Introduction to reactions — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Balancing chemical equations by inspection, writing molecular, full and net ionic equations, identifying spectator ions, and classifying common chemical reaction types per the AP CED framework.

You should already know: Chemical formula writing for ionic and covalent compounds, solubility rules for common ionic compounds, the law of conservation of mass.

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 Introduction to reactions?

Introduction to chemical reactions is the foundational opening topic of AP Chemistry Unit 4 (Chemical Reactions), covering the core conventions and rules for representing and classifying chemical change. A chemical reaction is defined as a process where reactant substances undergo bond rearrangement to form new product substances, distinct from the starting materials. Standard notation places reactants on the left of an arrow, products on the right, with state symbols $(s),(l),(g),(aq)$ indicating the physical state of every species. This topic contributes approximately 2-3% of the total AP Chemistry exam score, with the full Unit 4 accounting for 7-9% of total exam weight. Questions appear in both MCQ and FRQ sections; balanced equations are almost always required as the first step for longer FRQs covering stoichiometry, titrations, or equilibrium. This topic anchors all problem-solving involving chemical change, as every calculation begins with a correctly written and balanced equation.

2. Balancing Chemical Equations by Inspection

Balancing chemical equations is the process of adjusting stoichiometric coefficients (whole numbers written in front of each species) to satisfy the law of conservation of mass: the number of atoms of every element must be equal on the reactant and product sides. The most critical rule for balancing is that you never change subscripts in chemical formulas to balance an equation, because changing subscripts alters the identity of the substance. The standard method for balancing by inspection follows a simple sequence: 1) Write the unbalanced equation with all correct chemical formulas and state symbols first, 2) Count atoms of each element on both sides, 3) Start with elements that appear in only one reactant and one product, 4) Leave pure elemental species (like $O_2$, $H_2$) for last, 5) Clear any fractions by multiplying all coefficients by the denominator, 6) Reduce coefficients to the lowest whole-number ratio. Intuitively, balancing ensures that every atom that enters the reaction exits as product, consistent with mass conservation for chemical (non-nuclear) change.

Worked Example

Problem: Write the balanced equation for the complete combustion of butane ($C_4{H}_{10}$), which produces carbon dioxide gas and liquid water.

1. Write the unbalanced equation with correct formulas: $$C_4{H}_{10}(g)+O_2(g)\to C{O}_2(g)+H_{2}O(l)$$
2. Count initial atoms: Reactants = 4 C, 10 H, 2 O; Products = 1 C, 2 H, 3 O.
3. Balance carbon first (only one reactant/product): Add a coefficient of 4 to $C{O}_2$: $$C_4{H}_{10}(g)+O_2(g)\to 4C{O}_2(g)+H_{2}O(l)$$
4. Next balance hydrogen: 10 H on the reactant side requires a coefficient of 5 for $H_{2}O$: $$C_4{H}_{10}(g)+O_2(g)\to 4C{O}_2(g)+5H_{2}O(l)$$
5. Balance oxygen: Total oxygen on the product side = $(4\times 2)+(5\times 1)=13$, so we need 13 O atoms on the reactant side, giving a coefficient of $\frac{13}{2}$ for $O_2$. Clear the fraction by multiplying all coefficients by 2: $$2C_4{H}_{10}(g)+13O_2(g)\to 8C{O}_2(g)+10H_{2}O(l)$$
6. Final check: C = 8=8, H=20=20, O=26= 16+10=26. All balanced, lowest whole numbers.

Exam tip: Always recount oxygen last after adjusting coefficients for $C{O}_2$ and $H_{2}O$ in combustion problems; it is the most common element to have an unbalanced count.

3. Full and Net Ionic Equations

For reactions occurring in aqueous solution, three levels of equation representation are used: molecular, full ionic, and net ionic. A molecular (full formula) equation shows all compounds as neutral molecules, even if they dissociate in water. A full ionic equation splits all strong electrolytes (soluble ionic compounds, strong acids, strong bases) into their constituent free ions, as these species exist as separated ions in solution. Spectator ions are ions that appear unchanged on both sides of the full ionic equation, meaning they do not participate in the actual chemical change. A net ionic equation is obtained by canceling spectator ions from the full ionic equation, leaving only the species that undergo chemical change. A key rule to remember: never dissociate weak electrolytes (weak acids, weak bases), insoluble ionic compounds (precipitates), gases, or pure substances like water, as these do not exist as free ions in solution. Net ionic equations must also balance total charge, not just atoms, since we are working with charged ions.

Worked Example

Problem: Aqueous barium chloride reacts with aqueous sodium sulfate to form solid barium sulfate and aqueous sodium chloride. Write the molecular, full ionic, and net ionic equations, and identify spectator ions.

1. First write the balanced molecular equation with correct formulas and state symbols: $$BaC{l}_2(aq)+N{a}_{2}S{O}_4(aq)\to BaS{O}_4(s)+2NaCl(aq)$$
2. Split all strong electrolytes into ions: $BaC{l}_2$, $N{a}_{2}S{O}_4$, and $NaCl$ are all soluble ionic compounds (strong electrolytes), while $BaS{O}_4$ is insoluble so it remains as a neutral solid: $$B{a}^{2+}(aq)+2C{l}^{-}(aq)+2N{a}^{+}(aq)+S{O}_4^{2-}(aq)\to BaS{O}_4(s)+2N{a}^{+}(aq)+2C{l}^{-}(aq)$$
3. Identify spectator ions: $N{a}^{+}$ and $C{l}^{-}$ are identical on both sides, so they are spectators.
4. Cancel spectator ions and write the net ionic equation: $$B{a}^{2+}(aq)+S{O}_4^{2-}(aq)\to BaS{O}_4(s)$$
5. Check balance: Atoms: 1 Ba, 1 S, 4 O on both sides. Charge: (+2) + (-2) = 0 on left, 0 on right. Correct.

Exam tip: Always confirm solubility before splitting into ions; if a compound is insoluble, leave it as a neutral solid in the net ionic equation to earn full credit.

4. Classifying Common Reaction Types

AP Chemistry requires you to classify reactions by their pattern of reactants and products, which allows you to predict products for unknown reaction scenarios, a common exam skill. The five core reaction types you need to master are:

1. Combination (Synthesis): Two or more reactants combine to form a single product, following the pattern $A+B\to AB$. Example: $2Mg(s)+O_2(g)\to 2MgO(s)$.
2. Decomposition: A single reactant breaks down into two or more smaller products, following the pattern $AB\to A+B$. Example: $2H_2{O}_2(aq)\to 2H_{2}O(l)+O_2(g)$.
3. Combustion: A fuel (usually a hydrocarbon) reacts with oxygen gas. For complete combustion of hydrocarbons, the only products are carbon dioxide and water.
4. Double Displacement (Metathesis): Two ionic compounds swap cations and anions to form two new compounds, following $AB+CD\to AD+CB$. This includes precipitation and acid-base neutralization reactions.
5. Single Displacement: An elemental species displaces another element from its compound, following $A+BC\to AC+B$. Example: $Zn(s)+CuS{O}_4(aq)\to ZnS{O}_4(aq)+Cu(s)$.

Worked Example

Problem: Classify the reaction below, fill in the missing product formula, and write the complete balanced equation: $$\_\_\_Mg(s)+\_\_\_O_2(g)\stackrel{\Delta }{\to }\_\_\_[]$$

1. Identify the pattern: two elemental reactants forming one product, which fits the combination (synthesis) reaction pattern.
2. Predict the product: magnesium is a group 2 metal that forms a +2 cation, oxygen forms a -2 anion, so the product formula is $MgO(s)$.
3. Write the unbalanced equation and balance: $Mg(s)+O_2(g)\to MgO(s)$. Balance oxygen by adding a coefficient of 2 to $MgO$, then balance magnesium by adding a coefficient of 2 to $Mg$: $$2Mg(s)+O_2(g)\stackrel{\Delta }{\to }2MgO(s)$$
4. Final check: 2 Mg and 2 O on both sides, all coefficients are lowest whole numbers. Result: balanced combination reaction.

Exam tip: For any reaction involving a hydrocarbon and oxygen, assume complete combustion (only $C{O}_2$ and $H_{2}O$ as products) unless the problem explicitly states incomplete combustion.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Changing subscripts in chemical formulas to balance an equation, e.g., writing $H_2{O}_2$ instead of $H_{2}O$ to balance oxygen. Why: Students confuse stoichiometric coefficients (which count number of molecules) with subscripts (which count number of atoms per molecule). Correct move: Always write correct formulas for all compounds first, then only adjust coefficients in front of formulas to balance.
• Wrong move: Dissociating insoluble ionic compounds, weak acids, or pure water into ions when writing net ionic equations. Why: Students remember that ionic compounds split into ions, so they split all ionic compounds regardless of solubility. Correct move: Always check solubility rules first; only dissociate soluble strong electrolytes (soluble ionic, strong acids, strong bases) into ions.
• Wrong move: Forgetting to add state symbols to every species when writing equations for FRQs. Why: Students think state symbols are trivial, so they skip them. Correct move: Add $(s),(l),(g),(aq)$ to every species in every equation you write to avoid losing unnecessary points.
• Wrong move: Leaving fractional coefficients in the final balanced equation. Why: Students stop balancing after getting the correct atom count without clearing fractions. Correct move: Multiply all coefficients by the denominator of any fraction to get whole numbers before finishing.
• Wrong move: Only checking atom balance for net ionic equations, not charge balance. Why: Students focus on atom count and forget that net ionic equations must have equal total charge on both sides. Correct move: Add the total charge of reactants and products after writing a net ionic equation to confirm they match.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following is the correct net ionic equation for the reaction between aqueous perchloric acid ($HCl{O}_4$, a strong acid) and aqueous barium hydroxide ($Ba(OH{)}_2$, a strong base)? A) $HCl{O}_4(aq)+Ba(OH{)}_2(aq)\to Ba(Cl{O}_4{)}_2(aq)+H_{2}O(l)$ B) $H^{+}(aq)+Cl{O}_4^{-}(aq)+B{a}^{2+}(aq)+2O{H}^{-}(aq)\to B{a}^{2+}(aq)+2Cl{O}_4^{-}(aq)+2H_{2}O(l)$ C) $H^{+}(aq)+O{H}^{-}(aq)\to H_{2}O(l)$ D) $B{a}^{2+}(aq)+2Cl{O}_4^{-}(aq)\to Ba(Cl{O}_4{)}_2(aq)$

Worked Solution: First, recall that net ionic equations require removing all spectator ions from the full ionic equation. Option A is a molecular equation, not a net ionic equation, so it is incorrect. Option B is the full ionic equation, but it still includes spectator ions ($B{a}^{2+}$ and $Cl{O}_4^{-}$), so it is not net ionic. Cancel the spectator ions from B, leaving $H^{+}$ and $O{H}^{-}$ reacting to form water. After balancing, the net ionic equation is $H^{+}(aq)+O{H}^{-}(aq)\to H_{2}O(l)$, which matches option C. Option D is incorrect because $Ba(Cl{O}_4{)}_2$ is soluble, so $B{a}^{2+}$ and $Cl{O}_4^{-}$ are spectators and do not react. The correct answer is C.

Question 2 (Free Response)

Aqueous silver nitrate reacts with aqueous sodium phosphate to form a yellow precipitate of silver phosphate and aqueous sodium nitrate. (a) Write the balanced molecular equation for this reaction, including all state symbols. (b) Write the full ionic equation for this reaction. (c) Write the net ionic equation for this reaction, and identify all spectator ions.

Worked Solution: (a) First, write correct formulas for all species: silver nitrate = $AgN{O}_3$, sodium phosphate = $N{a}_{3}P{O}_4$, silver phosphate = $A{g}_{3}P{O}_4$, sodium nitrate = $NaN{O}_3$. Silver phosphate is insoluble, so it is solid; all other species are aqueous. Balancing gives: $$3AgN{O}_3(aq)+N{a}_{3}P{O}_4(aq)\to A{g}_{3}P{O}_4(s)+3NaN{O}_3(aq)$$ Check: 3 Ag, 3 N, 13 O, 3 Na, 1 P on both sides, balanced.

(b) Split all soluble strong electrolytes into ions, leave silver phosphate as a solid: $$3A{g}^{+}(aq)+3N{O}_3^{-}(aq)+3N{a}^{+}(aq)+P{O}_4^{3-}(aq)\to A{g}_{3}P{O}_4(s)+3N{a}^{+}(aq)+3N{O}_3^{-}(aq)$$

(c) Cancel spectator ions ($N{a}^{+}$ and $N{O}_3^{-}$) to get the net ionic equation: $$3A{g}^{+}(aq)+P{O}_4^{3-}(aq)\to A{g}_{3}P{O}_4(s)$$ Spectator ions are sodium ions ($N{a}^{+}$) and nitrate ions ($N{O}_3^{-}$). Charge balance: $3(+1)+(-3)=0$ on left, 0 on right, correct.

Question 3 (Application / Real-World Style)

Rust forms when solid iron metal reacts with oxygen gas and liquid water to form solid iron(III) hydroxide. Balance the reaction, then calculate how many moles of oxygen gas are consumed per mole of iron that reacts to form rust.

Worked Solution:

1. Write the unbalanced equation with correct formulas: $$Fe(s)+O_2(g)+H_{2}O(l)\to Fe(OH{)}_3(s)$$
2. Balance Fe first: 1 Fe on each side, so no adjustment needed yet.
3. Balance H: 2 H on left, 3 H on right, so add coefficients 2 to $Fe(OH{)}_3$ and 3 to $H_{2}O$: $Fe+O_2+3H_{2}O\to 2Fe(OH{)}_3$
4. Balance Fe: 2 Fe on right, add coefficient 2 to Fe: $2Fe+O_2+3H_{2}O\to 2Fe(OH{)}_3$
5. Balance O: Total O on right = $2\times 3=6$, O on left = $2+3=5$, so need 1 more O from $O_2$, giving a coefficient of $\frac{3}{2}$ for $O_2$. Multiply all coefficients by 2 to clear the fraction: $$4Fe(s)+3O_2(g)+6H_{2}O(l)\to 4Fe(OH{)}_3(s)$$
6. Check balance: 4 Fe, 12 O, 12 H on both sides, correct. Per 4 moles of Fe, 3 moles of $O_2$ are consumed, so per 1 mole of Fe, $\frac{3}{4}=0.75$ moles of oxygen are consumed. In context, this means 0.75 moles of oxygen react with every mole of iron metal to form solid iron(III) hydroxide (rust).

7. Quick Reference Cheatsheet

8. What's Next

This chapter is the foundational prerequisite for all subsequent topics in Unit 4 and the entire AP Chemistry course. Next, you will apply the balancing and net ionic equation skills you learned here to stoichiometry of chemical reactions, where you will calculate the amount of product formed or reactant consumed from a balanced equation. Without a correctly written and balanced equation, all stoichiometric calculations will be incorrect, so mastering this topic is non-negotiable for success on the exam. This topic also feeds into larger core concepts across the course: all reaction types (acid-base, oxidation-reduction, precipitation) require correctly written equations, and problems in equilibrium, kinetics, and thermodynamics all start with a balanced net ionic equation as the first step.

• Stoichiometry of chemical reactions
• Net ionic equations
• Oxidation-reduction reactions