Ten lessons covering physical, inorganic and organic chemistry for A-level, with practical and exam skills and interactive explorers.
A study guide to A-level Chemistry, not a full course or textbook. The paper structure follows the AQA 7405 page; other boards (Edexcel, OCR, WJEC) organize topics differently. Chemistry content is standard A-level knowledge for study, and you should check your own specification and data booklet. Lab numbers are practice values.
['GCSE Chemistry or Combined Science at higher tier']
Course outline
Atomic structure, amount of substance and the mole
Use relative masses, moles and concentration in calculations.
Bonding, structure and intermolecular forces
Link bonding type to structure and properties.
Energetics and thermodynamics
Calculate enthalpy changes and explain spontaneity.
Kinetics and equilibria
Explain rates, rate equations and the equilibrium constant.
Arrhenius, Kp and equilibrium calculations
Use the Arrhenius equation and equilibrium constants for gases.
Acids, bases and pH
Calculate pH and describe buffers and titrations.
Redox, electrode potentials and electrochemistry
Balance redox equations and use cell potentials.
Inorganic chemistry: periodicity, groups and transition metals
Explain trends in the periodic table and properties of transition elements.
Organic chemistry: structure, reactions and mechanisms
Name organic compounds and describe key mechanisms.
Analysis, practical skills and exam technique
Plan investigations and use analytical methods and the AQA paper structure.
Sources and curriculum note
Reviewed October 7, 2026. Confirm the specification for your board and exam year.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Atomic structure, amount of substance and the mole
Learning goal: Use relative masses, moles and concentration in calculations.
An atom has protons and neutrons in the nucleus and electrons in shells. Isotopes have the same number of protons but different numbers of neutrons. Relative atomic mass is the weighted mean mass of an atom compared with one twelfth of carbon-12. A mass spectrometer measures isotope abundances.
The mole is the amount of substance containing 6.02 x 10^23 particles (the Avogadro constant). Moles = mass / molar mass. For solutions, moles = concentration x volume in dm3. For gases, the ideal gas equation is pV = nRT with R = 8.31 J per K per mol, p in pascals, V in m3 and T in kelvin.
Empirical formula comes from the ratio of moles of each element. Percentage yield = actual yield / theoretical yield x 100. Atom economy = mass of desired product / total mass of products x 100. Work from a balanced equation and use the limiting reagent.
Write units at every step, keep significant figures consistent with the data, and round only at the end.
Worked example
Do a titration calc.
Moles of one reagent
Mole ratio
Moles of other
Concentration
Practice problem and solution
How many moles are in 36 g of water (M = 18 g per mol)? Enter a number.
36 / 18 = 2.
Mental model: Moles link mass, concentration and gas volume; use the balanced equation.
Common trap: Using grams where moles are needed.
2. Bonding, structure and intermolecular forces
Learning goal: Link bonding type to structure and properties.
Ionic bonding is the electrostatic attraction between oppositely charged ions in a giant lattice. Ionic compounds have high melting points and conduct when molten or dissolved. Covalent bonding is a shared pair of electrons. Giant covalent structures such as diamond have very high melting points, and simple molecules have low ones.
Metallic bonding is the attraction between positive ions and delocalized electrons, which explains conduction and malleability. Shapes of molecules follow electron-pair repulsion theory: 4 bonding pairs give tetrahedral at 109.5 degrees, 3 give trigonal planar at 120 degrees, and lone pairs reduce angles.
Electronegativity is the power of an atom to attract the bonding pair. A polar bond has a dipole, and a molecule is polar only if the dipoles do not cancel. Intermolecular forces are London forces, permanent dipole-dipole forces and hydrogen bonding, which needs a hydrogen bonded to N, O or F and a lone pair on the attracted atom.
Stronger intermolecular forces give higher boiling points. Compare size first for London forces, then polarity, then hydrogen bonding.
Worked example
Explain a property.
Name the structure
Name the particles
Name the force
Link to energy
Practice problem and solution
A molecule has 4 bonding pairs and no lone pairs. The tetrahedral angle is 109.5 degrees. What is it to the nearest 10 degrees? Enter a number.
109.5 rounds to 110.
Mental model: Match structure and bonding to properties; rank intermolecular forces.
Common trap: Saying boiling breaks covalent bonds in water.
3. Energetics and thermodynamics
Learning goal: Calculate enthalpy changes and explain spontaneity.
Enthalpy change is the heat change at constant pressure. Exothermic reactions have negative values. Standard conditions are 100 kPa and a stated temperature, usually 298 K. Standard enthalpy of formation is the change when one mole forms from its elements in standard states.
In calorimetry, q = m c delta T, with c about 4.18 J per g per K for water. Divide q by moles to get the enthalpy change per mole and give the sign. Heat loss to the surroundings makes measured values less exothermic than true values.
Hess's law says the enthalpy change is independent of route. Use formation data: delta H = sum of formation values for products minus sum for reactants. Bond enthalpies give estimates using bonds broken minus bonds made.
Entropy S measures disorder. The free-energy change delta G = delta H - T delta S decides feasibility: if delta G is negative, the reaction can occur spontaneously. A reaction with positive delta S becomes more feasible at higher temperature.
Worked example
Check feasibility.
Find delta H
Find delta S
Convert units
Compute delta G
Practice problem and solution
50 g of water rises by 10 K. Using c = 4.2 J per g per K, what is q in joules? Enter a number.
50 x 4.2 x 10 = 2100.
Mental model: Calorimetry, Hess's law and delta G link energy changes to feasibility.
Common trap: Mixing joules and kilojoules in delta G.
4. Kinetics and equilibria
Learning goal: Explain rates, rate equations and the equilibrium constant.
Reaction rate is the change in concentration per unit time. Collision theory says particles must collide with at least the activation energy and the right orientation. A catalyst gives an alternative route with lower activation energy. The Maxwell-Boltzmann distribution shows why a small temperature rise greatly increases the share of particles above the activation energy.
A rate equation has the form rate = k [A]^m [B]^n. The orders come from experiments, not the stoichiometry. The overall order is the sum. The rate constant k increases with temperature, and the Arrhenius equation ln k = ln A - Ea / RT gives a straight line when ln k is plotted against 1/T.
At dynamic equilibrium forward and reverse rates are equal. For aA + bB <-> cC + dD, Kc = [C]^c [D]^d / ([A]^a [B]^b). Kc changes only with temperature. Le Chatelier's principle predicts the direction of shift when concentration, pressure or temperature changes. A catalyst does not change the position of equilibrium.
Be careful to say 'rate' or 'yield' correctly. A change can speed a reaction without improving yield.
Worked example
Predict a shift.
Identify the change
Find the opposing direction
State the shift
Say what happens to Kc
Practice problem and solution
A reaction is first order in A with k = 0.5 per s. If [A] = 0.2 mol per dm3, what is the rate in mol per dm3 per s? Enter a number.
rate = k[A] = 0.5 x 0.2 = 0.1.
Mental model: Orders come from experiment; Kc depends only on temperature; catalysts change rate, not equilibrium.
Common trap: Taking orders from the balanced equation.
5. Arrhenius, Kp and equilibrium calculations
Learning goal: Use the Arrhenius equation and equilibrium constants for gases.
The Arrhenius equation k = A e^(-Ea/RT) links the rate constant to temperature. A is a constant for the reaction, Ea is the activation energy in joules per mole, R is 8.31 J per K per mol and T is in kelvin. Taking logs gives ln k = ln A - Ea/RT, so a plot of ln k against 1/T is a straight line with gradient -Ea/R.
A small rise in temperature can raise k a lot, because T sits in an exponent. Raising T from 300 K to 310 K roughly doubles k for an activation energy near 50 kJ per mole, which matches the old rule of thumb. A catalyst lowers Ea, which raises k at the same temperature.
For gases, equilibrium is often written with partial pressures. Partial pressure = mole fraction x total pressure. For N2O4 <-> 2NO2, Kp = p(NO2)^2 / p(N2O4). Kp has units that depend on the equation, here pressure. Like Kc, Kp changes only with temperature.
For an equilibrium calculation, write an ICE table: initial moles, change and equilibrium moles. Convert to concentrations or partial pressures and then substitute into the expression. Check that your answer makes sense, for example that no amount is negative.
Worked example
Find Kp.
Mole fractions
Partial pressures
Expression
Units
Practice problem and solution
A plot of ln k against 1/T has gradient -6000 K. What is Ea in J per mol using R = 8.31? Enter a number to the nearest 100.
6000 x 8.31 = 49860, which rounds to 49900.
Mental model: Arrhenius links k to T through Ea; Kp uses partial pressures; ICE tables organize equilibrium maths.
Common trap: Using Celsius instead of kelvin in the Arrhenius equation.
6. Acids, bases and pH
Learning goal: Calculate pH and describe buffers and titrations.
A Bronsted-Lowry acid donates a proton and a base accepts one. Strong acids dissociate fully and weak acids only partly. pH = -log10 [H+]. For a strong monoprotic acid [H+] equals the acid concentration. The ionic product of water is Kw = [H+][OH-] = 1.0 x 10^-14 at 298 K.
For a weak acid HA, Ka = [H+][A-] / [HA]. If dissociation is small, [H+] = square root of (Ka x [HA]). pKa = -log10 Ka. A smaller pKa shows a stronger acid.
A buffer resists pH change. It contains a weak acid and its conjugate base. Added acid reacts with the base, and added alkali reacts with the acid. The Henderson-Hasselbalch relation pH = pKa + log([A-]/[HA]) gives its pH.
A titration curve shows the equivalence point. Choose an indicator whose color change lies in the steep section. Strong acid with strong base gives a pH of 7 at equivalence, and weak acid with strong base gives a pH above 7.
Worked example
Plan a titration.
Choose indicator
Add slowly
Find end point
Calculate
Practice problem and solution
A buffer has pKa 4.8 and equal concentrations of acid and base. What is its pH? Enter a number.
log(1) = 0, so pH = pKa.
Mental model: pH is minus the log of [H+]; weak acids use Ka; buffers contain a weak acid and its conjugate base.
Common trap: Using the strong acid formula for a weak acid.
7. Redox, electrode potentials and electrochemistry
Learning goal: Balance redox equations and use cell potentials.
Oxidation is loss of electrons and reduction is gain. Oxidation numbers track this: an increase is oxidation, a decrease is reduction. A reducing agent is oxidized and an oxidizing agent is reduced. Write half-equations, balance atoms and charge, then combine them so electrons cancel.
The standard electrode potential is measured against the standard hydrogen electrode under standard conditions. The more positive the potential, the stronger the oxidizing agent. Cell potential is the more positive electrode potential minus the less positive one for a spontaneous cell.
A feasible reaction has a positive cell potential, although the rate may be very slow. Electrode potentials change with concentration and temperature away from standard conditions. Fuel cells react a fuel with oxygen to give electrical energy directly, and rechargeable cells reverse reactions on charging.
Transition metals show variable oxidation states and form colored complex ions, and many act as catalysts. These links connect redox ideas to inorganic chemistry.
Worked example
Compute a cell potential.
Choose electrodes
Higher minus lower
Check sign
State units
Practice problem and solution
Electrode potentials are +0.80 V and +0.34 V. What is the cell potential in volts? Enter a number.
0.80 - 0.34 = 0.46.
Mental model: Track electrons; the more positive electrode is reduced; positive cell potential means feasible.
Common trap: Reading feasible as fast.
8. Inorganic chemistry: periodicity, groups and transition metals
Learning goal: Explain trends in the periodic table and properties of transition elements.
Across period 3 atomic radius falls, first ionization energy rises with small dips, and metals change to non-metals. Down a group, radius grows and ionization energy falls. Explain trends with nuclear charge, shielding and distance.
In Group 2, reactivity rises down the group as ionization energy falls. Hydroxide solubility rises down the group and sulfate solubility falls. In Group 7, oxidizing power falls down the group, and halide ions are reducing agents with strength increasing down the group.
Transition metals form ions with partly filled d orbitals. They show variable oxidation states, form colored complex ions and act as catalysts. A complex has a central metal ion with ligands bonded by coordinate bonds. Colors arise when d electrons absorb part of visible light and move between split d levels.
Ligand exchange changes color, and some complexes show shape such as octahedral or tetrahedral. Learn typical tests for ions and the observations expected.
Worked example
Explain a trend.
State the trend
Name factor
Compare values
Link to energy
Practice problem and solution
A complex has 6 ligands, each bonded by one coordinate bond. How many coordinate bonds does it have? Enter a number.
6 x 1 = 6.
Mental model: Explain trends with charge, shielding and distance; transition metals have d-block properties.
Common trap: Writing trends without a reason.
9. Organic chemistry: structure, reactions and mechanisms
Learning goal: Name organic compounds and describe key mechanisms.
Organic compounds are named by chain length, functional group and position. Isomerism includes structural isomers and stereoisomers, such as E-Z isomers around a double bond and optical isomers with a chiral carbon.
Alkenes undergo electrophilic addition. The double bond attacks an electrophile, giving a carbocation. The more stable carbocation forms the major product. Halogenoalkanes undergo nucleophilic substitution, SN1 through a carbocation or SN2 in one step, and elimination with hot ethanolic hydroxide.
Alcohols can be oxidized: primary to aldehyde then carboxylic acid, secondary to ketone, and tertiary resist. Carbonyls undergo nucleophilic addition. Carboxylic acids form esters with alcohols. Aromatic compounds such as benzene undergo electrophilic substitution that keeps the delocalized ring.
Spectroscopy identifies structures: mass spectra give the molecular ion, infrared shows functional groups, and NMR shows hydrogen environments. Always draw curly arrows from lone pairs or bonds to the atom they attack.
Worked example
Draw a mechanism.
Lone pair or bond
Arrow to atom
Show charge
Final product
Practice problem and solution
How many hydrogen atoms are in a molecule of butane, C4H10? Enter a number.
The formula gives 10.
Mental model: Match functional groups to mechanisms; use curly arrows to show electron movement.
Common trap: Drawing arrows from atoms instead of electrons.
10. Analysis, practical skills and exam technique
Learning goal: Plan investigations and use analytical methods and the AQA paper structure.
Thin-layer and gas chromatography separate mixtures. Rf = distance moved by spot / distance moved by solvent. Gas chromatography with mass spectrometry identifies components. Titrations need a standard solution, a pipette, a burette and concordant results within 0.10 cm3.
Know qualitative tests: silver nitrate for halides, barium chloride for sulfate, and tests for carbonates and ammonium ions. Percentage uncertainty = absolute uncertainty / measurement x 100. Larger measured values reduce percentage uncertainty.
AQA A-level Chemistry (7405) has three papers, each of 2 hours. Paper 1 (105 marks, 35 percent) covers relevant physical chemistry and inorganic chemistry. Paper 2 (105 marks, 35 percent) covers relevant physical chemistry and organic chemistry. Paper 3 (90 marks, 30 percent) can assess any content and includes practical techniques and data analysis, a synoptic section and 30 multiple-choice questions.
Show working, include units, and use the data given. For extended answers, plan before you write. Check against the AQA page for the current format.
Worked example
Plan an analysis.
Choose method
Control variables
Collect data
Evaluate
Practice problem and solution
A spot moves 3.0 cm and the solvent moves 6.0 cm. What is the Rf? Enter a number.
3.0 / 6.0 = 0.5.
Mental model: Know the three papers, the practical methods and how to show full working.
Common trap: Giving a number without units or working.