Introduction to Reactions

AP Chemistry· AP Chemistry CED — Chemical Reactions· 14 min read

1. Core Definition of Chemical Reactions★☆☆☆☆⏱ 2 min

Introduction to chemical reactions is the opening topic of AP Chemistry Unit 4, covering core conventions for representing chemical change. This topic contributes approximately 2-3% of your total AP Chemistry exam score, with the full Unit 4 accounting for 7-9% of total weight. Balanced equations are almost always required as the first step for longer FRQs covering stoichiometry, titrations, or equilibrium.

📘 Definition

Chemical Reaction

Reactants \rightarrow Products

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 $(s), (l), (g), (a q)$ indicating physical state of each species.

2. Balancing Chemical Equations by Inspection★★☆☆☆⏱ 4 min

Balancing chemical equations adjusts stoichiometric coefficients (whole numbers 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: never change subscripts in chemical formulas to balance an equation, because this changes the identity of the substance.

Write the unbalanced equation with all correct chemical formulas and state symbols first
Count atoms of each element on both sides
Start with elements that appear in only one reactant and one product
Leave pure elemental species (like $O_{2}$ , $H_{2}$ ) for last
Clear any fractions by multiplying all coefficients by the denominator
Reduce coefficients to the lowest whole-number ratio

📐 Worked Example

Write the balanced equation for the complete combustion of butane ( $C_{4} H_{10}$ ), which produces carbon dioxide gas and liquid water.

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1. Write the unbalanced equation with correct formulas:
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$C_4H_{10}(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l)$
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1. Count initial atoms: Reactants = 4 C, 10 H, 2 O; Products = 1 C, 2 H, 3 O.
4
1. Balance carbon first (only one reactant/product): Add a coefficient of 4 to $C O_{2}$ :
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$C_4H_{10}(g) + O_2(g) \rightarrow 4CO_2(g) + H_2O(l)$
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1. Next balance hydrogen: 10 H on the reactant side requires a coefficient of 5 for $H_{2} O$ :
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$C_4H_{10}(g) + O_2(g) \rightarrow 4CO_2(g) + 5H_2O(l)$
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1. Balance oxygen: Total product oxygen = $(4 \times 2) + (5 \times 1) = 13$ , so $O_{2}$ gets a coefficient of $\frac{13}{2}$ . Clear the fraction by multiplying all coefficients by 2:
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$2C_4H_{10}(g) + 13O_2(g) \rightarrow 8CO_2(g) + 10H_2O(l)$
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1. 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. Molecular, Full, and Net Ionic Equations★★★☆☆⏱ 4 min

For reactions in aqueous solution, three levels of representation are used: (1) Molecular equations show all compounds as neutral molecules, even if they dissociate in water. (2) Full ionic equations split all strong electrolytes into free ions, since these exist as separated ions in solution. (3) Net ionic equations remove spectator ions to show only the species that actually react. Net ionic equations must balance both atoms and total charge.

📘 Definition

Spectator Ion

An ion that appears unchanged on both sides of a full ionic equation, meaning it does not participate in the actual chemical change. Spectator ions are canceled out to get a net ionic equation.

📐 Worked Example

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.

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1. First write the balanced molecular equation with correct formulas and state symbols:
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$BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)$
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1. Split all strong electrolytes into ions: $B a C l_{2}$ , $N a_{2} S O_{4}$ , and $N a C l$ are soluble ionic compounds, while $B a S O_{4}$ is insoluble so it remains solid:
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$Ba^{2+}(aq) + 2Cl^-(aq) + 2Na^+(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) + 2Na^+(aq) + 2Cl^-(aq)$
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1. Identify spectator ions: $N a^{+}$ and $C l^{-}$ are identical on both sides, so they are spectators.
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1. Cancel spectator ions and write the net ionic equation:
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$Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$
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1. Check balance: Atoms balance, and total 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 Chemical Reaction Types★★☆☆☆⏱ 3 min

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

Combination (Synthesis): Two or more reactants combine to form a single product: $A + B \to A B$ . Example: $2 M g (s) + O_{2} (g) \to 2 M g O (s)$ .
Decomposition: A single reactant breaks down into two or more smaller products: $A B \to A + B$ . Example: $2 H_{2} O_{2} (a q) \to 2 H_{2} O (l) + O_{2} (g)$ .
Combustion: A fuel (usually a hydrocarbon) reacts with oxygen gas. Complete combustion of hydrocarbons produces only carbon dioxide and water.
Double Displacement (Metathesis): Two ionic compounds swap cations and anions to form two new compounds: $A B + C D \to A D + C B$ . Includes precipitation and acid-base neutralization.
Single Displacement: An elemental species displaces another element from its compound: $A + B C \to A C + B$ . Example: $Z n (s) + C u S O_{4} (a q) \to Z n S O_{4} (a q) + C u (s)$ .

📐 Worked Example

Classify the reaction below, fill in the missing product formula, and write the complete balanced equation: $___M g (s) +___O_{2} (g) Δ___[]$

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1. Identify the pattern: two elemental reactants forming one product, which fits the combination (synthesis) reaction pattern.
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1. Predict the product: magnesium is a group 2 metal that forms a +2 cation, oxygen forms a -2 anion, so the product formula is $M g O (s)$ .
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1. Write the unbalanced equation and balance: Balance oxygen by adding a coefficient of 2 to $M g O$ , then balance magnesium by adding a coefficient of 2 to $M g$ :
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$2Mg(s) + O_2(g) \xrightarrow{\Delta} 2MgO(s)$
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1. 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. AP Style Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

Which of the following is the correct net ionic equation for the reaction between aqueous perchloric acid ( $H C l O_{4}$ , a strong acid) and aqueous barium hydroxide ( $B a (O H)_{2}$ , a strong base)?
- $H C l O_{4} (a q) + B a (O H)_{2} (a q) \to B a (C l O_{4})_{2} (a q) + H_{2} O (l)$
- $H^{+} (a q) + C l O_{4}^{-} (a q) + B a^{2 +} (a q) + 2 O H^{-} (a q) \to B a^{2 +} (a q) + 2 C l O_{4}^{-} (a q) + 2 H_{2} O (l)$
- $H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l)$
- $B a^{2 +} (a q) + 2 C l O_{4}^{-} (a q) \to B a (C l O_{4})_{2} (a q)$
Reveal answer
2 —
Option A is a molecular equation, not net ionic. Option B is the full ionic equation, still including spectators. Option D is incorrect because $B a^{2 +}$ and $C l O_{4}^{-}$ are spectators and do not react. After canceling spectators, the correct net ionic equation is option C.

📐 Worked Example

Aqueous silver nitrate reacts with aqueous sodium phosphate to form a yellow precipitate of silver phosphate and aqueous sodium nitrate. Write (a) balanced molecular, (b) full ionic, (c) net ionic equations, and identify spectator ions.

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(a) Balanced molecular equation with state symbols:
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$3AgNO_3(aq) + Na_3PO_4(aq) \rightarrow Ag_3PO_4(s) + 3NaNO_3(aq)$
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(b) Full ionic equation (split soluble strong electrolytes only):
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$3Ag^+(aq) + 3NO_3^-(aq) + 3Na^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) + 3Na^+(aq) + 3NO_3^-(aq)$
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(c) Cancel spectator ions ( $N a^{+}$ and $N O_{3}^{-}$ ) to get the net ionic equation:
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$3Ag^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s)$
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Charge balance: $3 (+ 1) + (- 3) = 0$ on left, 0 on right, which is correct. Spectator ions are $N a^{+}$ and $N O_{3}^{-}$ .

6. Common Pitfalls

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 $(s), (l), (g), (a q)$ 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.

7. Quick Reference Cheatsheet

Category	Rule/Formula	Notes
Law of Conservation of Mass	\sum \text{Mass of Reactants} = \sum \text{Mass of Products}	Applies to all chemical reactions; does not apply to nuclear reactions
Balancing Rule	Only adjust stoichiometric coefficients, never subscripts	Changing subscripts changes the identity of the chemical substance
Combination Reaction	A + B \rightarrow AB	Two or more reactants form one product
Decomposition Reaction	AB \rightarrow A + B	One reactant breaks into two or more products
Complete Combustion of Hydrocarbon	C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O	Assume complete combustion unless stated otherwise
Full Ionic Equation Rule	Split all strong electrolytes into ions	Strong electrolytes = soluble ionic compounds, strong acids, strong bases
Net Ionic Equation	Cancel spectator ions from full ionic equation	Must balance both atoms and total charge on both sides
Spectator Ions	Ions unchanged on both sides of full ionic equation	Do not participate in reaction, removed from net ionic equation

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AP Chem
MCQ balancing net ionic equation
2022 · AP Chem
FRQ write balanced equation
2021 · AP Chem
MCQ classify reaction type

What's Next

This foundational topic is a prerequisite for all subsequent units and problem-solving in AP Chemistry. You will next apply the balancing and net ionic equation skills you mastered here to stoichiometry, where you will calculate the amount of product formed or reactant consumed from a balanced equation. A correctly written and balanced equation is required for all stoichiometric calculations, so mastering this topic is critical for exam success. This topic also underpins all future work on acid-base reactions, oxidation-reduction, equilibrium, kinetics, and thermodynamics, all of which start with a correct balanced equation as the first step.