Chemical Reactions
A chemical reaction rearranges atoms from reactants into new products, breaking old bonds and forming new ones. Mass is conserved (atoms are neither created nor destroyed), so every equation must be balanced. The eight fundamental reaction types below cover the vast majority of chemistry encountered from high school through undergraduate courses.
| # | Type | General Form | Key Clue |
|---|---|---|---|
| 1 | Synthesis | A + B → AB | Two or more reactants → one product |
| 2 | Decomposition | AB → A + B | One reactant → two or more products |
| 3 | Single Displacement | A + BC → AC + B | Free element displaces one ion in compound |
| 4 | Double Displacement | AB + CD → AD + CB | Ions swap partners; often precipitate or gas |
| 5 | Combustion | fuel + O₂ → CO₂ + H₂O | Fuel reacts with oxygen; releases heat & light |
| 6 | Acid-Base | acid + base → salt + H₂O | Proton transfer (neutralisation) |
| 7 | Redox | oxidiser + reducer → products | Electrons transferred between species |
| 8 | Precipitation | ions(aq) → solid↓ + ions(aq) | Insoluble solid forms from two solutions |
Synthesis (Combination) A + B → AB
In a synthesis reaction two or more substances combine to form a single, more complex product. The reactants can be elements or compounds.
| Balanced Equation | Notes |
|---|---|
2H₂ + O₂ → 2H₂O | Formation of water; extremely exothermic |
2Na + Cl₂ → 2NaCl | Formation of table salt; vigorous reaction |
CaO + H₂O → Ca(OH)₂ | Slaked lime; used in construction |
3H₂ + N₂ → 2NH₃ | Haber-Bosch process; industrial ammonia synthesis |
SO₃ + H₂O → H₂SO₄ | Formation of sulfuric acid |
Problem Set
Decomposition AB → A + B
A single compound breaks down into two or more simpler substances. Energy input (heat, light, or electricity) is usually required.
| Balanced Equation | Notes |
|---|---|
2H₂O → 2H₂ + O₂ | Electrolysis of water |
2HgO → 2Hg + O₂ | Heat mercury(II) oxide; historic experiment |
CaCO₃ → CaO + CO₂ | Calcination of limestone at ~840 °C |
2H₂O₂ → 2H₂O + O₂ | Catalytic decomposition of hydrogen peroxide |
2KClO₃ → 2KCl + 3O₂ | Potassium chlorate heated; lab O₂ source |
Problem Set
Single Displacement A + BC → AC + B
A more reactive free element displaces a less reactive element from a compound. Reactivity (activity) series governs whether the reaction occurs.
| Balanced Equation | Notes |
|---|---|
Zn + 2HCl → ZnCl₂ + H₂ | Zinc displaces hydrogen from hydrochloric acid |
Fe + CuSO₄ → FeSO₄ + Cu | Iron displaces copper; copper deposits on nail |
2Al + 6HCl → 2AlCl₃ + 3H₂ | Aluminium dissolves in hydrochloric acid |
Mg + 2HCl → MgCl₂ + H₂↑ | Vigorous; magnesium more reactive than zinc |
Cl₂ + 2NaBr → 2NaCl + Br₂ | Halogen displacement; chlorine above bromine |
Problem Set
Double Displacement (Metathesis) AB + CD → AD + CB
Ions from two compounds exchange partners. The reaction is driven by formation of a precipitate, a gas, or a weakly ionised compound (like water).
| Balanced Equation | Notes |
|---|---|
NaCl + AgNO₃ → AgCl↓ + NaNO₃ | Classic precipitation test for chloride ions |
BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl | Barium sulfate precipitate; qualitative test |
HCl + NaOH → NaCl + H₂O | Strong acid-base neutralisation |
Pb(NO₃)₂ + 2KI → PbI₂↓ + 2KNO₃ | Vivid yellow lead(II) iodide precipitate |
Na₂CO₃ + CaCl₂ → CaCO₃↓ + 2NaCl | White calcium carbonate precipitate |
Problem Set
Combustion fuel + O₂ → CO₂ + H₂O
Complete combustion occurs with excess oxygen producing CO₂ and H₂O. Incomplete combustion (limited O₂) yields toxic CO and particulate soot (C).
| Balanced Equation | Notes |
|---|---|
CH₄ + 2O₂ → CO₂ + 2H₂O | Methane (natural gas); complete combustion |
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O | Propane; camping stove fuel |
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O | Acetylene; oxy-acetylene welding |
C + O₂ → CO₂ | Complete combustion of carbon |
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O | Octane (petrol); complete combustion |
Problem Set
Acid-Base (Neutralisation) acid + base → salt + H₂O
An acid donates H⁺ to a base (Brønsted-Lowry definition). The resulting ionic compound is a salt, and water is usually produced. Net ionic equation: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$.
| Balanced Equation | Notes |
|---|---|
HCl + NaOH → NaCl + H₂O | Strong acid + strong base; salt solution neutral |
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O | Diprotic acid requires 2 mol base |
CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂ | Acetic acid + baking soda; gas produced |
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O | Triprotic acid requires 3 mol base |
2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O | Nitric acid + calcium hydroxide |
Problem Set
Redox (Oxidation-Reduction) electron transfer
Oxidation is loss of electrons (OIL); reduction is gain of electrons (RIG). The species that loses electrons is the reducing agent; the one that gains electrons is the oxidising agent. Oxidation numbers track electron bookkeeping.
| Balanced Equation | Oxidation change |
|---|---|
2Fe + 3Cl₂ → 2FeCl₃ | Fe: 0 → +3 (oxidised); Cl: 0 → −1 (reduced) |
Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag | Cu: 0 → +2; Ag: +1 → 0 |
4Fe + 3O₂ → 2Fe₂O₃ | Rusting; Fe: 0 → +3; O: 0 → −2 |
2Mg + O₂ → 2MgO | Mg: 0 → +2; bright white flame |
Zn + CuSO₄ → ZnSO₄ + Cu | Zn: 0 → +2 (oxidised); Cu: +2 → 0 (reduced) |
Problem Set
Precipitation ions → solid↓
When two aqueous solutions are mixed, ions combine to form an insoluble ionic compound that separates from solution as a solid precipitate. Solubility rules determine which combinations precipitate.
| Balanced Equation | Precipitate colour |
|---|---|
Pb(NO₃)₂ + 2NaI → PbI₂↓ + 2NaNO₃ | Bright yellow |
AgNO₃ + NaCl → AgCl↓ + NaNO₃ | White (turns grey in light) |
BaCl₂ + K₂SO₄ → BaSO₄↓ + 2KCl | White; insoluble even in acid |
FeCl₃ + 3NaOH → Fe(OH)₃↓ + 3NaCl | Rust-brown |
Cu(NO₃)₂ + 2NaOH → Cu(OH)₂↓ + 2NaNO₃ | Pale blue |
Problem Set
Master Balancing Guide
A balanced equation has equal numbers of each atom on both sides and equal total charge. Use coefficients only — never change subscripts.
Step-by-Step Algorithm
- Write the unbalanced skeleton — correct formulae for all reactants and products with an arrow.
- Count atoms of each element on each side. List the imbalances.
- Balance metals first, then non-metals, then hydrogen, and save oxygen for last.
- Use the smallest integer coefficients that balance every element simultaneously. Start with the most complex formula and work outwards.
- Verify — recount all atoms and charge. Reduce coefficients by their GCF if needed.
Worked Examples
| Difficulty | Skeleton | Balanced |
|---|---|---|
| Easy | H₂ + O₂ → H₂O | 2H₂ + O₂ → 2H₂O |
| Easy | Na + H₂O → NaOH + H₂ | 2Na + 2H₂O → 2NaOH + H₂ |
| Medium | Fe + O₂ → Fe₂O₃ | 4Fe + 3O₂ → 2Fe₂O₃ |
| Medium | Al + H₂SO₄ → Al₂(SO₄)₃ + H₂ | 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂ |
| Hard | C₆H₁₂O₆ + O₂ → CO₂ + H₂O | C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O |
Oxidation Number Method (for Redox)
- Assign oxidation numbers to every atom using standard rules (pure element = 0; O usually −2; H usually +1; sum of compound = 0).
- Identify which atoms change oxidation number. Calculate the magnitude of change for each.
- Multiply coefficients so that the total electrons lost (oxidation) equals total electrons gained (reduction) — this is the electron balance.
- Balance the remaining atoms (not involved in electron transfer) by inspection. Check charge balance in ionic equations.
- Verify atom counts and charge on both sides.
Example: Balance $\text{KMnO}_4 + \text{HCl} \rightarrow \text{KCl} + \text{MnCl}_2 + \text{Cl}_2 + \text{H}_2\text{O}$
Mn goes +7 → +2 (gain 5e⁻, reduced). Cl goes −1 → 0 (loses 1e⁻ per Cl atom, oxidised). To equalise: 1 Mn × 5e⁻ = 5 Cl × 1e⁻. Coefficient 2 KMnO₄ and 10 HCl for the redox Cl; then balance remaining Cl and H by inspection:
2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O
Reaction Energy
Every reaction involves breaking existing bonds (requires energy) and forming new bonds (releases energy). The net difference is the enthalpy change $\Delta H$.
- Exothermic ($\Delta H < 0$): products have lower energy than reactants; heat is released to surroundings. Examples: combustion, neutralisation, rusting.
- Endothermic ($\Delta H > 0$): products have higher energy than reactants; heat is absorbed from surroundings. Examples: photosynthesis, dissolving ammonium nitrate, cooking.
- Activation energy ($E_a$): minimum energy needed to initiate the reaction — the "energy hill" that must be surmounted regardless of whether the reaction is exo- or endothermic.
- Catalysts lower $E_a$ without changing $\Delta H$, speeding up reactions by providing an alternative pathway.
The enthalpy change can be calculated from bond energies: $$\Delta H = \sum E_\text{bonds broken} - \sum E_\text{bonds formed}$$ or from Hess's Law by combining known reaction enthalpies. Standard enthalpy of formation values ($\Delta H_f^\circ$) allow calculation of any reaction enthalpy: $$\Delta H_\text{rxn}^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$$