[{"data":1,"prerenderedAt":206},["ShallowReactive",2],{"topic-intermolecular-forces-and-phase-changes":3},{"topic":4,"prev":191,"next":199},{"id":5,"slug":6,"title":7,"shortTitle":8,"lesson":5,"category":9,"description":10,"metaDescription":11,"objectives":12,"conceptSections":34,"workedExample":99,"oerSources":107,"selfStudyQuestions":113,"relatedTopicSlugs":140,"relatedElements":144,"relatedReferences":149,"keyEquations":152,"howTo":157,"conceptProbe":166,"practiceTeaser":170,"gatedContent":178,"seoKeywords":181,"lastAlignmentAudit":189,"objectiveCount":190},18,"intermolecular-forces-and-phase-changes","Intermolecular Forces and Phase Changes","IMFs & Phases","states-of-matter","Identify and rank intermolecular forces, interpret phase diagrams and heating/cooling curves, classify crystalline solids, calculate the energy required for temperature and phase changes, and compute density and atomic radius from cubic unit-cell parameters.","Learn intermolecular forces (London dispersion, dipole-dipole, hydrogen bonding, ion-dipole), phase diagrams, heating curves, crystalline-solid classification, phase-change energy calculations, and cubic unit-cell density and radius problems.",[13,16,19,22,25,28,31],{"id":14,"text":15},"18.1","Identify the types of intermolecular forces (London dispersion, dipole-dipole, hydrogen bonding, ion-dipole) present in a substance",{"id":17,"text":18},"18.2","Rank substances by the strength of their intermolecular forces and relate IMF strength to boiling point, vapor pressure, viscosity, and surface tension",{"id":20,"text":21},"18.3","Interpret and use phase diagrams to identify phases, phase transitions, triple point, and critical point",{"id":23,"text":24},"18.4","Classify crystalline solids as ionic, metallic, molecular, or covalent-network and relate type to properties",{"id":26,"text":27},"18.5","Interpret heating and cooling curves and identify phase-change regions",{"id":29,"text":30},"18.6","Calculate the total energy required for a temperature change that includes one or more phase changes",{"id":32,"text":33},"18.7","Calculate the density or atomic radius of a metallic crystalline solid from cubic unit-cell parameters",[35,44,52,61,69,77,85,91,95],{"heading":36,"content":37,"relatedObjectives":38,"deepDive":40},"Types of Intermolecular Forces","\u003Cp>\u003Cstrong>Intermolecular forces (IMFs)\u003C/strong> are attractions \u003Cem>between\u003C/em> molecules — distinct from the covalent bonds within them. Three main types, listed weakest to strongest:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>London dispersion forces\u003C/strong> — Present in \u003Cem>all\u003C/em> molecules. Temporary, instantaneous dipoles from electron motion induce dipoles in neighbours. Dispersion forces increase with \u003Cstrong>molar mass\u003C/strong> and \u003Cstrong>surface area\u003C/strong> (more electrons → greater polarisability). I\u003Csub>2\u003C/sub> is a solid while F\u003Csub>2\u003C/sub> is a gas at room temperature.\u003C/li>\u003Cli>\u003Cstrong>Dipole–dipole forces\u003C/strong> — Act between \u003Cem>polar\u003C/em> molecules. The δ+ end of one molecule attracts the δ− end of another. Stronger than London forces alone for molecules of comparable size.\u003C/li>\u003Cli>\u003Cstrong>Hydrogen bonding\u003C/strong> — A special, strong dipole–dipole force occurring when H is bonded to F, O, or N. The high electronegativity difference and tiny size of H create exceptionally strong attractions, explaining water’s high boiling point (100 °C) vs. H\u003Csub>2\u003C/sub>S (−60 °C).\u003C/li>\u003C/ul>\u003Cp>\u003Cstrong>Ion–dipole forces\u003C/strong> (between ions and polar molecules) are the strongest of all and drive dissolution of ionic compounds in water.\u003C/p>",[39],144,[41],{"label":42,"body":43},"Why this matters: geckos climb glass on the weakest force","\u003Cp>London dispersion forces are the feeblest attraction on this list, yet a gecko walks up a vertical pane of glass using nothing else. Each foot carries around half a million microscopic hairs, each splitting into hundreds of even finer tips that press so close to the surface that dispersion forces, useless at any visible distance, engage by the billions. Multiplied that many times, the weakest force in chemistry holds up a lizard, and inspired the adhesives industry to build gecko-mimicking tapes that stick without glue. Weak times enormous numbers is a recurring theme in this topic: it is also how dispersion forces hold liquids and molecular solids together at all.\u003C/p>",{"heading":45,"content":46,"relatedObjectives":47,"deepDive":48},"How IMFs Determine Physical Properties","\u003Cp>The strength of intermolecular forces directly controls a substance’s macroscopic behaviour:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Boiling and melting points\u003C/strong> — Stronger IMFs require more energy to separate molecules, raising both values. Substances with only London forces (noble gases, hydrocarbons) have low boiling points; those with hydrogen bonding (water, ethanol) have much higher boiling points.\u003C/li>\u003Cli>\u003Cstrong>Vapour pressure\u003C/strong> — Weaker IMFs allow molecules to escape the liquid more easily, giving \u003Cem>higher\u003C/em> vapour pressure at a given temperature. Diethyl ether (weak IMFs) is far more volatile than water at 25 °C.\u003C/li>\u003Cli>\u003Cstrong>Viscosity\u003C/strong> — Stronger IMFs and larger, more entangled molecules lead to greater resistance to flow. Glycerol, with three –OH groups for hydrogen bonding, is much more viscous than water.\u003C/li>\u003Cli>\u003Cstrong>Surface tension\u003C/strong> — Molecules at the surface feel a net inward pull from neighbours. Stronger IMFs mean higher surface tension, explaining why water forms nearly spherical droplets and supports small insects on its surface.\u003C/li>\u003C/ul>\u003Cp>To predict properties: identify the \u003Cem>strongest\u003C/em> IMF present and compare. Among molecules with the same IMF type, larger molecules generally have higher boiling points due to stronger dispersion forces.\u003C/p>",[39],[49],{"label":50,"body":51},"Hydrogen bonding is why Earth is habitable","\u003Cp>Judge water by its molar mass (18 g/mol) and its group neighbors, and it should boil somewhere far below zero: hydrogen sulfide, its heavier cousin at 34 g/mol, boils at −60 °C. Instead water boils at 100 °C, because each molecule hydrogen-bonds to several neighbors, and separating them costs extraordinary energy. No hydrogen bonding, no liquid water at Earth’s temperatures.\u003C/p>\u003Cp>The same force explains ice floating: the hydrogen-bonded lattice is unusually open, making solid water \u003Cem>less\u003C/em> dense than liquid, so lakes freeze from the top down and life survives underneath. Nearly every “anomaly” of water on this page is hydrogen bonding wearing a different hat.\u003C/p>",{"heading":53,"content":54,"relatedObjectives":55,"deepDive":57},"Vapour Pressure and the Clausius–Clapeyron Equation","\u003Cp>\u003Cstrong>Vapour pressure\u003C/strong> is the pressure exerted by a vapour in dynamic equilibrium with its liquid (or solid) at a given temperature. As temperature rises, more molecules have enough kinetic energy to escape the liquid surface, so vapour pressure increases exponentially.\u003C/p>\u003Cp>The \u003Cstrong>Clausius–Clapeyron equation\u003C/strong> quantifies this relationship:\u003C/p>\u003Cp>\u003Cstrong>ln(P\u003Csub>2\u003C/sub>/P\u003Csub>1\u003C/sub>) = −ΔH\u003Csub>vap\u003C/sub>/R · (1/T\u003Csub>2\u003C/sub> − 1/T\u003Csub>1\u003C/sub>)\u003C/strong>\u003C/p>\u003Cul>\u003Cli>P\u003Csub>1\u003C/sub>, P\u003Csub>2\u003C/sub> = vapour pressures at temperatures T\u003Csub>1\u003C/sub>, T\u003Csub>2\u003C/sub> (in K)\u003C/li>\u003Cli>ΔH\u003Csub>vap\u003C/sub> = enthalpy of vaporisation (J/mol)\u003C/li>\u003Cli>R = 8.314 J/(mol·K)\u003C/li>\u003C/ul>\u003Cp>A liquid boils when its vapour pressure equals the external atmospheric pressure. This is why water boils below 100 °C at high altitude (lower atmospheric pressure) and above 100 °C inside a pressure cooker.\u003C/p>",[56],143,[58],{"label":59,"body":60},"Why this matters: cooking times change with the phase diagram","\u003Cp>Water boils when its vapour pressure reaches the surrounding pressure, so “100 °C” is a sea-level fact, not a property of water. In Denver, water boils near 95 °C; on Everest, around 71 °C, too cool to cook pasta properly no matter how long it bubbles. A pressure cooker runs the trick in reverse: sealing in steam raises the pressure to roughly 2 atm, pushing the boiling point near 120 °C and cutting cooking times dramatically.\u003C/p>\u003Cp>Same liquid, same recipe, different pressure: the exponential vapour-pressure curve in this section is quietly rewriting cookbook instructions everywhere on Earth.\u003C/p>",{"heading":62,"content":63,"relatedObjectives":64,"deepDive":65},"Phase Transitions and Heating Curves","\u003Cp>Substances undergo six \u003Cstrong>phase transitions\u003C/strong>: melting, freezing, vaporisation, condensation, sublimation, and deposition. Each endothermic transition (melting, vaporisation, sublimation) absorbs energy to overcome intermolecular forces; the reverse transitions release energy.\u003C/p>\u003Cp>A \u003Cstrong>heating curve\u003C/strong> plots temperature vs. heat added for a substance. Diagonal segments show temperature increasing within a single phase; \u003Cem>flat\u003C/em> (plateau) segments show phase changes at constant temperature — added heat goes entirely into breaking IMFs rather than raising temperature. The length of each plateau is proportional to the enthalpy of that transition (ΔH\u003Csub>fus\u003C/sub> for melting, ΔH\u003Csub>vap\u003C/sub> for boiling).\u003C/p>\u003Cp>For water, ΔH\u003Csub>vap\u003C/sub> (40.7 kJ/mol) is much larger than ΔH\u003Csub>fus\u003C/sub> (6.02 kJ/mol) because vaporisation completely separates molecules, while melting only loosens the crystal lattice.\u003C/p>",[56],[66],{"label":67,"body":68},"Common mistake: expecting the thermometer to rise during melting","\u003Cp>The flats on a heating curve confuse everyone at first: you keep adding heat, yet the temperature refuses to move. During a phase change, every joule goes into dismantling intermolecular attractions, none into speeding molecules up, so the thermometer stalls until the transition finishes. Ice water stays at 0 °C until the last chip of ice is gone.\u003C/p>\u003Cp>The plateaus also explain why steam burns are worse than boiling-water burns at the same 100 °C: condensing steam first dumps its entire heat of vaporisation (about 2,260 J per gram, several times more than cooling the resulting water) into your skin before any cooling even begins.\u003C/p>",{"heading":70,"content":71,"relatedObjectives":72,"deepDive":73},"Phase Diagrams","\u003Cp>A \u003Cstrong>phase diagram\u003C/strong> plots pressure vs. temperature and shows which phase is stable under any combination of conditions:\u003C/p>\u003Cul>\u003Cli>\u003Cstrong>Phase boundaries\u003C/strong> — Curves where two phases coexist in equilibrium. The solid–liquid line shows melting points at various pressures; the liquid–gas line shows boiling points; the solid–gas line shows sublimation points.\u003C/li>\u003Cli>\u003Cstrong>Triple point\u003C/strong> — The unique T and P where all three phases coexist (for water: 0.01 °C, 0.00604 atm).\u003C/li>\u003Cli>\u003Cstrong>Critical point\u003C/strong> — Above this T and P, liquid and gas become indistinguishable as a supercritical fluid (for water: 374 °C, 218 atm).\u003C/li>\u003Cli>\u003Cstrong>Normal boiling/melting points\u003C/strong> — Found where a horizontal line at 1 atm crosses the boundaries.\u003C/li>\u003C/ul>\u003Cp>Water’s phase diagram is unusual: the solid–liquid line slopes to the \u003Cem>left\u003C/em>, meaning increasing pressure \u003Cem>lowers\u003C/em> the melting point. This occurs because ice is less dense than liquid water — a consequence of the open hydrogen-bonded crystal structure of ice. Most other substances have a positive slope.\u003C/p>",[56],[74],{"label":75,"body":76},"Why this matters: freeze-drying is a walk across the phase diagram","\u003Cp>Instant coffee, astronaut meals, and many injectable medicines are made by deliberately routing water around its liquid phase. Freeze the product, drop the chamber pressure below water’s triple point, and add gentle heat: the ice sublimes straight to vapour, leaving structure and flavour intact where boiling would have wrecked them.\u003C/p>\u003Cp>Dry ice performs the same maneuver in your freezer aisle for the opposite reason: carbon dioxide’s triple point sits at 5.1 atm, far above atmospheric pressure, so at 1 atm there is simply no temperature at which liquid CO\u003Csub>2\u003C/sub> can exist. It has no choice but to sublime, which is why dry ice never leaves a puddle.\u003C/p>",{"heading":78,"content":79,"relatedObjectives":80,"deepDive":81},"Types of Crystalline Solids","\u003Cp>Crystalline solids have particles arranged in a repeating 3-D pattern called a \u003Cstrong>crystal lattice\u003C/strong>. They are classified by particle type and bonding:\u003C/p>\u003Ctable>\u003Cthead>\u003Ctr>\u003Cth>Type\u003C/th>\u003Cth>Particles\u003C/th>\u003Cth>Forces\u003C/th>\u003Cth>Properties\u003C/th>\u003Cth>Examples\u003C/th>\u003C/tr>\u003C/thead>\u003Ctbody>\u003Ctr>\u003Ctd>\u003Cstrong>Ionic\u003C/strong>\u003C/td>\u003Ctd>Cations & anions\u003C/td>\u003Ctd>Electrostatic\u003C/td>\u003Ctd>Hard, brittle, high mp, conduct when molten/dissolved\u003C/td>\u003Ctd>NaCl, CaF\u003Csub>2\u003C/sub>\u003C/td>\u003C/tr>\u003Ctr>\u003Ctd>\u003Cstrong>Molecular\u003C/strong>\u003C/td>\u003Ctd>Molecules\u003C/td>\u003Ctd>London, dipole–dipole, H-bonds\u003C/td>\u003Ctd>Soft, low mp, poor conductors\u003C/td>\u003Ctd>Ice, dry ice, sugar\u003C/td>\u003C/tr>\u003Ctr>\u003Ctd>\u003Cstrong>Covalent network\u003C/strong>\u003C/td>\u003Ctd>Atoms (covalent bonds throughout)\u003C/td>\u003Ctd>Covalent bonds\u003C/td>\u003Ctd>Very hard, very high mp, poor conductors\u003C/td>\u003Ctd>Diamond, SiO\u003Csub>2\u003C/sub>\u003C/td>\u003C/tr>\u003Ctr>\u003Ctd>\u003Cstrong>Metallic\u003C/strong>\u003C/td>\u003Ctd>Metal cations in electron sea\u003C/td>\u003Ctd>Metallic bonding\u003C/td>\u003Ctd>Variable hardness, conduct heat & electricity, malleable\u003C/td>\u003Ctd>Fe, Cu, Au\u003C/td>\u003C/tr>\u003C/tbody>\u003C/table>\u003Cp>\u003Cstrong>Amorphous solids\u003C/strong> (glass, rubber, many plastics) lack long-range order and soften gradually over a temperature range instead of having a sharp melting point.\u003C/p>",[39],[82],{"label":83,"body":84},"Why this matters: diamond and pencil lead are the same element","\u003Cp>Diamond and graphite are both pure carbon; the difference is entirely in the lattice. Diamond bonds every atom to four neighbors in a rigid three-dimensional covalent network: the hardest natural material known. Graphite bonds each atom to three neighbors in flat sheets; the sheets themselves are strong, but only weak dispersion forces hold sheet to sheet, so they slide apart, onto your paper, with every pencil stroke.\u003C/p>\u003Cp>That one structural difference flips almost every property: hardness, feel, electrical conductivity (graphite conducts along its sheets; diamond insulates), and price. The lattice, not just the atom, decides what a solid is like.\u003C/p>",{"heading":86,"content":87,"relatedObjectives":88},"Cubic Unit Cells and Crystal-Lattice Calculations","\u003Cp>The repeating arrangement of particles in a crystalline solid is built from a smallest representative block called the \u003Cstrong>unit cell\u003C/strong>. Many metals (and some ionic solids) crystallize in one of three cubic unit cells:\u003C/p>\u003Ctable>\u003Cthead>\u003Ctr>\u003Cth>Unit cell\u003C/th>\u003Cth>Particles per cell (n)\u003C/th>\u003Cth>Edge-to-radius (metals)\u003C/th>\u003Cth>Coordination number\u003C/th>\u003C/tr>\u003C/thead>\u003Ctbody>\u003Ctr>\u003Ctd>\u003Cstrong>Simple cubic (SC)\u003C/strong>\u003C/td>\u003Ctd>1\u003C/td>\u003Ctd>r = a/2\u003C/td>\u003Ctd>6\u003C/td>\u003C/tr>\u003Ctr>\u003Ctd>\u003Cstrong>Body-centered cubic (BCC)\u003C/strong>\u003C/td>\u003Ctd>2\u003C/td>\u003Ctd>r = a√3 / 4\u003C/td>\u003Ctd>8\u003C/td>\u003C/tr>\u003Ctr>\u003Ctd>\u003Cstrong>Face-centered cubic (FCC)\u003C/strong>\u003C/td>\u003Ctd>4\u003C/td>\u003Ctd>r = a√2 / 4\u003C/td>\u003Ctd>12\u003C/td>\u003C/tr>\u003C/tbody>\u003C/table>\u003Cp>The particles-per-cell value n comes from counting fractional contributions: an atom at a corner is shared among 8 cells (counts as ⅛), one on a face among 2 cells (counts as ½), and one fully inside the cell counts as 1. SC has 8 corners × ⅛ = 1; BCC adds one full atom in the center (1 + 1 = 2); FCC adds 6 face atoms × ½ (1 + 3 = 4).\u003C/p>\u003Cp>\u003Cstrong>Density of a unit cell.\u003C/strong> Because a unit cell is microscopically small but representative, its density equals the density of the bulk solid:\u003C/p>\u003Cp style=\"text-align:center;\">d = (n · \u003Cem>M\u003C/em>) / (N\u003Csub>A\u003C/sub> · a\u003Csup>3\u003C/sup>)\u003C/p>\u003Cp>where n is particles per cell, \u003Cem>M\u003C/em> is molar mass (g/mol), N\u003Csub>A\u003C/sub> = 6.022 × 10\u003Csup>23\u003C/sup> /mol, and a is the edge length. To get density in g/cm\u003Csup>3\u003C/sup>, convert the edge length to centimeters before cubing it (1 pm = 10\u003Csup>−10\u003C/sup> cm, so 350 pm = 3.50 × 10\u003Csup>−8\u003C/sup> cm).\u003C/p>\u003Cp>\u003Cstrong>Atomic radius from density.\u003C/strong> If the density and the unit cell type are known, rearrange the same equation to solve for the edge length, then apply the cell's edge-to-radius relation. For BCC: a = ∛(n·\u003Cem>M\u003C/em> / (N\u003Csub>A\u003C/sub>·d)), then r = a√3 / 4. The result comes out in the same length units as a; convert to picometers via 1 cm = 10\u003Csup>10\u003C/sup> pm if needed.\u003C/p>\u003Cp>The same equation is the bridge that lets you back out any of the four quantities (n, M, a, d) when the other three are known — useful for identifying an unknown metal from its measured density and lattice parameter, or for confirming a unit-cell assignment.\u003C/p>",[89,90],141,142,{"heading":92,"content":93,"relatedObjectives":94},"Identifying IMFs: A Decision Strategy","\u003Cp>A systematic approach for identifying the intermolecular forces in any substance:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>Is it ionic?\u003C/strong> If the substance contains metal–nonmetal or polyatomic-ion combinations, the dominant forces are \u003Cem>ionic bonds\u003C/em> (not IMFs in the traditional sense).\u003C/li>\u003Cli>\u003Cstrong>Is it a network covalent solid?\u003C/strong> If atoms are linked by continuous covalent bonds throughout (diamond, SiO\u003Csub>2\u003C/sub>), the forces are covalent bonds.\u003C/li>\u003Cli>\u003Cstrong>Is the molecule polar or nonpolar?\u003C/strong> Draw the Lewis structure, determine the geometry (VSEPR), and check for a net dipole moment.\u003C/li>\u003Cli>\u003Cstrong>Does it have N–H, O–H, or F–H bonds?\u003C/strong> If yes, hydrogen bonding is present (in addition to dipole–dipole and London forces).\u003C/li>\u003Cli>\u003Cstrong>All molecules\u003C/strong> experience London dispersion forces, so include these last.\u003C/li>\u003C/ol>\u003Cp>List \u003Cem>all\u003C/em> IMFs present, then identify the \u003Cstrong>strongest\u003C/strong> one — it dominates the physical properties. When comparing two substances, the one with the stronger dominant IMF will generally have the higher boiling point, lower vapour pressure, and higher surface tension.\u003C/p>",[39],{"heading":96,"content":97,"relatedObjectives":98},"IMF Identification and Boiling-Point Prediction: Common Mistakes","\u003Cp>A reliable workflow for identifying intermolecular forces and predicting physical properties:\u003C/p>\u003Col>\u003Cli>\u003Cstrong>All molecules have London dispersion forces\u003C/strong>. Start here and note that LDF strength increases with molar mass and surface area.\u003C/li>\u003Cli>\u003Cstrong>Is the molecule polar?\u003C/strong> If yes, add dipole–dipole interactions.\u003C/li>\u003Cli>\u003Cstrong>Does the molecule have H bonded to N, O, or F?\u003C/strong> If yes, add hydrogen bonding (the strongest common IMF).\u003C/li>\u003Cli>\u003Cstrong>Rank the dominant IMF\u003C/strong>: hydrogen bonding > dipole–dipole > London dispersion (all else equal).\u003C/li>\u003Cli>\u003Cstrong>Predict properties\u003C/strong>: stronger IMFs → higher boiling point, lower vapour pressure, higher viscosity.\u003C/li>\u003C/ol>\u003Cp>Common mistakes: confusing intermolecular forces with intramolecular (covalent) bonds, thinking all polar molecules have hydrogen bonds (only those with H–N, H–O, or H–F qualify), ignoring London dispersion forces for large polar molecules where LDF may actually dominate, and assuming that ionic compounds have intermolecular forces (they have ionic bonds, not IMFs).\u003C/p>",[56,39],{"title":100,"problem":101,"steps":102,"answer":106},"Identifying IMFs and Predicting Boiling Points","Rank the following substances from lowest to highest boiling point and explain your reasoning: CH₄, CH₃OH, CH₃Cl.",[103,104,105],"Identify the IMFs for each substance. CH₄ is nonpolar → only London dispersion forces. CH₃Cl is polar → dipole–dipole forces + London forces. CH₃OH has an O–H bond → hydrogen bonding + dipole–dipole + London forces.","Compare IMF strengths. London only \u003C dipole–dipole \u003C hydrogen bonding. So the ranking from weakest to strongest total IMFs is: CH₄ \u003C CH₃Cl \u003C CH₃OH.","The substance with the weakest IMFs boils at the lowest temperature: CH₄ (–161 °C) \u003C CH₃Cl (–24 °C) \u003C CH₃OH (65 °C).","CH₄ \u003C CH₃Cl \u003C CH₃OH. Methane has only London forces (lowest bp), chloromethane adds dipole–dipole forces, and methanol has hydrogen bonding (highest bp).",[108,109,110,111,112],"OpenStax Chemistry 2e, Ch 10.1: Intermolecular Forces (CC BY 4.0)","OpenStax Chemistry 2e, Ch 10.2: Properties of Liquids (CC BY 4.0)","OpenStax Chemistry 2e, Ch 10.4: Phase Diagrams (CC BY 4.0)","OpenStax Chemistry 2e, Ch 10.5: The Solid State of Matter (CC BY 4.0)","OpenStax Chemistry 2e, Ch 10.6: Lattice Structures in Crystalline Solids (CC BY 4.0)",[114,116,119,121,123,125,127,129,131,133,135,138],{"question":115},"What are intermolecular forces and how do they differ from intramolecular bonds?",{"question":117,"hint":118},"What are London dispersion forces and what determines their strength?","Think about polarizability and molecular size.",{"question":120},"What is a dipole–dipole interaction?",{"question":122},"What is hydrogen bonding and what conditions are required for it?",{"question":124},"How do intermolecular force strengths relate to boiling points and melting points?",{"question":126},"What is vapour pressure and how is it related to intermolecular force strength?",{"question":128},"What does a phase diagram show and what are its key features?",{"question":130},"What is the difference between a phase transition and a chemical reaction?",{"question":132},"What are the main types of crystalline solids and how do they differ?",{"question":134},"How do you determine which type of intermolecular force is dominant for a given molecule?",{"question":136,"hint":137},"How many atoms per unit cell are there in a simple cubic, body-centered cubic, and face-centered cubic lattice?","1, 2, and 4 respectively, after counting fractional contributions of corner and face atoms.",{"question":139},"How is atomic radius related to edge length in BCC and FCC unit cells?",[141,142,143],"molecular-structure","gas-laws","solutions-colligative-properties",[145,146,147,148],"H","O","N","F",[150,151],"water-vapor-pressure","phase-diagrams",[153],{"label":154,"equation":155,"note":156},"Clausius-Clapeyron Equation","ln(P₂/P₁) = −ΔH_vap / R × (1/T₂ − 1/T₁)","R = 8.314 J/(mol·K); T in kelvin; P in any consistent unit",{"title":158,"steps":159},"How to Identify Intermolecular Forces in a Substance",[160,161,162,163,164,165],"Determine whether the substance is ionic, network covalent, metallic, or molecular. Only molecular substances have traditional IMFs.","If molecular, draw the Lewis structure and use VSEPR to decide if the molecule is polar or nonpolar.","Check for hydrogen bonding: does the molecule have H directly bonded to N, O, or F? If yes, hydrogen bonding is present.","If the molecule is polar but lacks H-N, H-O, or H-F bonds, dipole-dipole forces are the strongest IMF.","All molecules (polar and nonpolar) also have London dispersion forces. Note that LDF strength increases with molar mass and surface area.","Rank the dominant IMF to predict relative boiling points, vapor pressures, and other physical properties.",{"question":167,"answer":168,"type":169},"Methanol (CH3OH, molar mass 32 g/mol) boils at 64.7 °C, while ethane (C2H6, molar mass 30 g/mol) boils at -89 °C. These molecules have nearly the same molar mass. Why is methanol's boiling point so much higher?","Despite their similar molar masses, methanol has an O-H bond that allows it to form hydrogen bonds with neighboring molecules. Hydrogen bonding is much stronger than the London dispersion forces that are the only intermolecular attraction between nonpolar ethane molecules. The extra energy needed to break these hydrogen bonds raises methanol's boiling point dramatically compared to ethane.","conceptual",[171,174],{"id":172,"problem":173,"type":169},"pt-18-1","List all intermolecular forces present in acetone (CH3COCH3). Is acetone capable of hydrogen bonding with itself? What about with water?",{"id":175,"problem":176,"type":177},"pt-18-2","The vapor pressure of ethanol is 44 mmHg at 20 °C and its enthalpy of vaporization is 38.6 kJ/mol. Estimate its vapor pressure at 35 °C using the Clausius-Clapeyron equation.","calculation",{"workedExampleCount":179,"hasWorksheets":180},6,true,[182,183,184,185,186,187,188],"intermolecular forces","hydrogen bonding","phase diagram","vapor pressure","London dispersion","unit cell","crystal lattice","2026-07-17",7,{"id":192,"slug":193,"lesson":192,"title":194,"shortTitle":195,"description":196,"category":9,"objectiveCount":197,"problemCount":198},17,"gases-and-gas-laws","Gases and Gas Laws","Gas Laws","Convert between pressure units, apply the individual and combined gas laws, use the ideal gas law for pressure/volume/amount/temperature calculations, find molar mass and density, work with Dalton's partial pressures, perform gas stoichiometry, apply the van der Waals equation for non-ideal gases, use Graham's law for effusion, and calculate RMS molecular speeds.",9,58,{"id":200,"slug":201,"lesson":200,"title":202,"shortTitle":202,"description":203,"category":9,"objectiveCount":204,"problemCount":205},19,"colligative-properties","Colligative Properties","Understand colligative properties: boiling point elevation, freezing point depression, osmotic pressure, and Raoult's law.",5,48,1785108608146]