6.3 States of Matter & Thermochemistry

Why Phase Changes & Thermochemistry Matter for the NAPT

Heat transfer is one of the most heavily emphasized topics across the entire NAPT — it connects this chemistry chapter directly to Chapter 4's physics coverage of thermodynamics. Shipboard steam plants boil and condense water on a massive scale, refrigeration and air-conditioning systems rely on controlled phase changes, and reactor plants (covered in Chapter 7) manage enormous heat loads through pressurized water that never actually boils. Understanding how energy moves during a phase change — and how much energy a reaction releases or absorbs — is core chemistry content tested on the exam.

States of Matter & Phase Changes

Matter commonly exists in three familiar states (phases): solid (fixed shape and volume; particles vibrate in place), liquid (definite volume but no fixed shape; particles slide past one another), and gas (no fixed shape or volume; particles move freely and spread to fill their container). A fourth state, plasma (ionized gas), exists at very high energies but is rarely tested at this level.

A phase change is a physical change between states — the substance's chemical identity never changes, only its physical arrangement. There are six named phase changes, and each name depends on its direction:

Sublimation skips the liquid phase entirely — dry ice (solid carbon dioxide) sublimates directly into CO2 gas at room temperature and pressure, which is why it never leaves a puddle.

The Heating Curve

A heating curve plots temperature (y-axis) against heat added (x-axis) as a substance is warmed at a steady rate from solid through liquid to gas. For water, the curve has five distinct segments:

1. Solid warming — ice below 0°C rises in temperature as heat is added.
2. Melting plateau at 0°C — temperature holds flat while ice absorbs heat and turns to liquid water.
3. Liquid warming — water between 0°C and 100°C rises in temperature.
4. Boiling plateau at 100°C — temperature holds flat (much longer than the melting plateau) while liquid water absorbs heat and turns to steam.
5. Gas warming — steam above 100°C continues rising in temperature.

The key concept tested repeatedly on STEM screening tests: during a plateau, all added heat goes into breaking or forming intermolecular bonds — not into raising temperature. A thermometer reading 100°C tells you nothing about whether the water is still boiling or has fully turned to steam; only the amount of heat added (or removed) reveals that.

Two formulas cover the two kinds of segments:

• Sloped segments (temperature changing): q = mcΔT, where q is heat, m is mass, c is specific heat capacity (the heat needed to raise 1 gram of a substance by 1°C), and ΔT is the temperature change. Water's specific heat differs by phase: ice ≈ 2.09 J/(g·°C), liquid water = 4.18 J/(g·°C), steam ≈ 2.01 J/(g·°C).
• Flat segments (phase changing): q = mL, where L is the latent heat of the phase change — the heat of fusion (Lf) for melting/freezing (334 J/g for water) or the heat of vaporization (Lv) for boiling/condensing (2,260 J/g for water).

Worked Examples

Worked Example 1 (phase change only): How much heat is needed to melt 25.0 g of ice at 0°C into liquid water at 0°C?

• This is a flat-segment (phase-change) calculation, so use q = mLf.
• q = (25.0 g)(334 J/g) = 8,350 J = 8.35 kJ

Worked Example 2 (temperature change only): How much heat is needed to raise the temperature of 50.0 g of liquid water from 20.0°C to 80.0°C?

• This is a sloped-segment calculation, so use q = mcΔT.
• ΔT = 80.0°C - 20.0°C = 60.0°C
• q = (50.0 g)(4.18 J/g·°C)(60.0°C) = 12,540 J = 12.5 kJ

Worked Example 3 (full heating curve, multi-step): How much total heat is required to convert 10.0 g of ice at 0°C completely into steam at 100°C?

This problem crosses one phase change, one temperature rise, and a second phase change — three separate calculations added together:

• Step 1 — melt the ice (q = mLf): (10.0 g)(334 J/g) = 3,340 J
• Step 2 — heat the liquid water from 0°C to 100°C (q = mcΔT): (10.0 g)(4.18 J/g·°C)(100°C) = 4,180 J
• Step 3 — vaporize the water (q = mLv): (10.0 g)(2,260 J/g) = 22,600 J
• Total: 3,340 J + 4,180 J + 22,600 J = 30,120 J = 30.1 kJ

Notice that vaporizing the water (22,600 J) takes nearly seven times more energy than melting it (3,340 J), even though both start and end at a defined temperature. Vaporization must fully separate water molecules from one another, while melting only needs to loosen their rigid crystal structure — this is exactly why the boiling plateau on a heating curve is so much wider than the melting plateau.

Exothermic vs. Endothermic Reactions

A chemical reaction's enthalpy change (ΔH) describes whether it releases or absorbs heat relative to its surroundings:

• Exothermic reactions release heat to the surroundings (ΔH is negative). The surroundings feel warmer. Examples: combustion (burning fuel), the iron-oxidation reaction inside a disposable hand warmer, and the reaction 2H2 + O2 → 2H2O.
• Endothermic reactions absorb heat from the surroundings (ΔH is positive). The surroundings feel colder. Examples: an instant cold pack (ammonium nitrate dissolving in water), photosynthesis, and evaporation.

Worked Example 4 (heat-of-reaction stoichiometry): The reaction 2H2(g) + O2(g) → 2H2O(g) releases 483.6 kJ of heat for every 2.00 mol of H2 consumed. How much heat is released when 0.500 mol of H2 reacts completely?

• Heat per mole of H2 = 483.6 kJ ÷ 2.00 mol = 241.8 kJ/mol
• Heat released for 0.500 mol = 0.500 mol × 241.8 kJ/mol = 120.9 kJ

This is the same stoichiometric ratio-scaling skill from Chapter 5's mole calculations, applied to energy instead of mass.

Exam Strategy

1. Never use q = mcΔT during a phase change — temperature isn't changing there, so ΔT would be zero and the calculation would be meaningless.
2. Multi-step heating-curve problems add every segment's heat together — melting, warming, and vaporizing are three separate calculations, not one formula.
3. "Feels warmer" means exothermic (releases heat); "feels colder" means endothermic (absorbs heat) — reason from the surroundings' temperature change, not the reaction's name.
4. Heat of reaction scales with moles, just like stoichiometry — divide by the given mole amount to get heat per mole, then multiply by the new mole amount.