Representations of reactions — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: This chapter covers balanced chemical equation construction, molecular/complete ionic/net ionic equation writing, particle-level (particulate) reaction diagram interpretation, and matching different types of reaction representations for all core AP reaction types.

You should already know: Law of conservation of mass, solubility rules for common ionic compounds, definition of spectator ions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Representations of reactions?

Representations of reactions is the standardized set of notations chemists use to communicate what reactants are consumed, what products are formed, the stoichiometric ratios between species, and the particle-level changes that occur during a chemical reaction. Per the AP Chemistry Course and Exam Description (CED), this is topic 4.1 in Unit 4 (Chemical Reactions), accounting for ~7-9% of the overall exam score, and it is tested across both multiple-choice (MCQ) and free-response (FRQ) sections. Unlike vague word equations, formal reaction representations convey both exact qualitative and quantitative information. This topic includes not just writing equations, but also interpreting given representations, matching different representation types to one another, and identifying and correcting flawed representations that violate conservation laws. It is a gateway skill for almost all other topics in Unit 4 and the rest of the course, as every calculation involving reactions starts with a correct representation.

2. Balanced Molecular and Ionic Equation Representations

Reaction equations can be written at three levels of detail for solution reactions, each with a specific purpose. Molecular equations show all reactant and product species as neutral compounds, even if they dissociate into ions in solution. This is useful for describing the overall reaction starting from bulk reactants. Complete ionic equations split all soluble strong electrolytes (soluble ionic salts, strong acids, strong bases) into their dissociated ions, leaving insoluble compounds, weak electrolytes, and gases in their neutral molecular form. This shows all ions present in the solution. Net ionic equations remove spectator ions (ions that appear unchanged in charge and state on both sides of the reaction) to leave only the species that actually undergo chemical change. All balanced reaction equations must satisfy two requirements: conservation of mass (total number of each atom is equal on reactant and product sides) and conservation of charge (total net charge is equal on both sides, which is especially critical for ionic reactions).

Worked Example

Write the balanced net ionic equation for the reaction between aqueous calcium hydroxide and aqueous nitric acid, which forms liquid water and aqueous calcium nitrate.

1. First write the balanced molecular equation, ensuring all atoms are balanced: $${\text{Ca(OH)}}_2(aq)+2{\text{HNO}}_3(aq)\to 2{\text{H}}_2\text{O}(l)+{\text{Ca(NO}}_3{\text{)}}_2(aq)$$
2. Split all soluble strong electrolytes into ions; leave covalent liquid water intact: $${\text{Ca}}^{2+}(aq)+2{\text{OH}}^{-}(aq)+2{\text{H}}^{+}(aq)+2{\text{NO}}_3^{-}(aq)\to 2{\text{H}}_2\text{O}(l)+{\text{Ca}}^{2+}(aq)+2{\text{NO}}_3^{-}(aq)$$
3. Identify and cancel spectator ions: ${\text{Ca}}^{2+}$ and ${\text{NO}}_3^{-}$ appear unchanged on both sides, so they are removed.
4. Check the remaining equation for mass and charge balance: Mass: 2 H, 2 O, 1 Ca (canceled) = 2 H, 1 O × 2 = 2 H₂O: balanced. Charge: $2(-1)+2(+1)=0$ (reactants), $0$ (products): balanced. Final net ionic equation: ${\text{H}}^{+}(aq)+{\text{OH}}^{-}(aq)\to {\text{H}}_2\text{O}(l)$

Exam tip: Always check both mass and charge balance for net ionic equations on the AP exam; MCQ answer choices almost always include one or more options with correct atoms but unbalanced charge, which is an easy point to miss.

3. Interpreting Particulate (Particle-Level) Reaction Diagrams

Particulate diagrams are graphical representations of reactions that show individual atoms, ions, or molecules as differently sized/shaded spheres, to illustrate what happens at the particle level before and after a reaction. AP Chemistry regularly tests the skill of using a given particulate diagram to write a balanced reaction equation, identify excess/limiting reactants, or match the diagram to a corresponding equation. The key rule for working with these diagrams is that they show only a representative sample of particles, not the total amount in the entire reaction vessel. This means you must simplify the ratio of reacted reactants to formed products to the lowest whole number stoichiometry. Unreacted particles left over after the reaction are excess and should not be included in the balanced equation coefficients.

Worked Example

A particulate reaction diagram shows 6 molecules of hydrogen gas (${\text{H}}_2$) and 2 molecules of nitrogen gas (${\text{N}}_2$) before reaction. After reaction completes, the diagram shows 4 molecules of ammonia (${\text{NH}}_3$), 2 unreacted ${\text{H}}_2$ molecules, and 0 unreacted ${\text{N}}_2$ molecules. Write the balanced equation for this reaction based on the diagram.

1. Calculate how many of each reactant were consumed: ${\text{H}}_2$ consumed = initial ${\text{H}}_2$ - unreacted ${\text{H}}_2$ = $6-2=4$ ${\text{N}}_2$ consumed = initial ${\text{N}}_2$ - unreacted ${\text{N}}_2$ = $2-0=2$
2. Count the number of product molecules formed: 4 molecules of ${\text{NH}}_3$.
3. Write the unbalanced equation with the counted values: $4{\text{H}}_2+2{\text{N}}_2\to 4{\text{NH}}_3$
4. Divide all coefficients by the greatest common factor (2) to get the lowest whole number ratio: $2{\text{H}}_2+{\text{N}}_2\to 2{\text{NH}}_3$
5. Confirm atom balance: Reactants: 4 H, 2 N; Products: 2 × (1 N + 3 H) = 2 N, 4 H: balanced.

Exam tip: Always subtract unreacted excess reactant particles from the initial count to get the number of particles that actually reacted; AP exam diagrams always include excess particles to test if you incorrectly count them as reacted.

4. Matching Different Types of Reaction Representations

A very common AP exam question gives you one type of reaction representation (e.g. a net ionic equation or a particulate diagram) and asks you to identify which other representation matches it. To solve these problems, follow three systematic steps: 1) Confirm the state of matter of each species, which tells you if it should be dissociated (aqueous) or shown as an intact unit (solid, gas, liquid covalent). 2) Confirm the stoichiometric ratio of dissociated ions: for example, 1 mole of ${\text{FeCl}}_3$ dissociates into 1 ${\text{Fe}}^{3+}$ and 3 ${\text{Cl}}^{-}$, so the ratio of ion spheres in the diagram must be 1:3. 3) Confirm that the ratio of reacted reactants to formed products matches across both representations. Checking ion ratios first is the fastest way to eliminate incorrect answer choices.

Worked Example

Given the balanced net ionic equation: $2{\text{Ag}}^{+}(aq)+{\text{CO}}_3^{2-}(aq)\to {\text{Ag}}_2{\text{CO}}_3(s)$, which of the following particulate diagrams (reactant side only, spectators not shown) matches this representation?

• Diagram A: 1 ${\text{Ag}}^{+}$ sphere, 1 ${\text{CO}}_3^{2-}$ sphere
• Diagram B: 2 ${\text{Ag}}^{+}$ spheres, 1 ${\text{CO}}_3^{2-}$ sphere
• Diagram C: 2 ${\text{Ag}}^{+}$ spheres, 2 ${\text{CO}}_3^{2-}$ spheres
• Diagram D: 1 solid ${\text{Ag}}_2{\text{CO}}_3$ unit, 2 ${\text{Ag}}^{+}$ ions and 1 ${\text{CO}}_3^{2-}$ ion

1. From the net ionic equation, the stoichiometric ratio of ${\text{Ag}}^{+}$ to ${\text{CO}}_3^{2-}$ reactants is 2:1. The question asks for the reactant side, spectators not shown (consistent with net ionic, which omits spectators).
2. Eliminate incorrect options: Diagram A has a 1:1 ratio, so incorrect. Diagram C has a 2:2 = 1:1 ratio, so incorrect. Diagram D shows the product plus unreacted ions, which does not match the reactant-side net ionic, so incorrect.
3. Confirm Diagram B has the correct 2:1 ratio of reactant ions, matching the net ionic equation.

Exam tip: When matching representations, always check the ion ratio for dissolved ionic compounds first; this will usually eliminate 2-3 wrong answer choices immediately on MCQ.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Splitting a solid insoluble ionic compound into ions when writing a complete ionic equation. Why: Students confuse soluble and insoluble compounds, or forget only aqueous strong electrolytes are split. Correct move: Always apply solubility rules before splitting any compound; only split aqueous soluble ionic compounds, strong acids, and strong bases, and leave solids, weak acids/bases, and gases intact.
• Wrong move: Counting unreacted excess particles in a particulate diagram as part of the stoichiometric coefficients. Why: Students count all particles shown on the reactant side instead of only those that reacted. Correct move: Always mark off unreacted particles that appear unchanged after the reaction, and only count reacted particles when calculating stoichiometric ratios.
• Wrong move: Forgetting to balance charge in a net ionic equation, only balancing atoms. Why: Students are used to balancing molecular equations only by mass, and carry that habit over. Correct move: After balancing atoms, add up the total charge on the reactant side and product side; adjust coefficients if the charges do not match.
• Wrong move: Writing the wrong coefficient for a dissociated ion, e.g. ${\text{CaCl}}_2(aq)\to {\text{Ca}}^{2+}(aq)+{\text{Cl}}^{-}(aq)$. Why: Students forget the subscript in the original compound applies to the number of dissociated ions. Correct move: For every soluble ionic compound, multiply the ion coefficient by the subscript of that ion in the original neutral compound when splitting.
• Wrong move: Including spectator ions in a net ionic equation. Why: Students forget the definition of spectator ions, or do not cancel them correctly. Correct move: After writing the complete ionic equation, cross out every ion that appears in identical form (same state, same charge, same number) on both sides before writing the net ionic.
• Wrong move: Assuming all same-sized spheres in a particulate diagram are the same species. Why: Diagrams often use shading to distinguish different ions of similar size. Correct move: Always check the provided key to match each sphere's size and shading to the corresponding species before counting.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

A student mixes aqueous barium hydroxide with aqueous sulfuric acid, forming solid barium sulfate and liquid water. Which of the following is the correct balanced net ionic equation for this reaction? A) ${\text{Ba}}^{2+}(aq)+{\text{OH}}^{-}(aq)+{\text{H}}^{+}(aq)+{\text{SO}}_4^{2-}(aq)\to {\text{BaSO}}_4(s)+{\text{H}}_2\text{O}(l)$ B) ${\text{Ba}}^{2+}(aq)+2{\text{OH}}^{-}(aq)+2{\text{H}}^{+}(aq)+{\text{SO}}_4^{2-}(aq)\to {\text{BaSO}}_4(s)+2{\text{H}}_2\text{O}(l)$ C) ${\text{Ba(OH)}}_2(aq)+{\text{H}}_2{\text{SO}}_4(aq)\to {\text{BaSO}}_4(s)+2{\text{H}}_2\text{O}(l)$ D) ${\text{Ba}}^{2+}(aq)+{\text{SO}}_4^{2-}(aq)\to {\text{BaSO}}_4(s)$

Worked Solution: The question asks for a net ionic equation, so we can immediately eliminate option C, which is a balanced molecular equation, not net ionic. Next, we build the net ionic step-by-step: Barium hydroxide is a soluble strong base, so it dissociates into 1 ${\text{Ba}}^{2+}$ and 2 ${\text{OH}}^{-}$. Sulfuric acid is a strong acid, so it dissociates into 2 ${\text{H}}^{+}$ and 1 ${\text{SO}}_4^{2-}$. Products are solid ${\text{BaSO}}_4$ (intact) and liquid water (intact, covalent). There are no spectator ions, so all ions are included in the net ionic equation. Option A is unbalanced for atoms and charge, and option D omits the ${\text{OH}}^{-}$ and ${\text{H}}^{+}$ that react to form water, so only option B is balanced for both mass and charge. Correct answer: B.

Question 2 (Free Response)

A student studies the reaction between aqueous lead(II) nitrate and aqueous potassium carbonate. The reaction forms solid lead(II) carbonate and aqueous potassium nitrate. (a) Write the balanced complete ionic equation for this reaction. (b) Write the balanced net ionic equation for this reaction. (c) The student draws a particulate diagram of reactants with 3 ${\text{Pb}}^{2+}$ ions, 6 ${\text{NO}}_3^{-}$ ions, 4 ${\text{CO}}_3^{2-}$ ions, and 8 ${\text{K}}^{+}$ ions. How many total dissolved ions remain in solution after the reaction goes to completion? Justify your answer.

Worked Solution: (a) First, write the balanced molecular equation, then split all soluble compounds into ions: $${\text{Pb}}^{2+}(aq)+2{\text{NO}}_3^{-}(aq)+2{\text{K}}^{+}(aq)+{\text{CO}}_3^{2-}(aq)\to {\text{PbCO}}_3(s)+2{\text{K}}^{+}(aq)+2{\text{NO}}_3^{-}(aq)$$

(b) Cancel spectator ions (${\text{K}}^{+}$ and ${\text{NO}}_3^{-}$) and check balance: $${\text{Pb}}^{2+}(aq)+{\text{CO}}_3^{2-}(aq)\to {\text{PbCO}}_3(s)$$

(c) The reaction ratio of ${\text{Pb}}^{2+}$ to ${\text{CO}}_3^{2-}$ is 1:1. We have 3 ${\text{Pb}}^{2+}$ and 4 ${\text{CO}}_3^{2-}$, so ${\text{Pb}}^{2+}$ is limiting. All 3 ${\text{Pb}}^{2+}$ react with 3 ${\text{CO}}_3^{2-}$, leaving 1 unreacted ${\text{CO}}_3^{2-}$ in solution. All 6 ${\text{NO}}_3^{-}$ and 8 ${\text{K}}^{+}$ are spectator ions and remain dissolved. The reacted ions form solid ${\text{PbCO}}_3$, which is not dissolved, so they are not counted. Total dissolved ions = $1+6+8=15$.

Question 3 (Application / Real-World Style)

Hard water contains dissolved ${\text{Ca}}^{2+}$ ions that form insoluble calcium carbonate scale, damaging water heaters. A water softener uses solid sodium zeolite (${\text{Na}}_2\text{Z}(s)$, ${\text{Z}}^{2-}$ = insoluble zeolite anion) to remove ${\text{Ca}}^{2+}$ via the reaction: $${\text{Ca}}^{2+}(aq)+{\text{Na}}_2\text{Z}(s)\rightleftharpoons \text{CaZ}(s)+2{\text{Na}}^{+}(aq)$$ At equilibrium, a particulate diagram of the softener system shows 4 unreacted ${\text{Ca}}^{2+}$ ions and 12 ${\text{Na}}^{+}$ ions in solution. If 10 total ${\text{Ca}}^{2+}$ ions entered the system from hard water, how many ${\text{Na}}^{+}$ ions were released into solution by the zeolite? Explain what this means for water softening.

Worked Solution: First, calculate the number of ${\text{Ca}}^{2+}$ ions that reacted with the zeolite: total ${\text{Ca}}^{2+}$ minus unreacted ${\text{Ca}}^{2+}$ = $10-4=6$ reacted ${\text{Ca}}^{2+}$ ions. From the balanced reaction, 1 mole of ${\text{Ca}}^{2+}$ reacts to release 2 moles of ${\text{Na}}^{+}$, so total ${\text{Na}}^{+}$ released = $6\times 2=12$. In context, this means that every calcium ion (which forms harmful scale) is removed from the water and replaced with two sodium ions, which do not form insoluble scale, so the water is softened for use.

7. Quick Reference Cheatsheet

8. What's Next

Representations of reactions are the foundational skill for all remaining topics in Unit 4 and the entire AP Chemistry course. Next, you will use balanced reaction representations to solve stoichiometry problems, calculate reaction yields, identify limiting reactants, and classify reaction types. Without correctly writing or interpreting reaction representations, all stoichiometric calculations will be based on an incorrect reaction equation, leading to wrong answers even if your arithmetic is correct. This topic also feeds into later topics including acid-base titrations, redox reactions, and chemical equilibrium, where you must constantly interpret and write reaction representations to solve problems.

• Reaction stoichiometry
• Types of chemical reactions
• Introduction to acid-base reactions
• Particle models of equilibrium