Chemistry Final Exam Review

States of Matter and Phase Transitions

  • Solid Phase Characteristics:

    • Arrangement: Particles are close packed.

    • Description: Possesses a definite shape.

    • Kinetic Energy (KE): Low KE.

  • Liquid Phase Characteristics:

    • Arrangement: Particles are less close than in a solid.

    • Description: Takes the form of its container.

    • Kinetic Energy (KE): Higher KE than solids.

  • Gas Phase Characteristics:

    • Arrangement: Particles are very far apart.

    • Description: Expands to fill the container.

    • Kinetic Energy (KE): Highest KE.

  • Phase Change Definitions:

    • Evaporation (or Vaporization): The transition from the liquid phase to the gas phase.

    • Condensation: The transition from the gas phase to the liquid phase.

    • Sublimation: When a solid becomes a gas without passing through the liquid phase.

    • Deposition: The transition from a gas directly to a solid.

    • Melting (Fusion): The transition from solid to liquid.

    • Solidification (Freezing): The transition from liquid to solid.

  • Phase Diagrams:

    • Triple Point: The specific temperature and pressure at which the solid, liquid, and gas phases all exist at equilibrium.

    • Boundary Lines:

      • The line separating solid and gas represents sublimation and deposition.

      • The line separating solid and liquid represents melting and solidification.

      • The line separating liquid and gas represents vaporization and condensation.

Gas Laws and Kinetic Molecular Theory

  • Kinetic Molecular Theory (KMT) - Ideal vs. Real Gases:

    • Ideal Gas: Modeled as small hard spheres with no volume. There are no forces of attraction or repulsion between particles. Collisions are perfectly elastic. Particles are in constant random motion. All gas laws are based on these ideal assumptions.

    • Real Gas: Particles do have actual volume and exert forces of attraction and repulsion on each other.

    • Departure from Ideality: Real gases behave like ideal gases except under conditions of very high pressure (where particles are pushed close together) or very low temperatures (approaching the condensation point where gases become liquid).

  • Avogadro’s Hypothesis: Gases of the same volume at the same temperature (TT) and pressure (PP) must have the same number of particles, regardless of the gas identity.

  • Kinetic Energy and Velocity:

    • Formula: KE=12mv2KE = \frac{1}{2}mv^2.

    • If temperature is the same, average KE is the same for all gases.

    • Lower mass particles must have a higher velocity to maintain the same KE. Example: Between Neon, Hydrogen, Nitrogen, Methane (CH4CH_4), and Ammonia (NH3NH_3) at STP, Hydrogen has the greatest velocity because it has the smallest molar mass.

  • Gas Law Relationships:

    • Pressure (PP) and Volume (VV): Inversely proportional (when TT and nn are constant). If PP \uparrow, then VV \downarrow.

    • Pressure (PP) and Number of Molecules (nn): Directly proportional (when VV and TT are constant). If PP \downarrow, then nn \downarrow.

    • Volume (VV) and Temperature (TT): Directly proportional (when PP and nn are constant). If VV \uparrow, then TT \uparrow.

    • Pressure (PP) and Temperature (TT): Directly proportional (when VV and nn are constant). If PP \downarrow, then TT \downarrow.

    • Volume (VV) and Number of Molecules (nn): Directly proportional (when PP and TT are constant). If VV \uparrow, then nn \uparrow.

  • Mathematical Gas Law Formulas:

    • Boyle’s Law: P1V1=P2V2P_1V_1 = P_2V_2

    • Charles’ Law: V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}

    • Gay-Lussac’s Law: P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}

    • Combined Gas Law: P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

    • Ideal Gas Law: PV=nRTPV = nRT

Matter Classification and Atomic Structure

  • Classifying Matter:

    • Element: A pure substance consisting of one type of atom (e.g., Titanium).

    • Compound: Two or more elements chemically combined (e.g., carbon dioxide, water, methane).

    • Homogeneous Mixture: Multiple substances in the same state, uniformly distributed (e.g., Steel/alloy, salt water, air).

    • Heterogeneous Mixture: Multiple substances with different states or non-uniform distribution (e.g., chunky vegetable soup).

  • Atomic Subatomic Particles:

    • Protons: Positive charge; define the atomic number.

    • Electrons: Negative charge; equal to protons in neutral atoms. Found in orbitals outside the nucleus.

    • Neutrons: No charge; contribute to the mass number (Mass Number=Protons+Neutrons\text{Mass Number} = \text{Protons} + \text{Neutrons}).

    • Example (Lithium): 3 protons, 3 electrons, 4 neutrons (Mass Number=7\text{Mass Number} = 7, Atomic Number=3\text{Atomic Number} = 3).

  • Standard Notation:

    • Calcium-40: 2040Ca^{40}_{20}\text{Ca}, where 40 is mass and 20 is atomic number. It has 20 neutrons (4020=2040 - 20 = 20).

  • Isotopes:

    • Definition: Atoms of the same element with the same number of protons but different numbers of neutrons (e.g., Sulfur-32, Sulfur-33, Sulfur-34).

    • Chemical Identity: Isotopes are chemically alike because they have the same number of electrons (especially valence electrons), which determine chemical properties.

    • Atomic Mass Calculation: Calculated as the weighted average of isotopic masses. For Bromine (79Br^{79}\text{Br} at 50.69%, 81Br^{81}\text{Br} at 49.31%): (78.92×0.5069)+(80.92×0.4931)=79.9g/mol(78.92 \times 0.5069) + (80.92 \times 0.4931) = 79.9\,g/mol.

The Periodic Table and Trends

  • Elemental Classifications:

    • Alkali Metals: Group 1A (e.g., Potassium).

    • Alkaline Earth Metals: Group 2A (e.g., Barium, Magnesium). They lose 2 electrons to form +2+2 ions.

    • Transition Metals: B-group elements (e.g., Chromium, Tungsten).

    • Halogens: Group 7A (e.g., Fluorine, Chlorine).

    • Noble Gases: Group 8A (e.g., Neon, Argon). Inert with full valence shells.

    • Metalloids: Share properties of metals and non-metals (e.g., Arsenic, Antimony, Germanium).

  • Periodic Trends:

    • Atomic Radius:

      • Across a Period: Decreases from left to right. Electrons are added to the same main energy level but are pulled closer by a more positive/stronger nucleus.

      • Down a Group: Increases from top to bottom as new energy levels (orbitals) are added.

      • Ions: Cations (loss of electrons) are smaller than their parent atoms because an entire orbital may be lost. Anions (gain of electrons) are larger because increased electron repulsion expands the cloud.

    • Ionization Energy: The energy required to remove an electron.

      • Decreases down a group (electrons are further from the nucleus).

      • Increases across a period (stronger nuclear charge holds electrons tighter).

      • Alkali metals have a massive jump between 1st and 2nd ionization energies because the second electron must come from a lower, stable energy level.

    • Electronegativity: The ability of an atom to attract electrons in a bond.

      • Increases across a period and decreases down a group.

      • Fluorine (FF) is the most electronegative. Noble gases have no electronegativity values.

Electron Configurations

  • Principles:

    • Aufbau Principle: Electrons fill the lowest energy orbitals first.

    • Pauli Exclusion Principle: An orbital holds a maximum of 2 electrons with opposite spins.

    • Notation Details: In 3d53d^5, "3" is the principal energy level, "d" is the orbital shape, and "5" is the number of electrons.

  • Examples:

    • Fluorine (Z=9Z=9): 1s22s22p51s^2 2s^2 2p^5

    • Calcium (Z=20Z=20): 1s22s22p63s23p64s21s^2 2s^2 2p^6 3s^2 3p^6 4s^2

    • Silicon (Z=14Z=14): [Ne]3s23p2[Ne] 3s^2 3p^2

    • Rubidium (Z=37Z=37): [Kr]5s1[Kr] 5s^1

Chemical Bonding and Nomenclature

  • Comparison of Ionic and Molecular Compounds:

    • Components: Ionic (Metal + Non-metal); Molecular (2+ Non-metals).

    • Forces: Ionic (Attraction between oppositely charged ions); Molecular (Sharing electrons/covalent bonds).

    • Representative Particle: Ionic (Formula unit); Molecular (Molecule).

    • Physical Properties: Ionic (High melting point, dissolves and dissociates in water); Molecular (Low melting point, may dissolve but does not dissociate into ions).

  • Bonding Types:

    • Ionic Bond: Complete "give and take" transfer of electrons.

    • Polar Covalent Bond: Uneven sharing of electrons between atoms within a molecule (e.g., HH and OO in water).

    • Non-polar Covalent Bond: Relatively even sharing of electrons.

    • Hydrogen Bond: A weak intermolecular force of attraction between the dipoles of polar molecules (e.g., between different water molecules).

  • Nomenclature Examples:

    • Cl2O5Cl_2O_5: Dichloride pentoxide (M).

    • H2SO4H_2SO_4: Sulfuric acid (A).

    • Ba3(PO4)2Ba_3(PO_4)_2: Barium phosphate (I).

    • CuSCuS: Copper (II) sulfide (I).

    • HNO2HNO_2: Nitrous acid (A).

    • NO3NO_3^-: Nitrate ion (ion).

    • N3N^{3-}: Nitride ion (ion).

Chemical Reactions

  • Reaction Types:

    • Combination (Synthesis): 4Al+3O22Al2O34Al + 3O_2 \rightarrow 2Al_2O_3

    • Decomposition: 2H2O22H2O+O22H_2O_2 \rightarrow 2H_2O + O_2

    • Single Replacement: Zn+Cu(NO3)2Zn(NO3)2+CuZn + Cu(NO_3)_2 \rightarrow Zn(NO_3)_2 + Cu

    • Double Replacement: Pb(NO3)2+2KI2KNO3+PbI2Pb(NO_3)_2 + 2KI \rightarrow 2KNO_3 + PbI_2

    • Combustion: CH4+2O2CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O

    • Acid-Base Neutralization: 2KOH+H2SO4K2SO4+2H2O2KOH + H_2SO_4 \rightarrow K_2SO_4 + 2H_2O

  • Reaction Predictability:

    • Activity Series: Iron (FeFe) + Magnesium Chloride (MgCl2MgCl_2) yields NO REACTION because FeFe is less reactive than MgMg.

    • Solubility Rules: In the reaction of Ammonium phosphate and Barium nitrate, Barium phosphate (Ba3(PO4)2Ba_3(PO_4)_2) forms a solid precipitate. The spectator ions are NO3NO_3^- and NH4+NH_4^+.

Stoichiometry

  • Molar Calculations:

    • 1 mole = 6.02×10236.02 \times 10^{23} particles = Molar Mass (grams) = 22.4L22.4\,L (for a gas @ STP).

    • Molar Mass Example: Mg3(PO4)2=262.9g/molMg_3(PO_4)_2 = 262.9\,g/mol.

  • Stoichiometry Problem Steps:

    1. Convert given units to moles.

    2. Use the mole ratio from the balanced equation.

    3. Convert moles to the desired unit.

    • Example: Producing NH3NH_3 from 10.5g H2H_2: 10.5gH2×(1molH22.01g)×(2molNH33molH2)×(17.03gNH31molNH3)=59.3gNH310.5\,g\,H_2 \times (\frac{1\,mol\,H_2}{2.01\,g}) \times (\frac{2\,mol\,NH_3}{3\,mol\,H_2}) \times (\frac{17.03\,g\,NH_3}{1\,mol\,NH_3}) = 59.3\,g\,NH_3.

  • Yields:

    • Theoretical Yield: The maximum amount of product that could be formed from given reactants.

    • Percent Yield: (Actual YieldTheoretical Yield)×100(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}) \times 100.

Thermochemistry

  • Specific Heat Calculations:

    • Formula: q=mCΔTq = mC\Delta T

    • qq = heat (Joules), mm = mass (grams), CC = specific heat capacity (J/gCJ/g^\circ C), ΔT\Delta T = change in temperature (TfinalTinitialT_{final} - T_{initial}).

    • Heating 100 mL of water from 15.0C15.0^\circ C to 80C80^\circ C (C=4.18J/gCC = 4.18\,J/g^\circ C): q=100g×4.18×65=27,170Jq = 100\,g \times 4.18 \times 65 = 27,170\,J.

  • Endothermic vs. Exothermic:

    • Endothermic (+ΔH+\Delta H): Heat is absorbed from surroundings; products have higher potential energy; biological/chemical cold packs; melting; boiling.

    • Exothermic (ΔH-\Delta H): Heat is released to surroundings; products have lower potential energy; rusting; freezing; condensation; combustion.

  • Calorimetry:

    • qreaction=qsurroundingsq_{reaction} = -q_{surroundings}.

    • Enthalpy (ΔHrxn\Delta H_{rxn}) is often expressed in kJ/molkJ/mol.

  • Hess's Law (Example - Formation of Ethane):

    • Target Reaction: 2C(s)+3H2(g)C2H6(g)2C(s) + 3H_2(g) \rightarrow C_2H_6(g).

    • Step 1: Reverse combustion of ethane: 2CO2(g)+3H2O(l)C2H6(g)+72O2(g)2CO_2(g) + 3H_2O(l) \rightarrow C_2H_6(g) + \frac{7}{2}O_2(g), ΔH=+1560.7kJ\Delta H = +1560.7\,kJ.

    • Step 2: Multiply Carbon combustion by 2: 2C(s)+2O2(g)2CO2(g)2C(s) + 2O_2(g) \rightarrow 2CO_2(g), ΔH=787.0kJ\Delta H = -787.0\,kJ.

    • Step 3: Multiply Hydrogen combustion by 3: 3H2(g)+32O2(g)3H2O(l)3H_2(g) + \frac{3}{2}O_2(g) \rightarrow 3H_2O(l), ΔH=857.4kJ\Delta H = -857.4\,kJ.

    • Summing Enthalpies: ΔHtotal=1560.7787.0857.4=83.7kJ\Delta H_{total} = 1560.7 - 787.0 - 857.4 = -83.7\,kJ.