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AP® Chemistry · Unit 4 · Topic 4.3

Representations of Reactions: Study Guide (Chemistry 4.3)

Learn to turn balanced equations into consistent particle diagrams, and to read before-and-after diagrams back into equations, for reactions, solutions and physical changes.

An equation drawn at the scale of atoms

A balanced equation is a set of symbols. A particulate representation (particle diagram) is the same information drawn as atoms, molecules and ions. You need to move both ways:

  • from an equation to a diagram, and
  • from a diagram back to an equation.

The exam can give you a diagram and ask what reaction it shows, or give you an equation and ask you to draw the particles. A free-response question may ask you to draw the particle model itself, not just write the equation. The diagram has to be consistent: it must obey the same rules as the balanced equation.

The rules a consistent diagram follows

  1. A key. Each kind of atom has its own size, shading or letter. The key says which is which. Never rely on colour alone; add letters or labels.
  2. Atoms are conserved. Count each kind of atom before and after. The totals must match, because a reaction only rearranges atoms into new combinations.
  3. The reacting ratio matches the coefficients. If the equation says 2 : 1 : 2, the numbers of particles that react and form must be in the ratio 2 : 1 : 2. The box itself can hold more or fewer.
  4. Unreacted particles stay. If one reactant is in excess, the leftover molecules appear unchanged in the “after” box.
  5. Molecules stay together; dissolved ions separate. A molecule is drawn as joined atoms. A soluble ionic compound in water is drawn as separate ions, in its formula ratio, with charges labelled.
  6. States look right. Solids are packed in a regular arrangement, liquids are close together but disordered, gases are far apart. A precipitate is a packed cluster of ions.
  7. Charge balances. In any solution diagram, the positive and negative charges add up to zero.

If a diagram breaks any of these rules, it does not represent the reaction, however tidy it looks.

Worked example 1: reading a before-and-after diagram

Sulfur dioxide reacts with oxygen to form sulfur trioxide. Figure 1 shows a sealed container before and after the reaction.

Particle diagram for the reaction of sulfur dioxide with oxygen Two boxes joined by an arrow. The Before box contains seven sulfur dioxide molecules, each a dark S circle with two white O circles joined to it, and three oxygen molecules, each two joined white O circles. The After box contains six sulfur trioxide molecules, each a dark S circle with three white O circles joined to it, and one sulfur dioxide molecule. No oxygen molecules are left. A key below shows sulfur as a dark circle with the letter S and oxygen as a white circle with the letter O. Before After OOS OOS OOS OOS OOS OOS OOS OO OO OO OOOS OOOS OOOS OOOS OOOS OOOS OOS Ssulfur atom (dark, letter S)Ooxygen atom (white, letter O)
Figure 1. Before: 7 SO₂ and 3 O₂. After: 6 SO₃ and 1 SO₂. Each circle carries its element letter, so the diagram does not depend on shading.

Question. Write the balanced equation for the reaction shown in Figure 1, and check that the diagram is consistent.

  1. Count the particles in each box. Before: 7 SO₂ molecules and 3 O₂ molecules. After: 6 SO₃ molecules and 1 SO₂ molecule.
  2. Count the atoms. Before: S = 7; O = 7 × 2 + 3 × 2 = 20. After: S = 6 + 1 = 7; O = 6 × 3 + 1 × 2 = 20. Both totals match, so atoms are conserved.
  3. Find what actually changed. All 3 O₂ molecules were used. Only 6 of the 7 SO₂ molecules were used (1 is left over). 6 SO₃ molecules formed.
  4. Write the ratio and reduce it. Used and formed: 6 SO₂ : 3 O₂ : 6 SO₃. Divide by 3 to get 2 : 1 : 2.
  5. Write the equation. 2 SO₂(g) + O₂(g) → 2 SO₃(g).

Answer. 2 SO₂(g) + O₂(g) → 2 SO₃(g). The leftover SO₂ molecule is not part of the equation. It is in the box because more sulfur dioxide was present than the oxygen could react with.

Check. The number of molecules fell from 10 to 7. That is fine: molecules are not conserved in a chemical reaction, only atoms are. (Two common traps here are writing “7 SO₂ + 3 O₂ → 6 SO₃ + SO₂”, which leaves the unreacted SO₂ on both sides, and writing “7 SO₂ + 3 O₂ → 6 SO₃”, which does not balance.)

Representing reactions in solution

Many reactions you meet in Unit 4 happen in water. The particle diagram must show what is really in the beaker:

  • A soluble ionic compound is split into its ions. Three formula units of CaCl₂(aq) are drawn as 3 Ca²⁺ ions and 6 Cl⁻ ions, spread out and not touching. Drawing “CaCl₂” units stuck together would mean the solid had not dissolved.
  • A polyatomic ion such as SO₄²⁻ or NO₃⁻ stays in one piece. It is one particle with one charge, not separate S and O atoms.
  • A molecular solute such as glucose stays as whole molecules.
  • An insoluble product (a precipitate) appears as a packed cluster of ions, usually at the bottom of the container.
  • Spectator ions appear in both pictures unchanged. Their number does not change, so in the net ionic equation they are left out, just as in Topic 4.2.

This is why the three forms of equation from Topic 4.2 all match one particle diagram. The molecular equation names the compounds, the complete ionic equation lists every ion in the beaker, and the net ionic equation describes only the particles whose arrangement changes.

Worked example 2: drawing the “after” picture for a precipitation

Question. A beaker contains a solution with 3 Ca²⁺ ions and 6 Cl⁻ ions (from CaCl₂). A second solution with 9 Na⁺ ions and 3 PO₄³⁻ ions (from Na₃PO₄) is poured in. Calcium phosphate is insoluble. Describe the particle diagram after the reaction, and give the net ionic equation.

  1. Write the net ionic equation. Only calcium and phosphate ions combine. The solid must be neutral, so 3 Ca²⁺ (charge +6) combine with 2 PO₄³⁻ (charge −6): 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) → Ca₃(PO₄)₂(s). The ratio is 3 : 2.
  2. Find what reacts. There are 3 Ca²⁺ and 3 PO₄³⁻. The 3 Ca²⁺ use up 2 PO₄³⁻. One PO₄³⁻ is left over.
  3. List the spectators. Na⁺ and Cl⁻ do not change: all 9 Na⁺ and all 6 Cl⁻ stay dissolved.
  4. Draw the after picture. A solid cluster of 3 Ca²⁺ and 2 PO₄³⁻ ions, packed together. Spread through the solution: 9 Na⁺, 6 Cl⁻ and 1 PO₄³⁻, all separate (Figure 2).
  5. Check the charges. In solution: 9 × (+1) + 6 × (−1) + 1 × (−3) = 0. The solid is also neutral: 3 × (+2) + 2 × (−3) = 0.
  6. Check the counts. Phosphate before: 3. Phosphate after: 2 in the solid + 1 dissolved = 3. Calcium before: 3; after: 3 in the solid. Every ion is accounted for.
Particle diagram after mixing calcium chloride and sodium phosphate solutions A beaker. Spread through the liquid are nine separate sodium ions labelled Na plus, six separate chloride ions labelled Cl minus with dashed outlines, and one separate phosphate ion labelled PO4 3 minus. At the bottom is a packed row of five ions labelled precipitate: three calcium ions, dark circles labelled Ca 2 plus, alternating with two phosphate ions. A key on the right names each ion and notes that water molecules are not drawn. Na⁺ Na⁺ Na⁺ Na⁺ Na⁺ Na⁺ Na⁺ Na⁺ Na⁺ Cl⁻ Cl⁻ Cl⁻ Cl⁻ Cl⁻ Cl⁻ PO₄³⁻ Ca²⁺PO₄³⁻Ca²⁺PO₄³⁻Ca²⁺ precipitate Key (labels inside each circle) Na⁺sodium ion Cl⁻chloride ion (dashed) Ca²⁺calcium ion (dark) PO₄³⁻phosphate ion (one particle) Water molecules are not drawn.
Figure 2. After mixing: the precipitate is a packed cluster of 3 Ca²⁺ and 2 PO₄³⁻ ions; the spectator ions (Na⁺ and Cl⁻) and the excess PO₄³⁻ stay separate in solution. Each ion is identified by its label and outline, not by shading alone.

Answer. Net ionic equation: 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) → Ca₃(PO₄)₂(s). After the reaction the diagram shows 3 Ca²⁺ and 2 PO₄³⁻ packed as one formula unit of solid, with 9 Na⁺, 6 Cl⁻ and 1 PO₄³⁻ separate in solution.

Why the details matter. If you drew Na⁺ paired with PO₄³⁻, or Na⁺ paired with Cl⁻, you would be showing solid sodium phosphate or sodium chloride, which are both soluble. If you drew the solid with Ca²⁺ and PO₄³⁻ in a 1 : 1 ratio, it would carry a charge, which is impossible for a neutral solid. If you drew the leftover PO₄³⁻ as P and O atoms, you would be breaking covalent bonds that do not break in this reaction.

Representing physical processes

The same rules apply when nothing new is made. In a physical process the particles keep their identity. Only how close they are and how they are arranged changes.

Process Before After What stays the same
Boiling water, H₂O(l) → H₂O(g) 5 H₂O molecules close together, disordered 5 H₂O molecules far apart Every molecule is still H₂O; no O–H bonds break
Freezing water, H₂O(l) → H₂O(s) Molecules close, disordered Molecules in a regular, repeating arrangement Same molecules, same number
Dissolving glucose, C₆H₁₂O₆(s) → C₆H₁₂O₆(aq) Molecules packed in a solid Whole glucose molecules spread among water molecules Each glucose molecule stays intact
Dissolving NaCl, NaCl(s) → Na⁺(aq) + Cl⁻(aq) Na⁺ and Cl⁻ in a lattice Separate Na⁺ and Cl⁻ ions, each surrounded by water The same ions, 1 : 1 ratio

A common error is to draw boiling water as H₂ and O₂ molecules. That would be a chemical change (decomposition), not a phase change. Bubbles in boiling water contain H₂O molecules in the gas phase. Topic 4.4 looks at why dissolving an ionic solid sits close to the line between physical and chemical change.

When you draw a phase change, keep the number of molecules the same in both boxes, and make the spacing tell the story: wider gaps for a gas, a regular pattern for a solid.

Common misconceptions

  • “Molecules are conserved.” Only atoms (and total charge) are conserved. The number of molecules can go up or down, as in Figure 1 (10 molecules became 7).
  • “The box must contain exactly the coefficients.” Any numbers work, as long as what reacts and forms is in the coefficient ratio.
  • Leaving excess reactant out of the after box. Unreacted particles do not vanish. Leave them in, unchanged.
  • Putting excess reactant into the equation. The equation shows only what changes. Remove leftovers before you reduce the ratio.
  • Drawing dissolved ionic compounds as formula units. NaCl(aq) is separate Na⁺ and Cl⁻ ions, not NaCl pairs.
  • Breaking up polyatomic ions. SO₄²⁻, NO₃⁻ and NH₄⁺ stay in one piece in solution.
  • Wrong ion ratio. K₂SO₄(aq) has twice as many K⁺ ions as SO₄²⁻ ions.
  • Drawing a phase change as a reaction. Boiling or melting never splits a molecule into atoms.

Where this leads

This topic links the equations of Topic 4.2, Net Ionic Equations to the particle level. Next, Topic 4.4, Physical and Chemical Changes, uses particle pictures to decide whether bonds or only intermolecular attractions change in a process. Reading a diagram for the reacting ratio and leftover reactant also prepares you for stoichiometry and limiting reactants in Topic 4.5. Try the practice questions now, then use the revision notes and the checklist to consolidate.

Quick questions

Do I have to draw the water molecules in a solution diagram?

Usually not. Most questions say water is omitted for clarity. Draw the dissolved particles spread through the container and state in a key or label that water is not shown.

Does the number of particles in the box have to equal the coefficients?

No. The box can hold any numbers, as long as the particles that react and form are in the coefficient ratio. Six SO₂ reacting with three O₂ is the same 2 : 1 ratio as the equation 2 SO₂ + O₂ → 2 SO₃.

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