1. Liquid, Solids and Intermolecular Forces
Interactions
Properties of Liquids and Solids
Intermolecular Forces
Changes in state
Water

2. LIQUID
condensation
melting
evaporation
freezing
GAS (vapor)
SOLID
deposition
sublimation

3. Intermolecular forces vs Intramolecular forces
Intramolecular forces are bonds within an atom
Intermolecular forces are interactions between molecules/formula units

4. Intermolecular forces are weaker than intramolecular forces
Consider H2O(l)  H2O(g) 100oC
Consider 2H2O(l)  H2(g) and O2(g) 1% 2000oC
Boiling points and melting points
Why are they different for different compounds/molecules?
Depends upon the intermolecular forces.
Stronger attractive forces mean more energy (heat) needed to break intermolecular interactions apart.

5. Properties of Liquids
Liquids, gases and solids all have attractive forces between molecules (intermolecular forces). Between gases, the forces are very, very minimal, thus the gas laws.
The interactive forces between molecules of liquid are responsible for surface tension, viscosity, and vapor pressure.

6. Properties of Liquids and Solids

7. Liquids
No mathematical relationship like ideal gas law works for liquids and solids. Depends solely on the interactions (forces of attraction) between the molecules.
High densities in comparison to gases
Indefinite shape (assume shape of container)
Definite volume

8. Surface Tension and Viscosity – Intermolecular Forces
Surface tension is defined as the tendency of liquids to minimize their surface area (a “skin” is produced).
Viscosity is defined as the resistance of a liquid to flow.
Both of these properties of liquids are due to intermolecular forces.

10. Evaporation (Vaporization) and Condensation
Vaporization – a physical change in which a substance is converted from a liquid to a gas
Rate of vaporization –
Increases with greater surface area
Increases with increase in temperature
Decreases with decreasing strength of intermolecular forces
Condensation – a physical change in which a substance is converted from a gas to a liquid

11. Vapor Pressure and Dynamic Equilibrium
Vapor Pressure of a liquid is the partial pressure of its vapor in dynamic equilibrium with its liquid.

12. Vapor Pressure
Think of it as the “escaping tendency” of molecules to go from a liquid to a gas.
Vapor pressure increases as temperature rises.
The higher the vapor pressure the more volatile the liquid is. The more volatile a liquid is the more readily it will evaporate.

13. Boiling Point
The temperature at which the vapor pressure of a liquid is equal to the pressure above it
Normal Boiling Point- boiling point at 1 atmosphere

14. Heating Curve

15. Endothermic and Exothermic
Evaporation is endothermic.
When a liquid is converted to a gas, it absorbs energy.
Condensation is exothermic.
When a gas is converted to a liquid, it releases heat.

18. Example 12.1 Using the Heat of Vaporization in Calculations
Calculate the amount of water in grams that can be vaporized at its boiling point with 155 kJ of heat.
SORT
You are given the number of kilojoules of heat energy and asked to find the mass of water that can be vaporized with the given amount of energy.
STRATEGIZE
Draw the solution map beginning with the energy in kilojoules and converting to moles of water and then to grams of water.
GIVEN: 155 kJ
FIND: g H2O
SOLUTION MAP
RELATIONSHIPS USED
ΔHvap = 40.7 kJ/mol at 100 °C (Table 12.2)
18.02 g H2O = 1 mol H2O (molar mass of water)

19. Example 12.1 Using the Heat of Vaporization in Calculations
Continued
SOLVE
Follow the solution map to solve the problem.
SOLUTION
CHECK
Check your answer. Are the units correct? Does the answer make physical sense?
The units (g) are correct. The magnitude of the answer makes sense because each mole of water absorbs about 40 kJ of energy upon vaporization. Therefore 155 kJ should vaporize close to 4 mol of water, which is consistent with the answer (4 mol of water has a mass of about 72 g).

20. Example 12.1 Using the Heat of Vaporization in Calculations
Continued
SKILLBUILDER 12.1 Using the Heat of Vaporization in Calculations
Calculate the amount of heat in kilojoules required to vaporize 2.58 kg of water at its boiling point.
Answer: 5.83  103 kJ
SKILLBUILDER PLUS
A drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25 C. If the
heat released during condensation goes only toward heating the metal, what is the final temperature in Celsius of the
metal block? (The specific heat capacity of aluminum is J/g C.)
Answer: 47 C
For More Practice Example 12.7; Problems 47, 48, 49, 50, 51, 52.

21. Melting Point
The point at which enough thermal energy is applied to cause a solid to become a liquid.

24. Types of Intermolecular Forces
Dipole-dipole = two polar molecules
Hydrogen bonding is a strong dipole-dipole
Ion-dipole = an ion and a polar molecule
Dispersion forces = between nonpolar compounds; temporary dipoles

25. Example 12.2 Using the Heat of Fusion in Calculations
Calculate the amount of ice in grams that, upon melting (at 0 °C), absorbs 237 kJ of heat.
SORT
You are given the number of kJ of heat energy and asked to find the mass of ice that absorbs the given amount of energy upon melting.
STRATEGIZE
Draw the solution map beginning with the energy in kilojoules and converting to moles of water and then to grams of water.
GIVEN: 237 kJ
FIND: g H2O (ice)
SOLUTION MAP
RELATIONSHIPS USED
ΔHfus = 6.02 kJ/mol
1 mol H2O = g H2O (molar mass of water)

26. Example 12.2 Using the Heat of Fusion in Calculations
Continued
SOLVE
Follow the solution map to solve the problem.
SOLUTION
CHECK
Check your answer. Are the units correct? Does the answer make physical sense?
The units (g) are correct. The magnitude of the answer makes sense because each mole of water absorbs about 6 kJ of energy upon melting. Therefore 237 kJ should melt close to 40 mol of water, which is consistent with the answer (40 mol of water has a mass of about 720 g).

27. Example 12.2 Using the Heat of Fusion in Calculations
Continued
SKILLBUILDER 12.2 Using the Heat of Fusion in Calculations
Calculate the amount of heat absorbed when a 15.5-g ice cube melts (at 0 C).
Answer: 5.18 kJ
SKILLBUILDER PLUS
A 5.6-g ice cube (at 0 C) is placed into 195 g of water initially at 25 C. If the heat absorbed for melting the ice comes only from the 195 g of water, what is the temperature change of the 195 g of water?
Answer: 2.3 C
For More Practice Example 12.8; Problems 55, 56, 57, 58.

28. Correlate i, ii, iii with solid, liquid and gas

29. Sublimation
A substance changes from a solid to a gas.

30. Types of Intermolecular Forces
Dispersion Forces
Dipole-dipole forces
Hydrogen bonding

31. Dispersion Forces Temporary dipoles (instantaneous dipoles)
Depends on location of electrons at any given time frame in atom’s orbitals

33. Example 12.3 Dispersion Forces
Which halogen, Cl2 or I2, has the higher boiling point?
SOLUTION
The molar mass of Cl2 is g/mol, and the molar mass of I2 is g/mol. Since I2 has the higher molar mass, it has stronger dispersion forces and therefore the higher boiling point.
SKILLBUILDER 12.3 Dispersion Forces
Which hydrocarbon, CH4 or C2H6, has the higher boiling point?
Answer: C2H6
For More Practice Problems 63, 64.

34. Dipole-Dipole Forces Exist in all polar molecules
Permanent dipoles in a molecule that will interact with a dipole in another molecule

35. Example 12.4 Dipole-Dipole Forces
Determine whether or not each molecule has dipole–dipole forces.
(a) CO (b) CH2Cl (c) CH4
SOLUTION
A molecule has dipole–dipole forces if it is polar. To determine whether a molecule is polar, you must:
1. determine whether the molecule contains polar bonds, and
2. determine whether the polar bonds add together to form a net dipole moment (Section 10.8).
(a) Since the electronegativities of carbon and oxygen are 2.5 and 3.5, respectively (Figure 10.2), CO2 has polar bonds. The geometry of CO2 is linear. Consequently, the polar bonds cancel; the molecule is not polar and does not have dipole–dipole forces.
(b) The electronegativities of C, H, and Cl are 2.5, 2.1, and 3.5, respectively. Consequently,CH2Cl2 has two polar bonds (CCl) and two bonds that are nearly nonpolar (C  H). The geometry of CH2Cl2 is tetrahedral. Since the bonds C  Cl and the C  H bonds are different, they do not cancel, but sum to a net dipole moment. Therefore the molecule is polar and has dipole–dipole forces.

36. Example 12.4 Dipole-Dipole Forces
Continued
(c) Since the electronegativities of C and H are 2.5 and 2.1, respectively, the  of the molecule is tetrahedral, any slight polarities that the bonds might have will cancel. CH4 is therefore nonpolar and does not have dipole–dipole forces.
SKILLBUILDER Dipole–Dipole Forces
Determine whether or not each molecule has dipole–dipole forces.
(a) CI (b) CH3Cl (c) HCl
Answers:
(a) no dipole–dipole forces
(b) yes, it has dipole–dipole forces
(c) yes, it has dipole–dipole forces
For More Practice Problems 59, 60, 61, 62.

37. Hydrogen Bonding
Polar molecules containing hydrogen atoms directly bond to fluorine, oxygen or nitrogen exhibit hydrogen bonding.
The H-bonding will occur the H bonded as described above and the F, O or N on a neighboring molecule.
H-bonding is a type of dipole-dipole force but is much stronger than a normal dipole-dipole force.

38. H-bonding in hydrogen flouride

40. hydrogen peroxide 34.02 g/mol
Example 12.5 Hydrogen Bonding
One of these compounds is a liquid at room temperature. Which one and why?
SOLUTION
The three compounds have similar molar masses.
formaldehyde g/mol
fluoromethane g/mol
hydrogen peroxide g/mol
Therefore, the strengths of their dispersion forces are similar. All three compounds are also polar, so they have dipole–dipole forces. Hydrogen peroxide, however, is the only compound that also contains H bonded directly to F, O, or N. Therefore it also has hydrogen bonding and is most likely to have the highest boiling point of the three. Since the problem stated that only one of the compounds was a liquid, we can safely assume that hydrogen peroxide is the liquid. Note that although fluoromethane contains both H and F, H is not directly bonded to F, so fluoromethane does not have hydrogen bonding as an intermolecular force. Similarly, although formaldehyde contains both H and O, H is not directly bonded to O, so formaldehyde does not have hydrogen bonding either.
SKILLBUILDER 12.5 Hydrogen Bonding
Which has the higher boiling point, HF or HCl? Why?
Answer: HF, because it has hydrogen bonding as an intermolecular force
For More Practice Examples 12.9, 12.10; Problems 65, 66, 67, 68, 69, 70.

41. Example 12.9 Determining the Types of Intermolecular Forces in a Compound
Determine the types of intermolecular forces present in each substance
(a) N2
(b) CO
(c) NH3
SOLUTION
(a) N2 is nonpolar and therefore has only dispersion forces.
(b) CO is polar and therefore has dipole–dipole forces (in addition to dispersion forces).
(c) NH3 has hydrogen bonding (in addition to dispersion forces and dipole–dipole forces).

42. Example 12.10 Using Intermolecular Forces to Determine Melting and/or Boiling Points
Arrange each group of compounds in order of increasing boiling point.
(a) F2, Cl2, Br (b) HF, HCl, HBr
SOLUTION
(a) Since these all have only dispersion forces, and since they are similar substances (all halogens), the strength of the dispersion force will increase with increasing molar mass. Therefore, the correct order is F2 < Cl2 < Br2.
(b) Since HF has hydrogen bonding, it has the highest boiling point. Between HCl and HBr, HBr (because of its higher molar mass) has a higher boiling point. Therefore the correct order is HCl < HBr < HF.

43. Summary

44. H-bonding in DNA

48. Water

49. Water Most common and important liquid on Earth
Fills lakes, streams, oceans
Our body mass is majority water
Low molecular mass, but liquid at room temperature
Polarity and bent geometry gives it unique qualities: high boiling point, good as a solvent, liquid is higher in density than solid