Worked examples — Physical vs chemical change

This page is a practice gauntlet. The parent note Physical vs chemical change told you what the two categories are. Here we hit every kind of case you could ever be asked to classify — so that no exam question, no kitchen mishap, no science-fair result can surprise you.

Before we begin, one plain-language reminder of our two words, because everything below rests on them:

The single tool we use to decide is one question: "After the change, is the molecular formula still the same?" If yes → physical. If no → chemical. Every example is just that question, asked carefully.

Two pieces of shorthand appear in every example, so let us define them now, before we use them:

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The scenario matrix

There are only so many shapes a "is this physical or chemical?" problem can take. If we solve one of each, we've solved them all. Here is the full list of cells we will cover:

Cell	Case class	What makes it tricky	Covered by
A	State change (solid↔liquid↔gas)	Looks dramatic, but molecule unchanged	Ex 1 (freezing), Ex 5 (dry ice edge)
B	Dissolving (mixing into solution)	"It disappeared!" ≠ new substance	Ex 2 (salt)
C	Gas produced — new molecule	Bubbles ≠ automatically chemical	Ex 3 (vinegar + soda)
D	Gas produced — same molecule	The trap partner of Cell C	Ex 5 (dry ice)
E	Colour / precipitate appears	Colour can lie both ways	Ex 4 (silver + salt)
F	Reversible-looking but chemical	Rust, tarnish — cannot un-do	Ex 6 (rusting)
G	Zero / degenerate input — nothing happens	Is "no change" a change?	Ex 7 (limiting case)
H	Energy sign flip — heat absorbed vs released	Both can be physical OR chemical	Ex 8 (word problem)
I	Exam twist — mixed process, must separate steps	Two things happen at once	Ex 9 (mixed)

We now walk each cell. Try to forecast each answer before reading the steps — that is where the learning lives.

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Example 1 — Cell A: water freezing

Forecast: Bonds broke? New molecule? Guess now.

1. Identify before and after. Before: ${\text{H}}_2\text{O}(l)$. After: ${\text{H}}_2\text{O}(s)$. Why this step? Our one question needs a "before formula" and an "after formula" to compare.
2. Compare molecular formulas. ${\text{H}}_2\text{O}$ vs ${\text{H}}_2\text{O}$ — identical. Why this step? Same formula ⇒ no bond inside a molecule broke; only the arrangement between molecules changed (they locked into a crystal lattice — see States of Matter).
3. Check reversibility. Warm it back up → liquid water returns, tastes and behaves identically. Why this step? Easy reversal by a physical means (heating) is a strong hint of physical change.

Verify: The energy involved is the heat of fusion, $\Delta H_{\text{fus}}=6.0\text{kJ/mol}$ — tiny, in the intermolecular range ($<40$ kJ/mol), not the covalent bond range ($>400$ kJ/mol). Consistent with physical change. ✅

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Example 2 — Cell B: salt dissolving

Forecast: The salt vanished from sight. Trap or truth?

1. Before / after. Before: solid $\text{NaCl}$ + water. After: a clear salt solution. Why this step? "Disappeared" is a visual claim, not a molecular one. We must check molecules, not eyes.
2. What actually happened to the particles? Water pulled the ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions apart and surrounded them. The ions still exist — nothing new was built. Why this step? Pulling ions apart breaks an ionic attraction (a between-particle force), not a covalent bond that would create a new molecule.
3. The recovery test. Boil the water away → solid $\text{NaCl}$ crystals return, identical. Why this step? Full recovery by evaporation confirms no new substance formed. Compare Identification of Substances.

Verify: Dissolving $\text{NaCl}$ has $\Delta H_{\text{soln}}\approx +3.9\text{kJ/mol}$ — small, within the physical (between-particle) range. Physical change. ✅

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Example 3 — Cell C: vinegar + baking soda (gas, NEW molecule)

Forecast: Bubbles! Everyone's instinct says chemical — but why, exactly?

1. Before / after. Before: acetic acid + sodium bicarbonate. After: fizzing, and the solid dissolves. Why this step? Our one question needs a fixed "before" list of substances to compare against whatever appears afterwards — without pinning down the starting materials we cannot tell if anything new was made.
2. Name the gas. The gas is carbon dioxide, ${\text{CO}}_2$ — a molecule that was not present before. Why this step? This is the whole game. Bubbles only prove chemical change if the gas is a new substance (Cell C), not a phase-changed old one (Cell D).
3. Write the reaction. (Recall $\uparrow$ = leaves as a gas.) ${\text{CH}}_3\text{COOH}+{\text{NaHCO}}_3\to {\text{CH}}_3\text{COONa}+{\text{H}}_2\text{O}+{\text{CO}}_2\uparrow$ Why this step? Seeing new formulas on the right (sodium acetate, ${\text{CO}}_2$) proves bonds re-glued into new molecules. See Chemical Reactions.
4. Reversibility test. Cool/compress the ${\text{CO}}_2$ → you get dry ice, not vinegar back. Why this step? Failure to recover the originals confirms chemical change.

Verify (mass check). By Conservation of Mass, mass in = mass out. Left side atom count: from ${\text{CH}}_3\text{COOH}$ (C2 H4 O2) + ${\text{NaHCO}}_3$ (Na1 H1 C1 O3) = C3 H5 O5 Na1. Right side: ${\text{CH}}_3\text{COONa}$ (C2 H3 O2 Na1) + ${\text{H}}_2\text{O}$ (H2 O1) + ${\text{CO}}_2$ (C1 O2) = C3 H5 O5 Na1. Atoms balance. Chemical change. ✅

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Example 4 — Cell E: a precipitate + colour appears

(If the figure does not display: it shows two clear beakers on the left — an ${\text{AgNO}}_3$ solution with teal ${\text{Ag}}^{+}$ and orange ${\text{NO}}_3^{-}$ ions, and an $\text{NaCl}$ solution with plum ${\text{Na}}^{+}$ and ink-coloured ${\text{Cl}}^{-}$ ions. An arrow labelled "mix" points right to the result, where the teal ${\text{Ag}}^{+}$ and ink ${\text{Cl}}^{-}$ have clumped into a white solid $\text{AgCl}$, while plum ${\text{Na}}^{+}$ and orange ${\text{NO}}_3^{-}$ remain floating, still dissolved.)

Figure guide (read before the steps): In the picture, teal circles are the silver ion ${\text{Ag}}^{+}$, orange circles are the nitrate ion ${\text{NO}}_3^{-}$, plum circles are the sodium ion ${\text{Na}}^{+}$, and dark ink circles are the chloride ion ${\text{Cl}}^{-}$. The left beakers show the two starting solutions; the right side shows the result after mixing, where the teal ${\text{Ag}}^{+}$ has clumped with the ink ${\text{Cl}}^{-}$ into a solid, while the plum ${\text{Na}}^{+}$ and orange ${\text{NO}}_3^{-}$ stay floating (dissolved).

Forecast: Two clear liquids, and a solid appears from nowhere. What is that solid?

1. Before / after. Before: two colourless clear solutions. After: milky white cloud (a precipitate — a new solid falling out of solution). Why this step? A solid appearing where there was none is Cell E's signature. Look at the figure: the teal silver ions and ink chloride ions clump together.
2. Track the ions. In solution we have ${\text{Ag}}^{+},{\text{NO}}_3^{-},{\text{Na}}^{+},{\text{Cl}}^{-}$. The ${\text{Ag}}^{+}$ grabs the ${\text{Cl}}^{-}$ to form solid $\text{AgCl}$. Why this step? $\text{AgCl}$ is a brand-new insoluble compound — not present in either starting bottle.
3. Write it. (Recall $(aq)$ = dissolved in water, $(s)$ = solid, $\downarrow$ = falls out as solid.) ${\text{AgNO}}_3(aq)+\text{NaCl}(aq)\to \text{AgCl}(s)\downarrow +{\text{NaNO}}_3(aq)$ Why this step? New solid $\text{AgCl}$ on the right = new molecular arrangement = chemical.

Verify (mass balance). Left atoms: Ag1 N1 O3 + Na1 Cl1 = Ag1 N1 O3 Na1 Cl1. Right: AgCl (Ag1 Cl1) + NaNO₃ (Na1 N1 O3) = Ag1 Cl1 Na1 N1 O3. Balanced. Chemical change. ✅

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Example 5 — Cell D (and Cell A edge): dry ice subliming

Forecast: This is the twin of Example 3 — both make ${\text{CO}}_2$ gas. But the answer flips. Why?

1. Before / after. Before: ${\text{CO}}_2(s)$. After: ${\text{CO}}_2(g)$. Why this step? Notice: same formula on both sides, unlike Example 3 where new formulas appeared.
2. Name what broke. Only the weak ${\text{CO}}_2\cdots {\text{CO}}_2$ dispersion forces between whole molecules broke; the $\text{C=O}$ bonds inside each molecule stayed intact. Why this step? Between-molecule forces breaking = physical; inside-molecule bonds breaking = chemical.
3. Reversibility test. Cool the gas → dry ice re-forms. Recover the original perfectly. Why this step? Recovering the identical starting substance by a purely physical means (cooling) confirms no new molecule was ever made — the final proof of a physical change.

Verify: $\Delta H_{\text{sub}}({\text{CO}}_2)=25\text{kJ/mol}$, far below the $\text{C=O}$ bond energy of $799\text{kJ/mol}$. Ratio $799/25\approx 32$ — a factor of tens, exactly the between-molecule vs inside-molecule signature. Physical change. ✅ This is why Cell C and Cell D must always be separated: same gas, opposite verdicts, decided only by whether that gas is a new molecule. See Energy in Chemistry.

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Example 6 — Cell F: rusting (irreversible sign)

Forecast: No flame, no fizz, slow and quiet. Does "quiet" mean physical?

1. Before / after. Before: $\text{Fe}$ (shiny, magnetic). After: rust, mainly ${\text{Fe}}_2{\text{O}}_3\cdot n{\text{H}}_2\text{O}$ (dull, brittle, non-magnetic). Why this step? Property changes (magnetism lost, colour, brittleness) hint that the identity changed — the Molecular Structure is different. Here $n$ is the hydration number: how many water molecules are loosely locked into the rust's crystal for each ${\text{Fe}}_2{\text{O}}_3$ unit. It is not fixed — in real rust $n$ typically ranges from about $1$ to $3$ depending on humidity — which is why we write the letter $n$ instead of a number.
2. Where did the oxygen go? Oxygen atoms bonded directly onto iron atoms, forming $\text{Fe–O}$ bonds. Why this step? New bonds forming between different atoms = new molecule = chemical.
3. Reversibility test. No wiping or gentle heating restores shiny iron. Why this step? Cell F's whole point: slowness and quietness fool you, but the irreversibility gives it away.

Verify (balance the overall equation): $4\text{Fe}+3{\text{O}}_2\to 2{\text{Fe}}_2{\text{O}}_3$ Left: Fe4 O6. Right: 2×(Fe2 O3) = Fe4 O6. Balanced. Chemical change. ✅

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Example 7 — Cell G: the degenerate "nothing happens"

Forecast: A sneaky exam favourite. What's the honest answer?

1. Before / after. Before: glass marble + water. After: the same marble, same water, same everything. Why this step? Our decision question compares a "before" formula against an "after" formula; we must write both out explicitly to see that here they are identical for every substance present, leaving nothing to classify.
2. Apply the definitions strictly. No molecule changed → not chemical. But also no state, form, or arrangement changed → not even a physical change. Why this step? A change must change something. Zero difference is the degenerate input: it is classified as no change at all.
3. Contrast the edge. If instead the water slowly dissolved a little glass (it does, extremely slowly), that trace dissolving would be chemical. But over a short observation with no measurable effect, we report no change. Why this step? Naming the neighbouring real case shows where the degenerate cell sits — it stops you from lazily labelling every "boring" result as "no change" when a slow but genuine reaction may be hiding; the deciding factor is whether any measurable difference appears in the observation window.

Verify (logic check): "physical" requires (same molecule) AND (form/state/arrangement differs). Here the second condition is false, so the process is not physical; and (new molecule) is also false, so it is not chemical either. Both categories fail their test → no change at all. ✅ This is the limiting/degenerate cell every classifier must handle.

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Example 8 — Cell H: heat-absorbing word problem

Forecast: Energy is absorbed, not released. Does the sign of heat decide the category?

1. Separate the two questions. (a) Is it physical/chemical? (b) Does absorbing heat prove anything about that? These are independent. Why this step? Cell H's trap is believing heat direction picks the category. It does not — both physical and chemical changes can absorb or release heat, so we must not let the temperature drop bias our verdict.
2. Classify the process. ${\text{NH}}_4{\text{NO}}_3(s)\to {\text{NH}}_4^{+}(aq)+{\text{NO}}_3^{-}(aq)$. The salt splits into its existing ions surrounded by water — same ions, no new molecule. Why this step? Dissolving that only separates existing ions is Cell B territory → physical (a dissolution process; see States of Matter).
3. Explain the coldness. Pulling the crystal apart costs more energy than the water gives back when surrounding the ions, so net energy is drawn from the surroundings → the pack cools. This is an endothermic process (absorbs heat) but still physical. Why this step? Showing why it cools without any new molecule proves the coldness is an energy-bookkeeping effect, not evidence of a reaction — which directly refutes the student.

Verify: $\Delta H_{\text{soln}}({\text{NH}}_4{\text{NO}}_3)=+25.7\text{kJ/mol}$ (positive = heat absorbed). The magnitude sits in the between-particle range, and no new molecular formula appears → physical, endothermic. The student was wrong: heat sign never decides physical vs chemical. ✅

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Example 9 — Cell I: the exam twist (two things at once)

Forecast: One object, but two different things happen. Which "wins"?

1. Refuse to give one label to two processes. Split them: melting (i) and combustion (ii). Why this step? Cell I's lesson: exam candles bundle a physical and a chemical change. Answering "chemical" without noting the melting loses marks.
2. Sub-process (i): melting. ${\text{C}}_{25}{\text{H}}_{52}(s)\to {\text{C}}_{25}{\text{H}}_{52}(l)$. Same formula → physical (a state change, Cell A). Why this step? The wax molecule is unchanged; only its state went solid → liquid. Our one question answers "same formula" → physical.
3. Sub-process (ii): burning. Balance the combustion: ${\text{C}}_{25}{\text{H}}_{52}+38{\text{O}}_2\to 25{\text{CO}}_2+26{\text{H}}_2\text{O}$ Why this step? New molecules (${\text{CO}}_2$, ${\text{H}}_2\text{O}$) appear on the right → bonds re-glued → chemical (Cell C-type; energy released as light and heat, Energy in Chemistry).
4. Overall verdict. The candle as a working object undergoes a chemical change (the combustion is what produces the light and heat), accompanied by a physical change (the melting that keeps supplying liquid fuel to the wick). On an exam you must state both: melting = physical, burning = chemical, overall = chemical driven by the reaction. Why this step? This closes the matrix: the last cell is not "pick one" but "separate, classify each, then name the dominant one." Every scenario the topic can throw is now covered.

Verify (balance the combustion): Left: C25, H52, O = 38×2 = 76. Right: 25 CO₂ → C25, O50; 26 H₂O → H52, O26; total O = 50 + 26 = 76. C: 25=25, H: 52=52, O: 76=76. Balanced. ✅

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Recall Quick self-test

Bubbles always mean chemical change. ::: False — only if the gas is a new molecule (Cell C). Boiling water and subliming dry ice bubble too but are physical (Cell D). A change that absorbs heat and turns cold must be chemical. ::: False — the cold pack (Cell H) is physical dissolving; heat sign never decides the category. A white solid appearing from two clear solutions signals what? ::: A precipitate — a new insoluble compound, so chemical change (Cell E). Rust forms slowly and quietly, so it's physical. ::: False — new $\text{Fe–O}$ bonds form a new compound; slowness doesn't make it physical (Cell F). Pouring water on a marble with no visible effect is which cell? ::: Cell G — the degenerate case: no change at all, neither physical nor chemical. What does the small $n$ in ${\text{Fe}}_2{\text{O}}_3\cdot n{\text{H}}_2\text{O}$ mean? ::: The hydration number — how many water molecules sit in the rust per ${\text{Fe}}_2{\text{O}}_3$ unit, typically about 1 to 3.

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Parent: Physical vs chemical change See also: Chemical Reactions · Conservation of Mass · States of Matter · Energy in Chemistry · Molecular Structure · Identification of Substances