GENERAL CHEMISTRY 2 REVIEWER – 3RD QUARTER
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The kinetic molecular theory of matter states that:
o All matter is made up of very small particles that are constantly moving.
o All particles have energy, but the energy varies depending on the temperature the sample of
matter is in. This in turn determines whether the substance exists in the solid, liquid, or gaseous
state. Molecules in the solid phase have the least amount of energy, while gas particles have
the greatest amount of energy.
o The temperature of a substance is a measure of the average kinetic energy of the particles.
o A change in phase may occur when the energy of the particles is changed.
o There are spaces between particles of matter. The average amount of empty space between
molecules gets progressively larger as a sample of matter moves from the solid to the liquid and
gas phases.
o There are attractive forces between atoms/molecules, and these become stronger as the
particles move closer together. These attractive forces are called intermolecular forces.
Intermolecular forces
Intermolecular forces are forces that exist between molecules. Intramolecular forces are the forces that hold
atoms together within a molecule.
Hydrogen Bond (Strongest)
The hydrogen bond is a special form of theDipole-Dipole force, it’s not really a true chemical
bond.
o However, it can ONLY exist when a Hydrogen atom is present in the molecule and it’s bonded to
either a N, O, or F atom also present in the same molecule.
o H-bonding is especially strong in biological systems — such as DNA.
o H2O, HF, NH3, CH3OH,
o Dipole-dipole interaction (Less Strong)
Occurs between polar covalent molecules
A polar molecule has a partial positive charge on one end and a partial negative charge on the
other, forming a dipole (having two electrically charged poles)
o PCl3, CO, CHCl3, HCl, HI, PH3, H2S
o Dispersion Forces (Weakest)
o Sometimes called LONDON Dispersion Forces or induced dipole forces
o This force is the predominant intermolecular force for non-polar compounds
o CO2, Cl2, F2, CCl4, CH4
Properties of Liquids
o Liquids are much denser than gases because of the stronger intermolecular forces holding the
particles together.
o Surface tension - the amount of energy required to stretch or increase the surface of a liquid by
a unit area.
o Viscosity - a measure of a fluid’s resistance to flow.
o Vapor pressure - The equilibrium vapor pressure is the vapor pressure measured when a dynamic
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equilibrium exists between condensation and evaporation
Cohesion - the intermolecular attraction between like molecules(Water molecules are attracted
to themselves)
o Adhesion - an attraction between unlike molecules (ability to “climb” structures)
Properties of Solids
o They are incompressible and rigid.
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They have definite characteristic geometrical shape.
Types of Solids
 Crystalline Solids
 Usually consists of a large number of small crystals, each of them having a definite
characteristic geometrical shape.
 Quartz, sodium chloride
 Amorphous Solids
 Consist of particlesof irregular shape.
 Glass, rubber, plastics
Classification of Crystalline Solids
 Molecular Solids - Molecules are the constituent particles of molecular solids.
 Non polar Molecular Solids
 Polar Molecular Solids
Hydrogen Bonded Molecular Solids
 Ionic Solids - Ions are the constituent particles of ionic solids. Such solids are formed by
the three dimensional arrangements of cations and anions bound by strong coulombic
(electrostatic) forces.
 Metallic Solids–They readily conduct electricity. It has luster and is easily deformed. This is
also due to the presence of free electrons in them.
 Network Solids - A wide variety of crystalline solids of non-metals result from the
formation of covalent bonds between adjacent atoms throughout the crystal.Covalent
bonds are strong and directional in nature, therefore atoms are held very strongly at
their positions. E.g. Silica and diamonds
Phase Changes
o Is a change from one state of matter (solid, liquid, gas) to another.
 Vaporization(Boiling) - liquid to gas, adding energy
 Sublimation – solid to gas, adding energy
 Melting(Fusion) – solid to liquid, adding energy
 Freezing(Solidification) – liquid to solid, removing energy
 Condensation – gas to liquid, removing energy
 Deposition - gas to solid, removing energy
o Adding or removing energy (heat) to a substance causes phase changes
o During a phase change, temperature does NOT change
o Molar heat of fusion - The amount of energy absorbed by one mole of a substance to change
from solid to liquid
o Molar heat of vaporization – The amount of energy absorbed by one mole of a substance to change
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from liquid to gas
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Heating Curve
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Cooling Curve
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Phase Diagrams
o It is a graph of pressure versus temperature that shows in which phase a substance will exist under
different conditions of temperature and pressure.
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Solution
o Homogeneous mixtures of two or more substances(variable compositions).
o Has the same uniform appearance and composition throughout; maintain one phase (solid, liquid, gas)
o Can be physically separated
o Composed of solutes(the substance in the smallest amount and the one that dissolves in the solvent) and
solvents(the substance in the larger amount that dissolves the solute)
o Examples: salt water, leaded gasoline, soda, brass, oxygen dissolved in water
o Non-examples: salad, oil, water, sand/soil(sand dissolved in water), mercury, calcium carbonate
The Process of Dissolving
o Intermolecular forces between solute particles need to be broken—true for both ionic and molecular
solutes. This is endothermic.
o Some intermolecular forces in solvent (liquid) also need to be broken. This is endothermic.
o Attraction between solute and solvent particles.The temperature of the solution increases when there is
dissolution. This is exothermic.
Solubility
o Many factors affect solubility, such as nature of the solute and solvent, temperature and, in some cases,
pressure.
o For most solids dissolved in liquids, solubility increases as temperature increases.
o For gases dissolved in liquids, as temperature increases, solubility decreases.
o The solubility of solutes is very temperature dependent.
Types of Solution
o Unsaturated – more solute dissolves
o Saturated – no more solute dissolves
o Supersaturated –becomes unstable, crystals form
o Comparison between Types of Solution
 A supersaturated solution is more concentrated than a saturated solution.
 An unsaturated solution is more dilute than a supersaturated solution.
 Asupersaturated solution is a very unstable solution.
 A saturated solution need not be a concentrated solution.
Concentration of Solutions
o Concentration - refers to how much solute is dissolved in the solvent
 Concentrated: solutions with a large amount of solute dissolved in solvent
 Dilute: solutions with a small amount of solute dissolved the solvent
Methods of Expressing Concentration of Solutions
o Percent by Mass
 Mass percentage is expressed using the equation:
 % Mass of solute = Mass of solute x100
Mass of solution
 Sample Problem:
 What is the percent by mass when 40 g of salt is added to 200 g of water?
 %mass of solute = 40g
= 16.67%
x 100
40g + 200g
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Percent by Volume
 % Volume of solute = Volume of solute x 100
Volume of solution
 Sample Problem:
 In a solution, there is 111.0 mL solvent and 5.24 mL solute present in a solution. Find
volume percent of the solute.
 %Volume of solute = = 5.24mL
= 4.51%
x
100
5.24mL + 111mL
 Proof number
o Ex. A liquor that is 100 proof is 50% ethyl alcohol by volume
o Mole Fraction
 Mole fraction = number of moles of solute
total number of moles in solution
or
 Mole fraction =number of moles of solvent
total number of moles in solution
 Sample Problem:
 What is the mole fraction of carbon tetrachloride (CCl4) in solution if 3.47 moles of CCl4 is
dissolved in 8.54 moles of benzene (C6H6)?
 Mole fraction =
3.47 mol
= 0.29
3.47 mol + 8.54 mol
o Molarity
 Molarity tells us the number of moles of solute in exactly one liter of a solution.
 Molarity=
moles of solute
Volume of solution in liters
 Sample Problem:
 What is the molarity when 0.75mol is dissolved in 2.50 L of solution?
o 0.3 M
o Molality
 Molality, m, tells us the number of moles of solute dissolved in exactly one kilogram of solvent.
 Molality = moles of solute
kilogram of solvent
 Sample Problem:
 Calculate the molality of 0.210 mol of KBr dissolved in0.750 kg pure water.
 Molality = 0.210 mol = 0.28 m
0.75 kg
Colligative Properties of Solution
o Depend only on the number of solute particles present, not on the identity of the solute particles.
 Vapor pressure lowering - As solute molecules are added to a solution, the solvent become less
volatile (=decreased vapor pressure).
 Boiling point elevation - The boiling point of a solution containing a nonvolatile solute is higher
than that of the pure solvent.
 Tb = Kb m
 Tb = normal boiling point of the solvent + Tb
 Sample Problem:
 Eugenol is the active ingredient in the oil of cloves used for relieving toothache. Calculate
the boiling point of a solution in which 0.175 g of eugenol, C10H12O2, is dissolved in 10.0
g of benzene. Kb = 2.53°C/m
 0.175 g of C10H12O2 x 1 mol C10H12O2 = 0.001067 molC10H12O2
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164 g C10H12O2
Molality = 0.001067 molC10H12O2 = 0.1067 m
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0.01 kg benzene
Tb = Kb m
 Tb= 2.53°C/m0.1067 m = 0.269°C
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Tb = normal boiling point of the solvent + Tb
 Tb =80.1°C + 0.269°C = 80.37°C
Freezing point depression - Solute-solvent interactions also cause solutions to have lower
freezing points than the pure solvent.
Tf = Kf m
Tf (solution)= Tf (solvent) –Tf
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Osmotic pressure - Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one. A semipermeable membrane
allows the passage of solvent molecules but blocks the passage of solute molecules.
o A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct
electricity.
 Sugar Solution
o An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct
electricity.
 Salt Solution
Thermochemistry
o Energy Changes in Chemical Reactions
 Some reactions cause the temperature of the reaction mixture to increase. This type of reaction
is called exothermic. BOND FORMING
 Heat energy is given out by the reaction mixture hence the surroundings increase in
temperature.
 Freezing
 Some reactions cause the temperature of the reaction mixture to decrease. This type of reaction
is called endothermic. BOND BREAKING
 Heat energy is taken in by the reaction mixture hence the surroundings decrease in
temperature.
o Systems
 An open system is a system that freely exchanges energy and matter with its surroundings. For
instance, when you are boiling soup in an open saucepan on a stove, energy and matter are being
transferred to the surroundings through steam. The saucepan is an open system because it
allows for the transfer of matter (for example adding spices in the saucepan) and for the transfer
of energy (for example heating the saucepan and allowing steam to leave the saucepan).
 A closed system is a system that exchanges only energy with its surroundings, not matter. By
putting a lid on the saucepan, matter can no longer transfer because the lid prevents matter from
entering the saucepan and leaving the saucepan. Still, the saucepan allows energy transfer.
Imagine putting the saucepan on a stove and heating it. The saucepan allows energy transfer as
the saucepan heats up and heats the contents inside it.
 A thermos is used to keep things either cold or hot. Thus, a thermos does not allow for energy
transfer. Additionally, the thermos, like any other closed container, does not allow matter
transfer because it has a lid that does not allow anything to enter or leave the container. As a
result, the thermos is what we call an isolated system. An isolated system does not exchange
energy or matter with its surroundings.
o First Law of Thermodynamics
 Itstates that the total energy absorbed by the system is equal to the energy lost by the
surroundings.
 ∆U = Q + W
 Where,
 ∆U, change in internal energy
 Q, represents the heat added to the gas
 W, how much work was done
 The change in internal energy is going to equal the amount of heat that’s added to the gas
 Sample Problem:
o 60J of work is done on a gas, and the gas loses 150J of heat to its surroundings.
What is the change in internal energy?
 ∆U = Q + W
 ∆U = -150J + 60J = -90J