1 / 33

CHAPTER 12 Liquids and Solids

CHAPTER 12 Liquids and Solids Note that the material on interparticle forces is from Chapter 7, Section 3 of Burdge. Properties of Solids, Liquids, and Gases

sobrien
Télécharger la présentation

CHAPTER 12 Liquids and Solids

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. CHAPTER 12 Liquids and Solids Note that the material on interparticle forces is from Chapter 7, Section 3 of Burdge.

  2. Properties of Solids, Liquids, and Gases Condensed phase – A phase where the particles of the substance are in close contact with one another. Because of this, the volume occupied by a condensed phase is approximately independent of pressure and temperature. Solids and liquids are condensed phases.

  3. Properties of Solids, Liquids, and Gases Fluid phase – A phase where the particles of the substance can move about. Because of this, the phase does not have a definite shape, but instead has a shape determined by the container. Liquids and gases are fluids. The three common states of matter are solids, liquids, and gases. Their properties can be summarized as follows: solid - definite volume and definite shape; liquid - definite volume but indefinite shape; gas - indefinite volume and indefinite shape

  4. Effect of Temperature and Pressure on Phase The phase of a substance depends on both temperature and pressure. Generally speaking, substances go from solid to liquid to gas as temperature increases. It is often possible to condense a gas into a liquid or solid by increasing pressure. We will discuss this further later in the chapter.

  5. Summary of the Types of Intermolecular Forces Intermolecular forces are the forces that exists between different molecules or particles. We are more concerned with long range attractive forces and will ignore short range repulsive forces. Ion – ion. Forces between cations and anions. Dipole – dipole. Forces between molecules with a permanent dipole moment. This category includes hydrogen bonding, a particularly strong type of dipole – dipole force. London dispersion forces. Due to random movement of electrons. All particles have this type of force, but it is most important in molecules with no permanent dipole moment.

  6. Ion-Ion Forces Ion-ion - The attractive force acting between cations and anions. These are strong, and are found in substances where ionic bonding occurs.

  7. Dipole-Dipole Forces Dipole-dipole - The attractive force acting between polar molecules. The attraction is between the partial positive charge (+) on one molecule and the partial negative charge (-) on a different molecule. Generally speaking, the larger the partial positive and negative charges the stronger the dipole-dipole attraction.

  8. Dipole-Dipole Forces and Boiling Point When molecules have strong intermolecular attractive forces it takes more energy to overcome those attractive forces. One way of seeing this is in the boiling point for a substance. Generally speaking, the stronger the dipole-dipole attraction between molecules the higher the boiling point, parti-cularly for substances with approximately the same molecular mass.

  9. Hydrogen Bonding Hydrogen bonding - A particularly strong form of dipole-dipole attractive force. It is the attractive force that exists between a hydrogen atom bonded to an N, O, or F atom and lone pair electrons on a different N, O, or F atom (often an atom in a different molecule).

  10. Evidence For Hydrogen Bonding One effect of hydrogen bonding is to raise the boiling point of a liquid. This occurs because it requires more energy (and so a higher temperature) to break apart strong attractive forces between molecules than it does to break apart weak attractive forces between molecules. substanceboiling pointhydrogen bonding? H2O 100.0 C yes H2S - 60.7 C no H2Se - 41.5 C no H2Te - 4.4 C no

  11. Boiling Points for Binary Hydrogen Compounds

  12. London Dispersion Forces London dispersion forces - The attractive force that is due to the formation of instantaneous dipoles in a molecule. These instantaneous dipoles arise from the random motion of the electrons in the molecule. Polarizability - The shift in electrons in an atom or molecule when approached by a charged particle. In general, larger atoms or molecules have larger polarizabilities.

  13. Strength of London Dispersion Forces London dispersion forces are present in all molecules, but are the only intermolecular force present in nonpolar molecules. The strength of London dispersion forces is approximately proportional to the number of electrons, and so to the size of the molecule. Therefore, as a general rule, the larger the molecule the stronger the London dispersion forces. substanceboiling point He - 268.6 C Ne - 245.9 C Ar - 185.7 C Kr - 152.3 C Xe - 107.1 C

  14. Trends in the Boiling Points For Liquids The boiling point for a liquid depends on the strength of the attractive forces acting between the particles making up the liquid. Based on that, there are several general statements we can make concerning the boiling points for liquids. 1) Ionic compounds have extremely high boiling points. 2) Molecules with a permanent dipole moment have intermediate boiling points. Boiling points are highest for molecules where hydrogen bonding can occur. 3) For molecules interacting only by dispersion forces the boiling point should be low. In this case, the larger the molecule the higher the boiling point. 4) For similar molecules (diatomic molecules, hydrocarbons, alcohols), the larger the molecule the higher the boiling point.

  15. Sample Problem For each of the following pairs of molecules, pick the molecule expected to have the higher value for normal boiling point. (Normal boiling point - the boiling temperature when p = 1.00 atm). a) F2 or Cl2 b) NH3 or PH3 c) CH3OH or CH3CH2OH

  16. For each of the following pairs of molecules, pick the molecule expected to have the higher value for normal boiling point. (Normal boiling point - the boiling temperature when p = 1.00 atm). a) F2 or Cl2 Cl2 F2 (- 188. °C) Cl2 (- 35. ºC) Similar molecules, so larger molecule has higher boiling point. b) NH3 or PH3 NH3 NH3 (- 33. ºC) PH3 (- 88. ºC) NH3 can hydrogen bond, PH3 does not hydrogen bond. c) CH3OH or CH3CH2OH CH3CH2OH CH3OH (65. ºC) CH3CH2OH (78. ºC) Similar molecules, so larger molecule has higher boiling point.

  17. Vapor Pressure Vapor pressure is defined as the equilibrium partial pressure of vapor above a solid or liquid. The vapor pressure of substance increases as temperature increases. The normal boiling point for a pure chemical substance corresponds to the temperature at which the liquid and vapor pressures are at equilibrium and the vapor pressure is 1.00 atm.

  18. Clausius-Clapeyron Equation Based on experiment, the vapor pressure above a pure liquid is found to obey a simple equation, called the Clausius-Clapeyron equation ln(p) = - Hvap + C RT Based on this equation we expect a plot of ln(p) vs 1/T to give a straight line with slope m = - Hvap/R. The above equation may be used to find a second useful expression ln(p2/p1) = - (Hvap/R) [ (1/T2) – (1/T1) ] where p1 is the vapor pressure at temperature T1 p2 is the vapor pressure at temperature T2

  19. Clausius-Clapeyron Equation (Example) slope = - Hvap/R

  20. Example: The normal boiling point for water occurs at T = 100.0 C. The enthalpy of vaporization for water is Hvap = 40.67 kJ/mol. Based on this information estimate the vapor pressure of water at T = 20.0 C.

  21. Example: The normal boiling point for water occurs at T = 100.0 C. The enthalpy of vaporization for water is Hvap = 40.67 kJ/mol. Based on this information estimate the vapor pressure of water at T = 20.0 C. Recall that one form of the Clausius-Clapyron equation is: ln(p2/p1)= - Hvap1 _ 1 R T2 T1

  22. Example: The normal boiling point for water occurs at T = 100.0 C. The enthalpy of vaporization for water is Hvap = 40.67 kJ/mol. Based on this information estimate the vapor pressure of water at T = 20.0 C. Let T1 = 100.0 C = 373. K ; p1 = 1.00 atm T2 = 20.0 C = 293. K ln(p2/p1) = - 40670. J/mol11 = - 3.581 (8.314 J/mol.K) 293. K 373. K

  23. ln(p2/p1) = - 3.581 If we take the inverse logarithm of both sides, we get (p2/p1) = e-3.581 = 0.0279 p2 = (0.0279) p1 = (0.0279) (1.00 atm) = 0.0279 atm = 21. torr

  24. Solids Solids can be divided into two general categories Crystalline solid - Has a regular arrangement of the particles making up the solid (a crystal structure). Four main types exist: ionic, covalent, molecular, and metallic solids. Amorphous solid - Does not have a regular arrangement of the parti-cles making up the solid (no regular crystal struct-ure).

  25. Crystal Structure For crystalline solids, the crystal structure of a solid substance refers to the arrangement of the particles making up the solid. This is often given in terms of the unit cell, the smallest part of the crystal that can be used to construct the entire crystal.

  26. Phase Transitions The conversion of a substance from one phase to another phase is called a phase transition. Transitions can be caused both by adding heat and by removing heat from a substance. adding heat (H > 0)removing heat (H < 0) s   fusion (melting)   s freezing   g vaporization g   condensation s  g sublimation g  s deposition Recall that the enthalpy change for a phase transition is usually reported at the normal transition temperature, that is, the temperature at which the phase transition occurs when p = 1.00 atm. Since enthalpy is a state function: Hfreez = - Hfus Hcond = - Hvap Hdep = - Hsub

  27. Relationships Among Phase Transitions Based on Hess’ law, we would expect Hsub = Hfus + Hvap

  28. Thermodynamics of Phase Transitions We can study the thermodynamics of phase transitions by finding the heating curve for a substance. This is simply a plot of temperature vs. amount of heat added, under conditions where the heat is added slowly enough to maintain equilibrium. Experimentally we ex-pect to see two regions in the heating curve. Normally the temperature of the substance will increase as heat is added. However, at the temperature where a phase transition occurs the added heat will be used to carry out the transi-tion, and so temperature will remain constant until the phase transition is complete.

  29. Sample Heating Curve

  30. Phase Diagram A phase diagram is a diagram indicating which phase or phases are present at equilibrium as a function of pressure and temperature. H2O

  31. Important Features in a Phase Diagram Phase boundaries - Indicate where two phases can exist simul-taneously at equilibrium. Triple point - Indicates a point where three phases can exist simultaneously at equilibrium. Normal melting point - Solid-liquid equilibrium at p = 1.00 atm. Normal boiling point - Liquid-gas equilibrium at p = 1.00 atm. Normal sublimation point - Solid-gas equilibrium at p = 1.00 atm. (Note that substances will have either a normal melting and normal boiling point, or a normal sublimation point, but not both.) Critical point - Point below which a gas will undergo a phase transition (g   or g  s) when compressed reversibly at constant temperature. Above the critical point no such phase transition occurs. In this region of the phase diagram a supercritical fluid is present.

  32. Phase Diagram For CO2 CO2

  33. End of Chapter 12

More Related