Guided example: identify the pieces

Step 1 · Identify the pieces

Guided example: identify the pieces

Step 1 · Identify the pieces

Guided example: identify the pieces

Step 1 · Identify the pieces

Guided example: identify the pieces

Step 1 · Identify the pieces

Guided example: name the forces

Step 2 · Name every force present

Dispersion is in every substance. Polarity adds dipole–dipole.

Guided example: name the forces

Guided example: name the forces

Guided example: the route on the chart

Worked example 1: solution

Step 1 · Identify the pieces

Methane is a molecule, not ions in a solvent. Its four C–H bonds sit in a symmetric tetrahedron, so the molecule is nonpolar. No H is bonded to N, O, or F.

Worked example 1: solution

Dispersion is present in every substance. Nothing here adds dipole–dipole, hydrogen bonding, or ion–dipole.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the chart

Worked example 2: forces present

A common first answer

Methanol is polar, so dipole–dipole is real. It is not the strongest, because the H bonded to oxygen does more.

Worked example 2: forces present

A molecule, not ions. Polar. And one H is bonded directly to O: the trigger for hydrogen bonding.

Worked example 2: forces present

Worked example 2: the strongest force

Step 3 · Pick the strongest

Worked example 2: the route on the chart

Practice 1 answer: B

A has fluorine, but every H sits on carbon; the F carries no H. C puts H on bromine, which is not N, O, or F. D has four H atoms, all on silicon.

Practice 1: the route on the chart

Practice 2 answer: C

B stopped one rung early: HF is polar, but the H bonded to fluorine upgrades the strongest force to hydrogen bonding, above ordinary dipole–dipole. A defaulted to the weakest force; dispersion is in every molecule but never the strongest when a stronger one is present. D named the H–F bond inside the molecule, which is intramolecular and does not act between molecules.

Practice 2: the route on the chart

Practice 3 answer: A

B treated PF₃ like BF₃: the lone pair on P makes it pyramidal, so the three P–F dipoles do not cancel. C left out dispersion, which every substance has. D counted the F as a hydrogen-bond trigger, but PF₃ has no H at all.

Practice 3: the route on the chart

Practice 4 answer: B

A has polar C=O bonds, but O=C=O is linear with matching ends: the dipoles cancel, leaving dispersion only. C is polar, but each H sits on N, so hydrogen bonding outranks dipole–dipole. D counted C–H as a polar bond; ethane is nonpolar, with dispersion only.

Practice 4: the route on the chart

Worked example 3: forces present

Step 1 · Identify the pieces

Propane is a nonpolar hydrocarbon. Dimethyl ether is polar, but every H sits on carbon. Ethanol is polar and has an O–H.

Worked example 3: forces present

Worked example 3: the ranking

Step 3 · Pick the strongest

Worked example 3: the route on the chart

Practice 5 answer: D

Argon is a lone nonpolar atom: dispersion only. CH₂O is polar, but both H atoms sit on carbon: dipole–dipole. CH₃NH₂ has H on N: hydrogen bonding. Na⁺ pulls on water's dipoles: ion–dipole.

B put hydrogen bonding at the top; ion–dipole outranks it. C ranked by molar mass, 22.99 < 30.03 < 31.06 < 39.95 g/mol, as if dispersion decided everything. A is the right order read strongest first.

Practice 5: the route on the chart

Guided example: solution

Step 1 · Put the outside pressure in atm

The pressure is already in atm: 0.57 atm. No conversion is needed.

Guided example: solution

Water boils at 85 °C at this camp, not at 100 °C.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Worked example 1: solution

Step 1 · Identify each liquid's strongest force

Hexane is a nonpolar hydrocarbon: dispersion only. 1-Pentanol carries an O–H, so it hydrogen bonds. Nearly equal size means nearly equal dispersion; the O–H is the difference.

Worked example 1: solution

Viscosity measures resistance to flow. To flow, each molecule must slide past its neighbors, and sliding means briefly pulling away from them.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the strip

Worked example 2: solution

The common first answer, tested

Sealed above any liquid, at any temperature, some molecules escape while others return. Both flasks hold vapor at a steady pressure, so both liquids show a vapor pressure. Boiling is not required.

Worked example 2: solution

Water hydrogen bonds through its O–H. Diethyl ether is polar, but every H sits on carbon: ordinary dipole–dipole, far weaker.

Worked example 2: solution

Vapor pressure measures escape: how much vapor builds before condensation catches up with evaporation.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the strip

Practice 1 answer: C

A reversed the direction: air pressure falls going up a mountain and rises going below sea level. B treated the boiling point as fixed; 100 °C holds only at exactly 1 atm. D changed the forces; water's hydrogen bonds are the same everywhere, and only the pressure it must match changes.

Practice 2 answer: A

B swaps the reads: a lower outside pressure pairs with the lower temperature, never the higher one. C read the 1.00 atm line twice, as if a boiling point were fixed. D read butane's curve, the one on the right.

Practice 2: the route on the curve

Practice 3 answer: A

B describes water: when adhesion wins, the edge is dragged up the wall, giving a concave dip and a climbing column. C invents a repulsion; glass attracts every liquid at least weakly, and the dome comes from the liquid's own inward pull. D cannot happen: every substance has at least dispersion forces.

Practice 4 answer: B

The masses sit close, so force type decides. 1-Propanol's O–H hydrogen bonds. Acetone is polar with no O–H. Pentane is nonpolar. Vapor pressure measures escape, so it rises as the forces weaken.

A ranked the attractions, not the vapor pressure; stronger forces lower it. C ranked by molar mass alone, but pentane's extra mass cannot outpull a dipole or a hydrogen bond. D read acetone as nonpolar; its C=O makes it polar, so it holds on harder than pentane.

Practice 5 answer: D

Practice 5 answer: D

A stopped at the boiling point: 115 is where the water boils, not how far it warms. B used the normal boiling point: 100 − 22 = 78. C gave the normal boiling point itself, 100, a temperature and not a change.

Practice 5 answer: D

Practice 5: the route on the map

Guided example: solution

Step 1 · Identify each substance's strongest force

N₂ joins two identical atoms, so it is nonpolar: dispersion only. CO joins carbon to oxygen, which pull on the shared electrons unequally, so CO is polar: dipole–dipole.

Guided example: solution

The types differ, so the type decides. Dipole–dipole is stronger than dispersion alone, so CO boils higher.

Guided example: solution

Guided example: the route on the map

Practice 1 answer: B

A ranked by mass alone; the force type comes first, and HF has the stronger type. C stopped at dipole–dipole; the H on F adds hydrogen bonding on top. D let chlorine qualify; only N, O, or F do.

Practice 1 answer: B

Worked example 1: solution

Step 1 · Identify each substance's strongest force

Ethanol has an O–H bond, so it hydrogen bonds. Dimethyl ether is polar but has no O–H, so its strongest force is ordinary dipole–dipole attraction.

Worked example 1: solution

Equal molar mass means dispersion is the same for both, so only the force type differs. Hydrogen bonding is stronger than dipole–dipole, so ethanol boils higher.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Identify each substance's strongest force

Both are nonpolar, so dispersion is the only force acting in each.

Worked example 2: solution

Same force type does not mean same strength. The molecule held by stronger dispersion boils higher.

Worked example 2: solution

Iodine is far larger than chlorine, with many more electrons, so its dispersion is much stronger. Iodine boils higher.

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 answer: A

Both are nonpolar, so dispersion is the only force. Heptane is the larger molecule, so its dispersion is stronger and it boils higher. Answer A.

B confused forces with bonds: boiling overcomes the attractions between molecules, never the covalent bonds inside them. C reversed the trend: weakly held light molecules escape more easily and boil lower, not higher. D forgot that the same force type can differ in strength, and dispersion grows with size.

Practice 2 answer: A

Practice 3 answer: C

A let polarity decide inside a tie; with the type tied, size decides, and HBr's much larger electron cloud outweighs HCl's slightly larger dipole. B and D let Br or Cl qualify for hydrogen bonding; only N, O, or F do.

Practice 3 answer: C

Worked example 3: solution

Step 1 · Identify each substance's strongest force

Butane is nonpolar: dispersion only. Acetone is polar with no O–H: dipole–dipole. 1-Propanol has an O–H: hydrogen bonding.

Worked example 3: solution

Mass is nearly equal, so dispersion ties and the force type sets the order. Hydrogen bonding beats dipole–dipole beats dispersion, so 1-propanol > acetone > butane.

Worked example 3: solution

Worked example 3: the route on the map

Practice 4 answer: D

Neon and krypton are nonpolar atoms: dispersion only, and the larger krypton attracts more. PH₃ is polar, but its H sits on P: dipole–dipole. NH₃ has H on N: hydrogen bonding.

A ranked by mass alone; krypton is the heaviest, yet it has only dispersion. B let PH₃ hydrogen bond, but an H on P does not qualify. C reversed the size tie; the bigger atom disperses more and boils higher.

Practice 4 answer: D

Practice 5 answer: E

A ranked by molar mass alone; all four sit within 4.03 g/mol, so the force type decides. B ranked the two alcohols by chain length; the second O–H outweighs one extra carbon. C is the right order read highest first. D read butanone as nonpolar; its C=O makes it polar, like acetone.

Practice 5 answer: E

Guided example: solution

Step 1 · Find the point

Move 1: across the bottom to 25 °C. Move 2: up to 80 kPa. Move 3: the spot lies between the fusion and vaporization curves, so the water is liquid.

Guided example: solution

Step 1 · Find the point
Same 25 °C, now down to 1 kPa. The point drops below the vaporization curve: gas.

Guided example: solution

Step 1 · Find the point

Guided example: the route on the map

Worked example 1: solution

Step 1 · Find the point

The start sits in the solid region, so the sample begins solid.

Worked example 1: solution

Heating at constant pressure is a horizontal path to the right.

Worked example 1: solution

Above the triple point, the path crosses the fusion curve first, then the vaporization curve.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Find the point

On the 1-atm line, 30 °C lies between the fusion crossing (−15 °C) and the vaporization crossing (60 °C). That is the liquid region.

Worked example 2: solution

Step 1 · Find the point
Step 2 · Choose the path

Constant pressure: move right along the 1-atm line, from 30 °C to 100 °C.

Worked example 2: solution

Step 1 · Find the point
Step 2 · Choose the path
Step 3 · Cross the curves

Worked example 2: the route on the map

Your turn: the route on the map

Practice 1 answer: B

A took the left-leaning fusion curve to mean cold water under pressure is liquid; at −10 °C that needs about 110,000 kPa (1,100 atm), and 50 kPa is half an atmosphere. C read any pressure below 101.3 kPa as gas; at −10 °C the gas region lies below 0.26 kPa.

Practice 1: the route on the map

Practice 2 answer: C

A put the triple point on the path, but it sits at 0.25 atm, 0.75 atm below it. B read the triple-point temperature, 10 °C, as the melting point: 4 °C too low. D took the critical point, 320 °C, as the boiling point: 202 °C too high.

Practice 2: the route on the map

Practice 3 answer: A

B used the 1-atm points: 0 °C and 100 °C hold only at 101.3 kPa. C sent the path through the liquid region, which exists only above 0.61 kPa. D named the change backward: deposition runs gas → solid.

Practice 3: the route on the map

Practice 4 answer: B

The horizontal path runs 16.8 atm above the triple point, so cooling crosses the vaporization curve, then the fusion curve. It ends at −60 °C, 32 °C below the triple-point temperature: solid.

A took the sublimation route, which runs only below 3.2 atm. D stopped at the liquid; the path continues past the fusion curve. C misread the critical point: the gas starts 24 °C above the critical temperature, but cooling carries it below 96 °C, and at 20 atm, 28 atm under the critical pressure, it condenses.

Practice 4: the route on the map

Practice 5 answer: D

The vertical path at 60 °C crosses the vaporization curve once. The steep fusion curve rises from −35 °C and never reaches 60 °C.

A compared the starting 0.20 atm with the 0.80-atm triple point; a vertical path is judged by its temperature. B went on across the fusion curve, as a cooling path would. C checked only the pressure: 50 atm is 12 atm past 38 atm, but a supercritical fluid also needs a temperature above 120 °C.

Practice 5: the route on the map

Guided example: solution

Step 1 · Name the phase change

Liquid water turns to steam at 100 °C. The change is boiling.

Guided example: solution

Boiling uses the heat of vaporization, 40.7 kJ/mol. The 6.02 kJ/mol melts ice; it has no part here.

Guided example: solution

Moles cancel; kilojoules remain.

Guided example: solution

Guided example: the route on the map

Practice 1 answer: D

A used the heat of fusion: 1.60 × 6.02 = 9.63, but nothing melts or freezes. B put the heat per mole upside down: 1.60 ÷ 40.7 = 0.0393, and moles do not cancel. C is the heat for one mole alone; the steam is 1.60 mol.

Practice 1: the route on the map

Worked example 1: solution

Step 1 · Name the phase change

The ice is melting: solid water turns to liquid, all at 0 °C.

Worked example 1: solution

Melting uses the heat of fusion, ΔH_fus = 335 J/g.

Worked example 1: solution

Write the factor so grams cancel. Only one orientation does:

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Name the phase change

The water is boiling: liquid turns to gas, all at 100 °C.

Worked example 2: solution

Boiling uses the heat of vaporization, ΔH_vap = 2259 J/g.

Worked example 2: solution

Grams cancel; joules remain.

Worked example 2: solution

Worked example 2: the route on the map

Practice 2 answer: A

The heat is given, so ΔH_vap enters upside down: J on the bottom, so joules cancel. B used the heat of fusion: 5.00 × 10⁴ ÷ 335 = 149 g is the ice this heat would melt, not the water it boils. C used water's specific heat: 5.00 × 10⁴ ÷ 4.184 = 1.20 × 10⁴, but a plateau has no ΔT, so c does not apply. D kept the factor right side up: 5.00 × 10⁴ × 2259 = 1.13 × 10⁸, in J²/g, nothing cancels.

Practice 2: the route on the map

Worked example 3: solution

A common first attempt

The steam condenses at 100 °C into water at 100 °C. ΔT = 0, so the m·c·ΔT term adds nothing.

Worked example 3: solution

A common first attempt

The steam condenses at 100 °C into water at 100 °C. ΔT = 0, so the m·c·ΔT term adds nothing.
Step 1 · Name the phase change

The steam is condensing: gas turns to liquid, all at 100 °C.

Worked example 3: solution

A common first attempt

The steam condenses at 100 °C into water at 100 °C. ΔT = 0, so the m·c·ΔT term adds nothing.
Step 1 · Name the phase change
Step 2 · Pick its ΔH

Condensing uses the heat of vaporization, ΔH_vap = 2259 J/g, the same value as boiling.

Worked example 3: heat released

Step 3 · Multiply

Worked example 3: heat released

Heat leaves the steam as it condenses, so q is negative: q = −18,072 J.

Worked example 3: the route on the map

Practice 3 answer: C

A dropped the sign: freezing releases heat, so q for the water is negative. B used the heat of vaporization: −(45.0 × 2259) = −1.02 × 10⁵, but nothing boils or condenses. D divided by the heat of fusion: 45.0 ÷ 335 = 0.134, and grams do not cancel.

Practice 3: the route on the map

Worked example 4: solution

Step 1 · Name the phase change

The ice is melting at 0 °C, the same change as ever. Only the units of its heat are new: kilojoules per mole.

Worked example 4: solution

Melting uses the heat of fusion: ΔH_fus = 6.02 kJ/mol.

Worked example 4: solution

Grams cancel into moles, moles cancel into kilojoules.

Worked example 4: solution

The per-gram route on the same sample: 60.0 g × 335 J/g = 20,100 J = 20.1 kJ. Two routes, one heat, agreeing to the rounding of the constants.

Worked example 4: the route on the map

Practice 4 answer: E

Two conversion factors are needed.

A used the heat of fusion: 4.99 × 6.02 = 30.1. B skipped the molar mass: 90.0 × 40.7 = 3.66 × 10³, as if each gram were a mole. C put the molar heat upside down: 4.99 ÷ 40.7 = 0.123. D stopped at moles: 4.99 mol, one factor short.

Practice 4: the route on the map

Practice 5 answer: B

The heat is given, so ΔH_fus enters upside down and kJ cancels. A picked the vaporization heat: 15.0 ÷ 40.7 × 18.02 = 6.64 g, but no boiling or condensing happens here. C divided by the molar mass: 2.49 ÷ 18.02 = 0.138 g. D stopped at moles: 2.49 mol of water, one factor short of grams.

Practice 5: the route on the map

Guided example: solution

Step 1 · Identify each segment

Two pieces. The liquid warms from 80.0 °C to 100 °C, a slope. Then it boils at 100 °C, a plateau.

Guided example: solution

The slope has a ΔT of 20.0 °C. The plateau has none, so ΔHvap alone gives its heat.

Guided example: solution

Guided example: the route on the curve

Practice 1 answer: A

B started at the melting point: the ice must first warm from −15 °C to 0 °C. C skipped the plateau: melting takes heat even though the temperature holds. D stopped at the melting point: the water still has to warm to 60 °C.

Practice 1: the route on the curve

Worked example 1: solution

Step 1 · Identify each segment

The ice sits at its melting point. Two pieces follow: melt at a constant 0 °C, then warm the liquid to 25 °C.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the curve

Worked example 2: the melting step

A common first attempt

Melting happens at a constant 0 °C. There is no temperature change to multiply, so ΔHfus already gives the whole melting heat.

Worked example 2: the melting step

Two pieces: melt at 0 °C, then warm the liquid from 0 °C to 40 °C. Only the warming piece has a ΔT.

Worked example 2: the two pieces

Step 2 · Compute each piece

Worked example 2: the two pieces

Worked example 2: the route on the curve

Worked example 3: warming and melting

Step 1 · Identify each segment

Five pieces: warm ice (−20 → 0), melt at 0 °C, warm water (0 → 100), boil at 100 °C, warm steam (100 → 120). Three slopes and two plateaus.

Worked example 3: warming and melting

Worked example 3: boiling, then the total

Step 2 · Compute each piece

Worked example 3: boiling, then the total

Worked example 3: the route on the curve

Worked example 4: solution

Step 1 · Identify each segment

Two pieces: boil at 100.0 °C, then warm the steam to 130.0 °C. The water starts at its boiling point, so no liquid warms.

Worked example 4: solution

Worked example 4: adding the pieces

A common first attempt

The two numbers carry different units. Kilojoules and joules do not add as they stand.

Worked example 4: adding the pieces

Worked example 4: the route on the curve

Practice 2 answer: C

A skipped the melting plateau: 20.0 × 4.184 × 30.0 = 2.51 × 10³ J warms the liquid but never melts the ice. B stopped at the melting point: 20.0 × 335 = 6.70 × 10³ J leaves the water at 0 °C. D multiplied the plateau by a ΔT: 20.0 × 335 × 30.0 = 2.01 × 10⁵ J.

Practice 2: the route on the curve

Practice 3 answer: B

A used grams as moles: 72.1 × 6.02 + 6.03 = 4.40 × 10². C stopped at the melting point: 24.1 kJ leaves the water at 0 °C. D added J to kJ as they stand: 24.1 + 6033 = 6.06 × 10³.

Practice 3: the route on the curve

Practice 4 answer: D

A lost the sign: a cooling path releases heat, so q is negative. B skipped the condensation plateau: −285 − 1255 = −1.54 × 10³ J. C stopped at 100 °C: −285 − 22,590 = −2.29 × 10⁴ J.

Practice 4: the route on the curve

Practice 5 answer: E

A skipped the ice leg: 9020 + 3954 = 1.30 × 10⁴. B skipped the melt: 658 + 3954 = 4.61 × 10³. C used c(water) on ice: 1356 + 9020 + 3954 = 1.43 × 10⁴. D stopped at 0 °C: 658 + 9020 = 9.68 × 10³.

Practice 5: the route on the curve

Guided example: name the particle

Step 1 · Name the particle

Guided example: name the particle

Step 1 · Name the particle

Guided example: name the particle

Step 1 · Name the particle

Guided example: name the particle

Step 1 · Name the particle

Guided example: force and properties

Step 2 · Name the force

Cl pulls the shared electrons harder than H does. The H–Cl bond is polar, so the molecule is polar.

Guided example: force and properties

Guided example: the route on the map

Worked example 1: solution

Step 1 · Name the particle

A metal with a nonmetal gives ions: one Ca²⁺ for every two Cl⁻, so +2 + 2(−1) = 0.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Name the particle

Shiny, bendable, and conducting as a solid: only a metal fits all three. Its lattice holds metal cations in a sea of mobile electrons.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 answer: A

B contains a metal, but Ni with O is a metal with a nonmetal: ionic. C is shiny and gray, but silicon is on the network list: a covalent network. D is a single element, but sulfur is a nonmetal: S₈ molecules make a molecular solid.

Practice 1: the route on the map

Practice 2 answer: C

A matched hardness and melting point but skipped the molten test: a melted ionic solid conducts. B ignored the conductivity; a metal conducts as a solid. D read only the conductivity; a molecular solid melts low and is soft, never near 2000 °C.

Practice 2: the route on the map

Practice 3 answer: B

A applied the nonmetals-only rule without looking for a polyatomic ion. NH₄⁺ and NO₃⁻ carry charges, so the solid is built of ions. C picked a network solid, but the network list is short: C, Si, SiO₂, SiC. D named the right type but let the ions move in the solid; in the lattice they are locked in place.

Practice 3: the route on the map

Practice 4 answer: E

A read each NH₂ group as ammonium. NH₄⁺ carries 4 H and a charge; urea holds 2 × 2 = 4 H in two neutral NH₂ groups and no ion. B named the covalent bonds inside each molecule; melting separates whole molecules and breaks none of those bonds. C stopped at dispersion, the force every molecule has. D saw a polar molecule but missed H bonded to N.

Practice 4: the route on the map

Practice 5 answer: D

Conductivity decides first. Softness and a 98 °C melting point fit a metal too: metallic melting points vary widely.

A let the ions move in the solid; an ionic solid conducts only once melted or dissolved. B let the softness and the low melting point outvote the conductivity. C read polarity as charge; polar molecules carry no net charge and never conduct. E reached for graphite, the one conducting network; graphite stays solid past 3500 °C.

Practice 5: the route on the map