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 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

Practice 1 — answer: A

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. C 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.

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

Practice 2 — answer: A

B put the polar molecule below the lone atom; HCl has dipole–dipole on top of dispersion, so it outranks argon. C claimed HCl hydrogen bonds, but its H sits on chlorine, not on N, O, or F — HCl stops at dipole–dipole and stays below water. D treated one attraction as all the same strength; the type sets the strength, and the three types differ.

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 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

Practice 1 — 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.

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

Practice 2 — answer: A

Water's O–H bonds let it hydrogen bond. Hydrogen sulfide has no H on a small, strongly electron-attracting atom, so it is held only by weaker dipole–dipole forces. Water boils higher — answer A.

B ranked by mass alone: hydrogen sulfide is heavier, but its forces are weaker, so it boils lower. C confused forces with bonds: boiling overcomes the attractions between molecules, not the covalent O–H bonds inside them. D treated the same force type as the same strength, when hydrogen bonding is a much stronger special case.

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 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

Practice 1 — answer: B

A used the heat of fusion by mistake: 18.0 × 335 = 6,030 J is the heat to melt, not boil. C left out the mass: 2,259 J boils one gram only. D slid the decimal: 18.0 × 2259 = 40,662, then divided by 1000 to get 40.7 and mislabeled it joules.

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.

Practice 2 — answer: B

A used the heat of vaporization: 30.0 × 2259 = 67,770 J is the heat to condense steam, not freeze water. C has the right size but the wrong sign: freezing releases heat, so q is negative. D left out the mass: 335 J freezes one gram only.

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 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 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

Practice 1 — answer: C

A skipped the melting plateau: 20 × 4.184 × 30 = 2510 J warms the liquid but never melts the ice. B stopped at the melting point: 20 × 335 = 6700 J leaves the water at 0 °C. D treated fusion like a specific heat: 20 × 335 × 30 = 201,000 J, multiplying a plateau by a ΔT it does not have.

Practice 2 — answer: C

A skipped the melting plateau: 836 + 2092 = 2928 J never melts the ice. B stopped at the melting point: 836 + 8375 = 9211 J leaves the water at 0 °C. D used water's c for the ice: 25 × 4.184 × 16 = 1674 J for the ice leg gives 12,141 J.