Chapter 12

1. Chapter 12 Liquids, Solids, and Intermolecular Forces

2. Homework • Assigned Problems (odd numbers only) • Sections 12.4 to 12.6 • “Problems” (41-49), (51-61), (63-69), (87-93), (95-101) • “Cumulative Problems” (87 to 93), (95-101) • “Challenge Questions” 105 only (page 442)

3. Interactions between Molecules • Kinetic molecular theory of matter states that the particles present in any phase of matter are in constant, random motion: thermal energy • Thermal (kinetic) energy is temperature dependent and increases with increasing temperature • The particles interact together through attractions and repulsions that creates potential energy within the particles

4. Interactions between Molecules • Kinetic energy gives the particles their motion and tends to move the particles away from each other (disruptive force) • Potential energy is stored energy that matter possesses mainly due to electrostatic interactions from positive and negatively charged particles (cohesive force) • These are attractions and repulsions in which positively and negatively charged particles attract and repel each other

5. Interactions between Molecules • The relative influence of kinetic and potential energy is the main consideration when KM theory is used to explain the general properties of the gas, liquid, or solid states of matter • The type of energy that dominates will influence a substance’s physical state

6. Interactions between Molecules • Molecules in a liquid do not all have the same kinetic energy • These differences in energy are due to collisions between the molecules • A certain minimum KE is required for molecules to escape from the attractions of nearby molecules • As the temperature of a liquid increases, more molecules have this needed minimum KE. So, the rate of evaporation always increases as the temperature increases.

7. Interactions between Molecules • Intermolecular Forces are forces that act between a molecule and another molecule • They are electrostatic forces that are similar to the forces involved in ionic and covalent bonding: intramolecular forces • Intermolecular forces are much weaker forces but strong enough to influence behavior of liquids

8. Properties of Liquids and Solids • Temperature influences the thermal energy of the particles and this plays an important role in determining the physical state of a system (i.e. solid, liquid, or gas) • There are observable differences among the three states of matter: volume and shape, density, expansion, and compressibility • It is the strength of the intermolecular forces that determine the physical state of molecular substances at room temperature

9. Properties of Liquids and Solids • The solid state is predominated by potential energy rather than by (thermal) kinetic energy • Particles are in a fixed position by strong electrostatic attractions but vibrate due to kinetic energy • Definite volume and shapeand do not assume the shape of their container • High density: The particles are located as close as possible

10. Properties of Liquids and Solids • The liquid state is not dominated by potential energy or by kinetic energy • Particles are not in a fixed position due to kinetic energy which provides just enough motion energy for the particles to slide over each other. • The potential energy (cohesive force) is strong enough prevent total separation and retain a fixed volume • Assumes the shape of the container it occupies • Definite volume and indefinite shape • High density: The particles are not widely separated but located relatively close together

11. Properties of Liquids and Solids • The gaseous state is dominated completely by kinetic energy • Particles of a gas are independent of each other and move in a totally random manner due to their kinetic energy • The attractive forces between particles have been overcome by kinetic energy which allows particles to travel in all directions • Assumes both the volume and shape of the container it occupies • Indefinite volume and indefinite shape • Low density: The particles are widely separated and relatively few particles per unit volume

12. Properties of Liquids and Solids • Liquid and solid phases have many similar characteristics but gases are very different • The average distance between the particles is only slightly different in the solid and liquid but vastly different in the gaseous state

13. Evaporation and Condensation • Evaporation is the process by which molecules escape from the liquid phase to the gas phase • According to KM theory, at any given instant, not all molecules will have the same kinetic energy • The molecules with above average KE can overcome the attractive forces that are holding them in the liquid’s surface and escape into the gas phase

14. Evaporation and Condensation • Liquids are constantly evaporating and molecules at the surface break away from the liquid and enter into the gas phase • The rate of evaporation is influenced by an increase in surface area (more molecules at the surface) • An increase in temperature will give molecules the minimum needed kinetic energy to escape from the intermolecular attractions of other molecules

15. Vapor Pressure • A substance that readily evaporates (at room temp.) is described as a volatile substance • The molecules that escape from an evaporating liquid are referred to as a vapor: The gaseous state of a substance at a temperature and pressure at which the substance is normally a liquid • Liquids are constantly evaporating and molecules at the surface break away from the liquid and enter the gas phase

16. Vapor Pressure • In a closed container, the molecules continue to escape the liquid phase increasing their concentration in the gas phase • As the concentration increases in the gas phase (vaporization) , some of the particles return back to the liquid phase (condensation) • Since vaporization and condensation are occurring simultaneously, eventually a condition of equilibrium is established

17. Vapor Pressure • At equilibrium, the rates of evaporation and condensation are the same • The concentration of molecules in the liquid phase and gas phases are no longer changing • The vapor pressure is the pressure exerted by a vapor above a liquid when the liquid and vapor are in equilibrium

18. Boiling • Boiling is a special form of evaporation in which conversion from the liquid to the vapor state occurs within the body of a liquid through bubble formation • It occurs when a liquid is heated and the vapor pressure of the liquid reaches a value of the outside pressure • The normal boiling point of a liquid is the temp. at which its vapor pressure equals atmospheric pressure of 1 atm

19. Energetics of Evaporation and Condensation • Evaporation is the process by which molecules escape from the liquid phase to the gas phase • It is an endothermic process and requires the absorption of heat energy • The rate of evaporation always increases as a liquid’s temperature increases • This increases the number of molecules that possess the minimum KE needed to overcome the attractive forces (escape to vapor phase)

20. Energetics of Evaporation and Condensation • Condensation is the process by which a gas is changed to a liquid • This change of state is the reverse process of evaporation • It is an exothermic process and requires the release of heat energy • Energy is released as the intermolecular forces increase in number

21. Heating Curve A heating curve is a plot of temperature versus time with a constant amount of heat added Illustrates the steps involved in changing a solid to a gas Heat added is shown on the x-axis Temperature is shown on the y-axis Energy required to undergo a series of phase changes depends on the (three) property values of the substance: Specific Heat Heat of Fusion Heat of Vaporization

22. Heating Curve No temp change No temp change

23. Heating Curve during Boiling • Boiling is a phase change and the process occurs at a constant temperature • The energy absorbed is only used to overcome the intermolecular forces • The energy is also released as intermolecular forces are increased Heating Curve for Water

24. Vaporization/Condensation Heat of Vaporization Heat energy required to vaporize one mole of a substance (e.g. water) Heat energy that must be removed to condense one moleof a substance (e.g. water)

25. Change of State Problem Calculate the heat needed (in Joules) to heat 15 g of water from 75 °C to 100 °C, and to convert it to steam at 100 °C Two parts to the problem Heat the water (use specific heat for water) Convert water to steam (useheat of vaporization for water)

26. Change of State Problem Heat the water Calculate the heat energy needed to warm the water from 75 °C to 100 °C Ti = 75 °C Tf = 100 °C Given: 15.0 g H2O Find: kJ ×25 °C

27. Change of State Problem Calculate the heat required to vaporize 15.0 g of liquid water to steam at 100 °C No change in temperature for a phase change Ti = 100 °C Tf = 100 °C Given: 15.0 g H2O Find: kJ

28. Change of State Problem Calculate the heat required to vaporize 15.0 g of liquid water to steam at 100 °C No change in temperature for a phase change Ti = 100 °C Tf = 100 °C Given: 15.0 g H2O Find: kJ 1 mol H2O = 18.02 g H2O

29. Change of State ProblemCombining Energy Calculations Calculate the total heat Heat the water (q1) Convert liquid water to steam (q2)

30. Energetics of Melting and Freezing • Melting is the conversion of a solid to a liquid • It is an endothermic process that requires the absorption of heat • The heat energy absorbed is used to partially disrupt the intermolecular attractions, allowing the solid to melt • Freezing is the reverse process of melting and converts a liquid to a solid • It is an exothermic process that requires the release of heat

31. Melting/Freezing Reversible, change of state processes Heat of Fusion Heat energy required to melt 1 moleof a substance (e.g. water) Heat energy that must be removed to freeze 1 mole of a substance (e.g. water) Heat energy (to melt) 1 g of ice (water at 0 °C):

32. Change of State Problem Two parts to the problem Melt ice (use heat of fusion for water) Heat water (use specific heat of water) • Calculate the heat required (in Joules) to melt 15 g of ice at 0°C, and to heat the water to 75 °C

33. Change of State Problem Calculate the heat required to melt 15.0 g of ice to liquid water at 0 °C No change in temperature for a phase change Ti = 0 °C Tf = 0 °C Given: 15.0 g H2O Find: kJ

34. Change of State Problem Calculate the heat required to melt 15.0 g of ice to liquid water at 0 °C No change in temperature for a phase change Ti = 0 °C Tf = 0 °C Given: 15.0 g H2O Find: kJ 1 mol H2O = 18.02 g H2O

35. Change of State Problem Heat the water Calculate the heat energy needed to warm the water from 0°C to 75 °C Ti = 0 °C Tf = 75 °C Given: 15.0 g H2O Find: kJ ×75 °C

36. Change of State ProblemCombining Energy Calculations Calculate the total heat Melt the ice (q1) Heat the liquid water (q2)

37. Sublimation Sublimation: A phase change from solid to gas without going through the liquid state For example, the sublimation of water and carbon dioxide Requires the absorption of heat No temperature change occurs during this process Deposition is the reverse process (heat is released)

38. Intermolecular Forces • The strength of the attractive forces that exist between the molecules when they are close together will determine whether a substance is a solid, liquid, or a gas (e.g. water is a liquid, methane is gas, and glucose is a solid) • Intermolecular forces are the attractive forces that act between a molecule and another molecule • The type and magnitude of these forces will determine the characteristics of a compound: Relative Strengths of the Intermolecular Forces: covalent bonds >>> hydrogen bonds > dipole-dipole forces > London Forces

39. Intermolecular Forces • Intermolecular forces influence: • The physical state of a substance at room temperature • The boiling and freezing points of a substance • There are two factors that cause intermolecular forces: • The polarity of the bonds within molecules • The unsymmetrical movement of the electrons about the nuclei

40. Dispersion Force Dispersion (London) forces: Short-lived dipoles caused by uneven shifts in electron density The uneven shift causes one end of the molecule to be slightly positive and one end slightly negative This induces the same electron shift in adjacent molecules which creates a network of attractive forces All molecules can form these instantaneous dipoles but it is the only type of intermolecular force possible in nonpolar substances The magnitude of this force increases with increasing molar mass

41. Dipole-dipole Force Dipole-dipole: Attractive forces between molecules that have permanent dipoles (polar molecules) The nonsymmetrical distribution of the charge causes the molecules to line up Positive end of one directed (attracted) toward negative end of other Only molecules that possess permanent dipoles can engage in this type of interaction The magnitude of this interaction increases as the polarity of the molecule increases

42. Hydrogen Bonding The strongest of the three attractive forces is hydrogen bonding: A special type of dipole-dipole interaction Occurs between molecules that have a H atom bonded to F, O, or N Highly polar, covalent bonds with a H atom bonded with a small, highly electronegative atom leaving a significant partial positive charge on hydrogen The small size of the hydrogen atom allows close proximity to lone pair electrons on an F, O, or N of another molecule H O H O H H

43. Heats of Fusion and Vaporization vs. Intermolecular Forces Dipole-dipole Ionic forces H-bonding Dispersion forces

44. End