Unit A2 1 is the quantitative heart of the A2 year: rate equations, Kp, pH and Ka, buffers, Born–Haber cycles, entropy and Gibbs free energy.
The organic half goes deeper too — optical isomerism, carbonyl chemistry and aromatic chemistry, where you must explain mechanisms rather than recite products.
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.
Units of k depend on the overall order: 1st → s⁻¹ · 2nd → mol⁻¹ dm³ s⁻¹ · 3rd → mol⁻² dm⁶ s⁻¹.
The rate equation is a window on the mechanism. Only species involved in or before the rate-determining (slowest) step appear in it, and the order in each equals the number of its molecules in that step. A species that is zero order is still consumed — just in a later, faster step.
Arrhenius: ln k = ln A − Ea/RT. Plot ln k against 1/T and the gradient is −Ea/R.
[A]² = 0.0400. [A]²[B] = 0.0400 × 0.100 = 4.00 × 10⁻³. k = 8.00 × 10⁻³ ÷ 4.00 × 10⁻³ = 2.00 mol⁻² dm⁶ s⁻¹.
For aA(g) + bB(g) ⇌ cC(g) + dD(g):
Only temperature changes K. Raising the pressure shifts the position of equilibrium so that Kp is restored — the constant itself does not change. For an exothermic forward reaction, raising the temperature decreases K.
Kp = 40² ÷ 60 = 1600 ÷ 60 = 26.7 kPa. (Units: kPa² ÷ kPa = kPa.)
A strong acid dissociates fully, so [H⁺] = [acid]. A weak acid only partially dissociates:
A buffer is a weak acid plus its conjugate base. It resists pH change because each component mops up added base or acid:
A buffer does not stop pH changing — it limits the change. Because both reservoirs are large, the ratio [HA]/[A⁻] shifts only slightly, so the pH moves only slightly. Add enough acid to exhaust the conjugate base and the buffer collapses.
[H⁺] = 0.0200 mol dm⁻³ → pH = −log₁₀(0.0200) = 1.70.
[H⁺] = 1.00 × 10⁻³ mol dm⁻³. Ka = (1.00 × 10⁻³)² ÷ 0.0500 = 1.00 × 10⁻⁶ ÷ 0.0500 = 2.00 × 10⁻⁵ mol dm⁻³. pKa = −log(2.00 × 10⁻⁵) = 4.70.
[H⁺] = 1.35 × 10⁻⁵ × 0.6667 = 9.00 × 10⁻⁶ mol dm⁻³. pH = −log(9.00 × 10⁻⁶) = 5.05. Note the pH is above pKa (4.87) because there is more conjugate base than acid.
The lattice enthalpy of formation is the enthalpy change when one mole of an ionic solid forms from its gaseous ions. It is always exothermic and cannot be measured directly, so it is found indirectly with a Born–Haber cycle — an application of Hess's law.
Rearranged: ΔHLE = ΔHf − (sum of all the other steps).
What makes a lattice enthalpy more exothermic: smaller ionic radius and higher ionic charge, because both increase the electrostatic attraction between the ions. That is why MgO (2+, 2−, small ions) has a lattice enthalpy several times more exothermic than NaCl.
Sum of the other steps = 89 + 419 + 122 + (−349) = +281 kJ mol⁻¹. ΔHLE = −437 − 281 = −718 kJ mol⁻¹.
Entropy (S) measures the dispersal of energy and matter. Gases have far higher entropy than liquids, which have more than solids. An increase in the number of moles of gas raises entropy sharply.
The unit trap: ΔH is in kJ mol⁻¹ and ΔS is in J K⁻¹ mol⁻¹. Divide ΔS by 1000 before substituting, or your answer will be out by a factor of 1000.
Four cases: ΔH negative and ΔS positive → feasible at all temperatures. ΔH positive and ΔS positive → feasible only at high T. ΔH negative and ΔS negative → feasible only at low T. ΔH positive and ΔS negative → never feasible.
TΔS = 298 × (−0.1987) = −59.2 kJ mol⁻¹. ΔG = −92.2 − (−59.2) = −33.0 kJ mol⁻¹. Negative, so feasible at 298 K — but only just, and at the high temperatures used industrially ΔG becomes positive.
Using ΔG = ΔH − TΔS, tap a reaction, then tap when ΔG is negative.
A carbon bonded to four different groups is a chiral centre. The molecule then exists as two enantiomers — non-superimposable mirror images that rotate plane-polarised light by equal amounts in opposite directions. A 50 : 50 mixture is a racemic mixture and is optically inactive.
The carbonyl group, C=O, is polar (Cδ+=Oδ−), so it is attacked by nucleophiles — nucleophilic addition.
Why HCN + propanone gives a racemate: the carbonyl carbon is trigonal planar, so CN⁻ can attack from either face with equal probability. Equal amounts of the two enantiomers form, and their optical rotations cancel exactly.
Distinguishing tests: 2,4-DNPH gives an orange precipitate with any carbonyl; Tollens' reagent gives a silver mirror only with an aldehyde.
Benzene is planar and regular hexagonal, with all six C–C bonds the same length — intermediate between a single and a double bond. Each carbon donates a p electron to a delocalised π system above and below the ring.
The evidence for delocalisation: hydrogenating cyclohexene releases −120 kJ mol⁻¹, so three isolated double bonds should release 3 × −120 = −360 kJ mol⁻¹. Benzene actually releases only about −208 kJ mol⁻¹ — it is roughly 152 kJ mol⁻¹ more stable than the Kekulé structure predicts.
Because the delocalised ring is so stable, benzene undergoes electrophilic substitution, which preserves the ring, rather than addition. It does not decolourise bromine water.
Tap a reaction on the left, then its mechanism on the right.
Rates: rate = k[A]^m[B]^n — orders from experiment; only species in or before the RDS appear; constant half-life = first order
Kp: partial pressure = mole fraction × total pressure; only temperature changes K
Acids: pH = −log[H⁺] · [H⁺] = √(Ka[HA]) · Kw = 1.00 × 10⁻¹⁴ · buffer [H⁺] = Ka[HA]/[A⁻]
Thermodynamics: Born–Haber for lattice enthalpy · ΔG = ΔH − TΔS (convert ΔS to kJ)
Optical isomerism: four different groups → non-superimposable mirror images; a racemate is optically inactive
Carbonyls: nucleophilic addition; HCN gives a racemate because the C=O is planar
Aromatic: delocalisation makes benzene substitute, not add
That is Unit A2 1 — the toughest calculations in the course, plus the mechanisms to explain them. Press Finish to see your score.
You've worked through A2 1: Further Physical & Organic Chemistry for CCEA GCE Chemistry (1110). 🎉
Your stars: 0 / 0
Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.