CAPE Chemistry Unit 1 study materials
CAPE Chemistry Unit 1 study resources

CAPE Chemistry Unit 1: CAPE Chemistry Unit 1: Structure, Periodicity, Energetics, Kinetics and Equilibrium

CAPE Chemistry Unit 1 · Work through your exam, one topic at a time.

All guide sections

CAPE Chemistry Unit 1: Structure, Periodicity, Energetics, Kinetics and Equilibrium

CAPE Chemistry Unit 1 tests your ability to move between three representations of matter - symbolic equations, particulate models and quantitative data - across atomic structure, bonding, periodicity, energetics, kinetics and equilibrium. Questions are usually structured: a data stem, then a calculation, then an explanation carrying the heaviest marks.

Structure and periodicity

Across a period, increasing nuclear charge with roughly constant shielding contracts atomic radius and raises first ionisation energy. Down a group, the outer electron occupies a higher shell, so shielding increases and ionisation energy falls. Two dips are routinely examined: Be→B (and Mg→Al), where the electron removed comes from a filled s orbital that penetrates closer to the nucleus, and N→O, where pairing in one 2p orbital increases electron–electron repulsion.

Bonding and shape

Use VSEPR: count σ bonds plus lone pairs, then let lone pairs compress angles - CH₄ 109.5°, NH₃ 107°, H₂O 104.5°. Separate intramolecular covalent bonding from intermolecular forces (London < permanent dipole–dipole < hydrogen bonding), and use lattice enthalpy and polarisation to explain melting points and solubility trends.

Quantitative work

n = m/M, c = n/V, and pV = nRT with p in Pa, V in m³ and T in kelvin. Identify the limiting reagent before calculating yield. For rates, derive orders from initial-rate data; the units of k follow from the rate equation. For equilibrium, Kc omits pure solids and liquids, and comparing Q with K predicts the direction of shift.

Worked example: 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵). Find the pH. [H⁺] = √(Ka × c) = √(1.8 × 10⁻⁵ × 0.100) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³, so pH = 2.87. The approximation holds because 1.34 × 10⁻³ is only 1.3% of 0.100, comfortably below the 5% test.

Traps that cost marks

  • Leaving volume in cm³ when using c = n/V or pV = nRT.
  • Using °C instead of K, or kPa without converting to Pa.
  • Writing Kc with pure solids or water included.
  • Quoting average bond enthalpies as if they were exact for a specific compound.
  • Confusing rate with position of equilibrium: a catalyst changes neither K nor the yield.