McGraw-Hill Education 500 Review Questions for the MCAT General Chemistry (Mcgraw-hill's 500 Questions) 2

🧭 Overview

🧠 One-sentence thesis

The excerpt contains only copyright notices, table of contents, author information, and introduction material without substantive chemistry content for Chapter 1.

📌 Key points (3–5)

• What this excerpt includes: legal/copyright information, book structure (table of contents), author credentials, and introductory remarks about the book's purpose.
• Intended audience: students preparing for the MCAT General Chemistry exam.
• Book structure: 12 chapters covering topics from basics through electrochemistry, with 500 total multiple-choice questions.
• Pedagogical approach: questions parallel MCAT content and difficulty; each question has an explained answer.
• Common confusion: this excerpt does not contain the actual Chapter 1 content—only front matter and a fragment of questions from later chapters (radioactivity problems appear at the end).

📚 Book structure and purpose

📚 What the book contains

• 500 multiple-choice questions covering essential General Chemistry material for the MCAT.
• 12 chapters organized by topic:
  • Chapter 1: The Basics (Questions 1–38)
  • Chapter 2: Atomic Structure (Questions 39–83)
  • Chapter 3: Bonding and the Periodic Table (Questions 84–143)
  • Chapter 4: Phases (Questions 144–182)
  • Chapter 5: Gases (Questions 183–217)
  • Chapter 6: Solutions (Questions 218–255)
  • Chapter 7: Kinetics (Questions 256–294)
  • Chapter 8: Equilibrium (Questions 295–332)
  • Chapter 9: Acids and Bases (Questions 333–378)
  • Chapter 10: Thermodynamics (Questions 379–411)
  • Chapter 11: Electrochemistry (Questions 412–450)
  • Chapter 12: Final Review (Questions 451–500)
• Each question includes a clear explanation in the answer key.

🎯 Intended use

The book provides "valuable independent practice to supplement your regular textbook and the ground you have already covered in your classes."

• Designed for MCAT preparation, not as a primary learning resource.
• Questions parallel the content and difficulty level of actual MCAT questions.
• Can be used for extra study weeks before the exam or as last-minute review.
• The introduction states: "If you practice with all the questions and answers in this book, we are certain you will build the skills and confidence needed to excel on the MCAT."

👥 Authors and credentials

👥 Who wrote the book

Author	Credentials	Experience
John T. Moore, EdD	Director of Teaching Excellence Center and Codirector of STEM Center	Over 40 years teaching chemistry at Stephen F. Austin State University
Richard H. Langley, PhD	Has written AP Chemistry exam questions	Over 30 years teaching university chemistry; taught General Chemistry and Organic Chemistry MCAT review courses

• The introduction emphasizes that the authors "know the MCAT inside and out and can identify crucial information as well as the kinds of questions that are most likely to appear on the exam."

📄 Legal and copyright information

📄 Copyright and usage terms

• The eBook version corresponds to print ISBN 978-0-07-183616-6.
• MCAT is a registered trademark of the Association of American Medical Colleges, which did not produce or endorse this product.
• The work is provided "AS IS" with no guarantees or warranties regarding accuracy, adequacy, or completeness.
• Users may store and retrieve one copy for noncommercial and personal use only.
• Prohibited actions include: decompiling, disassembling, reverse engineering, reproducing, modifying, creating derivative works, transmitting, distributing, selling, or sublicensing without McGraw-Hill Education's prior consent.

📄 Disclaimer

• McGraw-Hill Education and its licensors disclaim all warranties, express or implied, including warranties of merchantability or fitness for a particular purpose.
• They are not liable for any inaccuracy, error, omission, or damages resulting from use of the work.
• This limitation applies to any claim in contract, tort, or otherwise.

🧪 Fragment of sample questions

🧪 Radioactivity questions (incomplete)

The excerpt ends with two incomplete questions about radioactive isotopes:

Question 42 (incomplete): Asks which isotope is most dangerous with brief skin contact, listing four options with half-lives, emission types, and particle energies:

• Lead-187 (half-life 15.2 s, alpha emission, 5.99 MeV)
• Actinium-230 (half-life 2.0 m, beta emission, 1.4 MeV)
• Astatine-211 (half-life 7.21 h, electron capture, 0 MeV)
• Thorium-232 (half-life 1.4 × 10^10 years, alpha emission, 2.83 MeV)

Question 43 (incomplete): Describes a radon-222 laboratory accident (half-life 3.8 days), initial radiation level 1 × 10^–5 μCi/mL, safe level below 1 × 10^–6 μCi/mL; asks for the minimum number of days before safe entry.

• These questions appear to be from a later chapter (likely Chapter 2: Atomic Structure, based on the topic).
• The excerpt does not include the actual Chapter 1 content or answers.

🧭 Overview

🧠 One-sentence thesis

Radioactive decay transforms unstable nuclei into more stable ones through several distinct emission or capture processes, each governed by half-life kinetics and nuclear stability principles.

📌 Key points (3–5)

• Main decay modes: alpha emission (helium nucleus), beta emission (electron), gamma emission (high-energy photon), plus less common modes like electron capture, positron emission, and spontaneous fission.
• Half-life calculations: the time required for half of a radioactive sample to decay determines how long radiation levels remain dangerous.
• Nuclear stability factors: "magic numbers" of protons or neutrons (2, 8, 20, 28, 50, 82, 126) confer extra stability; neutron-to-proton ratios determine decay mode.
• Common confusion: distinguishing emission (something leaves the nucleus) from electron capture (something enters); also, mass number changes differ by decay type.
• Practical implications: decay products and half-lives determine radiation hazards and safety protocols.

☢️ Types of radioactive decay

⚛️ Alpha emission

Alpha emission is the radioactive decay mode in which a nucleus emits an alpha particle.

• An alpha particle is essentially a helium nucleus (2 protons + 2 neutrons).
• Mass change: emitting two alpha particles changes mass by 8 units (each alpha = 4 mass units).
• Example: If a container with an unknown isotope shows high helium gas concentration, the isotope undergoes alpha decay.
• Penetration: Alpha particles have relatively low penetration but high energy (e.g., 5.99 MeV for lead-187).

🔵 Beta emission

Beta emission is the radioactive decay mode in which a nucleus emits a beta particle.

• A beta particle is an electron emitted from the nucleus.
• Example: Carbon-14 is a beta emitter that produces nitrogen-14.
• Why it happens: The neutron-to-proton ratio is too high; a neutron converts to a proton plus an electron.
• Electrical field behavior: Beta particles show the greatest deviation in an electrical field (compared to alpha or gamma) because of their small mass and charge.

🌟 Gamma emission

Gamma emission is the radioactive decay mode in which a nucleus emits a gamma ray.

• Gamma rays are high-energy photons, not particles.
• Often follows other decay modes when the resulting nucleus is in an excited (metastable) state.
• Example: Technetium-99m emits gamma rays to reach the ground state.
• No mass or charge change: Gamma emission does not change the element or mass number.

🔄 Less common modes

Decay mode	What happens	Key feature
Electron capture	Nucleus absorbs an electron	Something enters the nucleus (unlike emission modes)
Positron emission	Nucleus emits a positron	Produces the same mass-number isotope of a lower-atomic-number element
Spontaneous fission	Nucleus splits into two or more nuclei	No specific symbol; nucleus breaks apart
Neutron emission	Nucleus emits a free neutron	Example: thorium-232 bombarded with alpha particles absorbs the particle and emits a neutron

Don't confuse: Electron capture is the only common mode where something enters rather than leaves the nucleus.

⏱️ Half-life and decay kinetics

⏳ What half-life means

• Half-life (t₁/₂): the time required for half of a radioactive sample to decay.
• Example: Radon-222 has a half-life of 3.8 days; hydrogen-3 has a half-life of about 14 years.

📉 Calculating decay time

• To drop from an initial radiation level to a target level, count how many half-lives are needed.
• Example: Initial level 1 × 10⁻⁵ μCi/mL, target 1 × 10⁻⁶ μCi/mL (one-tenth the original).
  • After one half-life: 5 × 10⁻⁶ (half remains).
  • After two half-lives: 2.5 × 10⁻⁶ (one-quarter remains).
  • After three half-lives: 1.25 × 10⁻⁶ (one-eighth remains).
  • Need slightly more than three half-lives; for radon-222 (3.8 days each), about 12 days minimum.
• Partial half-lives: After 8 years of a 14-year half-life, approximately 67% remains (not yet one full half-life).

☣️ Safety implications

• Protective gear is insufficient until radiation drops below safe thresholds.
• Half-life determines how long an area remains hazardous after a release.

🧲 Nuclear stability principles

🎯 Magic numbers

Magic numbers (2, 8, 20, 28, 50, 82, 126) are certain numbers of protons or neutrons that impart additional stability to a nucleus.

• Nuclei with a magic number of protons or neutrons are less likely to be radioactive.
• Doubly magic: both protons and neutrons are magic numbers → even more stable.
• Example: Lead-208 is doubly magic (82 protons, 126 neutrons), making it highly stable.
• Least radioactive: Among light nuclei, those closest to magic numbers (e.g., beryllium-8 has 4 protons and 4 neutrons, not magic) are more stable.

⚖️ Neutron-to-proton ratios

• Light isotopes: stable ones have approximately equal numbers of protons and neutrons.
• Heavy isotopes: stable ones have more neutrons than protons.
• Isodiapheres: isotopes of different elements with the same number of excess neutrons over protons.
• Example: If the neutron-to-proton ratio is too high, the nucleus is a beta emitter (converts a neutron to a proton).

🔗 Decay series

• An isotope may undergo multiple decay steps to reach stability.
• Example: Thorium-232 decay series ends with lead-204; counting mass and atomic number changes reveals the number of alpha and beta particles emitted.
  • Each alpha decay: mass decreases by 4, atomic number by 2.
  • Each beta decay: mass unchanged, atomic number increases by 1.
• Same-element result: If a series starts and ends with the same element, alpha decays (which lower atomic number) must be balanced by beta decays (which raise it).

🔬 Electron configuration and atomic structure

🌀 Quantum numbers and orbitals

• Each electron is described by four quantum numbers: principal (n), angular momentum (ℓ), magnetic (mℓ), and spin (±½).
• Example: A valence electron on sodium (3s¹) might have quantum numbers 3, 0, 0, +½.
• Unique sets: No two electrons on the same atom can have identical sets of all four quantum numbers.
• Degenerate orbitals: orbitals with the same energy (e.g., the three p orbitals in a nitrogen atom).

📊 Electron shells and capacity

• The n = 4 shell can hold a maximum of 32 electrons (formula: 2n²).
• Subshells: s (2 electrons), p (6), d (10), f (14).

🎨 Ground state vs excited state

• Ground state: lowest-energy electron configuration.
• Excited state: one or more electrons occupy higher-energy orbitals than in the ground state.
• Example: Silicon ground state is 1s² 2s² 2p⁶ 3s² 3p²; an excited state might be 1s² 2s² 2p⁶ 3s² 3p¹ 4s¹.
• Don't confuse: Excited-state configurations still obey electron-filling rules but have electrons promoted to higher orbitals.

🧪 Isoelectronic species

• Species with the same number of electrons.
• Example: The iron ion in Fe₂O₃ is Fe³⁺; isoelectronic species have the same electron count (e.g., Mn²⁺ has 23 electrons, same as Fe³⁺).

🔥 Flame tests and spectra

• Bright bands: Energy is emitted when electrons drop to lower energy levels, producing characteristic colors.
• Example: Lithium produces a brilliant crimson flame because energy is emitted and there is a bright band in the atomic spectrum.

⚡ Mass-energy relationships

⚖️ Mass defect

Mass defect is the amount by which the actual mass of an isotope is less than the sum of the masses of its constituents.

• The "missing" mass is converted to binding energy that holds the nucleus together.
• Fusion and fission: When nuclei fuse or split and release energy, the exact mass of the products is less than the reactants.

💥 Energy release calculations

• Energy released from mass conversion uses the relationship: energy equals mass times the speed of light squared (E = mc²).
• Example: If 2.0 × 10⁻²⁸ g is converted to energy, the energy released is 1.8 × 10⁻¹¹ J.
• Fusion products: The nucleus formed has less mass than the sum of the two fusing nuclei because energy is released.

🛡️ Practical applications and hazards

☢️ Radiation hazards

• Particle energy and half-life: Both determine danger level.
• Example: Lead-187 (half-life 15.2 s, alpha energy 5.99 MeV) vs thorium-232 (half-life 1.4 × 10¹⁰ years, alpha energy 2.83 MeV).
• Shorter half-lives mean faster decay but higher initial intensity; longer half-lives mean prolonged exposure.

🩺 Medical and environmental concerns

• Iodine-131: A beta emitter released in nuclear accidents; humans absorb it along with natural iodine.
• Treatment strategy: Ingest large amounts of nonradioactive iodine to saturate the body's iodine uptake, reducing absorption of the radioactive isotope.
• Don't confuse: Boiling does not accelerate radioactive decay (decay rate is independent of temperature); precipitation or diuretics do not selectively remove radioactive isotopes.

🧬 Bohr atoms

• A Bohr atom is a hydrogen-like species with only one electron.
• Example: He⁺ (helium with one electron removed) qualifies; H⁻ (hydrogen with an extra electron) does not.

🧭 Overview

🧠 One-sentence thesis

Lewis structures provide a systematic way to track valence electrons and predict bonding behavior, stoichiometry, and molecular geometry across representative elements.

📌 Key points (3–5)

• Lewis structures use symbols (dots, circles) around element symbols to represent valence electrons and reveal bonding information.
• Coordinate covalent bonds form when one atom donates both electrons in a bond (e.g., oxygen binding to iron in hemoglobin).
• Periodic trends include atomic radius, ionization energy, electron affinity, and electronegativity, which vary predictably across periods and groups.
• Common confusion: ionic vs. covalent bonding—MgCl₂ shares zero electrons (ionic), while coordinate covalent bonds involve electron-pair donation to a metal.
• Molecular geometry depends on hybridization and electron arrangement, affecting polarity, intermolecular forces, and physical properties.

🔬 Lewis structures and bonding fundamentals

🔬 What Lewis symbols represent

Lewis symbol: the chemical symbol of an element with an indication of the valence electrons.

• Any symbol (dots, circles, etc.) can be placed around the central element symbol to show valence electrons.
• Lewis structures can be written for atoms, ions, or compounds.
• They immediately reveal valence electron count and bonding patterns.

🔗 Types of bonds in Lewis structures

• Shared electrons: covalent bonds where atoms share electron pairs.
• Zero shared electrons: ionic compounds like MgCl₂ involve electron transfer, not sharing.
• Coordinate covalent bonds: one atom donates both electrons in the bond.
  • Example: Oxygen (O₂) binding to iron in hemoglobin—oxygen acts as a Lewis base (electron-pair donor), forming a coordinate covalent bond.
• Don't confuse: coordinate covalent bonds are still covalent, but the electron pair comes from one atom only.

🧲 Lewis acids and bases

• Lewis base: donates an electron pair (e.g., oxygen molecule binding to iron).
• Lewis acid: accepts an electron pair (e.g., boron halides reacting with halide ions: BX₃ + X⁻ → BX₄⁻).
• Example: Chelating agents in chelation therapy act as Lewis bases, donating electron pairs to heavy metal ions.

🔄 Periodic trends and atomic properties

📏 Atomic radius trends

• Across a period: radius decreases as effective nuclear charge increases (more protons pull electrons closer).
• Down a group: radius increases as additional electron shells are added.
• Example: Potassium is larger than arsenic because potassium has a smaller effective nuclear charge despite being in the same period.

Comparison	Explanation
K vs. As	K is larger; smaller effective nuclear charge
Isoelectronic series (La³⁺, Ba²⁺, Xe, Te²⁻)	More positive charge → smaller radius; order: La³⁺ < Ba²⁺ < Xe < Te²⁻

⚡ Ionization energy

Ionization energy: the energy required to remove an electron from an atom.

• Trend: increases across a period (stronger nuclear attraction); decreases down a group (electrons farther from nucleus).
• Example: Sodium has lower ionization energy than chlorine because sodium's valence electron is farther from the nucleus and experiences lower effective nuclear charge.
• Second and third ionization energies: removing additional electrons requires more energy; Ca's third ionization energy is much higher than Sc's because Ca must remove an electron from a stable inner shell.

🧲 Electron affinity

Electron affinity: the energy change when an atom gains an electron.

• First electron affinity: usually exothermic (energy released).
• Second electron affinity: significantly greater (less favorable) due to repulsion between like charges.
• Example: Neon has the lowest electron affinity because it has a stable, complete outer shell.

🔋 Electronegativity trends

• General trend: increases across a period and up a group.
• Affects bond polarity: greater electronegativity difference → more polar bond.
• Example: Bond polarity order: F–F < N–F < C–F (increasing polarity as electronegativity difference increases).

🏗️ Molecular geometry and hybridization

🏗️ Hybridization and geometry

• sp³ hybridization: tetrahedral geometry (e.g., four bonds around carbon).
• sp² hybridization: trigonal planar geometry (e.g., three bonds, 120° angles).
• sp hybridization: linear geometry.
• Example: In nitrous acid, the oxygen atoms have sp³ and sp² hybridization (left to right).

📐 Common geometries

Geometry	Example	Bond angles
Trigonal planar	BF₃, NO₃⁻	120°
Tetrahedral	CH₄	109.5°
Trigonal bipyramidal	PF₅	90° and 120°
Octahedral	Six-coordinate iron in hemoglobin	90°
Square planar	ClF₄⁻	90°

• Don't confuse: NH₃ is not trigonal planar—it has trigonal pyramidal geometry due to a lone pair.

🔺 Formal charge and resonance

• Formal charge: helps determine the most stable Lewis structure.
• Example: In the cyanate ion (OCN⁻), the nitrogen has a formal charge of 0 in one resonance form.
• Resonance stabilization: multiple valid Lewis structures distribute charge; fulminate ion (CNO⁻) is less stable than cyanate due to unfavorable formal charge distribution and less resonance.

🌀 Geometry changes with bonding

• Example: Sulfur trioxide (SO₃) is trigonal planar; when it dissolves in water to form sulfuric acid, the geometry around sulfur changes from trigonal planar to tetrahedral.
• Example: Boron halides (BX₃) are trigonal planar; when reacting with halide ions (BX₃ + X⁻ → BX₄⁻), geometry changes to tetrahedral.

🧪 Intermolecular forces and physical properties

🧪 Types of intermolecular forces

• Hydrogen bonding: strong intermolecular force; important for water, DNA double helix, and protein secondary structure.
  • Requires H bonded to N, O, or F.
  • Example: CH₃OH and H₂SO₄ can participate; CHF₃ cannot (H not bonded to F).
• Dipole-dipole forces: between polar molecules.
• London dispersion forces: weakest; present in all molecules, including nonpolar ones.
  • Example: Helium solidifies at 1 K under 26 atm due to temporary dipole-dipole attractions.

❄️ Melting and boiling points

• Stronger intermolecular forces → higher melting/boiling points.
• Order of increasing melting point: Cl₂ < HCl < H₂O < NaCl.
  • Cl₂: only dispersion forces.
  • HCl: dipole-dipole forces.
  • H₂O: hydrogen bonding.
  • NaCl: ionic bonding (strongest).

🔬 Polarity and solubility

• Polar molecules: have asymmetric charge distribution.
• Nonpolar molecules: symmetric charge distribution cancels dipoles.
  • Example: CF₄ is nonpolar (tetrahedral, symmetrical); HCl is polar.
• Solubility: nonpolar compounds (e.g., dimethyl mercury) cross hydrophobic cell membranes more easily than ionic compounds, making them more toxic.

Geometry	Can be nonpolar?
Linear	Yes
Tetrahedral	Yes (if symmetrical)
Octahedral	Yes (if symmetrical)
Bent	No (always polar)

🔍 Special topics in bonding

🔍 Free radicals

Free radical: a molecule with an unpaired electron.

• Example: NO is a free radical; the unpaired electron in the methyl radical (CH₃) is in an unhybridized p-orbital.

🔍 Spectroscopy and electronic transitions

• Electronic transitions: require UV photons (high energy).
• Vibrational transitions: require infrared photons (lower energy).
• Example: Water molecules have vibrational spectra in the infrared region.
• No vibrational spectrum: Monoatomic gases like Xe have no vibrational modes.

🔍 Capillary action and meniscus

• Strong intermolecular forces between glass and solution cause capillary action.
• To measure liquid volume accurately in a burette, always read at the same point (top or bottom of meniscus) for consistency.

🧭 Overview

🧠 One-sentence thesis

Water's unusually high specific heat, heat of vaporization, and hydrogen-bonding structure enable it to moderate Earth's temperature, form ice less dense than liquid water, and exhibit unique phase-transition behaviors that distinguish it from typical liquids.

📌 Key points (3–5)

• Water's unusual properties stem from hydrogen bonding: water forms two hydrogen bonds per molecule (via two lone pairs), more than most liquids like ammonia (one lone pair).
• High specific heat and heat of vaporization moderate temperature: water requires large amounts of energy to heat or vaporize, stabilizing Earth's climate; the moon lacks liquid water and experiences extreme temperature swings.
• Ice is less dense than liquid water: strong hydrogen bonding creates an open structure in ice, causing ~10% volume increase upon freezing.
• Common confusion—energy at phase transitions: at the same temperature (e.g., boiling point), steam contains far more energy than liquid water due to the high heat of vaporization (~10× more joules per mass).
• Phase diagrams and pressure effects: water boils at lower temperatures at higher altitudes because atmospheric pressure is lower; phase changes (sublimation, deposition, melting) depend on pressure and temperature conditions.

💧 Water's hydrogen bonding and thermal properties

💧 Why water is unusual

Water is "not always best to think of … as a typical liquid" because its highly polar molecules form extensive hydrogen bonds.

• Hydrogen bonding capacity:
  • Water has two lone pairs on oxygen → can attract hydrogen from two separate water molecules → forms two hydrogen bonds per molecule.
  • Ammonia has one lone pair → forms only one hydrogen bond per molecule.
  • This extensive network makes water's properties (specific heat, heat of vaporization, viscosity, surface tension) unusually high.

🔥 High specific heat (4.18 J/g°C)

• What it means: a large amount of heat is needed to raise water's temperature; conversely, removing heat cools water slowly.
• Why it matters:
  • Oceans (large volume of water) absorb/release heat slowly → moderate Earth's temperature.
  • Example: the Moon receives the same solar energy as Earth but has almost no liquid/gaseous water → temperature swings from ~–150°C to above 100°C; Earth's temperature range is much narrower.
  • Water vapor in the atmosphere also moderates temperature to a lesser extent.

♨️ High heat of vaporization (~40 kJ/mol)

• What it means: evaporating water requires enormous energy; water vapor carries a large amount of heat energy.
• Why it matters:
  • Evaporation of massive amounts of water supplies water vapor for precipitation.
  • Evaporation helps moderate Earth's temperature by absorbing heat.
• Don't confuse liquid and vapor energy:
  • At the boiling point, liquid water and steam are at the same temperature, but steam contains ~10× more energy per mass.
  • Example: one teaspoon of boiling water (~1,000 J) vs. the same mass of steam (~10,000 J) → steam burns are far more severe.

🧊 Ice density and hydrogen bonding

• Unusual property: ice is less dense than liquid water (most substances have denser solids than liquids; bismuth is another exception).
• Mechanism: strong hydrogen bonding in ice creates an open, ordered structure → ~10% volume increase when water freezes.
• Consequence: ice floats; freezing water can burst pipes or crack sidewalks.
• Preventing freezing damage (colligative property): adding solutes (e.g., NaCl, (NH₄)₂SO₄, Na₃PO₄) lowers the freezing point; effectiveness depends on the number of particles produced per mole (more ions → more effective).

🌊 Viscosity and surface tension

• High viscosity: not easily observed in everyday life, but noticeable when walking through shallow water in a pool (resistance).
• High surface tension: water droplets on a surface draw into rounded beads; other liquids (e.g., gasoline) spread out more.
• Both properties result from strong hydrogen bonding between water molecules.

🔄 Phase transitions and energy changes

🔄 Heats of fusion, vaporization, and sublimation

Phase change	Energy direction	Typical magnitude
Fusion (solid → liquid)	Absorbs heat	~330 J/g for water (~6 kJ/mol)
Vaporization (liquid → gas)	Absorbs heat	~2,260 J/g for water (~40 kJ/mol)
Sublimation (solid → gas)	Absorbs heat	Sum of fusion + vaporization
Deposition (gas → solid)	Releases heat	Negative of sublimation
Condensation (gas → liquid)	Releases heat	Negative of vaporization
Freezing (liquid → solid)	Releases heat	Negative of fusion

• Key relationship: heat of sublimation ≈ heat of fusion + heat of vaporization.
• Example: heat of deposition of water = –(heat of fusion + heat of vaporization) = –(330 + 2,260) J/g ≈ –2,590 J/g.
• Spontaneous deposition: releases heat and decreases entropy (gas → solid is more ordered).

📈 Heating curves and calculations

• Heating curve: temperature vs. heat added; shows phase transitions as horizontal plateaus (temperature constant while phase changes).
• Slanting regions (single phase): use specific heat (or molar heat capacity) to calculate temperature change.
  • Formula (in words): heat = mass × specific heat × temperature change.
  • Example: heating liquid water from melting point to boiling point requires knowing the mass, specific heat of liquid water, and the two temperatures (T₂ and T₃).
• Horizontal regions (phase change): use heat of fusion or heat of vaporization; temperature does not change.
  • Example: melting ice at 0°C requires heat of fusion; boiling water at 100°C requires heat of vaporization.
• Heat capacity vs. specific heat:
  • Specific heat: energy per gram per degree (J/g°C).
  • Heat capacity: energy per degree for a given sample (J/°C) = specific heat × mass.
  • Example: 200 g water with specific heat 4.2 J/g°C → heat capacity = 200 × 4.2 = 840 J/°C.

🧪 Calorimetry and enthalpy calculations

• Bomb calorimeter: measures heat of combustion; total heat released = heat capacity of calorimeter × temperature change.
  • Example: 0.46 g ethanol burned, calorimeter heat capacity 1.2 kJ/°C, ΔT = 10°C → total heat = 1.2 × 10 = 12 kJ; convert to per mole using molar mass.
• Coffee-cup calorimeter: measures heat of dissolution or reaction in solution; assumes specific heat of solution ≈ water (4.2 J/g·K).
  • Example: dissolving NaNO₃ in water causes temperature drop → endothermic (positive ΔH); calculate using mass of solution, specific heat, and ΔT.
• Hess's Law: combine thermochemical equations to find ΔH for a target reaction.
  • Example: given ΔH for formation reactions, calculate ΔH for a synthesis by reversing/scaling equations and summing.

📊 Phase diagrams and pressure effects

📊 Reading phase diagrams

• Axes: pressure (y-axis) vs. temperature (x-axis).
• Regions: solid, liquid, gas.
• Lines: phase boundaries (coexistence curves).
• Triple point: all three phases coexist.
• Critical point: liquid and gas phases become indistinguishable; beyond this, no distinct liquid/gas boundary exists.
• Normal boiling point: temperature at which vapor pressure = 1 atm (standard atmospheric pressure); on the diagram, this is at P = 1 atm on the liquid-gas boundary.
• Normal melting point: temperature at which solid and liquid coexist at 1 atm.

🏔️ Pressure and boiling point

• Why water boils at lower temperature on a mountain: atmospheric pressure is lower at higher altitude.
  • Boiling occurs when vapor pressure of liquid = external pressure.
  • Lower external pressure → lower temperature needed to reach that vapor pressure → lower boiling point.
• Boiling above normal boiling point: possible if pressure is increased above 1 atm (e.g., pressure cooker).
  • Example: liquid nitrogen (normal boiling point 77 K) can exist as liquid above 77 K if pressure > 1 atm.

🧊 Phase changes at constant temperature or pressure

• Compressing water vapor at constant temperature (e.g., T₁ on diagram):
  • Sequence depends on phase boundaries crossed.
  • Example: gas → liquid → solid (if pressure increases enough to cross both boundaries).
• Heating ice at low pressure (e.g., 4 × 10⁻³ atm):
  • If pressure is below the triple point, ice sublimates directly to gas (no liquid phase).
  • Example: heating from –10°C to +10°C at 4 × 10⁻³ atm → sublimation (solid → gas).
• Don't confuse:
  • Sublimation (solid → gas) vs. deposition (gas → solid).
  • Melting (solid → liquid) vs. boiling (liquid → gas).

🔬 Intermolecular forces and phase properties

🔬 Why diamond vs. graphite hardness differs

• Diamond: 3D network of sp³-hybridized carbon; each carbon bonded to four others via strong covalent bonds → extremely hard.
• Graphite: planar layers of sp²-hybridized carbon; each carbon bonded to three others in-plane via resonating covalent bonds; layers held together only by weak London dispersion forces → soft (layers slide easily).
• Don't confuse: hardness (resistance to deformation) vs. heat of sublimation (energy to convert solid → gas).
  • Both diamond and graphite have very high heats of sublimation because sublimation requires breaking strong covalent bonds (within the network or layers).

🧲 Predicting heats of vaporization and fusion

• General trend: stronger intermolecular forces → higher heat of vaporization/fusion.
• Hydrogen bonding (e.g., water, methanol) > dipole-dipole (e.g., HCl) > London dispersion (e.g., nonpolar molecules).
• Ionic compounds (e.g., NaCl, Al₂O₃) have very high heats of fusion due to strong ionic bonds.
• Example comparison (heats of fusion): Al₂O₃ > NaCl > H₂O > CCl₄ (network ionic > simple ionic > hydrogen bonding > nonpolar).
• Methanol vs. methyl chloride: methanol has higher heat of vaporization because hydrogen bonding (O–H···O) is stronger than dipole-dipole forces in methyl chloride, even though methyl chloride has higher molecular weight.

💨 Vapor pressure and partial pressure

• Vapor pressure: pressure exerted by a vapor in equilibrium with its liquid/solid phase.
• Partial pressure: pressure contributed by one component in a gas mixture.
• Example: in the stratosphere (total pressure ~8 torr), ozone is one component → partial pressure of ozone < 8 torr (the rest is other gases like O₂, N₂).
• Boiling a solution (e.g., NaCl in water): the vapor phase is predominantly H₂O(g); nonvolatile solutes (NaCl) remain in the liquid phase and do not vaporize significantly.

🧪 Energy and entropy in phase changes

• Exothermic (release heat): freezing, condensation, deposition (gas/liquid → more ordered solid).
• Endothermic (absorb heat): melting, vaporization, sublimation (solid/liquid → less ordered gas).
• Entropy changes:
  • Increases: melting, vaporization, sublimation (more disorder).
  • Decreases: freezing, condensation, deposition (more order).
• Example: spontaneous deposition releases heat (exothermic) and decreases entropy (gas → solid).

🔢 Calculating energy for multi-step heating

• Strategy: break the process into segments (single-phase heating + phase changes); sum the energies.
• Example: converting solid HI at melting point to vapor at 0°C:
  1. Melt solid → liquid (heat of fusion).
  2. Heat liquid from melting point (–50°C) to boiling point (–35°C) (specific heat of liquid).
  3. Vaporize liquid → gas (heat of vaporization).
  4. Heat gas from boiling point (–35°C) to 0°C (specific heat of gas).
  • Total energy = (moles × heat of fusion) + (moles × specific heat of liquid × ΔT) + (moles × heat of vaporization) + (moles × specific heat of gas × ΔT).

🧭 Overview

🧠 One-sentence thesis

Gases behave predictably according to gas laws that relate temperature, pressure, volume, and amount, because gas particles are far apart with high kinetic energy and negligible intermolecular forces.

📌 Key points (3–5)

• Gases vs vapors: gases exist in the gaseous state at room temperature; vapors are gaseous forms of substances normally liquid or solid at room temperature.
• Microscopic properties: gas particles have low mass, large distances between them (explaining compressibility), and enough kinetic energy to overcome intermolecular forces.
• Gas laws: relationships between temperature, pressure, volume, and moles can be determined when two variables are held constant.
• Ideal vs real behavior: gases approach ideal behavior at low pressure and high temperature; deviations occur when particle volume or intermolecular forces become significant.
• Common confusion: kinetic energy vs velocity—at the same temperature, all gases have the same average kinetic energy, but lighter molecules move faster.

🔬 Macroscopic vs microscopic properties

🔬 Macroscopic characteristics

• Compressibility: gases are highly compressible (unlike solids and liquids).
• Expansion: gases expand to fill the volume of their container.
• Homogeneity: gases always form homogeneous mixtures when mixed.

🔬 Microscopic structure

• Particle mass: gas particles are normally low atomic or molecular mass atoms or molecules.
• Spacing: relatively large distance between gas particles explains their compressibility.
• Kinetic energy: particles possess enough kinetic energy to overcome attractive forces (intermolecular forces), so particles are relatively independent of each other.

Condensed phases: solids and liquids, which have much less space between particles than gases.

Intermolecular forces: attractive forces between particles that gas particles have enough kinetic energy to overcome.

⚖️ Gas laws and relationships

⚖️ Core principle

• All gases behave similarly with respect to temperature, pressure, volume, and amount (moles).
• Relationships between pairs of these conditions can be determined if the other two are held constant.
• These relationships were discovered between the mid-1600s and mid-1800s and are named after their discoverers.

🌡️ Temperature-volume relationship (constant pressure)

• Heating a gas at constant pressure causes volume to increase.
• Example: A 0.50 mole sample of argon at 0°C (273 K) and 1.00 atm with initial volume 11 L, heated to 273°C (546 K), doubles in volume to 22 L.
• The absolute temperature (in Kelvin) is proportional to volume when pressure is constant.

🔒 Temperature-pressure relationship (constant volume)

• Heating a gas at constant volume (locked piston) causes pressure to increase.
• Example: Same argon sample heated from 0°C to 273°C with locked piston results in final pressure = 2 × initial pressure, while volume remains constant.

📏 Pressure-volume relationship (constant temperature)

• At constant temperature, if pressure decreases, volume increases (and vice versa).
• Example: After heating and unlocking the piston at constant temperature, final pressure < initial pressure and final volume > initial volume.

🧮 Partial pressures and mixtures

🧮 Dalton's law concept

• In a gas mixture, each gas exerts a partial pressure.
• Total pressure = sum of all partial pressures.

🧮 Calculating partial pressures

• Example: Mixture with total pressure 1,000 torr containing 25% methane, 40% oxygen, 35% hydrogen → partial pressure of hydrogen = 350 torr (35% of 1,000).
• Mole fraction = partial pressure / total pressure.

Gas component	Relationship to total
Partial pressure	Proportional to mole fraction
Mole fraction	Partial pressure ÷ total pressure

🧮 Unit conversions

• Pressures can be expressed in atm, torr, or kPa; must convert to same units before adding.
• Example: He = 0.50 atm, Ne = 76 torr (≈0.10 atm), Ar = 10 kPa (≈0.10 atm) → total ≈ 0.70 atm.

🚀 Kinetic molecular theory

🚀 Key postulates

• Gas is composed of particles (atoms or molecules).
• Volumes of gas particles are negligible compared to container volume.
• No attractive or repulsive interactions between gas particles (ideal assumption).
• Particles are constantly moving.
• Average kinetic energy depends on absolute temperature.

🚀 Kinetic energy vs velocity

Kinetic energy formula: KE = 1/2 mv² (one-half mass times velocity squared).

• At a given temperature, all gas molecules have the same average kinetic energy.
• Don't confuse: same kinetic energy ≠ same velocity.
• Lighter molecules must move faster to have the same kinetic energy as heavier molecules.
• Example: Hydrogen and oxygen at the same temperature and pressure have the same average kinetic energy, but hydrogen molecules move faster because they are lighter.

🚀 Diffusion and effusion

• The basis for explaining diffusion of gases is the kinetic energy equation KE = 1/2 mv².
• Rate of effusion is inversely related to molecular mass: lighter gases effuse faster.
• Example: If an unknown gas effuses at half the rate of methane (molecular weight 16), the unknown likely has molecular weight around 64 (e.g., SO₂).
• Example: Helium effuses out of a balloon faster than air molecules effuse in, because helium is lighter.

🎯 Ideal vs real gas behavior

🎯 When gases are ideal

• Gases approach ideal behavior at low pressure and high temperature.
• Under these conditions, the assumptions of kinetic molecular theory hold well.

🎯 Deviations at high pressure

• Problem postulate: "The volumes of the gas particles are negligible."
• At high pressure, particles are forced closer together, so particle volume becomes significant compared to container volume.

🎯 Deviations at low temperature

• Problem postulate: "There are no attractive or repulsive interactions between the gas particles."
• At low temperature, particles move slower, so intermolecular attractions become more significant.
• Example: Water vapor shows lower pressure than expected from PV = nRT because of strong attraction between water molecules.

🎯 Van der Waals constants

• Gases with larger van der Waals constants deviate more from ideal behavior.
• Larger constants indicate stronger intermolecular forces or larger particle volumes.
• Example: Among O₂, CO₂, and SO₂, the one with the largest van der Waals constants is the least ideal.

🎯 Which gases are most ideal

• Gases most likely to be ideal under normal conditions: small, nonpolar, low-mass molecules.
• Example: Helium (He) is more ideal than Xe (heavy), HF (polar with hydrogen bonding), or C₂H₆ (larger molecule).

📐 Practical calculations and applications

📐 Standard conditions (STP)

• STP = 0°C (273 K) and 1 atm pressure.
• At STP, 1 mole of an ideal gas occupies 22.4 L.
• Example: 56 g of N₂ = 2 moles (molecular weight 28) → volume at STP = 2 × 22.4 = 44.8 L.

📐 Ideal gas equation

• PV = nRT relates pressure, volume, moles, and temperature.
• Can be used to find any one variable if the others are known.
• Example: 2.0 moles at 2.0 atm and 249°C (522 K) → volume = (2.0 × R × 522) / 2.0 ≈ 42.8 L.

📐 Density and molecular weight

• Density of a gas can be used to identify it.
• Example: 1.0 mole at 1.0 atm and 0°C with density 5.8 g/L → molar mass ≈ 5.8 × 22.4 ≈ 130 g/mol, suggesting Xe.

📐 Temperature-volume problems

• Must use absolute temperature (Kelvin).
• Example: Gas at 127°C (400 K) occupying 1.0 L, cooled to 500 mL → new temperature = 400 × (500/1000) = 200 K = –73°C.

📐 Water displacement caution

• When collecting gas over water, the measured pressure includes water vapor pressure.
• Don't confuse: total pressure = gas pressure + water vapor pressure.
• Example: If calculated pressure is 1.0 atm but measured pressure is higher, the student likely did not account for water vapor pressure.

🔧 Applications and real-world examples

🔧 Shock absorbers

• Shock absorbers rely on air because air is compressible.
• Water could not replace air because water is nearly incompressible.
• Don't confuse: density or freezing point are not the main reasons; compressibility is key.

🔧 Gas-phase reactions

• Volatile compounds can react in the gas phase to form nonvolatile products, visible as a cloud.
• The location of the cloud depends on the relative speeds (and thus molecular weights) of the reactants.
• Example: If a cloud forms near the center with dimethylamine, the other reactant has similar molecular weight; candidates include acids like HCOOH.

🔧 Balloon comparisons

• Two balloons at the same volume, pressure, and temperature contain the same number of moles (from PV = nRT).
• The balloon with heavier molecules (e.g., oxygen vs hydrogen) has greater density.
• Both have the same average kinetic energy (temperature-dependent), but lighter molecules move faster.

🧭 Overview

🧠 One-sentence thesis

Colligative properties—freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure—depend only on the number of solute particles in solution, not their identity, making the van't Hoff factor critical for calculating these effects when electrolytes dissociate into ions.

📌 Key points (3–5)

• What colligative properties are: solution properties that depend only on the number of solute particles, not their chemical identity.
• How electrolytes differ from non-electrolytes: strong electrolytes like NaCl dissociate completely into ions, so a 1.0 M NaCl solution behaves like a 2.0 M solution for colligative properties.
• The van't Hoff factor (i): the number of particles (ions) produced per formula unit of solute; for NaCl, i = 2; for Ca(NO₃)₂, i = 3.
• Common confusion: molarity vs. molality and particle concentration—a 1.0 M NaCl solution is 1.0 M in Na⁺ and 1.0 M in Cl⁻, totaling 2.0 M particles.
• Why it matters: colligative properties explain freezing point depression (road salt), boiling point elevation (cooking), vapor pressure lowering, and osmotic pressure in biological and industrial systems.

🧪 What colligative properties are

🧪 Definition and core principle

Colligative properties: solution properties that depend only upon the number of solute particles present.

• The key word is "only"—the chemical nature of the solute does not matter, only how many particles are dissolved.
• Four main colligative properties are mentioned or implied:
  • Freezing point depression
  • Boiling point elevation
  • Vapor pressure lowering
  • Osmotic pressure

🔍 Why solutes change freezing and boiling points

• The presence of a solute lowers the vapor pressure of the solvent.
• Because vapor pressure is lowered:
  • The freezing point of the solution is lower than that of the pure solvent.
  • The boiling point of the solution is higher than that of the pure solvent.
• Example: pure water freezes at 0°C, but adding salt lowers the freezing point below 0°C.

🔢 The van't Hoff factor and electrolytes

🔢 What the van't Hoff factor is

van't Hoff factor (i): the number of ions (particles) produced by an electrolyte in solution.

• For non-electrolytes (e.g., sugar, ethanol), i = 1 (the molecule does not dissociate).
• For strong electrolytes, i equals the number of ions per formula unit:
  • NaCl → Na⁺ + Cl⁻, so i = 2
  • Ca(NO₃)₂ → Ca²⁺ + 2 NO₃⁻, so i = 3
  • (NH₄)₂SO₄ → 2 NH₄⁺ + SO₄²⁻, so i = 3

⚡ How to count particles for strong electrolytes

• Table salt (NaCl) is a strong electrolyte and dissociates completely.
• A 1.00 M (or 1.00 m) NaCl solution is not 1.00 M in particles; it is:
  • 1.00 M in Na⁺
  • 1.00 M in Cl⁻
  • Total: 2.00 M in particles
• Therefore, the freezing point depression of a 1.00 m NaCl solution is twice that of a 1.00 m sugar solution.
• Don't confuse: the molarity of the compound vs. the total particle concentration—always convert strong electrolytes to individual ions before calculating colligative effects.

🧮 Two methods for calculating temperature change

1. Direct particle counting: convert the electrolyte to ions, add up the separate concentrations, then use the ΔT equation.
2. Using the van't Hoff factor: multiply the formula concentration by i in the equation ΔT = i K m, where K is the freezing point depression constant (Kf) or boiling point elevation constant (Kb), and m is molality.

🌡️ When the van't Hoff factor deviates from ideal

• The van't Hoff factor may vary due to interactions between particles (ion pairing, incomplete dissociation).
• At low concentrations, ions are farther apart, interactions are weaker, and i is closest to the ideal value.
• At high concentrations, ions interact more, and the effective number of particles is lower than ideal.
• Example: the observed freezing point of a 0.1 m sodium nitrate solution may be higher than the theoretical (less depression) because the actual number of free particles is lower than ideal.

❄️ Freezing point depression

❄️ The freezing point depression equation

• ΔT = i Kf m
  • ΔT: the change in freezing point (negative, indicating a decrease)
  • i: van't Hoff factor
  • Kf: freezing point depression constant (specific to the solvent)
  • m: molality of the solution

🧊 Key insight: concentration matters, not total amount

• If a solution has ΔT = –0.80°C and you spill half of it, the freezing point of the remaining solution is still –0.80°C.
• Why? Colligative properties depend on concentration (molality), not the total volume or amount of solution.
• Spilling half does not change the ratio of solute particles to solvent molecules.

🧂 Why salts lower freezing points

• Dissolved ions (e.g., Na⁺, Cl⁻, Ca²⁺) interfere with the formation of ice crystals.
• This is why road salt (NaCl, CaCl₂) is used to melt ice: it lowers the freezing point of water below 0°C.
• Don't confuse: the mechanism is not "compensating for heat of solution" or "increasing density"—it is disruption of crystal formation.

🔥 Boiling point elevation

🔥 The boiling point elevation equation

• ΔT = i Kb m
  • Kb: boiling point elevation constant (specific to the solvent)
  • For water, Kb is about 0.5°C per molal (0.5°C/mol).

🧪 Example: potassium phosphate solution

• Potassium phosphate is K₃PO₄.
• It dissociates into 3 K⁺ + PO₄³⁻, so i = 4.
• For a 1 molal solution:
  • ΔT = 4 × 0.5°C = 2.0°C
  • Boiling point ≈ 100°C + 2.0°C = 102.0°C

🔍 Comparing solutions: which freezes or boils at the extreme?

Solution	Molarity	Van't Hoff factor (i)	Effective particle concentration	Effect
1.0 M NaNO₃	1.0 M	2	2.0 M	Lower freezing point than non-electrolyte
1.0 M Ca(NO₃)₂	1.0 M	3	3.0 M	Even lower freezing point
1.0 M CaCl₂	1.0 M	3	3.0 M	Highest boiling point among 1 M solutions
2.0 M ethanol	2.0 M	1	2.0 M	Non-electrolyte, fewer particles per mole

• The solution with the highest particle concentration will have the lowest freezing point and the highest boiling point.
• Example: among 1 M solutions, 1.0 M CaCl₂ (i = 3) will boil at a higher temperature than 1.0 M NaCl (i = 2) or 1.0 M ethanol (i = 1).

💧 Vapor pressure lowering

💧 How solutes lower vapor pressure

• Adding a non-volatile solute (e.g., salt, sugar) to a solvent (e.g., water) lowers the vapor pressure of the solvent.
• Fewer solvent molecules are at the surface to evaporate because solute particles occupy space.
• Example: pure water at 30°C has a vapor pressure of 32 torr; adding 10.0 g of magnesium chloride to 100.0 mL of water will lower the vapor pressure.

🌫️ Collecting gas over water

• When hydrogen gas is collected over water, the measured pressure includes:
  • The pressure of the hydrogen gas
  • The vapor pressure of water
• If a student calculates pressure using only the ideal gas law and ignores water vapor pressure, the measured pressure will be greater than calculated.

🔬 Why vapor pressure depends on particle number

• Vapor pressure lowering is a colligative property.
• The more solute particles, the greater the reduction in vapor pressure.
• Does not depend on: external pressure (that affects boiling point, not vapor pressure at a given temperature).
• Does depend on: temperature, hydrogen bonding (affects pure solvent vapor pressure), and the presence of solutes.

🧬 Osmosis and osmotic pressure

🧬 What osmosis is

Osmosis: the flow of solvent through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration.

• The membrane allows solvent (e.g., water) to pass but blocks solute particles.
• Solvent flows to equalize concentrations on both sides.

🔄 Key principles

• If concentrations are the same on both sides, the solvent does not flow (no net osmosis).
• The solvent always flows into the solution of higher concentration (more solute particles).
• Osmotic pressure is the pressure needed to stop osmosis; it is proportional to the particle concentration (another colligative property).
• Don't confuse: osmotic pressure does not "force" the solvent through—it is the pressure required to prevent the natural flow.

🧪 Comparing osmotic pressures

• A 1.0 m NaCl solution (i = 2, effective 2.0 m particles) has a higher osmotic pressure than a 1.0 m acetic acid solution (weak acid, i ≈ 1).
• The statement "osmotic pressure of 1.0 m NaCl = 1.0 m acetic acid" is false.

🔬 Reverse osmosis

• Applying pressure greater than the osmotic pressure forces solvent to flow from high solute concentration to low concentration, purifying the solvent.
• Used in water purification to remove ions.
• A reducing agent is often used beforehand to convert certain ions (e.g., Hg²⁺) to less harmful forms.

🧪 Solubility and solution behavior

🧪 Factors affecting solubility

• Polarity: polar solutes (e.g., NH₃, HCl) are more soluble in polar solvents (water) than non-polar solutes (e.g., O₂).
  • NH₃ and HCl are polar; O₂ is not, so NH₃ and HCl are much more soluble in water.
• Hydrogen bonding: molecules with more –OH groups (hydroxyl groups) are more soluble in water.
  • Example: a compound with four –OH groups is more soluble than one with two –OH groups and a long hydrocarbon chain.
• Acid-base reactions: acids are more soluble in basic solutions because they react to form ions.
  • Decanoic acid (C₉H₁₉COOH) is more soluble in a basic solution than in pure water because the acid reacts with the base to produce a hydrophilic (water-loving) ion (the carboxylate anion).

🧂 Solubility of salts

• Most salts are soluble in water, but some (e.g., Ag₂S, CaCO₃, AlPO₄) have very low solubility.
• If 2 grams of an unknown salt dissolves completely in 100 mL of water, it is likely a soluble salt such as Ca(C₂H₃O₂)₂ (calcium acetate).

🔬 Precipitate formation and ionic bonding

• Mixing certain salt solutions produces a precipitate (insoluble solid).
• Example: mixing BaCl₂ with Na₂C₂O₄, NaIO₃, Na₂CO₃, or Na₃PO₄ all produce barium precipitates.
• The precipitate with the strongest ionic bonding forms from ions with the greatest charges.
  • Ba³⁺ and PO₄³⁻ (from Na₃PO₄) have higher charges than Ba²⁺ and CO₃²⁻, so Ba₃(PO₄)₂ has the strongest ionic bonding.

🧪 Redissolving precipitates

• Many transition metal carbonate precipitates can be redissolved by adding an acid.
• Example: a precipitate formed by mixing Na₂CO₃ with a transition metal ion solution can be redissolved with 1 M HNO₃ (the acid reacts with the carbonate to form CO₂ and water, removing the anion and dissolving the precipitate).

🔬 Non-colligative properties and special cases

🔬 What is NOT a colligative property

• Density does not depend on the number of particles in the same way; it depends on mass and volume.
• Vapor pressure, freezing point, boiling point, and osmotic pressure are all colligative.

🧪 Ideal vs. non-ideal solutions

• An ideal solution is one where the van't Hoff factor equals the theoretical value and interactions between particles are negligible.
• Solutions behave closest to ideal at low concentrations and with low-charge ions.
• Example: 0.01 M NaCl (i = 2, low charge) behaves more ideally than 0.01 M Al₂(SO₄)₃ (i = 5, high charges, strong interactions).

🔬 Ion interactions and the van't Hoff factor

• In a CuSO₄ solution, interactions between Cu²⁺ and SO₄²⁻ are greater than between K⁺ and Cl⁻ in a KCl solution.
• Why? The charges of the ions are greater (Cu²⁺ and SO₄²⁻ are both doubly charged; K⁺ and Cl⁻ are singly charged).
• Higher charges lead to stronger electrostatic attraction (ion pairing), reducing the effective number of free particles.

🧪 Distillation and volatile contaminants

• Distillation purifies water by boiling and condensing the vapor.
• It works best when the contaminant is non-volatile (does not evaporate easily).
• Example: HgCl₂ (mercury(II) chloride) is less volatile than organic compounds like CH₂Cl₂ or C₂H₂Br₂, so distillation is most effective for removing HgCl₂.

🔬 Mixing acids and bases: effect on freezing point

• Mixing 50 mL of 4.00 M NaOH with 50 mL of 4.00 M HNO₃ produces a neutralization reaction:
  • NaOH + HNO₃ → NaNO₃ + H₂O
• The final solution is a salt solution (NaNO₃) with a lower freezing point than either of the original solutions (because the salt dissociates into ions, increasing particle concentration).

🧪 Ion-exchange resins and water softeners

• Water softeners use resins with Na⁺ ions to remove "hard" ions (Ca²⁺, Mg²⁺) from water.
• Ions in solution displace Na⁺ on the resin if they have a greater affinity for the resin.
• A sodium ion water softener is least effective at removing Li⁺ (lithium is chemically similar to sodium and has low affinity for the resin compared to larger or doubly charged ions).

🔬 Identifying unknowns by precipitation

• If an unknown solution produces:
  • A white precipitate with NaCl
  • No reaction with NaNO₃
  • A yellow precipitate with NaI
  • A black precipitate with Na₂S
• The unknown is likely AgNO₃ (silver nitrate):
  • AgCl is white
  • AgI is yellow
  • Ag₂S is black
  • Silver salts are known for forming colored precipitates with halides and sulfides.

🧭 Overview

🧠 One-sentence thesis

Chemical reaction rates depend on multiple factors—concentration, physical state, and temperature—and studying these rates through rate laws allows us to predict when reactions complete, when they started, and how to control them in practical applications.

📌 Key points (3–5)

• What affects reaction rate: concentration of reactants (higher = faster contact), physical state (gases mix faster than solids), and temperature (higher = faster molecular motion).
• Why physical state matters: solids react only at contact surfaces, so powdering increases surface area and rate; gases mix readily and react faster.
• Rate laws and their uses: mathematical relationships that predict reaction completion (chemical engineering), starting time (archaeology/dating), or dosing schedules (medicine).
• Common confusion: rate constant vs. rate of reaction—the rate constant depends on temperature and activation energy but is independent of concentration and enthalpy change.
• Order of reaction: determined experimentally from concentration data; tells how each reactant's concentration affects the overall rate.

🔬 Factors controlling reaction speed

🧪 Concentration of reactants

• Core idea: Higher concentration → more frequent collisions → faster reaction.
• Atoms or molecules must come into contact to react; more particles per unit volume increases the likelihood of contact.
• Example: Doubling the concentration of a reactant may double, quadruple, or have no effect on the rate, depending on the reaction order for that substance.

🧊 Physical state and surface area

Physical state: whether reactants are solid, liquid, or gas; determines how easily particles can contact each other.

• Solid-solid reactions: Very slow because contact occurs only at the interface.
  • Example: A block of calcium and a block of sulfur touching react only along the contact plane; once those atoms react, the reaction stops.
  • Grinding into powder increases surface area dramatically, allowing many more atoms to contact and react.
• Gas-phase reactions: Much faster because gases mix readily.
  • Example: Hydrogen gas and chlorine gas mix quickly, so molecules contact and react rapidly.
• General rule: Gases react more rapidly than solids.

🌡️ Temperature

• Mechanism: Higher temperature → molecules move faster → more frequent and energetic collisions → faster reaction.
• The excerpt notes that increasing temperature by 30°C can increase the rate by a factor of 2 or 8 (questions 256, 281 suggest typical factors).
• Don't confuse: Temperature affects the rate constant itself, not just the rate; lowering temperature from 37°C to 27°C roughly halves the rate (question 281).

📐 Rate laws and reaction order

📊 What is a rate law?

Rate law: a mathematical relationship showing how the rate of a reaction depends on the concentrations of reactants.

• Developed by studying reaction rates experimentally.
• Form: Rate = k [A]^m [B]^n, where k is the rate constant, [A] and [B] are concentrations, and m and n are the orders with respect to each reactant.
• The exponents (orders) are determined from experimental data, not from the balanced equation.

🔢 Determining reaction order from data

• Method: Compare experiments where one concentration changes while others stay constant; see how the rate changes.
• Example (question 257 table): If doubling [A] quadruples the rate while [B] is constant, the reaction is second order in A.
• Example (question 259): The order for each substance (I⁻, ClO⁻, OH⁻) is found by comparing trials; if [OH⁻] appears in the rate law, it affects the rate even if it's not a "reactant" in the overall equation.

🧮 Using rate laws for calculations

• Second-order reaction (question 261): Rate = k [A]², so k = Rate / [A]².
  • Example: If rate = 1.6 × 10⁻² M/s and [acetaldehyde] = 0.40 M, then k = (1.6 × 10⁻²) / (0.40)² = 0.10 M⁻¹ s⁻¹.
• Effect of concentration changes (question 258): If Rate = k [A]² [B], halving [A] reduces the rate by (1/2)² = 1/4, and doubling [B] increases it by 2; net effect = (1/4) × 2 = 1/2.

⚙️ Reaction mechanisms and energy profiles

🧩 Rate-determining step

Rate-determining step: the slowest step in a multi-step mechanism; it controls the overall reaction rate.

• In a mechanism with multiple steps, the slowest step is the bottleneck.
• Example (question 266): A mechanism with three steps (fast, slow, fast) has the slow step as rate-determining.
• Energy profile: The rate-determining step corresponds to the highest activation energy peak (question 272).

🏔️ Activation energy and energy profiles

• Activation energy: the energy barrier that must be overcome for reactants to form products.
• Higher activation energy → slower reaction (question 263: H₂ + N₂ is slower than H₂ + O₂ because activation energy is higher, even though both are exothermic).
• Energy profile features:
  • Peak height = activation energy.
  • Difference between start and end = enthalpy change (ΔH).
  • Exothermic: products lower than reactants (question 270).
  • Endothermic: products higher than reactants.
• Reverse reaction activation energy (question 273): For an exothermic reaction with forward activation energy 285 kJ/mol and ΔH = –145 kJ/mol, reverse activation energy = 285 + 145 = 430 kJ/mol.

🧬 Intermediates and catalysts in mechanisms

Role	Definition	Example from excerpt
Intermediate	Formed in one step, consumed in a later step; does not appear in overall equation	•OH in Cl₂ + H₂O mechanism (question 277)
Catalyst	Speeds up reaction without being consumed; appears in mechanism but not overall equation	NO in ozone destruction (question 266); hemoglobin in H₂O₂ decomposition (question 274)
Inhibitor	Slows down reaction	Hydroxide ion (OH⁻) in I⁻ + ClO⁻ reaction (question 260)

• How to identify: An intermediate is produced then consumed within the mechanism; a catalyst is present at the start and regenerated at the end.
• Example (question 283): Br(g) is formed in the first step and consumed in the second, so it is an intermediate.

🧪 Catalysts and reaction control

🔑 What catalysts do

• Effect: Lower the activation energy → increase the rate constant → speed up the reaction.
• Catalysts do not change the enthalpy (ΔH) or the equilibrium constant; they only change the path (question 278: a catalyst changes the height of the peak, not the start or end positions).
• Example (question 274): Hemoglobin increases the rate of H₂O₂ decomposition from 1.0 × 10⁻⁸ M/s to 2.0 × 10⁻¹ M/s; the rate of O₂ production is half the rate of H₂O₂ disappearance (stoichiometry 2:1), so 1.0 × 10⁻¹ M/s.

🛑 Inhibitors and promoters

• Inhibitor: Substance that slows the reaction (e.g., OH⁻ in question 260).
• Promoter: Substance that speeds up the reaction (question 277 context).
• Free radical scavenger: Removes reactive intermediates, inhibiting the reaction (question 268).

🌟 Light and other factors

• Some mechanisms are promoted by light (photochemical reactions, question 269).
• Adding an inert gas (e.g., argon) at constant volume has the least effect on rate because it doesn't change concentrations or collisions (question 269).

🧮 Practical applications of rate laws

🏭 Chemical engineering

• Engineers need to know when a reaction is complete to design plants efficiently.
• Rate laws predict how long a reaction takes under given conditions.

🏺 Archaeology and dating

• Rate laws allow archaeologists to determine when a reaction started (e.g., radioactive decay for carbon dating).
• Example: First-order kinetics with known half-life can calculate time elapsed.

💊 Medicine and pharmacology

• Rate laws predict when a patient needs another dose of a drug.
• Example (question 294): A drug with first-order metabolism and 4-hour half-life drops to 10% of initial concentration after about 13.3 hours (between 12 and 24 hours; closest answer is 12 hours).
  • Calculation: After 1 half-life (4 h) → 50%; after 2 (8 h) → 25%; after 3 (12 h) → 12.5%; after 4 (16 h) → 6.25%. So ~13 hours to reach 10%.

🔍 Common confusions and distinctions

⚖️ Rate constant vs. rate of reaction

Property	Rate constant (k)	Rate of reaction
Depends on	Temperature, activation energy, presence of catalyst	Concentration of reactants, rate constant
Independent of	Concentration, enthalpy change (ΔH)	(Varies with all factors)
Changes when	Temperature changes, catalyst added/removed	Concentration changes, temperature changes, catalyst added

• Don't confuse (question 264): The rate constant is independent of enthalpy change and concentration.
• Don't confuse (question 287): The rate of reaction is independent of the overall enthalpy change (ΔH), but depends on activation energy.

🔄 Forward vs. reverse reaction rates

• Energy profiles show both directions (question 275).
• The relative rates depend on activation energies: lower activation energy → faster reaction.
• At equilibrium, forward and reverse rates are equal, but before equilibrium, the direction with lower activation energy is faster.

📉 What does NOT change the rate?

• Concentration of products (question 276): Adding CaCl₂ to the reaction CaCO₃ + HCl → CaCl₂ + … does not change the rate (only reactant concentrations matter for rate).
• Enthalpy change (question 282): Reactions with different ΔH values cannot be compared for rate without knowing activation energies; ΔH alone does not determine rate.

🧪 Stoichiometry and rate expressions

• The rate of disappearance/appearance depends on stoichiometric coefficients.
• Example (question 288): For 2 KMnO₄ → 2 MnSO₄, Rate = –(1/2) Δ[KMnO₄]/Δt (the 1/2 accounts for the coefficient 2).
• Example (question 293): For 2 MnO₄⁻ → 5 I₂, if MnO₄⁻ disappears at 2.0 × 10⁻³ M/s, then I₂ appears at (5/2) × 2.0 × 10⁻³ = 5.0 × 10⁻³ M/s.

🧭 Overview

🧠 One-sentence thesis

Chemical equilibrium is a dynamic state where forward and reverse reactions proceed at equal rates, and Le Châtelier's principle predicts how systems respond to stresses such as concentration, pressure, temperature changes, and catalysts.

📌 Key points (3–5)

• Dynamic equilibrium: both forward and reverse reactions continue simultaneously at equal rates, so concentrations remain constant even though reactions never stop.
• Le Châtelier's principle: applying stress (concentration, volume/pressure, temperature, or catalyst) causes the system to shift to counter that stress and establish a new equilibrium.
• Catalysts do not change equilibrium position: they speed up both forward and reverse reactions equally, so equilibrium constants remain unchanged.
• Common confusion: catalysts vs. concentration changes—catalysts affect rate but not position; concentration changes shift equilibrium position.
• Solubility equilibria (Ksp): apply the same principles; common ion effect decreases solubility, while removing ions (e.g., via acid) can increase solubility.

⚖️ What is chemical equilibrium

⚖️ Reversibility and the equilibrium state

Chemical equilibrium: the state when the forward reaction proceeds at the same rate as the reverse reaction, so no apparent change in concentration occurs.

• All reactions are, in principle, reversible—products can reform reactants.
• Initially, one reaction (forward or reverse) is faster because activation energies differ.
• As the faster reaction consumes its reactants, it slows down; the slower reaction speeds up as its reactants accumulate.
• Eventually, both rates become equal → equilibrium is established.

🔄 Dynamic nature

• Equilibrium is dynamic, not static: molecules continue reacting in both directions.
• There is no net change in concentrations, but individual molecules are constantly converting between reactants and products.
• The double arrow symbol (⇌) indicates equilibrium.
• Example: In a weak acid equilibrium like HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq), the solution contains all three species simultaneously.

📊 Maximum stability

• At equilibrium, the system's stability is at a maximum.
• No net change occurs unless an external stress is applied.

🧪 Le Châtelier's principle and stress responses

🧪 The principle

Le Châtelier's principle: applying a stress to a system in equilibrium causes the system to respond in order to counter the stress, leading to a new equilibrium.

• A "stress" is virtually any change imposed on an equilibrium system.
• The system shifts to partially undo the stress.

🔢 Concentration changes

• Adding a reactant or product: the equilibrium shifts to consume the added substance.
• Removing a substance: the equilibrium shifts to produce more of it.
• Example: Adding more reactant shifts equilibrium toward products; removing a product shifts equilibrium to produce more product.
• Don't confuse: adding a solid reactant or product in a heterogeneous equilibrium has no effect because solids do not appear in the equilibrium expression.

📏 Volume and pressure changes

• Applies to gaseous equilibria.
• Decreasing volume (increasing pressure): equilibrium shifts toward the side with fewer moles of gas.
• Increasing volume (decreasing pressure): equilibrium shifts toward the side with more moles of gas.
• Example: For CH₄(g) + H₂O(g) ⇌ CO(g) + 3 H₂(g), increasing pressure favors the left (2 moles) over the right (4 moles).
• If both sides have equal moles of gas, pressure changes have minimal effect.

🌡️ Temperature changes

• Temperature is the only stress that changes the value of the equilibrium constant.
• Exothermic reactions (ΔH < 0): increasing temperature shifts equilibrium toward reactants (left); decreasing temperature shifts toward products (right).
• Endothermic reactions (ΔH > 0): increasing temperature shifts equilibrium toward products (right).
• Example: For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) with ΔH = –17 kJ/mol, increasing temperature shifts left, decreasing ammonia concentration.

⚗️ Catalysts

• Catalysts speed up both forward and reverse reactions equally.
• They do not change the equilibrium position or the equilibrium constant.
• They only help the system reach equilibrium faster.
• Don't confuse: catalysts with concentration changes—catalysts affect kinetics (speed), not thermodynamics (position).

🧮 Equilibrium constants and calculations

🧮 Equilibrium constant expressions

• For a general reaction, the equilibrium constant (K) is the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient.
• Kc: uses molar concentrations.
• Kp: uses partial pressures (for gases).
• Example: For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), Kp = [SO₃]² / ([SO₂]² [O₂]).
• Solids and pure liquids do not appear in the expression.

🔍 Solubility product constant (Ksp)

Ksp: the equilibrium constant for the dissolution of a sparingly soluble ionic compound.

• Smaller Ksp → less soluble.
• Example: AgCN (Ksp = 2.2 × 10⁻¹⁶) is much less soluble than AgOCN (Ksp = 2.3 × 10⁻⁷).
• For a compound like CaF₂, Ksp = [Ca²⁺][F⁻]².

🧂 Common ion effect

• Adding an ion that is already part of the equilibrium decreases solubility.
• Example: CaSO₄ is less soluble in 0.1 M Na₂SO₄ than in pure water because the sulfate ion is common to both.
• Conversely, removing an ion (e.g., by forming a weak acid) can increase solubility.

📉 Reaction quotient (Q)

• Q has the same form as K but uses current concentrations, not equilibrium concentrations.
• Q < K: reaction shifts right (toward products).
• Q > K: reaction shifts left (toward reactants).
• Q = K: system is at equilibrium.

🧬 Special equilibria and applications

🧬 Autoionization of water

• Water undergoes a reversible reaction: 2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq).
• This is an endothermic process.
• Increasing temperature shifts equilibrium right, increasing [H₃O⁺] and decreasing pH.
• Don't confuse: pH of pure water at higher temperatures is not 7, but the solution is still neutral because [H₃O⁺] = [OH⁻].

🩸 Dynamic equilibria with isotopes

• In a dynamic equilibrium, adding a soluble compound with a different isotope leads to isotope exchange.
• Example: Adding nonradioactive Co-59 as CoCl₂ to a solution in equilibrium with radioactive Co-60 in solid CoS will result in both isotopes appearing in the solid over time, because dissolution and precipitation continue.

🧪 Precipitation and dissolution

• Precipitation occurs when the ion product exceeds Ksp.
• Adding a common ion shifts equilibrium toward the solid (precipitation).
• Adding an acid can shift equilibrium toward dissolution if it reacts with one of the ions (e.g., forming a weak acid or gas).

Stress type	Effect on equilibrium	Effect on K
Concentration change	Shifts position	No change
Pressure/volume change	Shifts position (if Δn ≠ 0)	No change
Temperature change	Shifts position	Changes K
Catalyst	No shift	No change

🧭 Overview

🧠 One-sentence thesis

The carbonic acid/bicarbonate buffer system maintains blood pH within the narrow 7.2–7.6 range by responding to hydrogen ion changes through equilibria involving carbon dioxide, carbonic acid, and bicarbonate ion.

📌 Key points (3–5)

• Why blood pH must be controlled: many body reactions are pH dependent; values outside 7.2–7.6 can cause acidosis or death.
• The key buffer system: carbonic acid (H₂CO₃) and bicarbonate ion (HCO₃⁻) work together, with carbon dioxide playing a critical role because carbonic acid decomposes to CO₂ and water.
• Buffer capacity principle: a 1:1 ratio of acid to base maximizes buffering capacity, making pH equal to pKₐ.
• Common confusion: the carbonic acid buffer's pKₐ is 6.1, far from blood pH 7.4, but the CO₂/bicarbonate ratio adjusts the effective pH.
• Acid-base fundamentals tested: conjugate acid-base pairs, pH calculations, polyprotic acids, and the relationship between Kₐ and Kᵦ.

🩸 The blood buffer system

🩸 Why blood pH control matters

• Normal blood pH is 7.4; the safe range is 7.2 to 7.6.
• When the body oxidizes glucose, it produces both carbon dioxide and hydrogen ions (H⁺).
• If H⁺ concentration increases unchecked, blood pH drops below 7.4, causing acidosis.
• Many reactions in the body are pH dependent, so extreme pH values can result in death.

🧪 The carbonic acid/bicarbonate equilibrium

The most important buffer system in the body: H₂CO₃(aq) ⇌ HCO₃⁻(aq) + H⁺(aq)

• Kₐ for this system is approximately 7.9 × 10⁻⁷ at body temperature.
• For maximum buffering capacity, the acid-to-base ratio should be 1:1, giving pH = pKₐ.
• For carbonic acid, pKₐ = 6.1, which is far outside the normal blood pH of 7.4.
• Don't confuse: the simple pKₐ calculation doesn't account for the full system.

🌬️ The role of carbon dioxide

• Carbonic acid is unstable in water and decomposes: H₂CO₃(aq) ⇌ CO₂(g) + H₂O(l)
• A more complete representation combines both equilibria:
  • CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) ⇌ HCO₃⁻(aq) + H⁺(aq)
• The amount of carbon dioxide affects the buffer system.
• An equilibrium expression relates hydrogen ion concentration to the CO₂ and bicarbonate concentrations.
• The pH depends on the ratio of carbon dioxide to bicarbonate, not just the pKₐ of carbonic acid alone.

Example: If CO₂ levels rise (e.g., from impaired breathing), the equilibrium shifts to produce more H⁺, lowering pH and potentially causing acidosis.

🧪 Buffer fundamentals

🧪 What makes a buffer

• A buffer resists pH changes when small amounts of acid or base are added.
• A buffer requires a conjugate acid-base pair: a weak acid and its conjugate base (or a weak base and its conjugate acid).
• Example combinations from the excerpt:
  • NaHCO₃ and Na₂CO₃ (bicarbonate and carbonate)
  • Nitrous acid (HNO₂) combined with a source of its conjugate base (e.g., NaOH to form NO₂⁻)

🔄 How buffers respond to added acid or base

• Adding base (e.g., NaOH) to a weak acid solution:
  • Shifts the equilibrium to the right (toward the conjugate base).
  • Increases the concentration of the conjugate base (e.g., formate ion from formic acid).
  • Increases the solution's ability to conduct electricity (more ions).
  • Does not decrease the pH; it increases pH.
• Adding acid to a buffer consumes the conjugate base, minimizing pH drop.

📊 Maximizing buffer capacity

Factor	What it means	Why it matters
High concentrations	More moles of acid and base	More capacity to neutralize added H⁺ or OH⁻
Equal concentrations	Acid-to-base ratio = 1:1	pH = pKₐ; balanced capacity for both directions
Appropriate pKₐ	pKₐ near the target pH	Buffer works best within ±1 pH unit of pKₐ

Don't confuse: "buffer capacity" (how much acid/base it can neutralize) with "buffer range" (the pH range over which it works well).

🧂 Common ion effect

Adding a salt containing the conjugate base to a weak acid solution shifts the equilibrium and changes the pH.

• Example: Adding sodium cyanate (NaOCN) to cyanic acid (HOCN) solution.
• The cyanate ion (OCN⁻) is the conjugate base of cyanic acid.
• The common ion effect shifts the equilibrium to the left, reducing ionization of the acid.
• This increases the pH (makes the solution less acidic).

🔢 pH and acid-base calculations

🔢 pH from strong acids

• For a strong acid like HNO₃, assume complete dissociation.
• A 10⁻² M HNO₃ solution has [H⁺] = 10⁻² M, so pH = 2.
• Don't confuse: diluting an acid raises pH but cannot make it basic; diluting HCl with pH 6 to 1000 mL will give pH < 7 (still acidic).

🔢 pH from weak acids and bases

• For a weak acid with Kₐ = 2 × 10⁻⁶ and equal concentrations of acid and conjugate base:
  • pH ≈ pKₐ = –log(2 × 10⁻⁶) ≈ 5.7 (between 5 and 6).
• For a weak base like pyridine (C₅H₅N) with Kᵦ = 2 × 10⁻⁹ in 1 M solution:
  • Calculate [OH⁻], then pOH, then pH = 14 – pOH.
  • Approximate pH ≈ 9.

🔢 Polyprotic acids

Polyprotic acids can donate more than one proton.

• Example: Arsenic acid (H₃AsO₄) is a weak triprotic acid with three ionization steps.
• Each step has its own Kₐ (Kₐ₁, Kₐ₂, Kₐ₃), with Kₐ₁ > Kₐ₂ > Kₐ₃.
• In a 0.10 M H₃AsO₄ solution:
  • [H⁺] is dominated by the first ionization.
  • [H⁺] > [H₂AsO₄⁻] because not all acid molecules ionize.
  • [AsO₄³⁻] is very small because the third ionization is weak.

Example: Phosphoric acid (PO(OH)₃, also written H₃PO₄) is polyprotic; acetic acid (CH₃COOH) is monoprotic.

🔢 Relating Kₐ and Kᵦ

• For a conjugate acid-base pair: Kₐ × Kᵦ = Kw (the ion product of water, 1.0 × 10⁻¹⁴ at 25°C).
• If a compound's solution has pH 4.3 (acidic), the compound is a weak acid, so Kₐ > Kᵦ.

🔢 pH comparisons and concentration ratios

• pH is a logarithmic scale: pH = –log[H⁺].
• Vinegar (pH 3) vs. ammonia (pH 11): [H⁺] in vinegar is 10⁸ times higher.
• Blood (pH 7.4) vs. stomach acid (pH 1): [H⁺] in stomach acid is 10⁶ times higher.

🧬 Acid-base strength and structure

🧬 What makes an acid strong or weak

• Strong acids have Kₐ > 1; they ionize nearly completely in water.
• Weak acids have Kₐ < 1; they ionize only partially.
• In water, all strong acids appear equally strong (leveling effect); use a different solvent (e.g., acetic acid, ammonia) to differentiate them.

🧬 Factors affecting acid strength

Factor	Effect on acidity	Example from excerpt
Electronegativity of X in H–X	Higher electronegativity → weaker H–X bond → stronger acid	Hydrogen halides: HI > HBr > HCl > HF (bond energy decreases down the group)
Bond energy	Lower bond energy → easier to break H–X → stronger acid	H₂X compounds: acidity increases as bond energy decreases
Electron-withdrawing groups	Stabilize the conjugate base → stronger acid	CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH (more Cl atoms withdraw electrons)

Don't confuse: HF is the weakest hydrogen halide acid despite F being the most electronegative, because the H–F bond is very strong.

🧬 Conjugate acid-base pairs

When an acid donates a proton, it forms its conjugate base; when a base accepts a proton, it forms its conjugate acid.

• Example: HCO₃⁻ is the conjugate base of H₂CO₃.
• Example: NH₄⁺ is the conjugate acid of NH₃.
• Incorrect pairing: HNO₃ is not the conjugate acid of NO₂⁻; it would be the conjugate acid of NO₃⁻.

🧬 Comparing base strength

• The weaker the acid, the stronger its conjugate base.
• Given Kₐ values, the conjugate base of the acid with the smallest Kₐ is the strongest base.
• Example: If Kₐ(HCN) < Kₐ(HC₂H₃O₂), then CN⁻ is a stronger base than C₂H₃O₂⁻.

🧪 Acid-base behavior in solution

🧪 Acidic, basic, and neutral salts

• Neutral salts (e.g., NaCl, KNO₃): neither cation nor anion hydrolyzes; pH ≈ 7.
• Basic salts (e.g., Na₂CO₃, K₃PO₄): the anion is the conjugate base of a weak acid and hydrolyzes to produce OH⁻; pH > 7.
• Acidic salts (e.g., NH₄Cl): the cation is the conjugate acid of a weak base and hydrolyzes to produce H⁺; pH < 7.

Example: Sodium fluoride (NaF) in water produces a basic solution (pH > 7) because F⁻ is the conjugate base of weak HF and hydrolyzes: F⁻ + H₂O ⇌ HF + OH⁻.

🧪 Amphoteric substances

Amphoteric substances can act as either an acid or a base.

• Example: Amino acids have both an amine group (–NH₂, basic) and a carboxylic acid group (–COOH, acidic).
• They can donate or accept protons depending on the pH of the solution.
• Bicarbonate ion (HCO₃⁻) is amphoteric: it can donate H⁺ (acting as an acid) or accept H⁺ (acting as a base).

🧪 Brønsted-Lowry acids and bases

Brønsted-Lowry acid: proton donor. Brønsted-Lowry base: proton acceptor.

• Example reaction: HNO₃(aq) + H₂SO₄(aq) → (NO₂⁺)(HSO₄⁻)(aq) + H₂O(l)
  • HNO₃ accepts a proton from H₂SO₄, so HNO₃ is the Brønsted-Lowry base.
  • H₂SO₄ donates a proton, so it is the Brønsted-Lowry acid.

🧪 Indicators and titrations

• Acid-base indicators change color at specific pH ranges.
• For a titration of a weak acid with a strong base, the equivalence point pH is > 7.
• Choose an indicator that changes color near the equivalence point pH.
• Example: For a weak acid titrated with strong base, use an indicator with a transition range above pH 7 (e.g., phenolphthalein).

🧪 Neutralization and pH after mixing

• Mixing equal volumes of 0.50 M HNO₃ (strong acid) and 0.50 M NH₃ (weak base):
  • They react 1:1 to form NH₄NO₃.
  • The resulting solution is 0.25 M NH₄NO₃ (because volume doubled).
  • NH₄⁺ is a weak acid, so pH < 7.
• After neutralizing 99% of 0.1 M HNO₃ with KOH, 1% remains: [H⁺] = 0.001 M, so pH = 3.

🌍 Applications and special cases

🌍 Solubility and pH

• Salts of weak acids (e.g., CaCO₃) are more soluble in acidic solutions because H⁺ reacts with the anion (CO₃²⁻), shifting the solubility equilibrium.
• Example: CaCO₃ is more soluble in blood (pH 7.4) than in pure water (pH 7) because blood is slightly more acidic than pure water? No—blood pH 7.4 is slightly basic. The excerpt states CaCO₃ is more soluble in blood than in water, implying other factors (e.g., buffering) affect solubility.

🌍 Oxides and acidity

• Nonmetal oxides (e.g., SO₂, CrO₃) react with water to form acids.
• Example: A chromium oxide dissolved in water gives pH 1.0 (very acidic) → likely CrO₃ (chromium in high oxidation state).
• Metal oxides (e.g., CaO) react with water to form bases.

🌍 Temperature and Kw

• The reaction H⁺ + OH⁻ → H₂O is very exothermic (releases heat).
• By Le Chatelier's principle, heating water shifts the equilibrium to the left, increasing [H⁺] and [OH⁻].
• Therefore, Kw increases with temperature.

🌍 Carbonate detection

• To detect carbonate ion (CO₃²⁻) in solution, add an acid.
• The acid reacts with carbonate to produce CO₂ gas: CO₃²⁻ + 2H⁺ → H₂O + CO₂(g).
• Observation: gas generation indicates carbonate presence.

🌍 Functional groups in indicators

• Acid-base indicators use functional groups that can donate or accept protons and change structure (and color) when protonated or deprotonated.
• Possible groups: sulfonic acid (–SO₃H), amine (–NH₂), carboxylic acid (–COOH).
• Alcohol (–OH) does not act as an acid-base indicator functional group because it is too weak an acid and not a base.

🌍 Methylamine buffer

• Methylammonium ion (CH₃NH₃⁺) has Kₐ = 2 × 10⁻¹².
• When [CH₃NH₂] = [CH₃NH₃⁺], pH = pKₐ = –log(2 × 10⁻¹²) ≈ 12.

🌍 Hydrogen phosphate dissociation

• Hydrogen phosphate ion (HPO₄²⁻) can act as an acid by donating a proton.
• Acidic dissociation: HPO₄²⁻ → H⁺ + PO₄³⁻.
• The result is phosphate ion (PO₄³⁻).

🌍 Hydrogen halide conjugate bases

• Hydrogen halides (HF, HCl, HBr, HI) are acids in water.
• Their conjugate bases are the halide ions: F⁻, Cl⁻, Br⁻, I⁻.
• Astatine (At) is also a halogen, so At⁻ (astatide ion) could be the conjugate base of HAt.

🧭 Overview

🧠 One-sentence thesis

Thermodynamics governs whether reactions are spontaneous by relating enthalpy, entropy, and Gibbs free energy, and calorimetry allows scientists to measure the heat changes that accompany chemical processes.

📌 Key points (3–5)

• Calorimetry measures heat changes: constant-pressure calorimeters (coffee cups) measure enthalpy changes in solution reactions; constant-volume bomb calorimeters measure heats of combustion.
• Spontaneity depends on ΔG: a positive ΔG means a reaction is nonspontaneous under those conditions; a negative ΔG means spontaneous.
• Entropy and phase changes: gases have higher entropy than solids; reactions that increase the number of gas molecules typically increase entropy.
• Common confusion—system vs surroundings: a process can have negative entropy change in the system (like freezing) but still be spontaneous if the surroundings gain enough entropy.
• Temperature affects spontaneity: changing temperature can shift a nonspontaneous endothermic reaction to spontaneous by altering the balance between ΔH and TΔS in the Gibbs equation.

🔬 Calorimetry and measuring heat

☕ Constant-pressure calorimeters (coffee cup)

Constant-pressure calorimeter: a simple device that holds pressure constant at atmospheric pressure by having the reaction chamber open to the atmosphere; used to measure enthalpy changes in solution reactions.

• How it works: two stacked polystyrene (Styrofoam) coffee cups insulate the system from surroundings; a thermometer measures temperature change; a stirrer circulates the solution.
• Why Styrofoam works: it has very low heat capacity, so almost no heat is lost to heating the cups—heat stays in the solution.
• Key principle: heat lost or gained by the process equals heat gained or lost by the solution; knowing the solution's heat capacity, mass, and temperature change allows calculation of heat of reaction (q_rxn).
• Adiabatic nature: the insulation prevents heat exchange with surroundings (question 379 confirms this is why a coffee cup is effective).
• Limitations: will not work at very high temperatures, very high pressure, or very low pressure (question 380).

Example: Mixing 25.00 mL of 1.5 M NaOH with 25.00 mL of 1.5 M HNO₃ causes a 15.0°C temperature increase; repeating with acetic acid (a weak acid) gives a smaller temperature change because the enthalpy change is smaller (question 381).

💣 Constant-volume calorimeters (bomb calorimeter)

Bomb calorimeter: a constant-volume calorimeter with a heavy-walled reaction vessel used to measure heats of combustion.

• How it works: sample placed in a cup inside the bomb; air removed and replaced with oxygen; electrical ignition burns the sample; heat absorbed by water and calorimeter itself.
• Setup: bomb sits in insulated container with known amount of water; thermometer measures temperature changes; stirrer circulates water around bomb.
• Use case: measuring heat associated with combustion (burning) of substances.

🧪 Interpreting calorimetry experiments

• Dissolving can be exothermic or endothermic: question 382 shows that when NaOH(s) dissolves, the process is exothermic, adding extra heat to the system beyond the neutralization reaction.
• Don't confuse: the heat measured includes both the reaction heat and any heat from dissolving solids.

🌡️ Entropy and the second law

📈 What entropy measures

Entropy: a measure of disorder or the number of available microstates; gases usually have greater entropy than solids.

• Phase and entropy: gases > liquids > solids in entropy (question 386 confirms this).
• Molecular changes: reactions that increase the number of gas molecules typically have the greatest increase in entropy (question 404).
• Absolute zero: cooling any substance to absolute zero reduces its entropy to zero (question 386).

Example: The reaction CaO(s) + CO₂(g) → CaCO₃(s) decreases entropy because a gas molecule is consumed and only a solid is produced (question 390).

🔄 Entropy changes in reactions

• Gas formation increases entropy: converting liquids or solids to gases increases disorder.
• Condensation decreases entropy: ammonia gas condensing to liquid ammonia has negative ΔS (question 396).
• Heating increases entropy: heating water from 25°C to 75°C increases entropy mainly because the average kinetic energy of water molecules increases (question 405).

🌍 Second law and spontaneity

• Second law of thermodynamics: for any spontaneous process, the total entropy change (system + surroundings) must be positive.
• Apparent paradox: a liquid can freeze spontaneously even though the system's entropy decreases—this is because the removal of heat during freezing causes a greater positive entropy change in the surroundings (question 395).
• Don't confuse: system entropy can decrease if surroundings entropy increases more; the total must be positive for spontaneity.

⚡ Gibbs free energy and spontaneity

🎯 What ΔG tells us

Gibbs free energy change (ΔG): determines spontaneity; positive ΔG means nonspontaneous, negative ΔG means spontaneous, ΔG = 0 means equilibrium.

• Nonspontaneous reactions: if a reaction is nonspontaneous, ΔG must be positive under those conditions (question 385).
• Relationship to equilibrium constant: positive ΔG corresponds to K < 1; negative ΔG corresponds to K > 1 (question 398).
• Standard conditions: ΔH°_f and ΔG°_f for any element in its standard state are zero (question 386).
• Note: STP and standard conditions are not the same (question 386).

🔥 Enthalpy and entropy balance

The Gibbs equation relates ΔG to enthalpy (ΔH) and entropy (ΔS): ΔG = ΔH – TΔS (in words: free energy change equals enthalpy change minus temperature times entropy change).

ΔH	ΔS	Spontaneity
Negative	Positive	Always spontaneous (question 393)
Positive	Negative	Never spontaneous (question 393)
Positive	Positive	Spontaneous at high temperature (question 393)
Negative	Negative	Spontaneous at low temperature; requires cooling to become spontaneous (question 393)

Example: The reaction COCl₂(g) → CO(g) + Cl₂(g) is nonspontaneous and endothermic (ΔH positive); increasing temperature can make it spontaneous by making the TΔS term larger (question 401).

🧮 Calculating ΔG and ΔH

• Standard Gibbs free energy of formation: for HF(g), ΔG°_f = –270 kJ/mol; for the reaction H₂(g) + F₂(g) → 2 HF(g), ΔG_rxn = 2 × (–270 kJ) = –540 kJ (question 387).
• At equilibrium: ΔG = 0 (question 409).
• Temperature dependence: changing temperature will most likely change ΔG° for a reaction; increasing reaction rate, adding a catalyst, or lowering activation energy do not change ΔG° (question 384).

Don't confuse: catalysts and activation energy affect reaction rate, not thermodynamic spontaneity (ΔG).

🔋 Energy, work, and the first law

⚙️ Internal energy change (ΔE)

First law of thermodynamics: ΔE = q + w, where q is heat and w is work.

• Sign conventions: heat lost by the system is negative; heat gained is positive; work done by the system (expansion) is negative; work done on the system (compression) is positive.
• Adiabatic system: no heat exchange between system and surroundings (q = 0) (question 389).
• Isothermal system: temperature remains constant (not the same as adiabatic).

Example: A helium gas sample expands, doing 1,475 J of work on surroundings (w = –1,475 J) and cooling through removal of 375 J of heat (q = –375 J); ΔE = –375 J + (–1,475 J) = –1,850 J (question 394).

🏋️ Work in gas reactions

• Work calculation: when a gas is produced at constant pressure, work = –PΔV (in words: negative pressure times volume change).
• Example: 4.0 g of calcium reacts with water to generate hydrogen gas at 0.0°C and 1.00 atm; the work involved is approximately –220 J (question 399).

🌐 Greenhouse effect and energy

• If a gas system absorbs more energy than it emits (greenhouse effect), q is positive (question 403).

🧊 Phase changes and intermolecular forces

🌡️ Enthalpy of phase changes

• Vaporization vs fusion: the enthalpy change for vaporization (liquid → gas) is higher than for fusion (solid → liquid) because it requires more energy to completely overcome intermolecular forces than to partially overcome them (question 397).
• Example: For CCl₄, vaporization requires more energy to completely overcome London dispersion forces than fusion does to partially overcome them.

❄️ Freezing and condensation

• Freezing (solidification): ΔH is negative (exothermic), ΔS is negative (entropy decreases).
• Condensation: ammonia gas condensing to liquid has ΔH negative and ΔS negative (question 396).
• Melting point estimation: melting point ≈ ΔH_fusion / ΔS_fusion; for aluminum, with ΔH_fusion = 10.0 kJ/mol and ΔS_fusion = 9.50 J/mol·K, the melting point is approximately 1,000 K (question 411).

🔥 Combustion and exothermic reactions

🔥 Combustion characteristics

• LP gas (liquid propane): combustion is very exothermic, meaning K_p for the combustion is positive and large (question 388).
• Methane combustion: 2 CH₄(g) + 3 O₂(g) → 2 CO(g) + 4 H₂O(g) has ΔH negative (exothermic) and ΔS positive (more gas molecules produced) (question 407).
• Glucose combustion: C₆H₁₂O₆(s) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(g) probably has the greatest increase in ΔS because a solid reactant produces many gas molecules (question 404).

⚡ Electrolysis and nonspontaneous reactions

• Fluorine production: electrolysis of liquid HF produces F₂ and H₂; the reaction is nonspontaneous because breaking hydrogen-hydrogen bonds requires energy (question 392).
• Don't confuse: bond breaking always requires energy; bond formation releases energy; the net determines spontaneity.

🧪 Reaction direction and equilibrium

⚖️ Le Chatelier's principle and solids

• Adding a pure solid: for the equilibrium SnCl₂(s) + Cl₂(g) ⇌ SnCl₄(l), adding more SnCl₂(s) causes no change in the reaction direction because pure solids do not appear in the equilibrium expression (question 391).

📉 Energy profiles

• Overall enthalpy: in an energy profile diagram, the overall enthalpy for the reaction is the difference between products and reactants (final minus initial energy) (question 400).

🔄 Oxidation and reduction in reactions

• Barium production: 4 BaO(s) + Si(s) → Ba₂SiO₄(s) + 2 Ba(g); silicon undergoes oxidation (question 406).
• Entropy in this reaction: there is an increase in entropy because gaseous barium is produced from solid reactants.

🧭 Overview

🧠 One-sentence thesis

Electrolytic cells use external electrical energy to drive nonspontaneous redox reactions, enabling the isolation of active metals like magnesium from their compounds by forcing oxidation at the positive anode and reduction at the negative cathode.

📌 Key points (3–5)

• Electrolytic vs galvanic cells: electrolytic cells have negative cell potentials and require external energy; galvanic cells are spontaneous with positive potentials.
• Electrode polarity reversal: in electrolytic cells, the anode is positive (oxidation still occurs there) and the cathode is negative (reduction still occurs there)—opposite polarity from galvanic cells.
• Competing half-reactions: when water or other substances are present, multiple half-reactions are possible; the actual products depend on which oxidation and reduction are energetically favored.
• Common confusion: oxidation always occurs at the anode and reduction at the cathode in both cell types, but the electrode charges flip between galvanic and electrolytic cells.
• Industrial application: electrolysis is the method of choice for isolating highly active metals (like magnesium and sodium) because their compounds are too stable for chemical reduction.

⚡ Electrolytic cell fundamentals

⚡ What makes electrolytic cells different

Electrolytic cell: a cell that uses external electrical energy to drive a nonspontaneous redox reaction; cell potentials are negative.

• Unlike galvanic (voltaic) cells, which produce electricity spontaneously, electrolytic cells consume electricity to force a reaction that would not occur on its own.
• The minimum energy needed depends on the cell potential—either a standard value from reduction potential tables or a nonstandard value calculated via the Nernst equation.
• Example: breaking down water or magnesium chloride requires an external power source because these reactions are not spontaneous.

🔄 Electrode definitions and polarity

Feature	Galvanic cell	Electrolytic cell
Anode	Oxidation occurs; negative	Oxidation occurs; positive
Cathode	Reduction occurs; positive	Reduction occurs; negative
Cell potential	Positive (spontaneous)	Negative (nonspontaneous)

• Don't confuse: the process at each electrode stays the same (anode = oxidation, cathode = reduction), but the charge reverses.
• In an electrolytic cell, electrons are forced into the cathode (making it negative) and pulled from the anode (making it positive).

🏭 Magnesium production by electrolysis

🏭 Why electrolysis is necessary for magnesium

• Magnesium is an active metal (high on the activity series), so its compounds are very stable.
• Isolating magnesium by chemical means is energetically impractical; electrolysis is the method of choice.
• Example: magnesium chloride is so stable that no cheaper chemical reaction can break it down efficiently.

🧪 The multi-step preparation process

1. Precipitate magnesium hydroxide: Add calcium oxide (CaO) to seawater. CaO reacts with water to form calcium hydroxide, Ca(OH)₂, a strong base. This base reacts with magnesium ions in seawater to precipitate magnesium hydroxide, Mg(OH)₂.
2. Convert to magnesium chloride: Separate the precipitate and add hydrochloric acid (HCl) to convert Mg(OH)₂ to MgCl₂.
3. Dry and melt: Dry the magnesium chloride and heat it in an electrolytic cell until it melts.
4. Electrolyze: Pass electric current through molten MgCl₂ to produce magnesium metal and chlorine gas.

⚙️ The electrolysis half-reactions

• Cathode (reduction): Mg²⁺ + 2 e⁻ → Mg
  • Magnesium ions gain electrons and are reduced to magnesium metal.
• Anode (oxidation): 2 Cl⁻ → Cl₂ + 2 e⁻
  • Chloride ions lose electrons and are oxidized to chlorine gas.
• Overall cell reaction: MgCl₂ → Mg + Cl₂
  • This is the sum of the two half-reactions.

💧 Why molten salt instead of aqueous solution

• If aqueous magnesium chloride were used instead of molten MgCl₂, the products would be hydrogen gas (H₂), chlorine gas (Cl₂), and solid magnesium hydroxide (Mg(OH)₂)—not magnesium metal.
• Reason: Water introduces additional possible half-reactions (oxidation and reduction of water itself).
• In molten MgCl₂, only two half-reactions are possible: reduction of Mg²⁺ and oxidation of Cl⁻.
• In aqueous solution, water can be reduced (forming H₂) or oxidized (forming O₂), and these may be energetically favored over magnesium reduction.
• Don't confuse: the presence of water changes which reactions actually occur, even though magnesium and chloride ions are still present.

🔍 Competing half-reactions in electrolysis

🔍 Why multiple half-reactions matter

• In any electrolytic cell, we must consider all possible half-reactions.
• For molten magnesium chloride, there are only two: Mg²⁺ reduction and Cl⁻ oxidation.
• When water is present (aqueous solutions), we must also consider the oxidation and reduction of water.
• Additional substances (like other ions or solvents) increase the number of possible half-reactions.

🎯 Selecting the actual half-reactions

• We must reduce all possible half-reactions to only one oxidation and one reduction.
• The actual reactions that occur depend on the standard reduction potentials (or nonstandard potentials from the Nernst equation).
• Generally, the reduction with the most positive (or least negative) potential will occur at the cathode, and the oxidation with the least positive (or most negative) potential will occur at the anode.
• Example: In aqueous potassium fluoride, water reduction (E° = –0.83 V) is favored over potassium reduction (E° = –2.93 V) at the cathode, so hydrogen gas forms instead of potassium metal.

🧂 Electrolytes in nonelectrolyte solutions

• To electrolyze a nonelectrolyte like pure water, an inert electrolyte must be added (e.g., a few drops of phosphoric acid).
• The electrolyte provides ions to carry current through the solution, but it does not participate in the redox reactions.
• Example: Adding phosphoric acid to water allows electrolysis to proceed, producing oxygen gas at the anode and hydrogen gas at the cathode.

🔋 Cell potential and energy requirements

🔋 Calculating minimum voltage

• The minimum voltage necessary for electrolysis is determined by the cell potential.
• For the magnesium example:
  • Reduction: Mg²⁺ + 2 e⁻ → Mg, E° = –2.37 V
  • Oxidation: 2 Cl⁻ → Cl₂ + 2 e⁻ (reverse of Cl₂ + 2 e⁻ → 2 Cl⁻, E° = +1.36 V)
  • Cell potential = (–2.37 V) – (+1.36 V) = –3.73 V
• The negative sign confirms the reaction is nonspontaneous; the magnitude (3.73 V) is the minimum voltage needed.
• Don't confuse: the cell potential is negative for electrolytic cells, but the applied voltage must be positive (and at least equal to the magnitude of the cell potential).

⚖️ Relationship to spontaneity and Gibbs free energy

• A spontaneous reaction (galvanic cell) has a positive cell potential and negative Gibbs free energy change (ΔG < 0).
• A nonspontaneous reaction (electrolytic cell) has a negative cell potential and positive Gibbs free energy change (ΔG > 0).
• External energy input overcomes the positive ΔG, forcing the reaction to proceed.

🧮 Quantitative electrolysis calculations

🧮 Faraday's laws and stoichiometry

• The amount of substance produced or consumed in electrolysis depends on the quantity of electric charge passed through the cell.
• Faraday constant: 1 F = 96,500 coulombs per mole of electrons.
• Charge (coulombs) = current (amperes) × time (seconds).
• Moles of electrons = charge / Faraday constant.
• Use stoichiometry from the half-reaction to relate moles of electrons to moles of product.

🔢 Example calculation framework

• Given: current, time, and half-reaction.
• Step 1: Calculate total charge: Q = current × time (convert time to seconds).
• Step 2: Calculate moles of electrons: moles e⁻ = Q / 96,500.
• Step 3: Use half-reaction stoichiometry to find moles of product.
• Step 4: Convert moles to grams using molar mass.
• Example: For the reaction Mg²⁺ + 2 e⁻ → Mg, 2 moles of electrons produce 1 mole of magnesium.

⚗️ Relating different products

• If a reaction produces multiple products (e.g., Mg and Cl₂ from MgCl₂), use the overall balanced equation to relate their amounts.
• Example: MgCl₂ → Mg + Cl₂ shows that 1 mole of Mg forms for every 1 mole of Cl₂.
• If 2.7 g of aluminum forms from AlCl₃, calculate moles of Al, then use stoichiometry (2 Al + 3 Cl₂ from 2 AlCl₃) to find moles of Cl₂.

🔬 Practical considerations and applications

🔬 Why some metals require electrolysis

• The most active metals (high on the activity series) form very stable compounds.
• Chemical reduction is impractical or impossible for these metals.
• Electrolysis is the only viable method for isolating elements like sodium, magnesium, and aluminum.
• Example: Sodium metal reacts violently with water, so it cannot be produced by electrolysis of aqueous sodium chloride—molten NaCl must be used instead.

🛠️ Electroplating and other applications

• Electrolysis can deposit a metal coating onto an object (electroplating).
• The object to be plated is the cathode (negatively charged), where reduction deposits metal atoms.
• Example: Silver plating uses a silver ion solution; silver ions are reduced at the cathode (the object), forming a silver coating.

🌊 Water electrolysis details

• Pure water is a poor conductor; an inert electrolyte (like phosphoric acid or sulfuric acid) is added.
• At the anode (positive), water is oxidized to oxygen gas: 2 H₂O → O₂ + 4 H⁺ + 4 e⁻.
• At the cathode (negative), water is reduced to hydrogen gas: 2 H₂O + 2 e⁻ → H₂ + 2 OH⁻.
• Don't confuse: the electrolyte provides conductivity but does not appear in the net reaction (it is "inert").

🧭 Overview

🧠 One-sentence thesis

This final review tests integrated understanding of chemistry topics including electrochemistry, stoichiometry, thermodynamics, kinetics, equilibrium, acid-base chemistry, and periodic trends through multi-step problem-solving.

📌 Key points (3–5)

• Electrochemistry applications: galvanic cells, electrolysis calculations, and biological redox processes (cytochromes) require understanding of reduction potentials and spontaneity.
• Stoichiometry and limiting reagents: theoretical yield depends on the limiting reagent amount, not excess reagent concentration or recovery efficiency.
• Intermolecular forces and physical properties: heat of vaporization, melting point, and solubility are determined by hydrogen bonding strength, ionic character, and molecular mass.
• Common confusion—activation energy vs. heat of reaction: activation energy (Eₐ) tells you about reaction rate, not whether a reaction is exothermic; heat of reaction depends only on reactants and products, not the transition state or catalyst.
• Equilibrium and Le Chatelier's principle: catalysts do not change equilibrium constants or heats of reaction; they only affect the rate of reaching equilibrium.

⚡ Electrochemistry and redox processes

⚡ Galvanic cell voltage conditions

• A galvanic cell uses two half-reactions; the cell with the greater difference in reduction potentials produces higher voltage.
• The excerpt shows: HNO₂ + H⁺ + e⁻ → NO + H₂O (E° = +1.00 V) and Ce⁴⁺ + e⁻ → Ce³⁺ (E° = +1.61 V).
• Greatest voltage condition: Low pH increases H⁺ concentration, shifting the first half-reaction forward and increasing cell voltage (answer: B).
• Don't confuse: equilibrium conditions yield zero voltage because the reaction stops driving electrons.

🔋 Electrolysis calculations

• Question 450: How many grams of chromium from Cr(NO₃)₃ electrolysis using 2.00 amps for 3.00 hrs?
• Chromium in Cr(NO₃)₃ is Cr³⁺, requiring 3 electrons per atom.
• Current × time gives total charge; divide by Faraday's constant to get moles of electrons, then divide by 3 to get moles of Cr.
• Convert moles to grams using chromium's molar mass (~52 g/mol).
• Answer: (A) 3.90 g.

🧬 Biological redox: cytochrome electron transfer

• Question 472: Cytochromes transfer electrons spontaneously in the sequence determined by reduction potentials.
• For spontaneous transfer, electrons move from lower (more negative) to higher (more positive) reduction potential.
• The table lists cytochromes a₃, a, c, and b with their standard reduction potentials.
• Sequence: electrons flow from the cytochrome with the lowest potential to the highest.
• Answer: (B) cytochrome b → c → a → a₃.

🔌 Why batteries stop before completion

• Question 471: Batteries stop producing electricity while limiting reagent remains.
• Reason: the system reaches equilibrium, not because the reaction is incomplete or the rate is too slow.
• At equilibrium, forward and reverse reaction rates are equal, so no net electron flow occurs.
• Don't confuse: normal reactions go to completion because products are removed or the equilibrium constant is very large; in a closed battery, equilibrium is reached.

🧪 Stoichiometry and solution chemistry

🧪 Theoretical yield and limiting reagents

• Question 473: For 2 HCl(aq) + Na₂CO₃(aq) → 2 NaCl(aq) + CO₂(g) + H₂O(l), theoretical yield of NaCl depends on:
• Answer: (A) the amount of the limiting reagent.
• Theoretical yield is the maximum product calculated from the limiting reagent, not the actual amount recovered or the concentration of excess reagent.

💧 Ion concentration after dilution

• Question 454: 100 mL of 0.50 M aluminum nitrate (Al(NO₃)₃) added to 0.900 L water.
• Aluminum nitrate dissociates: Al(NO₃)₃ → Al³⁺ + 3 NO₃⁻.
• Each mole of Al(NO₃)₃ produces 3 moles of NO₃⁻.
• Initial moles of NO₃⁻: 0.100 L × 0.50 M × 3 = 0.15 mol.
• Total volume: 0.100 L + 0.900 L = 1.0 L.
• Concentration: 0.15 mol / 1.0 L = 0.15 M.
• Answer: (D) 0.15 M.

🧂 Moles of ions in solution

• Question 458: Which produces the most moles of ions?
• Compare total ion moles from each compound:
  • 1.0 mol CHCl₃ (molecular, non-ionic): 0 ions.
  • 0.30 mol CaCl₂ → 0.30 mol Ca²⁺ + 0.60 mol Cl⁻ = 0.90 mol ions.
  • 0.45 mol KCl → 0.45 mol K⁺ + 0.45 mol Cl⁻ = 0.90 mol ions.
  • 0.25 mol CeCl₃ → 0.25 mol Ce³⁺ + 0.75 mol Cl⁻ = 1.00 mol ions.
• Answer: (D) 0.25 mol CeCl₃.

💉 Why use saline solution for dehydrated patients

• Question 459: Why not pure water?
• Answer: (A) The osmotic pressure of water is too low.
• Pure water would cause cells to swell and burst due to osmotic imbalance.
• Also valid: (B) dehydrated patients may have low salt levels.

🔥 Thermodynamics and physical properties

🔥 Heat of vaporization factors

• Question 455: Which has the highest heat of vaporization?
• Heat of vaporization depends on intermolecular force strength.
• Compare:
  • (A) CH₃CH₂CH₂OH: hydrogen bonding (one –OH group).
  • (B) HOCH₂CH₂OH: hydrogen bonding (two –OH groups, strongest).
  • (C) CH₃CH₂CH₂CH₃: only London dispersion forces.
  • (D) CH₃CH₂CH₂F: dipole-dipole, weaker than hydrogen bonding.
• Answer: (B) HOCH₂CH₂OH.

🔥 What heat of vaporization does NOT depend on

• Question 456: Heat of vaporization does not depend on:
• Answer: (C) the mass of the molecules.
• It depends on intermolecular force strength (hydrogen bonding, dipole-dipole), external pressure, but not molecular mass directly.

🧊 Highest melting point

• Question 457: Which has the highest melting point?
• Ionic compounds have higher melting points than molecular compounds.
• Compare ionic compounds:
  • (B) CaO: Ca²⁺ and O²⁻ (both doubly charged, very strong ionic bonding).
  • (D) NaCl: Na⁺ and Cl⁻ (singly charged).
• Answer: (B) CaO (highest charge product → strongest ionic bonding).

🌡️ Entropy change for phosgene formation

• Question 494: CO(g) + Cl₂(g) → COCl₂(g).
• Entropy change ΔS = S(products) – S(reactants).
• The reaction goes from 2 moles of gas to 1 mole of gas, so entropy decreases (ΔS is negative).
• Using the provided thermodynamic data, calculate ΔS.
• Answer: (A) –131 J/K.

⚙️ Kinetics and reaction mechanisms

⚙️ Activation energy and reaction rate

• Question 461: Which reaction proceeds at the greatest rate?
• Lower activation energy (Eₐ) → faster reaction.
• Compare:
  • (A) Eₐ = 100 kJ (lowest).
  • (B) Eₐ = 150 kJ.
  • (C) Eₐ = 250 kJ.
• Answer: (A) A + B → C with Eₐ = 100 kJ.

⚙️ Identifying intermediates in a mechanism

• Question 462: Cl₂(g) + H₂O(g) → HCl(g) + HClO(g).
• Proposed mechanism:
  1. Cl₂ → 2 Cl• (fast).
  2. Cl• + H₂O → HCl + •OH (slow).
  3. Cl• + •OH → HClO (fast).
• Intermediates: species produced in one step and consumed in another.
• Both Cl• and •OH are intermediates.
• Answer: (C) Cl• and •OH.

⚙️ Catalyst effects on equilibrium and heat of reaction

• Question 466: Adding a catalyst to a reaction:

• Answer: (C) does not change the equilibrium constant.

• Catalysts speed up both forward and reverse reactions equally.

• Question 468: Catalyst effect on heat of reaction:

• Answer: (A) does not change the heat of reaction.

• Heat of reaction depends only on reactants and products, not the pathway.

🔬 What determines heat of reaction

• Question 469: Heat of reaction depends upon:
• Answer: (D) the reactants and the products.
• The transition state and activation energy affect the rate, not the heat of reaction.
• Don't confuse: Eₐ (kinetics) vs. ΔH (thermodynamics).

🔬 Determining if a reaction is exothermic

• Question 467: Which reaction is exothermic?
• Answer: (D) It is impossible to determine.
• Activation energy (Eₐ) does not tell you if a reaction is exothermic or endothermic; you need ΔH or the energy difference between reactants and products.
• Note: (B) shows Eₐ = –150 kJ, which is physically impossible (activation energy cannot be negative).

⚖️ Equilibrium and Le Chatelier's principle

⚖️ Increasing drug yield by shifting equilibrium

• Question 463: Pt(NH₃)₂Cl₂ formation is an equilibrium process.
• To increase yield, shift equilibrium toward products.
• Answer: (D) Increasing the ammonia concentration.
• Adding a reactant shifts equilibrium to the right (toward products).
• Don't choose: (A) CHCl₃ does not directly increase Cl⁻ in a useful way; (B) and (C) would shift equilibrium away from the product.

⚖️ Volume change effect on gas equilibrium

• Question 489: 2 NO₂(g) ⇌ 2 NO(g) + O₂(g).
• Decreasing volume increases pressure.
• Le Chatelier: system shifts toward the side with fewer moles of gas.
• Left side: 2 moles; right side: 3 moles.
• Answer: (A) reaction shifts to the left.

⚖️ Reaction quotient vs. Kₛₚ

• Question 465: If the reaction quotient exceeds Kₛₚ:
• Answer: (A) a precipitate will form.
• When Q > Kₛₚ, the solution is supersaturated, and solid will precipitate to restore equilibrium.

⚖️ Increasing solubility of ammonium perchlorate

• Question 464: NH₄ClO₄ is sparingly soluble.
• NH₄ClO₄ ⇌ NH₄⁺ + ClO₄⁻.
• NH₄⁺ is a weak acid; lowering pH (adding H⁺) suppresses its formation, but increasing pH (adding OH⁻) reacts with NH₄⁺, shifting equilibrium to the right and increasing solubility.
• Answer: (B) Increasing the pH.

🧬 Acid-base chemistry

🧬 Conjugate bases

• Question 491: Conjugate bases of H₂PO₄⁻, CH₃OH, and NH₃.
• Remove one H⁺ from each:
  • H₂PO₄⁻ → HPO₄²⁻.
  • CH₃OH → CH₃O⁻.
  • NH₃ → NH₂⁻.
• Answer: (C) HPO₄²⁻, CH₃O⁻, and NH₂⁻.

🧬 Relationship between Ka and Kb

• Question 492: Ka of phenol is 1 × 10⁻¹⁰; find pKb of phenolate ion.
• For a conjugate acid-base pair: Ka × Kb = Kw = 1 × 10⁻¹⁴.
• Kb = Kw / Ka = 10⁻¹⁴ / 10⁻¹⁰ = 10⁻⁴.
• pKb = –log(Kb) = –log(10⁻⁴) = 4.
• Rewrite: pKb = 14 – pKa = 14 – (–log(10⁻¹⁰)) = 14 – 10 = 4.
• Answer: (C) 14 + log(1 × 10⁻¹⁰) [since log(10⁻¹⁰) = –10, this gives 14 – 10 = 4].

🧬 Calculating pKa from Ka

• Question 493: Ka for HNO₂ is 5.1 × 10⁻⁴.
• pKa = –log(Ka) = –log(5.1 × 10⁻⁴) ≈ –(log 5.1 + log 10⁻⁴) ≈ –(0.7 – 4) ≈ 3.3.
• Answer: (A) 3.3.

🧬 Choosing a buffer

• Question 490: Best acid for a buffer at pH 8.6.
• A buffer works best when pH ≈ pKa of the weak acid.
• Choose the acid with pKa closest to 8.6.
• (The excerpt does not provide the table, so the specific answer cannot be determined, but the principle is clear.)

🧬 pH change during aluminum dissolution

• Question 495: 2 Al(s) + 6 H⁺(aq) → 2 Al³⁺(aq) + 3 H₂(g).
• The reaction consumes H⁺, so [H⁺] decreases.
• Lower [H⁺] → higher pH.
• Answer: (A) The pH will increase.

🔬 Atomic structure and periodic trends

🔬 Ground-state electron configuration of gallium

• Question 475: Gallium (Ga) is element 31.
• Configuration: [Ar] 4s² 3d¹⁰ 4p¹.
• Answer: (A) [Ar]4s² 3d¹⁰ 4p¹.

🔬 Excited-state electron configuration of sodium

• Question 477: Sodium ground state: 1s² 2s² 2p⁶ 3s¹.
• Excited state: an electron is promoted to a higher orbital.
• Answer: (A) 1s² 2s² 2p⁶ 3p¹ (the 3s electron is promoted to 3p).

🔬 Periodic trends: ionization energy, electronegativity, atomic radii

• Question 478: Ionization energy increases left to right.
• Other trends in the same direction:
  • Electronegativity increases (atoms hold electrons more tightly).
  • Atomic radii decrease (more protons pull electrons closer).
• Answer: (A) Increasing electronegativity and decreasing atomic radii.

🔬 Least reactive element

• Question 481: Among Ca, Be, Sr, Ra (all alkaline earth metals).
• Reactivity increases down the group (larger atoms lose electrons more easily).
• Answer: (B) Be (smallest, least reactive).

🔬 Lowest melting point ionic compound

• Question 482: Melting point of ionic compounds depends on charge and size.
• Larger ions and lower charges → weaker ionic bonding → lower melting point.
• Compare:
  • (A) RbI: large cation and anion, singly charged.
  • (B) AlN: Al³⁺ and N³⁻, highly charged, very strong bonding.
  • (C) CaO: Ca²⁺ and O²⁻, doubly charged.
  • (D) LiF: small ions, strong bonding.
• Answer: (A) RbI.

🧪 Gas laws and behavior

🧪 Partial pressure calculation

• Question 483: Total pressure 1,000 torr; 3 mol O₂, 1 mol H₂, 4 mol N₂.
• Partial pressure of H₂: (moles of H₂ / total moles) × total pressure.
• Total moles: 3 + 1 + 4 = 8 mol.
• Partial pressure of H₂: (1 / 8) × 1,000 torr = 125 torr.
• Answer: (C) 125 torr.

🧪 Effusion rate ratio

• Question 484: Rate of effusion of methane (CH₄, molar mass 16 g/mol) vs. oxygen (O₂, 32 g/mol).
• Graham's law: rate ratio = square root of (M₂ / M₁) = √(32 / 16) = √2 ≈ 1.4.
• Answer: (A) 1.4.

🧪 Volume change with temperature (Charles's law)

• Question 485: Helium at 500 cm³ at 227°C; find volume at 727°C (constant pressure).
• Convert to Kelvin: 227°C = 500 K; 727°C = 1000 K.
• V₁ / T₁ = V₂ / T₂ → V₂ = V₁ × (T₂ / T₁) = 500 × (1000 / 500) = 1,000 cm³.
• Answer: (A) 1,000 cm³.

🧪 Conditions for ideal gas behavior

• Question 486: Water vapor approaches ideal behavior when:
• Answer: (A) the pressure is low and the temperature is high.
• Low pressure → molecules are far apart, intermolecular forces negligible.
• High temperature → kinetic energy overcomes intermolecular attractions.

🧬 Molecular structure and bonding

🧬 Polar vs. nonpolar molecules

• Question 479: Which is NOT polar?
• (A) HCl: polar (linear, different atoms).
• (B) CH₃Cl: polar (tetrahedral, asymmetric).
• (C) H₂S: polar (bent, like water).
• (D) CF₄: nonpolar (tetrahedral, symmetric, dipoles cancel).
• Answer: (D) CF₄.

🧬 Orbital geometry of sp hybridized carbon

• Question 480: sp hybridization → 2 hybrid orbitals.
• Geometry: linear (180° bond angle).
• Answer: (B) Linear.

🧬 Lewis structure of chlorine dioxide

• Question 497: ClO₂ is a free radical (odd number of electrons).
• The correct Lewis structure shows Cl with one unpaired electron and two oxygen atoms.
• (The excerpt does not display the structures clearly, so the specific answer cannot be determined from the text alone.)

🧬 Molecular geometry of CO₂ and CH₂O

• Question 498: CO₂(g) + 2 H₂(g) → CH₂O(g) + H₂O(g).
• CO₂: linear (O=C=O, no lone pairs on C).
• CH₂O (formaldehyde): trigonal planar (C double-bonded to O, single bonds to two H, no lone pairs on C).
• Answer: (D) CO₂ is linear, and CH₂O is trigonal planar.

🧬 Hydrogen bonding to water

• Question 460: Which CANNOT hydrogen bond to water?
• Hydrogen bonding requires H bonded to N, O, or F, or a lone pair on N, O, or F.
• (A) CH₃OH: has –OH, can hydrogen bond.
• (B) CH₃Cl: no H on N/O/F, Cl is not effective for hydrogen bonding.
• (C) C₆H₁₂O₆ (glucose): has –OH groups, can hydrogen bond.
• (D) CH₃NH₂: has –NH₂, can hydrogen bond.
• Answer: (B) CH₃Cl.

☢️ Nuclear chemistry

☢️ Nuclear fission product identification

• Question 474: U-235 + neutron → Cs-140 + 3 neutrons + ?
• Conserve mass number: 235 + 1 = 140 + 3×1 + A → A = 93.
• Conserve atomic number: 92 + 0 = 55 + 0 + Z → Z = 37 (rubidium).
• Answer: (A) Rubidium-93.

☢️ Mirror nuclei

• Question 476: Mirror nuclei swap proton and neutron numbers.
• Check each pair:
  • (A) ²⁷Al (13 p, 14 n) and ²⁷Si (14 p, 13 n): mirror nuclei.
• Answer: (A) ²⁷Al and ²⁷Si.

☢️ Half-life calculation

• Question 499: 6.25% remains.
• Each half-life reduces the amount by half: 100% → 50% → 25% → 12.5% → 6.25%.
• Answer: (D) 4 half-lives.

🧪 Miscellaneous applications

🧪 Barium ion mimicry in the body

• Question 451: Barium ions (Ba²⁺) are retained because they mimic:
• Answer: (C) Ca²⁺.
• Ba²⁺ and Ca²⁺ are both group 2 ions with similar size and charge, so biological systems mistake Ba²⁺ for Ca²⁺.

🧪 Molar mass of hydrated calcium sulfate

• Question 452: CaSO₄·2H₂O.
• Molar mass: Ca (40) + S (32) + 4×O (16) + 2×[2×H (1) + O (16)] = 40 + 32 + 64 + 36 = 172 g/mol ≈ 170 g/mol.
• Answer: (A) 170 g/mol.

🧪 Same oxidation state halogens

• Question 453: Compare oxidation states of halogens in each pair.
• (The excerpt does not provide enough detail to calculate oxidation states for all compounds, but the method is: assign oxidation numbers and compare.)

🧪 Respiration as a redox process

• Question 470: C₆H₁₂O₆(aq) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(l).
• Glucose is oxidized (C goes from 0 to +4 in CO₂); oxygen is reduced (O₂ goes from 0 to –2 in H₂O and CO₂).
• Answer: (A) Redox process.

🧪 Effect of removing air from ethanol container

• Question 496: Connecting a vacuum pump removes air.
• Removing vapor above the liquid does not change the vapor pressure (an intrinsic property at a given temperature).
• Lower external pressure → boiling point decreases.
• Answer: (C) The boiling point of ethanol decreases.