 C

  35. Which species has the greater surface tension 25 C ? CH3CH2CH2OH () or HOCH2CH2OH ()

  36. Vapor Pressure – the partial pressure vapor molecules exert at equilibrium h liquid h = height of the Hg = vapor pressure, VP VPwater = 17.5 mm HgVPmethanol = 46.0 mm Hg at 20 C

  37. VP H2O < VP CH3OH VP increases as intermolecular forces decrease VP increases as temperature increases Which species has the higher vapor pressure at 25 C ?CH3CH2CH2OH () or HOCH2CH2OH () Higher VP

  38. Phase Changes • Conversion from one state of matter to another is called a phase change. • Energy is either added or released in a phase change. • Phase changes: melting/freezing, vaporizing/condensing, subliming/depositing.

  39. Energy Change and Change of State • The heat of fusionis the energy required to change a solid at its melting point to a liquid. • The heat of vaporizationis the energy required to change a liquid at its boiling point to a gas. • The heat of sublimationis the energy required to change a solid directly to a gas.

  40. Heating Curves • A graph of temperature vs. heat added is called a heating curve. • Within a phase, heat is the product of specific heat, sample mass, and temperature change. • The temperature of the substance does not rise during a phase change. • For the phase changes, the product of mass or moles and heat of fusion or vaporization is heat.

  41. Heating Curve of a Solid • As you heat a solid, its temperature increases linearly until it reaches the melting point. • q = mass ×Cs×DT • Once the temperature reaches the melting point, all the added heat goes into melting the solid. • The temperature stays constant. • Once all the solid has been turned into liquid, the temperature can again start to rise. • Ice/water will always have a temperature of 0 °C at 1 atm.

  42. The amount of heat energy required to melt one mole of the solid is called the heat of fusion, DHfus. Sometimes called the enthalpy of fusion It is always endothermic; therefore, DHfus is +. DHcrystallization= −DHfusion Generally much less than DHvap DHsublimation = DHfusion + DHvaporization Heat of Fusion

  43. Heating Curve of a Liquid • As you heat a liquid, its temperature increases linearly until it reaches the boiling point. • q = mass ×Cs×DT • Once the temperature reaches the boiling point, all the added heat goes into boiling the liquid; the temperature stays constant. • Once all the liquid has been turned into gas, the temperature can again start to rise.

  44. Heat of Vaporization • The amount of heat energy required to vaporize one mole of the liquid is called the heat of vaporization, DHvap. • Sometimes called the enthalpy of vaporization • It is always endothermic; therefore, DHvap is +. • DHcondensation= −DHvaporization

  45. Heating Curve of Water How much heat is involved in converting 1 mol of ice at -25oC to steam at 125oC?

  46. Heating 1.00 mole of ice at −25.0 °C up to the melting point, 0.0 °C q = mass ×Cs×DT Mass of 1.00 mole of ice = 18.0 g Cs(ice) = 2.09 J/g∙ °C Segment 1

  47. Melting 1.00 mole of ice at the melting point, 0.0 °C q = n ∙ DHfus n = 1.00 mole of ice DHfus= 6.02 kJ/mol Segment 2

  48. Heating 1.00 mole of water at 0.0 °C up to the boiling point, 100.0 °C q = mass ×Cs×DT Mass of 1.00 mole of water = 18.0 g Cs(water)= 4.18 J/g∙ °C Segment 3

  49. Boiling 1.00 mole of water at the boiling point, 100.0 °C q = n ∙ DHvap n = 1.00 mole of ice DHvap= 40.7 kJ/mol Segment 4

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