OCR A-level Chemistry A (H432) · Module 3: Periodic table and energy
Mini-Lesson
Periodic Table & Energy
Module 3 joins the inorganic half of A-level (periodicity, Group 2, the halogens, ion tests) to the energy half (enthalpy, Hess, rates and equilibrium).
You will calculate enthalpy changes three different ways — from calorimetry, from bond enthalpies and from Hess cycles — then handle rates, the Boltzmann distribution and Kc.
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.
3.1.1 Periodicity
Ionisation energy across a period
Across Period 3, the first ionisation energy generally increases: the nuclear charge rises, the atomic radius falls and the shielding is essentially unchanged, so the outer electron is held more tightly.
Dip at Al — the outer electron of Al is in a 3p orbital, which is higher in energy and more shielded than Mg's 3s.
Dip at S — in sulfur, two electrons occupy the same 3p orbital and repel each other, making one easier to remove than in phosphorus.
Down a group, first ionisation energy falls: more shells, larger radius and more shielding outweigh the greater nuclear charge.
Successive ionisation energies give away the group: a big jump tells you an electron has been pulled from a closer shell. A jump after the 2nd electron means Group 2.
Quick check
Think it through
?The first ionisation energy of sulfur is lower than that of phosphorus. Why?
3.1.2 & 3.1.3 Groups 2 and 7
Group 2 metals and the halogens
Group 2 — reactivity increases down the group: the atomic radius grows and shielding increases, so the two outer electrons are lost more easily (first ionisation energy falls).
With oxygen: 2Mg + O₂ → 2MgO. With water: Ca + 2H₂O → Ca(OH)₂ + H₂.
Group 2 hydroxides get more soluble down the group — Mg(OH)₂ is sparingly soluble (milk of magnesia); Ca(OH)₂ (limewater) is more so.
Group 2 sulfates get less soluble down the group — BaSO₄ is insoluble, which is why Ba²⁺ is the test for sulfate.
Group 7 — reactivity decreases down the group. As an oxidising agent, a halogen displaces any halide below it:
Cl₂ + 2KBr → 2KCl + Br₂chlorine is the stronger oxidising agent, so it takes the electrons from Br⁻
Disproportionation — one element is both oxidised and reduced. Chlorine with water: Cl₂ + H₂O → HClO + HCl. Chlorine goes from 0 to +1 (in HClO) and 0 to −1 (in HCl) in the same reaction.
Quick check
Think it through
?Chlorine water is added to aqueous potassium iodide. What is observed, and why?
Match it
Match the test to the observation
Tap an ion test on the left, then the observation it gives on the right.
Test
Observation
3.2.1 Enthalpy changes
Enthalpy, calorimetry and q = mcΔT
An exothermic reaction releases energy to the surroundings: ΔH is negative. An endothermic reaction takes energy in: ΔH is positive.
q = mcΔTq in J · m = mass of the solution in g · c = 4.18 J g⁻¹ K⁻¹ · ΔT in K or °C
To get ΔH in kJ mol⁻¹: divide q (converted to kJ) by the moles of the limiting reactant, then attach the sign — negative if the temperature rose.
Standard conditions for a standard enthalpy change (⦵): 100 kPa, a stated temperature (usually 298 K) and all substances in their standard states. Standard enthalpy of formation of an element in its standard state is zero by definition.
Calculate
Your turn — calculation 1
125.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 25.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises by 6.5 °C. Take the total mass as 50.0 g and c = 4.18 J g⁻¹ K⁻¹. Calculate the enthalpy change of neutralisation in kJ mol⁻¹ (include the sign), to 3 significant figures.
Breaking bonds is endothermic; making bonds is exothermic. So:
ΔH = Σ(bonds broken) − Σ(bonds made)using mean bond enthalpies, which are averages over many compounds
Why bond-enthalpy answers never quite match the data book: mean bond enthalpies are averaged across different molecules, and they apply to substances in the gaseous state. Any species that is a liquid at 298 K adds an extra enthalpy change that the calculation ignores.
Calculate
Your turn — calculation 2
2For H₂(g) + Cl₂(g) → 2HCl(g), use mean bond enthalpies: H–H = +436, Cl–Cl = +242, H–Cl = +431 kJ mol⁻¹. Calculate ΔH in kJ mol⁻¹ (include the sign).
3Use mean bond enthalpies to find ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). C–H = +412, O=O = +496, C=O = +805, O–H = +463 kJ mol⁻¹. Give your answer in kJ mol⁻¹ (include the sign).
Hess's law: the enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same. Two standard cycles:
ΔHr = Σ ΔHf(products) − Σ ΔHf(reactants)use with enthalpies of formation — arrows point up into the reaction
ΔHr = Σ ΔHc(reactants) − Σ ΔHc(products)use with enthalpies of combustion — arrows point down out of the reaction
Remember the direction: formation — products minus reactants. Combustion — reactants minus products. Getting these the wrong way round flips the sign and loses every mark.
Calculate
Your turn — calculation 4
4Calculate ΔH for CaCO₃(s) → CaO(s) + CO₂(g) using enthalpies of formation: CaCO₃ = −1207, CaO = −635, CO₂ = −394 kJ mol⁻¹. Include the sign.
Collision theory, the Boltzmann distribution and catalysts
A reaction happens only when particles collide with energy ≥ the activation energy, Ea, and with the correct orientation.
The Boltzmann distribution shows how molecular energies are spread. Only the molecules in the tail to the right of Ea can react.
Raise the temperature — the curve flattens and shifts right; a much larger proportion of molecules exceed Ea. This, not the extra collision frequency, is the main reason rate rises so steeply.
Increase concentration or pressure — more particles per unit volume → more frequent collisions. The curve itself does not change shape.
Add a catalyst — provides an alternative route with a lower Ea. The Boltzmann curve is unchanged, but Ea moves left, so more molecules can react.
Two catalyst facts: the curve never changes shape when you add a catalyst — only the position of Ea moves. And a catalyst does not change ΔH or the position of equilibrium; it only gets you there faster.
Calculate
Your turn — calculation 5
5In a rates experiment, 48 cm³ of gas is collected in 40 s. Calculate the mean rate of reaction in cm³ s⁻¹.
cm³ s⁻¹
Hint: Mean rate = change in volume ÷ time = 48 ÷ 40.
Method
Rate = 48 ÷ 40 = 1.2 cm³ s⁻¹. Note this is a mean rate — the initial rate, from the gradient of the tangent at t = 0, is higher.
Quick check
Think it through
?Which statement about a catalyst and the Boltzmann distribution is correct?
3.2.3 Chemical equilibrium
Dynamic equilibrium, Le Chatelier and Kc
At dynamic equilibrium in a closed system, the forward and reverse rates are equal and the concentrations stay constant — but both reactions are still happening.
Le Chatelier's principle: if a change is imposed, the equilibrium shifts to minimise that change.
For the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹:
Higher pressure → shifts to the side with fewer gas moles (4 → 2), so more NH₃.
Higher temperature → shifts in the endothermic (reverse) direction, so less NH₃.
Catalyst → no change in position or yield; equilibrium is simply reached faster.
Kc = [products]coefficients ÷ [reactants]coefficientsKc changes only with temperature — never with pressure, concentration or a catalyst
Calculate
Your turn — calculation 6
6For H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium mixture in a 1.00 dm³ vessel contains 0.20 mol H₂, 0.20 mol I₂ and 1.60 mol HI. Calculate Kc.