Chemical Reactions
Welcome to Chemical Reactions & Equations — the heart of chemistry. While Module 1 taught you about individual atoms and elements, this module is all about what happens when substances interact, transform and create something new.
You will learn how to write and balance chemical equations, identify different types of reactions (combustion, neutralisation, decomposition and more), understand energy changes in reactions, work with rates of reaction and chemical equilibrium, identify redox reactions, and explore acids, bases and salts. These topics feed three of the four modules of the CSCA Chemistry syllabus.
By the end of this module you will be able to predict products, balance equations efficiently, and explain why some reactions are fast and others slow. Let's get mixing!
What the Exam Tests
The official chemistry syllabus has four modules: Basic Chemical Concepts and Calculations, Properties and Reactions of Substances, Chemical Theories and Laws, and Chemical Experiments and Applications. This module feeds three of them. Balancing Equations, Stoichiometry and Amount of Substance & Gas Calculations are Basic Concepts and Calculations (chemical equations, the amount of substance, solution concentration and the ideal gas law). Reaction Types belongs to Properties and Reactions of Substances, and Reaction Rates and Chemical Equilibrium to Chemical Theories and Laws.
Questions can ask for a mass or a gas volume from an equation, the concentration after a dilution, the limiting reagent, the effect of temperature or a catalyst on a rate, or the direction in which an equilibrium shifts.
In our mock papers Concepts and Calculations gets 12 of the 48 questions, Substances and their Reactions 12, and Theories and Laws 14. The CSCA publishes the topics but not how many questions each gets, so this split is our own reading of the syllabus.
A chemical equation is a shorthand way of describing a chemical reaction. Reactants go on the left, products on the right, and an arrow (→) means "reacts to form".
Word Equations
Start simple: sodium + water → sodium hydroxide + hydrogen
Symbol Equations
Replace names with formulas: Na + H₂O → NaOH + H₂
Balancing
The law of conservation of mass says atoms cannot be created or destroyed. Every atom on the left must appear on the right. We balance by adding coefficients (big numbers in front).
Steps to Balance
- Write the unbalanced equation with correct formulas.
- Count atoms of each element on both sides.
- Adjust coefficients one element at a time. Start with the element that appears in the fewest formulas.
- Check: every element must have the same count on both sides.
- Make sure all coefficients are the smallest possible whole numbers.
Example: Na + H₂O → NaOH + H₂
Na: 1 left, 1 right ✓
H: 2 left, 3 right ✗ → multiply: 2Na + 2H₂O → 2NaOH + H₂
Na: 2=2 ✓, H: 4=4 ✓, O: 2=2 ✓
State Symbols
Show the physical state of each substance: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water).
Chinese convention (化学方程式): Chinese textbooks usually write a balanced equation with "=" (or →) and put the reaction condition over it — 点燃 (ignite), △ (heat), 高温 (high temperature), 催化剂 (catalyst), 通电 (electrolysis). Instead of state symbols they mark a gas product with ↑ (only if no reactant is a gas) and a precipitate in solution with ↓ (only if no reactant is a solid): 2Na + 2H₂O = 2NaOH + H₂↑; CuSO₄ + 2NaOH = Cu(OH)₂↓ + Na₂SO₄. A "which equation is written correctly" option can be wrong just because a condition, ↑ or ↓ is missing or misplaced.
Final balanced equation: 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
Key takeaway: Balance equations by adjusting coefficients to ensure the same number of each type of atom on both sides. Never change the subscripts in formulas — only add coefficients.
Key Formulas
Balanced equation: same atoms on both sides | State symbols: (s) solid, (l) liquid, (g) gas, (aq) dissolved | Conservation of mass: total mass of reactants = total mass of products
Worked Example 1
Example 1: Balance: Fe + O₂ → Fe₂O₃
Start with O, the hard element: 2 atoms in O₂, 3 in Fe₂O₃. The lowest common multiple is 6, so write 3O₂ and 2Fe₂O₃. Now there are 4 Fe on the right → put 4Fe on the left. Recount: 4Fe + 3O₂ → 2Fe₂O₃. Fe: 4=4 ✓, O: 6=6 ✓.
Note: this is only a balancing exercise (a simplified equation for rusting, which also needs water). Iron burning in pure oxygen gives black Fe₃O₄: 3Fe + 2O₂ → Fe₃O₄ (点燃).
Worked Example 2
Example 2: Balance: CH₄ + O₂ → CO₂ + H₂O
C: 1=1 ✓. H: 4 left, 2 right → put 2H₂O. O: 2 left, 4 right → put 2O₂. CH₄ + 2O₂ → CO₂ + 2H₂O. C:1=1, H:4=4, O:4=4 ✓.
Worked Example 3
Example 3: Balance: Al + HCl → AlCl₃ + H₂
Al: 1=1 ✓. Cl: 1 left, 3 right → 3HCl. H: 3 left, 2 right — not even. Try: 2Al + 6HCl → 2AlCl₃ + 3H₂. Al:2=2, H:6=6, Cl:6=6 ✓.
Exam tip: If you struggle with balancing, try the "odd-even trick" (奇数配偶法): if an element has an odd number of atoms in a formula on one side and an even number on the other, put a 2 in front of the formula with the odd count first (Fe₂O₃ has 3 O, O₂ has 2, so write 2Fe₂O₃ first).
Exam tip: When state symbols are shown, they are part of the answer — two options can differ only in an (aq) against an (s), or in a missing ↑ or ↓.
Common mistake: Changing subscripts to balance: Writing H₃O instead of putting a coefficient in front of H₂O changes the substance entirely.
Common mistake: Forgetting to re-count after adjusting: Changing one coefficient may unbalance another element. Always do a final check.
Try This Yourself
Balance: Mg + HCl → MgCl₂ + H₂
Module Summary
Open and read all sections to complete this module