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Honors Chemistry Final Exam Review: States of Matter, Solutions, Thermochemistry, Reaction Rates, Equilibrium, Acids & Bases, Nuclear Chemistry, Redox Reactions

States of Matter

  • Endothermic Processes:
    • Melting: Solid to liquid.
    • Boiling: Liquid to gas.
    • Evaporation: Liquid to gas (occurs at the surface).
    • Sublimation: Solid directly to gas.
  • Exothermic Processes:
    • Freezing: Liquid to solid.
    • Condensation: Gas to liquid.
    • Deposition: Gas directly to solid.
  • Helium Balloon in Cold Weather:
    • The balloon will shrink because the decrease in temperature reduces the kinetic energy of the gas particles inside the balloon. According to the ideal gas law, as temperature decreases, volume decreases if pressure is constant.
  • Sealed Syringe in Hot Water:
    • The syringe piston will slide outwards. This is because increasing the water temperature increases the temperature of the air inside the syringe, leading to a higher kinetic energy of the air particles. According to the ideal gas law, if the amount of gas and pressure are kept constant, the volume increases with temperature.
  • Effusion of Gases:
    • Hydrogen (H2) would effuse fastest. According to Graham's law of effusion, the rate of effusion is inversely proportional to the square root of the molar mass of the gas. The gas with the lowest molar mass effuses fastest.
  • Gas Temperature, Volume, and Pressure Relationship:
    • If the temperature of a gas is quadrupled (multiplied by 4) and the volume is doubled (multiplied by 2), the new pressure will be twice the original pressure. This can be derived from the combined gas law: \frac{P1V1}{T1} = \frac{P2V2}{T2}. If T2 = 4T1 and V2 = 2V1, then P2 = 2P1.
  • Ideal Gas Law Calculation:
    • To find the temperature of 10.0 g of hydrogen gas at a pressure of 633 kPa and a volume of 0.020 kL, use the ideal gas law: PV = nRT.
    • Convert pressure to atm: 633 \text{ kPa} = 6.25 \text{ atm}
    • Convert volume to L: 0.020 \text{ kL} = 20.0 \text{ L}
    • Calculate moles of H2: n = \frac{10.0 \text{ g}}{2.016 \text{ g/mol}} = 4.96 \text{ mol}
    • Solve for T: T = \frac{PV}{nR} = \frac{(6.25 \text{ atm})(20.0 \text{ L})}{(4.96 \text{ mol})(0.0821 \text{ L atm / (mol K)})} = 307 \text{ K}
  • Gas at Constant Volume:
    • To find the temperature at which a gas occupies the same volume but has a different pressure, use the formula: \frac{P1}{T1} = \frac{P2}{T2}.
    • Given: P1 = 106.3 \text{ kPa}, T1 = 25 + 273.15 = 298.15 \text{ K}, P_2 = 100.0 \text{ kPa}
    • T2 = \frac{P2T1}{P1} = \frac{(100.0 \text{ kPa})(298.15 \text{ K})}{106.3 \text{ kPa}} = 280.5 \text{ K}

Solutions

  • Types of Solutions:
    • Saturated: On the line of a solubility curve; contains the maximum amount of solute that can dissolve at a given temperature.
    • Unsaturated: Below the line on a solubility curve; contains less solute than the maximum amount that can dissolve at a given temperature.
    • Supersaturated: Contains more solute than the maximum amount that can dissolve at a given temperature; unstable, and excess solute may precipitate out.
    • Undissolved crystals may appear at the bottom of a beaker for saturated or supersaturated solutions.
    • A solid crystal may come out of solution when agitated in a supersaturated solution.
  • Solubility Curve:
    • To find the solubility of calcium chloride (CaCl2) at 5°C, read the value directly from the solubility curve at 5°C. Actual value depends on the specific curve provided.
    • If 100 g of water contains 30 g of KCl, determine how many more grams of KCl could be dissolved at 40°C to make a saturated solution. Read the solubility of KCl at 40°C from the curve, and subtract the current amount (30 g) from the saturated amount.
    • At 90°C, 10 g of potassium chlorate (KClO3) is dissolved in 100 g of water. To determine if this solution is saturated, unsaturated, or supersaturated, compare the amount dissolved (10 g) with the solubility of KClO3 at 90°C from the curve.
  • Molarity Calculation:
    • Molarity (M) is defined as moles of solute per liter of solution. To find the molarity of a solution containing 212.5 g of sodium nitrate in 3.0 L of solution:
      • Convert mass of sodium nitrate to moles: n = \frac{212.5 \text{ g}}{84.99 \text{ g/mol}} = 2.5 \text{ mol}
      • Calculate molarity: M = \frac{2.5 \text{ mol}}{3.0 \text{ L}} = 0.833 \text{ M}
  • Dilution Calculation:
    • To prepare 300.0 mL of 0.750 M NaBr using a 2.00 M NaBr stock solution, use the dilution formula:M1V1 = M2V2
      • (2.00 \text{ M})V_1 = (0.750 \text{ M})(300.0 \text{ mL})
      • V_1 = \frac{(0.750 \text{ M})(300.0 \text{ mL})}{2.00 \text{ M}} = 112.5 \text{ mL}
  • Percent Volume (% v/v):
    • % v/v is defined as (volume of solute / volume of solution) x 100.
      • \frac{60.0 \text{ mL}}{500.0 \text{ mL}} \times 100 = 12.0 \%
  • Percent Mass/Volume (% m/v):
    • % m/v is defined as (mass of solute / volume of solution) x 100.
      • 15 \% = \frac{\text{mass of NaCl}}{2.25 \text{ L}} \times 100
      • Convert volume to mL: 2.25 \text{ L} = 2250 \text{ mL}
      • \text{mass of NaCl} = \frac{15 \times 2250}{100} = 337.5 \text{ g}

Colligative Properties

  • Particles in Solution:
    • Potassium carbonate (K2CO3) produces three particles in solution: 2 K+ ions and 1 CO32- ion.
  • Freezing Point Depression:
    • CaCl2 will lower the freezing point more because it produces three ions (1 Ca2+ and 2 Cl-) in solution per formula unit, while NaCl produces two ions (1 Na+ and 1 Cl-). The more particles in solution, the greater the effect on colligative properties like freezing point depression.
  • Molality Calculation:
    • Molality (m) is defined as moles of solute per kilogram of solvent.
      • Convert mass of NaCl to moles: n = \frac{\text{mass}}{58.44 \text{ g/mol}}
      • 0.50 \text{ molal} = \frac{n}{0.750 \text{ kg}}
      • n = 0.50 \times 0.750 = 0.375 \text{ mol}
      • \text{mass of NaCl} = 0.375 \text{ mol} \times 58.44 \text{ g/mol} = 21.9 \text{ g}
  • Boiling Point Elevation:
    • The boiling point elevation is calculated using the formula: \Delta Tb = iKbm, where i is the van't Hoff factor, Kb is the ebullioscopic constant, and m is the molality.
      • For NaCl, i = 2. \Delta T_b = 2 \times 0.512 \times 2.00 = 2.044 ^\circ \text{C}
      • Boiling point = 100 + 2.044 = 102.044 ^\circ \text{C}
  • Freezing Point Depression:
    • The freezing point depression is calculated using the formula: \Delta Tf = iKfm, where i is the van't Hoff factor, Kf is the cryoscopic constant, and m is the molality.
      • Find the molality of the solution: Convert mass of NaCl to moles: n = \frac{25.0 \text{ g}}{58.44 \text{ g/mol}} = 0.428 \text{ mol}
      • Convert volume of water to mass: 900.0 mL of water = 900.0 g = 0.900 kg
      • Calculate molality: m = \frac{0.428 \text{ mol}}{0.900 \text{ kg}} = 0.476 \text{ m}
      • \Delta T_f = 2 \times 1.86 \times 0.476 = 1.77 ^\circ \text{C}
      • Freezing point = 0 - 1.77 = -1.77 ^\circ \text{C}
  • Nonpolar Molecules in Water:
    • Nonpolar molecules do not dissolve in water because water is a polar solvent. Water molecules are attracted to each other through hydrogen bonds due to their polarity. Nonpolar molecules cannot form strong intermolecular interactions with water, so they do not dissolve.
  • Conductivity of Solutions:
    • Nonelectrolyte: The light bulb would not light up because the solution does not conduct electricity.
    • Weak electrolyte: The light bulb would light up dimly because the solution partially ionizes and conducts electricity weakly.
    • Strong electrolyte: The light bulb would light up brightly because the solution fully ionizes and conducts electricity strongly.

Thermochemistry

  • Enthalpy Curve:
    • Sketch a curve showing the enthalpy changes during phase transitions. The curve should increase as it moves from solid to liquid (melting), then liquid to gas (boiling/evaporation). Freezing and condensation would be represented moving down the curve.
  • Energy and Phase Changes:
    • Moving up the curve (solid to liquid to gas), energy is absorbed (endothermic).
    • Moving down the curve (gas to liquid to solid), energy is released (exothermic).
  • Endothermic Phase Changes:
    • Melting (solid to liquid).
    • Boiling/Evaporation (liquid to gas).
    • Sublimation (solid to gas).
  • Energy Released by Iron:
    • To calculate the energy released, use the formula q = mc\Delta T, where m is mass, c is specific heat capacity, and \Delta T is the change in temperature.
      • The specific heat capacity of iron is approximately 0.450 \text{ J/g}^\circ \text{C}.
      • \Delta T = 25.3 - 200.0 = -174.7 ^\circ \text{C}
      • q = (34.12 \text{ g})(0.450 \text{ J/g}^\circ \text{C})(-174.7 ^\circ \text{C}) = -2678 \text{ J} = -2.678 \text{ kJ}
  • Energy to Convert Ice to Steam:
    • This involves several steps:
      • Heating ice from -45.9°C to 0°C: q_1 = mc\Delta T (using specific heat of ice, which is approx. 2.09 \text{ J/g}^\circ \text{C}).
      • Melting ice at 0°C: q2 = m \Delta H{fus}, where \Delta H_{fus} is the heat of fusion of water (334 J/g).
      • Heating water from 0°C to 100°C: q_3 = mc\Delta T (using specific heat of water, which is approx. 4.186 \text{ J/g}^\circ \text{C}).
      • Evaporating water at 100°C: q4 = m \Delta H{vap}, where \Delta H_{vap} is the heat of vaporization of water (2260 J/g).
      • Total energy = q1 + q2 + q3 + q4.
      • q_1 = (15.0 \text{ g})(2.09 \text{ J/g}^\circ \text{C})(0 - (-45.9)) ^\circ \text{C} = 1436 \text{ J}
      • q_2 = (15.0 \text{ g})(334 \text{ J/g}) = 5010 \text{ J}
      • q_3 = (15.0 \text{ g})(4.186 \text{ J/g}^\circ \text{C})(100 - 0) ^\circ \text{C} = 6279 \text{ J}
      • q_4 = (15.0 \text{ g})(2260 \text{ J/g}) = 33900 \text{ J}
      • q_{\text{total}} = 1436 + 5010 + 6279 + 33900 = 46625 \text{ J} = 46.625 \text{ kJ}
  • Heat Evolved in Reaction:
    • First, find the number of moles of SO2: n = \frac{87.9 \text{ g}}{64.07 \text{ g/mol}} = 1.372 \text{ mol}
    • From the balanced equation, 2 moles of SO2 produce -198.2 kJ of heat.
    • q = 1.372 \text{ mol} \times \frac{-198.2 \text{ kJ}}{2 \text{ mol}} = -135.9 \text{ kJ}

Reaction Rates & Equilibrium

  • Ways to Increase Reaction Rate:
    • Increase the concentration of reactants.
    • Increase the temperature.
    • Add a catalyst.
    • Increase the surface area of solid reactants.
    • Increase pressure (for gaseous reactants).
  • Equilibrium Expression:
    • For the reaction 2 N2O5 (g) \leftrightarrow 4 NO2 (g) + O2 (g), the equilibrium expression is: K{eq} = \frac{[NO2]^4[O2]}{[N2O_5]^2}
  • Equilibrium Constant Calculation:
    • Given [NO2] = 0.80 M, [N2O5] = 0.50 M, [O2] = 0.20 M:
      • K_{eq} = \frac{(0.80)^4(0.20)}{(0.50)^2} = \frac{0.08192}{0.25} = 0.328
  • Reactant or Product Favoritism:
    • Since Keq < 1, reactants are favored.
  • Le Chatelier's Principle:
    • Addition of oxygen to the vessel: The equilibrium will shift to the left, favoring the reverse reaction to consume the added oxygen and re-establish equilibrium.
    • Decrease in pressure: The equilibrium will shift to the side with more moles of gas. In this reaction, there are 2 moles of gas on the reactant side and 5 moles of gas on the product side. Therefore, the equilibrium will shift to the right (favoring products).
  • Rate Law:
    • For the reaction A + B ↔ C + D, where the rate is zero order with respect to A and second order with respect to B, the rate law is: Rate = k[B]^2.
    • If the concentration of A is quadrupled, the rate remains unchanged because the reaction is zero order with respect to A.
    • If the concentration of B is tripled, the rate increases by a factor of 9 because the reaction is second order with respect to B. (3)^2=9

Acids & Bases

  • Nomenclature:
    • H2CO3 (aq): Carbonic acid
    • H2S (aq): Hydrosulfuric acid
  • Formulas:
    • Ammonia: NH3
    • Nitrous acid: HNO2
  • pH Calculations:
    • a. [H+] = 4.4 x 10-11 M: pH = -log[H+] = -log(4.4 x 10-11) = 10.36
    • b. [OH-] = 2.2 x 10-7: pOH = -log[OH-] = -log(2.2 x 10-7) = 6.66; pH = 14 - pOH = 14 - 6.66 = 7.34
    • c. pOH = 1.4: pH = 14 - pOH = 14 - 1.4 = 12.6
  • Classification of Solutions:
    • pH = 10.36: Basic
    • pH = 7.34: Basic
    • pH = 12.6: Basic
  • Acid Dissociation:
    • H2SO4 dissociates in water to give 2 H+ ions so it is a diprotic acid. HCl dissociates in water to give 1 H+ ions so it is a monoprotic acid.
    • A 1 M HCl solution will have a higher pH than a 1 M H2SO4 solution. Since H2SO4 is diprotic, it produces twice as many H+ ions as HCl, resulting in a lower pH (more acidic).
  • Conjugate Acid-Base Pairs:
    • C6H5NH2 (aq) + H2O (l) ↔ C6H5NH3+ (aq) + OH- (aq)
      • Acid: H2O
      • Base: C6H5NH2
      • Conjugate Acid: C6H5NH3+
      • Conjugate Base: OH-
  • Base Dissociation Expression:
    • For the reaction C6H5NH2 (aq) + H2O (l) ↔ C6H5NH3+ (aq) + OH- (aq), the base dissociation expression is:
      • Kb = \frac{[C6H5NH3^+][OH^-]}{[C6H5NH_2]}
  • Acid Dissociation of Formic Acid:
    • Acid dissociation equation: HCOOH (aq) ↔ H+ (aq) + HCOO- (aq)
    • Expression for the acid dissociation:
      • K_a = \frac{[H^+][HCOO^-]}{[HCOOH]}
    • Given Ka = 1.764 x 10-4 and [HCOOH] = 0.10 M, use an ICE table.
      • K_a = \frac{x^2}{0.10 - x} = 1.764 \times 10^{-4}
      • Assuming x is small compared to 0.10, x^2 = 1.764 \times 10^{-5}
      • x = \sqrt{1.764 \times 10^{-5}} = 0.0042, so [H+] = 0.0042 M
      • pH = -log[H+] = -log(0.0042) = 2.38
  • Titration Calculation:
    • To find the molarity of a sodium hydroxide solution, use the equation: M1V1 = n M2V2, where n is the number of H+ ions from the acid.
      • M_{NaOH}(38 \text{ mL}) = 2(0.75 \text{ M})(14 \text{ mL})
      • M_{NaOH} = \frac{2 \times 0.75 \text{ M} \times 14 \text{ mL}}{38 \text{ mL}} = 0.553 \text{ M}
  • Neutralization Calculation:
    • First, find the number of moles of aluminum hydroxide: n = \frac{0.055 \text{ g}}{78.00 \text{ g/mol}} = 7.05 \times 10^{-4} \text{ mol}
    • The reaction is: Al(OH)3 + 3HCl \rightarrow AlCl3 + 3H_2O, so 3 moles of HCl are needed to neutralize 1 mole of Al(OH)3.
      • n_{HCl} = 3 \times 7.05 \times 10^{-4} \text{ mol} = 2.115 \times 10^{-4} \text{ mol}
      • V = \frac{n}{M} = \frac{2.115 \times 10^{-3} \text{ mol}}{0.200 \text{ M}} = 0.0106 \text{ L} = 10.6 \text{ mL}
  • Function of Acid/Base Indicator:
    • An acid/base indicator is a substance that changes color depending on the pH of the solution. It is used to visually determine the endpoint of a titration.

Nuclear Chemistry

  • Fission vs. Fusion:
    • Fission: The splitting of a heavy nucleus into lighter nuclei, releasing energy.
    • Fusion: The combining of light nuclei to form a heavier nucleus, releasing energy.
  • Difference Between Nuclear and Chemical Reactions:
    • Nuclear reactions involve changes in the nucleus of an atom and can result in the formation of different elements. Chemical reactions involve the rearrangement of electrons and do not change the nucleus.
  • Completing Nuclear Equations:
    • b. 146C → 0-1e + 147N
    • c. 24195Am → 42He + 23793Np
    • d. 167N → 168O + 0-1e
  • Writing Nuclear Equations:
    • a. Alpha decay of francium-208: ^{208}{87}Fr \rightarrow ^{4}{2}He + ^{204}_{85}At
    • b. Electron capture by beryllium-7: ^{7}{4}Be + ^{-1}{0}e \rightarrow ^{7}_{3}Li
    • c. Beta emission by argon-37: ^{37}{18}Ar \rightarrow ^{0}{-1}e + ^{37}_{19}K
    • d. Positron emission by fluorine-17: ^{17}{9}F \rightarrow ^{0}{+1}e + ^{17}_{8}O
  • Completing Transmutation Reactions:
    • b. ^{6}{3}Li + ^{1}{0}n \rightarrow ^{4}{2}He + ^{3}{1}H
    • c. ^{235}{92}U + ^{1}{0}n \rightarrow ^{93}{36}Kr + ^{141}{56}Ba + 3 \, ^{1}_{0}n
    • d. ^{27}{13}Al + ^{4}{2}He \rightarrow ^{30}{14}Si + ^{1}{0}n
    • e. ^{1}{0}n + ^{235}{92}U \rightarrow ^{144}{58}Cs + 6 \, ^{1}{0}n + 1 \, ^{0}_{-1}e
  • Half-Life Calculation:
    • Polonium-214 has a half-life of 164 s.
      • \frac{8.0}{0.25} = 32. The number of half-lives = 5. (2^5=32)
      • 5 \times 164 \text{ s} = 820 \text{ s}
  • Half-Life Calculation:
    • Palladium-103 has a half-life of 17 days.
      • \frac{16}{1.0} = 16. The number of half-lives = 4. (2^4=16)
      • 4 \times 17 \text{ days} = 68 \text{ days}
  • Half-Life Calculation:
    • Argon-35 decays from 1.20 g to 0.15 g in 5.49 seconds.
      • \frac{1.20}{0.15} = 8. The number of half-lives = 3. (2^3=8)
      • \frac{5.49 \text{ s}}{3} = 1.83 \text{ s}

Redox Reactions

  • Oxidation Numbers:
    • b. Sn: 0
    • c. K+: +1
    • d. S2-: -2
    • e. Fe3+: +3
  • Oxidation Numbers of Chromium:
    • b. Cr2O3: +3
    • c. H2Cr2O7: +6
    • d. CrSO4: +2
    • e. CrO42-: +6
  • Determining Oxidation/Reduction:
    • b. 2 Sr + O2 → 2 SrO
      • Sr: 0 → +2 (oxidized, reducing agent)
      • O: 0 → -2 (reduced, oxidizing agent)
    • c. 4 Fe + 3 O2 → 2 Fe2O3
      • Fe: 0 → +3 (oxidized, reducing agent)
      • O: 0 → -2 (reduced, oxidizing agent)
    • d. Mg + 2 HCl → MgCl2 + H2
      • Mg: 0 → +2 (oxidized, reducing agent)
      • H: +1 → 0 (reduced, oxidizing agent)
  • Balancing Redox Reactions (Half-Reaction Method):
    • a. C + H2SO4 → CO2 + SO2 + H2O
      • Oxidation: C → CO2 + 4e- + 4H+
      • Reduction: H2SO4 + 2e- → SO2 + H2O
      • Balanced: C + 2H2SO4 → CO2 + 2SO2 + 2H2O
    • b. Sb + HNO3 → Sb2O5 + NO + H2O
      *Oxidation: Sb + 5 H2O → Sb2O5 + 10e- + 10H+
      *Reduction: HNO3 + 3e- → NO + H2O
      *Balanced: Sb + HNO3 → Sb2O5 + NO + H2O
    • c. KMnO4 + HCl → MnCl2 + Cl2 + H2O + KCl
      *Oxidation: HCl → Cl2 + 2e-
      *Reduction: KMnO4 + 5e- → MnCl2
      *Balanced: KMnO4 + HCl → MnCl2 + Cl2 + H2O + KCl
    • d. Zn + Cr2O72- + H+ → Zn2+ + Cr3+ + H2O
      *Oxidation: Zn → Zn2+ + 2e-
      *Reduction: Cr2O72- + 6e- + H+ → Cr3+ + H2O
      *Balanced: Zn + Cr2O72- + H+ → Zn2+ + Cr3+ + H2O