1. Liquids and Solids
Chapter 10

2. Solids and Liquids
the “condensed” states of matter. Forces of attraction (“cohesive” forces) are stronger than the repulsive forces that make the Ideal Gas Law.

3. Cohesive Forces
def forces between particles in a lattice or a liquid. There are five types of cohesive forces.
1. Ionic – two ions connected together, one atom gives up electrons while the other atom gains the electrons
2. Covalent – two atoms sharing electons

4. Covalent Forces 3. hydrogen bonding 4. dipole-dipole
5. Van der Waals (London dispersion forces)
Ionic and covalent forces are intramolecular forces (inside the molecules)
Hydrogen, dipole-dipole and van der Waals forces are intermolecular forces (between molecules)

5. Relative Magnitudes of Forces
The types of bonding forces vary in their strength as measured by average bond energy.
Strongest Ionic bonds
Covalent bonds (400 kcal/mol)
Hydrogen bonding ( kcal/mol )
Dipole-dipole interactions ( kcal/mol)
London forces (less than kcal/mol)
Weakest

6. Hydrogen bonding Two consequences of hydrogen bonding
Gives substances high boiling points
Substances with hydrogen bonding tend to be more viscous
Both are explained by increased attractions making them more difficult to separate

7. Dipole-Dipole
result when the partial positive (d+) and partial negative (d-) charges of neighboring POLAR covalent molecules attract (roughly 1% as strong as a “typical” covalent bond)
Arrange themselves so the negative and positive ends attract one another
Eventually arrange in order to find the best compromise between attraction and repulsion

8. London Dispersion (van der Waals)
Small electrostatic forces are caused by the movement of electrons in the covalent bonds of molecules
London forces increase with the size of the molecules and with surface area.

9. London Dispersion (van der Waals)
As one approaches another, the electrons of one or both are temporarily displaced
This movement causes small, temporary dipoles to be set up which attract one another
Large molecules with big surface area have more electrons and more dispersion forces and greater attraction and, therefore, higher boiling points

10. The stronger the cohesive forces
THE HIGHER THE MELTING POINT
THE HIGHER THE BOILING POINT
This is because you gotta add more energy to overcome the stronger cohesive forces

11. Solids Ionic structures
Lattice structures – a three dimensional system of points; held together by strong electrostatic interactions
Strong bonds give them high melting points
Can only conduct electricity when molten or in solution

12. Solids Molecular structures
Held in lattice structures held together by weak dispersion forces
Fairly low melting points
A nonconductor
Example - iodine

13. Liquids
Substances that have insufficient energy to overcome intermolecular forces but enough energy to break away from the ordered state of solids

14. Liquids
Capillary action – spontaneous rising of a liquid in a narrow tube
Cohesive forces – between the molecules
Adhesive forces – between molecules and container

15. VAPOR PRESSURE AND PHASE DIAGRAMS
2 important vapor pressure factors
1. the lower the cohesive forces in the condensed state, the higher the vapor pressure
2. the higher the system temperature, the higher the vapor pressure

16. Phase Diagrams
A way of combining most of this chapter (info on condensed states) into one diagram: a graph of Pressure versus Temperature under all conditions, that then shows physical state at that point.

17. Phase Diagrams
Represents phases as a function of temperature and pressure.
Critical temperature: temperature above which the vapor can not be liquefied.

18. Phase Diagrams

19. Colligative Properties: General Solutions Properties
Chapter 11

20. Colligative Properties
Those properties that depend only on the number of solute particles present, not their nature

21. Colligative Properties
Non-electrolyte solutions
Covalent solute
No dissociation
Electrolyte solutions
Ionic solute
Dissocation
Van’t Hoff factor
Freezing point depression
Osmotic Pressure
Freezing point depression
Vapor pressure
Raoult’s law
Osmotic pressure
Boiling point elevation
Boiling point elevation
2 non-volatile components
Deviations
Non-volatile solute

22. More Solution Concentration Units
Since volumes of solutions are temperature dependent, molarity of solutions changes slightly with temp. Chemists have three other solutions concentration units that are NOT temp dependent, since they are related only to MASS and NOT VOLUME. They are mass percent, mole fraction, and molality.

23. Mass Percent and Mole Fraction
Mass Percent Equation
Mass percent = (mass of solute/mass of solution) x 100
Mole Fraction Equation
Mole fraction of A = cA = nA/(nA+nB)

24. Molality
Molality = m = moles solute/kilograms solvent

25. Ex You add 5. 84 g formaldehyde, H2CO, to 100. 0 g H2O
Ex You add 5.84 g formaldehyde, H2CO, to g H2O. The final volume is ml. Find molarity, molality, mass %,and  H2CO?

26. Solubility Trends
The solubility of MOST solids increases with temperature.
The rate at which solids dissolve increases with increasing surface area of the solid.
The solubility of gases decreases with increases in temperature.
The solubility of gases increases with the pressure above the solution.

27. Therefore… Solids tend to dissolve best when:
Heated
Stirred
Ground into small particles
Gases tend to dissolve best when:
The solution is cold
Pressure is high

28. Pressure/Temperature Effects on Solubility of Solutes
Pressure has does not effect how well a solid or liquid dissolve into a solution.
Pressure INCREASES gas solubility. There are more collision on a given surface of the solvent with increased gas pressure. The pressure –concentration relationship is linear (at LOW gas concentrations) and only applies to gases that do not chemically react with the solvent to make new products.

29. Pressure/Temperature Effects on Solubility of Solutes
This is related mathematically by HENRY’S LAW (this relates gas pressure and solubility).
c = KP
c = molar concentration (mol/L)
K = a constant that depends on temperature (mol/L atm)
P = pressure in atm (atmospheres)

30. Ex The solubility of O2 is 2. 2 X 10-4 M at 0 oC and 0. 10 atm
Ex The solubility of O2 is 2.2 X 10-4 M at 0 oC and 0.10 atm. What is the solubility if you increase pressure to 0.35 atm?

31. General Temperature Effects
As a GENERAL RULE, increased temperature…
INCREASES solid/liquid solubility into solvents
DECREASES gas solubility because this increases vapor pressure

32. Pressure
Is caused by the collisions of molecules with the walls of a container
is equal to force/unit area
SI units = Newton/meter2 = 1 Pascal (Pa)
1 atmosphere = 101,325 Pa
1 atmosphere = 1 atm = 760 mm Hg = 760 torr

33. Pressure - examples Convert 5.6 atm to torr. Convert 920mmHg to atm.
Convert 99.8kPa to torr.

34. Vapor Pressure Lowering
Two ways to explain why the vapor pressure is lowered when a solute is added
The process of solvent molecules escaping is partially blocked by the interactions with solute particles at the surface, causing vapor pressure to fall
When a solute is added to a pure solvent, the entropy (disorder) dramatically increases. This is the preferred situation. Prior to the solute being added, the solvent could only increase the disorder by evaporating. Adding a solute created disorder and there is less incentive for the solvent to evaporate.

35. Raoult’s Law (Non-volatile Solute)
explains nonvolatile solute effects on vapor pressure.
Non-volatile solutes do not have their own vapor pressure

36. Raoult’s Law (Non-volatile Solute)
Vapor pressure will be lower than that of the pure solvent
Vp = cPo
Vp = vapor pressure
c = mole fraction of the solvent
Po = vapor pressure of the pure solvent

37. Raoult’s Law
Ex What is the vapor pressure of a glycerin (nonvolatile), C3H8O3, solution made by adding 164 grams of it to 338 ml of H2O at 39.8 °C? The PH2O at that temp is torr and the density is g/mL

38. Raoult’s Law
Raoult’s Law also holds for electrolytes (soluble salts), but you actually have more solute present because of ionization.

39. Raoult’s Law
Ex What would be the vapor pressure effect of adding grams of NaCl to the water in the previous example?

40. Raoult’s Law
ex At 29.6 oC, PoH2O = 31.1 torr. You make a solution using 86.7 grams of unknown “M” in grams of H2O, and determine the resulting vapor pressure to be 28.6 torr. What is the molar mass of M?

41. Raoult’s Law for volatile components
If a solute that has its OWN vapor pressure (is volatile), you must do a separate Raoult’s calculation on each volatile component.
Ptotal = cAPAo + cBPBo
Ex The Po of hexane is 573 torr at 60.0°C (C6H14). The Po of benzene (C6H6) is 391 torr. What would be the vapor pressure if you mixed 58.9 grams hexane with 44.0 grams benzene?

42. Boiling Point Elevation
Boiling point of a solution is reached when the external atmospheric pressure is equal to its vapor pressure allowing a change from liquid to gas
Since the addition of a solute lowers the vapor pressure of a solution it must raise the boiling point

43. Boiling Point Elevation
DTb = Kbm
DTb = change in boiling point of the solvent
Kb = molal boiling point constant for the solvent
m = molality (mol/kg) of the solute

44. Examples
Ex What is the boiling point of a solution made by adding g of NaCl to grams of H2O at 34.0°C? Kb H2O 0.51 oC kg/mol.

45. Freezing Point Depression
Freezing points of a solvent are lowered by the addition of solutes
Can be explained in two ways
Solute particles get in the way of the solid structure being formed. If the solidification process is hindered, more heat must be removed from the solution and the freezing point is lowered.
The solution has more disorder than the pure solvent so it needs more energy removed from it in order to create the more ordered solid

46. Freezing Point Depression
Change in freezing point for non-electrolyte solutions can be summarized as
DTf = Kfm
DTf = change in freezing point of the solvent
Kf = molal freezing point constant for the solvent
m = molality of the solute

47. Examples
Ex How many grams of glycerin, C3H8O3, must be added to grams of water in order to lower the freezing point to –3.84oC? Kf H2O = 1.86 oC kg/mol.

48. Osmotic Pressure
Osmotic pressure () is the pressure needed to stop osmosis of a solvent across a semipermeable membrane. It can also be used to determine molar mass.
 = MRT
M = molarity
R = L atm/K mol (constant)
T = temperature in Kelvin

49. Osmotic Pressure
ex The osmotic pressure of a 26.5 mg/L solution of aspartame is 1.70 torr at 30 oC. What is the molar mass (formula weight) of aspartame?

50. Electrolyte Solutions and the van’t Hoff Factor
Van’t Hoff factor (i) corrects for the dissociation of ions in solution
Added to the formulas previously used
DTb = iKbm
DTf = iKfm
 = iMRT
Given as
i = actual number of particles (ions or molecules) in solution after dissociation

51. Gases
Chapter Five

52. General Properties and Kinetic Theory
Gases are made up of particles that have (relatively) large amounts of energy
A gas has no definite shape or volume and will expand to fill as much space as possible
As a result of the large amount of empty space in a volume of gas, gases are easily compressed

53. The Nature of Gases Gases expand to fill their containers
Gases are fluid – they flow
Gases have low density
1/1000 the density of the equivalent liquid or solid
Gases are compressible
Gases effuse and diffuse

54. Standard Temperature and Pressure
P = 1 atmosphere, 760 torr
T = 0°C, 273 Kelvins
The molar volume of an ideal gas is liters at STP

55. Ideal Gases
Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory.
Gases consist of tiny particles that are far apart relative to their size.
Collisions between gas particles and between particles and the walls of the container are elastic collisions
No kinetic energy is lost in elastic collisions

56. Ideal Gases
Gas particles are in constant, rapid motion. They therefore possess kinetic energy, the energy of motion.
There are no forces of attraction between gas particles
The average kinetic energy of gas particles depends on temperature, not on the identity of the particle.

57. Kinetic Molecular Theory
Particles of matter are ALWAYS in motion
Volume of individual particles is  zero.
Collisions of particles with container walls cause pressure exerted by gas.
Particles exert no forces on each other.
Average kinetic energy µ Kelvin temperature of a gas.

58. Boyle’s Law
Pressure is inversely proportional to volume when temperature is held constant.
As pressure increases, volume decreases
P1V1 = P2V2

59. Boyle’s Law - Example
In an automobile engine the gaseous fuel-air mixture enters the cylinder and is compressed by a moving piston before it is ignited. In a certain engine the initial cylinder volume is 0.725L. After the piston moves up, the volume is 0.075L. The fuel-air mixture initially has a pressure of 1.00atm. Calculate the pressure of the compressed fuel-air mixture, assuming that both the temperature and the amount of gas remain constant.

60. Charles’s Law
The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin (P = constant)
As temperature increases, volume increases
V1 = V2
T1 T2

61. Charles’s Law - example
In former times, gas volume was used as a way to measure temperature using devices called gas thermometers. Consider a gas that has a volume of 0.675L at 35oC and 1atm pressure. What is the temperature (in units of oC) of a room where this gas has a volume of 0.535L at 1atm of pressure.

62. Avogadro’s Law
For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).
As the number of moles increases, volume increases
V1 = V2
n1 n2
n = number of moles of gas

63. Avogadro’s Law - example
Suppose we have a 12.2L sample containing 0.50mol of oxygen gas at a pressure of 1atm and a temperature of 25oC. If all of this oxygen gas is converted to ozone, O3, at the same temperature and pressure, what will be the volume of the ozone formed?

64. Gay Lussac’s Law
The pressure and temperature of a gas are directly related, provided that the volume remains constant.
As temperature increases, pressure increases
P1 = P2
T1 T2

65. Gay Lussac’s Law Example
A gas at 25oC in a closed container has its pressure raised from 150. atm to 160. atm. What is the final temperature of the gas?

66. Combined Gas Law
The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.
P1V1 = P2V2
T T2
Boyle’s law, Gay-Lussac’s law, and Charles’ law are all derived from this by holding a variable constant.

67. Combined Gas Law
A sample of diborane gas, B2H6, a substance that bursts into flames when exposed to air, has a pressure of 0.454atm at a temperature of -15oC and a volume of 3.48L. If conditions are changed so that the temperature is 36oC and the pressure is 0.616atm, what will the new volume be?

68. Ideal Gas Law PV = nRT P= pressure in atm V = volume in liters
n = moles
R = proportionality constant
= L atm/ mol·K
T = temperature in Kelvins
Holds closely at P < 1 atm and T > 273 K

69. Gas Laws Example
Anesthetic gas is normally given to a patient when the room temperature is 20oC and the patient’s body temperature is 37oC. What would this temperature change do to 1.60L of gas if the pressure remains constant?

70. Gas Laws Example 2
What volume is occupied by 0.250mol of carbon dioxide gas at 25oC and 371torr?

71. Gas Density Density = mass = molar mass volume molar volume
… so at STP …
Density = molar mass
22.4 L

72. Gas Stoichiometry #1
If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios.
3 H2(g) N2(g)  NH3(g)
3 moles H mole N2  moles NH3
3 liters H liter N  liters NH3

73. Gas Stoichiometry #2
How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen?
3 H2(g) N2(g)  NH3(g)

74. Gas Stoichiometry #3
How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)

75. Gas Stoichiometry #4
How many liters of oxygen gas, at 37.0C and atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)

76. Mole Fractions
Partial Pressures are determined by the mole of each gas present.
Find the number of moles of each gas present – total moles creates total pressure
Pgas = Xgas*Ptotal
Xgas = mole gas / total moles

77. Dalton’s Law of Partial Pressure
For a mixture of gases in a container,
PTotal = P1 + P2 + P
This is particularly useful in calculating the pressure of gases collected over water.
Can also get the following equation
Ptotal = ntotal RT V

78. Real Gases (van der Waal’s equation)
Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).
[Pobs + a(n/V)2] x (V-nb) = nRT
corrected pressure corrected volume

79. Kinetic Energy of Gas Particles
At the same conditions of temperature, all gases have the same average kinetic energy.
KE = (1/2)mv2

80. The Meaning of Temperature
(KE)avg = (3/2)RT
Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

81. Root Mean Square Velocity
The average velocity (speed) of a gas particle
R = J/K.mol or kg.m2/s2
M = molar mass in kg/mol

82. Root Mean Square Velocity
Example: High above the stratosphere, the temperature of the atmosphere may reach 1000oC. However, spacecraft and astronauts do not burn up because the concentration of molecules is very low and the impacts transfer very little energy. What is the rms velocity of nitrogen molecules at that temperature?

83. Diffusion
Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

84. Effusion
Effusion: describes the passage of gas into an evacuated chamber.

85. Graham’s Law of Effusion