MATSEC AM Chemistry study materials
MATSEC AM Chemistry study resources

MATSEC AM Chemistry: Chemistry (AM syllabus content): Physical, Inorganic and Organic Chemistry at Advanced Level

MATSEC AM Chemistry · Work through your exam, one topic at a time.

All guide sections

Chemistry (AM syllabus content): Physical, Inorganic and Organic Chemistry at Advanced Level

Domain overview

The MATSEC Advanced Matriculation Chemistry syllabus tests physical, inorganic and organic chemistry to a depth beyond Intermediate level: quantitative equilibrium and thermodynamics, reaction mechanisms with curly-arrow reasoning, and structured inorganic chemistry of period 3 and the d-block. Examiners reward correct units, significant figures, and mechanistic justification as much as the final number.

Core rules to command

  • Thermodynamics: ΔG = ΔH − TΔS; a reaction is feasible when ΔG < 0. ΔH is independent of route (Hess's law), so lattice enthalpy from a Born–Haber cycle must balance formation, ionisation, atomisation and electron affinity terms with their signs.
  • Equilibria: Kc and Kp expressions exclude pure solids and liquids. For a weak acid, Ka ≈ [H+]²/[HA] only when dissociation is small (<5%); otherwise solve the quadratic.
  • Buffers: pH = pKa + log([A−]/[HA]); maximum buffer capacity occurs at pH = pKa.
  • Kinetics: rate = k[A]^m[B]^n; orders come from experiment, not stoichiometry. The Arrhenius plot of ln k against 1/T has gradient −Ea/R.
  • Organic: SN1 proceeds via a planar carbocation (racemisation, tertiary > primary); SN2 is backside attack (inversion, primary > tertiary). Electrophilic substitution of benzene needs a strong electrophile generated in situ, and substituents direct ortho/para if electron-donating.
  • Inorganic: d-block ions are coloured by d–d transitions split by ligands; complex charge and oxidation state must be shown, e.g. [Cu(NH3)4]2+.

Worked example

25.0 cm³ of 0.100 M ethanoic acid (Ka = 1.8 × 10−5) is titrated. Find the pH before any base is added. Set Ka = x²/(0.100 − x); assume x is small: x ≈ √(1.8 × 10−6) = 1.34 × 10−3 M, so pH = −log(1.34 × 10−3) = 2.87. Check the assumption: 1.34 × 10−3/0.100 = 1.3%, acceptable, so the approximation holds.

Marks lost

Writing ΔG = ΔH + TΔS; confusing the exothermic lattice enthalpy of formation from gaseous ions with the endothermic lattice-dissociation step; quoting pH to four decimals from two-significant-figure data; stating a mechanism without curly arrows or the intermediate; omitting units and state symbols in enthalpy cycles.