Module 5 is the most mathematically demanding part of the A-level. Everything here is quantitative: rate equations, Kc and Kp, pH and Ka, Born–Haber cycles, entropy, Gibbs and electrode potentials.
Then transition elements pull it together — complexes, ligand substitution, colour, catalysis and redox titrations.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
The overall order is the sum of the individual orders. The units of k depend on the overall order: 1st order → s⁻¹; 2nd order → mol⁻¹ dm³ s⁻¹; 3rd order → mol⁻² dm⁶ s⁻¹.
The rate-determining step (RDS) is the slowest step. Only species involved in or before the RDS appear in the rate equation — and the order in each species equals the number of its molecules in the RDS. This is how a rate equation reveals a mechanism.
Arrhenius: k = Ae−Ea/RT, or ln k = −Ea/RT + ln A. A plot of ln k against 1/T gives a straight line of gradient −Ea/R.
[B]² = 0.0400. [A][B]² = 0.100 × 0.0400 = 4.00 × 10⁻³. k = 2.00 × 10⁻³ ÷ 4.00 × 10⁻³ = 0.500 mol⁻² dm⁶ s⁻¹. (Overall order 3 → units mol⁻² dm⁶ s⁻¹.)
Tap an observation, then tap the order in X that it indicates.
Kc uses equilibrium concentrations; Kp uses partial pressures for gases.
For aA + bB ⇌ cC + dD, Kp = (pCc × pDd) ÷ (pAa × pBb).
Only temperature changes K. Changing pressure or concentration shifts the position of equilibrium so that K stays the same. A catalyst changes neither. For an exothermic forward reaction, raising the temperature decreases K.
A Brønsted–Lowry acid is a proton donor; a base is a proton acceptor. Every acid has a conjugate base.
For a strong monobasic acid, dissociation is complete, so [H⁺] = [acid].
For a weak acid HA ⇌ H⁺ + A⁻:
Two approximations are built into that: [H⁺] = [A⁻] (we ignore water's own dissociation) and [HA]eqm ≈ [HA]initial (little dissociates).
[H⁺] = 0.0500 mol dm⁻³. pH = −log₁₀(0.0500) = 1.30.
[H⁺] = √(1.74 × 10⁻⁶) = 1.319 × 10⁻³ mol dm⁻³. pH = −log(1.319 × 10⁻³) = 2.88. Compare with 1.00 for 0.100 mol dm⁻³ HCl — the weak acid is far less dissociated.
A buffer resists a change in pH when small amounts of acid or base are added. An acidic buffer is a weak acid plus its conjugate base (e.g. CH₃COOH + CH₃COONa).
Choosing an indicator: the indicator must change colour entirely within the vertical section of the titration curve. Strong acid–strong base: any (both work). Strong acid–weak base: methyl orange. Weak acid–strong base: phenolphthalein. Weak acid–weak base: no suitable indicator — there is no vertical section.
[H⁺] = 1.74 × 10⁻⁵ × 2.00 = 3.48 × 10⁻⁵ mol dm⁻³. pH = −log(3.48 × 10⁻⁵) = 4.46.
Lattice enthalpy of formation is the enthalpy change when one mole of an ionic compound is formed from its gaseous ions. It is always exothermic and cannot be measured directly — hence the Born–Haber cycle.
By Hess's law, going round the cycle:
Rearranged: ΔHLE = ΔHf − (all the other steps).
What makes a lattice enthalpy more exothermic: smaller ionic radii and greater ionic charges, both of which increase the electrostatic attraction. MgO (2+ and 2−, small ions) has a far more exothermic lattice enthalpy than NaCl. The same two factors make hydration enthalpies more exothermic.
Sum of the other steps = 107 + 496 + 122 + (−349) = +376 kJ mol⁻¹. ΔHLE = −411 − 376 = −787 kJ mol⁻¹.
Entropy (S) measures the number of ways energy and particles can be arranged — the dispersal of energy. Solids < liquids << gases. Producing more moles of gas raises entropy sharply.
The unit trap that ruins Gibbs calculations: ΔH is in kJ mol⁻¹ but ΔS is in J K⁻¹ mol⁻¹. Convert ΔS to kJ (divide by 1000) before substituting, or you will be out by a factor of 1000.
At the temperature where ΔG = 0, the reaction is just becoming feasible, so T = ΔH ÷ ΔS.
ΔS = (39.7 + 213.8) − 92.9 = 253.5 − 92.9 = +160.6 J K⁻¹ mol⁻¹. It is strongly positive because a gas is produced from a solid.
T = ΔH ÷ ΔS = 178 ÷ 0.1606 = 1108 K (about 835 °C). Below this, ΔG is positive and limestone does not decompose — which is why lime kilns run so hot.
A standard electrode potential, E°, is measured against the standard hydrogen electrode (E° = 0.00 V) under standard conditions: 298 K, 100 kPa, all solutions 1.00 mol dm⁻³.
Two limitations of E° feasibility that carry marks: (1) E° values apply only under standard conditions — change the concentrations and the potential changes. (2) A positive E°cell shows the reaction is thermodynamically feasible, but says nothing about the rate. A very high activation energy can leave a feasible reaction not happening at all.
Fuel cells: H₂ + ½O₂ → H₂O, producing only water. They convert chemical energy directly to electrical energy and do not need recharging while fuel is supplied — but the H₂ has to be produced and stored, which usually consumes energy.
E°cell = (+0.34) − (−0.76) = +1.10 V. Positive, so the reaction Zn + Cu²⁺ → Zn²⁺ + Cu is feasible; zinc is the negative electrode.
A transition element forms at least one stable ion with a partially filled d sub-shell. (Sc and Zn are therefore d-block but not transition elements: Sc³⁺ is d⁰ and Zn²⁺ is d¹⁰.)
Their characteristic properties all flow from that partly-filled d sub-shell:
A ligand is a species that donates a lone pair into a vacant orbital on the metal — a dative covalent (coordinate) bond. The coordination number is the number of coordinate bonds.
Ligand substitution to remember: [Cu(H₂O)₆]²⁺ (pale blue) + excess concentrated HCl → [CuCl₄]²⁻ (yellow-green) — note the coordination number falls from 6 to 4 because Cl⁻ is much larger. With excess NH₃ you get the deep blue [Cu(NH₃)₄(H₂O)₂]²⁺, where only four water ligands are replaced.
Redox titration with MnO₄⁻ is self-indicating — the end-point is the first permanent pale pink from a trace of excess MnO₄⁻:
n(MnO₄⁻) = 0.02240 × 0.0200 = 4.48 × 10⁻⁴ mol. n(Fe²⁺) = 5 × 4.48 × 10⁻⁴ = 2.24 × 10⁻³ mol. c = 2.24 × 10⁻³ ÷ 0.0250 = 0.0896 mol dm⁻³.
Tap a species on the left, then its colour on the right.
Rates: rate = k[A]^m[B]^n — orders come from experiment; only species in or before the RDS appear; ln k vs 1/T has gradient −Ea/R
Kc and Kp: partial pressure = mole fraction × total pressure; only temperature changes K
Acids: pH = −log[H⁺] · Ka = [H⁺]²/[HA] for a weak acid · Kw = 1.00 × 10⁻¹⁴ · buffer [H⁺] = Ka[HA]/[A⁻]
Thermodynamics: Born–Haber for lattice enthalpy · ΔS = ΣS(products) − ΣS(reactants) · ΔG = ΔH − TΔS (convert ΔS to kJ!)
Electrode potentials: E°cell = E°(positive) − E°(negative); positive = feasible, but says nothing about rate
Transition elements: partially filled d sub-shell → variable oxidation states, colour (d–d transitions), catalysis, complex formation
That is Module 5 — the quantitative core of the A-level. Press Finish to see your score.
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